Transcript for:
Comprehensive Edexcel IGCSE Chemistry Guide

hello lovelies and this we're gonna go over everything that you need for your Edexcel eye level chemistry to go with this so the revision guide which is free on my website where you can tick things off as you go along if there's anything that you're not sure about you can link through to the teaching video or go and do the free multiple choice questions which are also over on my website and then there are the predictive papers for this year's exams written and thoroughly checked and the video walkthroughs to go with so many questions good luck guys I'm gonna be here with you every single step of the way foreign [Music] structure and as atoms are made up from three different subatomic particles protons neutrons and electrons protons are in the nucleus and have a mass of one and a charge of plus one neutrons are also in the nucleus have a mass of one but a charge of zero electrons are found in the outer shells that mass is very small it's 1 1036 the mass of a proton and they have charge of minus one when we say the charge is plus one zero and minus one this is the relative charge compared to other things the actual charger on this we can measure in coulombs and is very very small but it is much easier to say plus one and minus one the mass of protons neutrons and electrons has been worked out based on common 12 as a reference standard and it is really important to remember that this drawing of an atom is not too scared it is mainly empty space the nucleus the diameter of the nucleus is basically 10 to the minus 50 meters whereas the whole spherical diameter of the atom is 10 to the minus 10 meters there is a big difference in those numbers the structure of the atom has changed over time as New Evidence has presented itself we started off with a blob basically an atom means uncuttable we then went to a solid sphere with negative bits inside it before we have ended up with today's structure on the periodic table you'll see boxes with numbers in now it doesn't matter where these numbers are but the larger number of the two will be the mass number this is going to be the average mass of the naturally occurring isotopes of that atom because we can't have half a neutron or half a proton the atomic number is the smaller number associated with an element this is the number of protons you will frequently see the atomic number with the symbol Z and the mass number with the symbol a now the mass number is the total number of protons and neutrons which is why having a mass number of 35.5 or whatever decimal is is a bit strange but it hand made decimal because it is an average number of the naturally occurring isotopes we can have isotopes of an element or different versions of an element here we have carbon 12 and carbon 14. they will have the same atomic number 6. they will have six protons but they will have a different mass number 12 and 14. this is due to a change in the number of neutrons we calculate the mass from this by looking at the number of protons plus the number of neutrons and an increase in neutrons from a six to eight will give us an increase in Mass they will have the same electronic structure the same number and Arrangements of electrons so they will have the same chemical properties but there are different masses means they might have different physical properties these are a couple of important definitions you will see these phrases used a lot and it is important that you know exactly what we are referring to when we say these phrases the relative molecular mass is the average mass of molecule compared to 112 the mass of one atom of carbon the relative atomic mass is the average mass of one atom compared to 1 12. the average mass of one atom of carbon you can also see these referred to as a r and m Are the mass number that we see on the periodic table is an average of all the naturally occurring isotopes of an element we can use the mass spec to work this out here we see that 20 of the naturally occur in boron has a mass of 10 and 80 of the natural occurring boron has a mass of 11. so we can do 20 times 10 plus 80 the percentage times 11 the mass divided by a hundred which is the total of the percentage this will give us 10.8 as the relative atomic mass for Boron and this is the number that you will see written on the periodic table these are some really important definitions to learn and to make sure you understand properly the first ionization energy is the energy that is required for the removal of one electron from each atom within one mole of atoms in a gaseous form to make one mole of gaseous plus one ions hydrogen and using our state symbols gas will be turned into hydrogen ion gas plus one electron sodium gaseous form is really important to get these right will be turned into sodium Plus in the gases form plus one electron the second ionization energy is the energy that is required for the removal of one electron from each ion within one mole of plus one ions in a gaseous form to make one mole of gaseous plus two ions our equations and again it is important to get the state symbols correct here gaseous helium plus will go to helium two plus plus one electron sodium plus we'll go to sodium two plus in a gaseous form plus one electron there are a number of factors that affect ionization energy the atomic radius where the larger the distance between the nucleus and the outer electrons the less the attraction will be electron shielding or electron repulsion electrons are all negative and the inner ones the inner electrons repel the outer electrons reducing the attraction nuclear charge the more protons in the nucleus the greater the attraction between it and the outer electrons if we look at Trends in successive ionization energies we can see some patterns the first seven electrons on an outer shell follow a different pattern to the two electrons on the inner shell we can see a big jump here between electrons seven and electrons eight there is a big jump in ionization energy it is always going to get harder to remove electrons and as electrons are removed the repulsion between the remaining ones is less when we are looking at Trends in ionization energies we will see that there is an increase across periods there is a sharp drop in the first ionization energy between the end of one period and the beginning of the next and these two bits of data give us the evidence for shells ionization energy can provide evidence for electron structure if we look at our graph here we have increasing atomic number here are the groups and this is the ionization energy there is a small drop in ionization energy between groups two and three for example beryllium to Boron and magnesium to aluminum this drop can be used as evidence for electron configuration if we look at beryllium and Boron the fifth electron is the first one in the 2p because a new subshell has been started the fifth electronic is easier to remove another drop can be seen between groups five and six if we look at nitrogen and oxygen with seven and eight electrons you can see that eighth electron is the first one to be paired in the 2p so the ionization energy is giving us evidence for electron configuration unfortunately the structure we have the picture that you're used to drawing of an atom is fake we need to look at the cells the subshells and the orbitals we can look at the periodic table and we can categorize things as d block s block F block or P block all based on the cell subshells and orbitals and we can draw them like this so for example if we look at calcium it has 20 electrons so we are going to start filling from the bottom each needs to be filled singly each electron and within each box they must have opposite spins two electrons three four five six seven eight nine ten eleven twelve thirteen fourteen fifteen sixteen Seventeen eighteen nineteen twenty that is the electronic structure of calcium but we can write it out a bit neater One S two two s two two P six three s two three p six four S two here we have a shell this is a sub shell and this is an orbital they are very easy to get these confused if you're not 100 clear on what is what the shapes of the atomic number tools can be looked at as spfd they can be spheres they can be dumbbells and all are present at once we start with One S and then 2s and three two P orbitals each orbital can hold two electrons so the total number of electrons in first shell is two in the second shell it is two plus six giving us eight the shorthand way of writing this literally looks at the shells and the orbitals a number of electrons so argon's 18 electrons or One S two two s two two P six three s two three p six and the shorthand for that would be argon and square brackets calcium has 20 electrons 1s2 2s2 gp6 3s2 3p64 S2 but that's a bit of a mouthful so we can say argon which covers that first bit there for S two because it's got the noble gas Arrangements the periodic table can be divided up into blocks the S block the D-Block the p-block and the F block period three goes across here on the periodic table this is an example period and similar Trends are seen in other periods for example period two and just above it as we move across the group we are going to see a decrease in atomic radius as the number of protons increases so does the positive charge in the nucleus further attracting the outermost electrons inwards if we look at the trends in first ionization energy there will be a decrease between magnesium and aluminum this is as the S shell gets full and we start filling up P orbitals there is also another drop between groups five and six June phosphora and sulfur as pairing starts repulsion increases slightly here sodium magnesium and aluminum will have strong bonds leading to high melting and boiling points silicon also has strong covalent bonds it is a giant structure it will have high melting and boiling points chlorine sulfur and phosphorus are simple covalents so they will have a weak intermolecular bonds leading to low boiling points and argon is monoatomic very low boiling points [Music] [Music] ionic bonding we need metals and non-metals and we can call this the transfer of electrons from a metal to a non-metal resulting in a positive metal ion and a negative non-metal ion between these two ions there will be an electrostatic attraction and this is what we call an ionic bond if we look at magnesium chloride magnesium has two electrons in the outer shell and chlorine has seven electrons in the outer shell magnesium will transfer one electron from its outer shell to each chlorine's outer shell we can draw this using square brackets and having the charge on the outside of these square brackets this will give us a formula mgcl2 if we compare the electronic configuration of the atoms and the ions of both magnesium and chlorine you will see that magnesium has lost these three S2 so it is bringing it down to full shelf whereas chlorine has 3p5 and it wants another one to make this the repeat six to make it more stable this formula can be confusing it is important to remember that the Magnesium the chlorine will not just be attracted to the iron it gained or lost electrons with all ions will feel this false not just single ones electrostatic traction is stronger if ions have higher charges or if they are smaller ions are atoms that have lost or gained electrons for example here we see the electronic configuration for calcium 1s2 2s2 2p6 3s2 3p64s2 now you can either remember this which you can do for the group 1 and group 2 as plus one and plus two or you can work it out group one will form plus one ions group two will form plus two ions the transition metals have a very block station State group 7 will form minus one ions and group six for minus two ions but they are all aiming for a noble gas configuration so here we have calcium's electronic structure it is here it has these electrons in the 4s2 and to get to noble gas configuration it is going to lose those electrons one at a time so it has now formed a plus one iron losing another one will give us a plus two iron and we can change the way we write the electronic structure to reflect that we can see some Trends in ionic radio as we go down a group the ionic radii will increase this is as we see an increase in the number of shells or the number of energy at levels so the outermost electrons are further away from the nucleus if we have isoelectric ions this is ions with the same electron extractor for example ions with the same electronic structure as Neon going from nitrogen 3 minus to aluminum three plus you'll see we increase the number of protons 7 8 9 10 11 12 13 protons whereas in a movie with electrons stays the same as we increase the number of protons the nucleus will have an increasing attractions to the electrons thus the radii gets smaller when we are thinking about the properties of ironic structures it is really important that we think of these as a lattice a large structure here we have sodium chloride and we write it as NaCl but it is massive it's not just one sodium and one chlorine each sodium plus iron is surrounded by six chlorine ions and each chloride ion is surrounded by six sodium ions all of these are attracted to each other and weakly attracted to the ones that are further away the attraction is lattice wide not just the one that donated or received the electrons Gio and ionic glasses will have high melting points and high boiling points this is because the strong electrostatic attraction requires a large amounts of energy to overcome that strong electrostatic attraction they will be soluble in water is polar it will interact with the ions on the outside and pull them off they will conduct electricity when molten or dissolved because the ions can move freely whereas in a solid they cannot the solids are hard and brittle as the ions are fixed in place and not free to move around covalent bonding occurs between non-metals it is the sharing of electrons between two non-metals when we are drawing it a single bond is one pair of electrons being shared a double bond is two pairs of electrons being shared so four in total a triple bond is three pairs of electrons being shared six electrons in total when we are drawing things at a level circles are optional and I will draw them from now on without circles so here we have oxygen X is from one O's from the other we have four electrons or two pairs in the middle making our double bond for nitrogen we will have six electrons in the middle or three pairs in dative covalent bonding one element gives both electrons for the bond here we have ammonia with its lone pair of electrons up here at the top a hydrogen ion has no electrons to share so when it bonds with it to form an ammonium ion all of the electrons will have been donated by this nitrogen here this hydrogen that joined it did not give any electrons to this and we draw that with an arrow here is another example in the bonding here this Bond here all of the electrons will have been provided by the nitrogen the Boron didn't provide any of these electrons so this Bond here is a dative covalent bond whereas the rest of them are traditional covalent bonds giving out nh3bcl3 you need to know the shapes of lots of different molecules or at the very least be able to work them out so here are some examples we have carbon dioxide hcn bef2 and these are linear that Bond angles are going to be 180 degrees and there are no lone pairs I'm going to be using Molly mods to show you these if you are confused about this topic I strongly suggest that you get yourself a set of Molly mods and actually sit there building these and playing with them you can see how the shape is linear how the bonding sets up and then we can draw the dot and cross diagram down here and relate it to the the stick diagram with the 180 degree Bond angle for SO3 bcl3 and alcl3 these are trigonal planar they will have Bond angles of 120 degrees and these have no lone pairs they are fat and we can see the bonding here for CH4 and nh4 plus these are tetrahedral they have Bond angles of 109.5 and there are no lone pairs when we are drawing 3D structures a line like this shows it is in plain with the paper here we have backwards and then forwards so we can have two Bonds in line with the paper one coming forwards out of the paper and one going backwards into the paper and it is much easier to see if you actually have this in your hands and you can hold it and you can align it so that you have the bonds in playing with the paper going backwards and going forwards if we just have the dot on Cross diagram over here it is really hard to see how the bonds are arranged for NH3 and clo3 these are trigonal pyramidal that Bond angles are 107 so roughly 2.5 less than that tetrahedral and they have one lone pair which is shown in pink here adding on the lone pair really helps to remind you that the bond angles are different compared to this structure here where it's a bit hard to see and you can see we have one bond in plane one forward and One backwards water H2O is a bent shape it has a bond angle of 104.5 that is 2.5 less than trigonal pyramidal and it has two lone pairs again shown here in pink it really helps to remind you if you add in the lone pairs compared to water without the lone pairs the bond angle is going to be reduced due to the valence electron Theory the lone pairs can be seen here on the dust and cross diagram pcl5 is trigonal by pyramidal it has two different Bond angles of 120 degrees and 90 degrees there are no lone pairs the bond angles are really easy to see the difference in in the 3D structure using molar mods but much harder to see when you are drawing it out in 3D way or even in the dot and cross diagram if this is something you struggle with remembering then either using One mods or flashcards because this is something that you need to be able to remember you will notice that around the phosphorus in the middle there are 10 electrons this is an expanded octet sf6 is octahedron all of the bond angles are 90 degrees and there are no lone hairs we can see this in the 3D model we can see this when we are actually holding the Molly mods in our hands that the bond angles are 90 degrees whereas on here it is much harder to visualize these being 90 degree Bond angles this is another one with an expanded octet around the central atom b-s-e-p-r theory is a bit of a mouthful but then so is valence shell electron pair repulsion Theory electrons are negative and they repel each other so the electrons on the outer shell arrange themselves so they are as far apart as possible here are some examples each of these has four pairs of electrons in the outer shell CH4 has four bonding pairs NH3 has three bonding Pairs and one lone pair H2O has two bonding Pairs and two lone pairs so while they all have four pairs of electrons in the outer shell they have different Bond angles CH4 will have a hundred and nine point five degrees in our bond angle and H3 will have 107 and H2O has 104.5 this is because lone pairs are more repulsive than bonding pairs when you increase the number of Lone pairs the bond angle decreases a lone pair loan pay is more repulsive than a lone pair bonding pair is more repulsive than a bonding pair bonding pair electronegativity is a measure of how much an element will attract electrons we have fluorine over here as the most electronegative elements as we move across a period the electronegativity will increase as the number of protons increases in a period the number of electron shells Remains the Same so the atomic radius is decreasing as the electrons are pulled in electronegativity decreases as we move down groups the increasing number of shells increases the shielding around the nucleus covalent bonding and ionic bonding are not completely different things there is a spectrum and that goes for increasing polarity when two elements are the same we can have pure covalent bonding with the electron cloud shared evenly if we have different elements in a covalent bond then the electron cloud may be shifted more towards one side and as the polarity increases the the shift of the electron cloud is going to increase as well until we get to the point where the electrons have been transferred and we have ionic bonding for lots of these partial dipoles will be set up within the bond and this can all be due to the difference in electronegativity if there is more than a two different in electronegativity then we're going to be seeing ionic bonding sharing electrons is a continuous Spectrum with covalent bonding at one end and ionic bonding at the other end if for example in hydrogen gas we have equal electronegativity between these two elements so they share electrons equally in HCL the chlorine is more electronegative so it's going to attract the electrons more and our electron cloud is going to be shifted setting up a dipole when we are talking about permanent dipole permanent dipole forces always use the full wording even though it might seem repetitive and a lot to write out in the exam this is what the examiners expect to see some molecules are made up from atoms with different electronegativities for example HCL and the electron cloud here is not going to be even it's going to be shifted more towards the chlorine which is more electronegative this is going to set up a bond polarity and we can have permanent dipoles now in a large collection of HCL molecules they will all have these dipoles and this will result in attractions between the Delta negative and the Delta positive Parts this will result in a higher melting and boiling point as the intermolecular bonding is stronger and we can see this as HCL is asymmetrical for a symmetrical molecule the forces are going to balance each other out and it will not be polar when we are looking at induced dipoles they can be referred to as dispersion forces London forces or London dispersion forces but because old-fashioned I will refer to them when I'm doing my work as induced dipoles instantaneous dipoles which is much more descriptive of what they actually are this traffic upper will occur in most things but not in ionic substances here we have a chlorine molecule and it has an evenly shared electron cloud the electrons are evenly distributed between each atom but they are always moving around the random movement means that at any point they can all be around one atom and not the other meaning a dipole will instantaneously form this will induce a dipole in a neighboring molecule the strength of these forces will depend on the number of electrons the more electrons involved the stronger it will be the shape of the molecule the more surface area that allows more contact the stronger the forces will be for example between these two linear compounds there are lots of opportunities for contact whereas in the branch isomer there are much fewer opportunities for contacts meaning the strength is going to be reduced straight isomers will have more contact points meaning the intermolecular forces will be stronger and they will have a higher boiling point hydrogen bonding is an area where we see a lot of crossover with Biology it occurs when hydrogen is bonded to either nitrogen oxygen or fluorine right over here in the very electronegative area of the periodic table between these two elements there was a large difference in electronegativity at the same time as hydrogen bonding you can have other forces occurring such as vandals but hydrogen bonding is stronger than the other intermolecular forces and this leads to a few properties they will have anomalously high boiling points and we will see in water that ice is less dense and water has a very high specific heat capacity you need to be aware on how intermolecular forces can have an effect on the properties for a long chain alkene the chain length will have an effect on the boiling point as we increase the chain length there is more opportunity for intermolecular forces to occur thus increasing the amount of energy that needs to be put in to break these thus increasing the boiling point the more branching we see on an alkene chain will decrease the opportunity to enter molecular bonds decreasing the boiling point compared to a chain of the same length that doesn't have branching alcohol groups have very electronegative oxygen in there so they are going to be polar slightly positive slightly negative leading to Stronger intermolecular bonds decreasing the boiling point they will have a low volatility compared to similar length alkanes for the hydrogen halides we can write them in order like this as you would expect to see them on the periodic table but from hydrogen chloride down to hydrogen iodide we will see an increase in boiling point due to the larger halogens an increasing strength in the intermolecular bonds however hydrogen and fluoride shows hydrogen bonding moving it down to the bottom of the list if we are going to put them in order of boiling point these are very strong intermolecular bonds and requires a lot of energy to break them when you are looking for a solvent there are a number of factors that will affect your choice of solvents mainly is it polar or nonpolar and is the thing you're trying to dissolve polar or nonpolar water is a good solvent as some ionic compounds due to the hydration of ions water can be used as a solvent alcohols due to hydrogen bonding however water is a poor solvent for some compounds due to the inability to form these hydrogen bonds to put this in context you need to aim from a similar intimate with bonds between the compound and the solvent for example the thing that is hot in chilies the capsid is a non-polar compound so if you want to remove that burning sensation from your mouth you will need a non-polar solvent to do that milk is a non-polosovan so it'll be suitable in this situation whereas water is a polar solvent so would not be suitable here we have our model of metallic bonding we have blue positive metal ions in green we have the delocalized electrons which are free and kind of floating around the electrostatic attraction between the delocalized electrons and the positive ions leads to bonding the stronger bonding will happen with smaller ions more delocalized electrons and a more positive nucleus giant metallic lattices have a few particular properties they have high melting and boiling points as their strong electrostatic attractions require lots of energy to overcome them they're insoluble in water they will conduct when solid or Molten as in both forms the electrons can freely move around they can conduct both electricity and heat they are ductile and malleable because the metal ions can slide over each other you need to know the properties of giant covalent macromolecular structures and we're going to use silicon dioxide as an example these giant covalent structures will have high melting and boiling points because the strong intramolecular bonds require large amounts of energy to overcome them they insoluble in water they are poor conductors as the electrons are not freely available to move around diamond and graphite are special examples of giant covalent structures they are both made from carbon in Diamond each carbon makes four carbon-carbon bonds whereas in graphite each carbon makes three carbon-carbon bonds diamond is very hard in a lattice structure graphite is soft as the atoms are arranged in layers that can slide over each other Diamond does not conduct electricity all the electrons are involved in bonding and there are no free electrons graphite does conduct electricity there is a free electron that is not involved in bonding meaning it can move around and conduct when we are talking about simple molecular structures covalent ones there are some common examples that you should be familiar with water H2O ammonia and H3 nitrogen gas N2 carbon dioxide CO2 and oxygen gas O2 obviously there were lots more examples but these are common ones that you should be very very familiar with simple molecular substances will have low melting and boiling points due to a weak inter molecular bonds they are insoluble in water they do not conduct there are two commonly confused because the spelling types of bonding here we are going to look at covalent intra molecular bonds which are the strong covalent bonds between atoms in a compound these are very very strong and then there are the weaker intermolecular bonds these ones are much easier to overcome and give rise to properties such as the low melting and boiling points [Music] foreign [Music] Topic in chemistry that I have the most videos on so if anything on this single slide is confusing go and check out my individual videos on this there are some rules Fox station states that we have to obey to make everything simple uncombined elements will have an oxidation state of zero the total oxidation States in a compound will add up to zero the total oxidation stays in an iron will add up to the charge on that ion group one elements will be plus one group two will be plus two hydrogen is plus one except in metal hydrides when it is minus one oxygen is minus two except in peroxides or when combined with fluorine chlorine ambromine are minus one accepting compounds with oxygen or fluorine here we have sulfuric acid hydrogen we know is plus one and we have two of those plus one times two gives us plus two in total oxygen we know is minus two we have four of those minus two times four gives us minus eight in total this adds up to zero meaning sulfur must be plus six for our carbon iron we have an overall charge of minus t oxygen we know is minus two and there are three and then minus two times three is minus sixth overall we need to get to minus two so what plus minus six makes minus two well that is plus four for the carbon in a metal hydride we know sodium in group one will be plus one this is an exception for hydrogen so hydrogen will be minus one giving us zero over all hydrogen peroxide H2O2 hydrogen is going to be plus one and we have two of those so plus one times two gives us plus two this needs to be zero overall which means the oxygen needs to in total add up to minus two meaning they need to be minus one each minus one times two gives us minus t we can look at the names and the Roman numerals and things to work out what they are copper 2 means it has a two plus and nitrate five means the nitrogen is going to be plus five and from copper two nitrate five we can work out the formula of this compound really useful if you have iron compounds we know the nitrate is going to be minus one oxygen is minus two we have three of those that gives us minus six we know nitrate is plus five giving us NO3 minus copper is plus two so we need two nitrate ions to go with the copper when we are looking at redox reactions we can use oil rig to help us remember what things are oxidation is loss of electrons reduction is gain of electrons if we have a decrease in the oxidation number it has gained electrons and been reduced an increase in the oxidation state we have lost electrons and been oxidized so chlorine on its own will have an oxidation state of zero sodium group one metal will be plus one oxygen minus two hydrogen plus one these are from our rules sodium plus one chlorine minus one we can see chlorine started off with zero and went to minus one and plus one so here we have seen a decrease in the oxidation state showing that he has gained electrons and been reduced here we see an increase in the oxidation state this has lost electrons and been oxidized this type of reaction where the same thing is oxidized and reduced in a single reaction is a disproportionation reaction and that's one of my favorite words sometimes with redox reactions we know the start point and we know the end point and we can work out the half equation that has happened again there are some rules for us to follow any oxygens are balanced with water hydrogens are balanced with hydrogen ions and any charges that are uneven are balanced with electrons here we have a reaction that we know occurs but it is not balanced adding inner four Waters will balance out the oxygens but now the hydrogens are not balanced so we need to add in eight hydrogen ions and then add in electrons to balance the charges the other half of this reaction we need to start by balancing the hydrogens because the oxygens are already charged and then add in electrons to balance out the charge then we can add these two together looking at the electrons there are not the same number of electrons and we can't just have electrons disappear so we need to change the bottom reaction multiplying it by five so these 10 electrons and the top one by two so there are ten electrons so I'm just going to write it out here with this reaction being multiplied by 2. and with the bottom reaction being multiplied by 5 we can now use algebra to start canceling things out 10 electrons on either side don't need to be written down they can cancel it out and so can sum of the hydrogens we can then write this as a complete overall reaction this may seem very complicated but with a bit of practice you'll get the hang of it no problem [Music] foreign [Music] there is structure and all the 2D periodic table and understanding it will help you immensely it is arranged by increasing atomic number not by math for example we have argon here and Krypton here argon has 18 protons and a mass of 40. there is potassium which comes after it in the periodic table has 19 protons and a mass of 39 a lower Mass which might make it seem like it's in the wrong order but it is arranged by atomic number everything within a group will have similar chemical properties they have the same number of electrons on the outer shell periods will go across the periodic table they will show a repeating trend this is periodicity they will have the same number of electron shells but an increasing number of electrons on the outer shell the structure can also help you remember how many electrons go into each shell for example in the first period period one there were two elements there are two electrons in the second period there are eight electrons and eight elements the same in the third period there are eight elements in that period in there are eight electrons in that shell for group two we are looking at the ones going down on the left hand side these are the alkaline earth metals and they would have two electrons in their outer shells this makes them all very reactive as you go down the group the atomic radius increases this is due to the increase in the number of electron shells the melting point will decrease this is due to the increased distance between the nucleus and the outermost electrons reducing the electrostatic attractions between atoms there is a decrease in the first ionization energy as the number of shells increases the shielding increases reactivity increases as you move down the group as outermost electrons are more easily lost if we have a group two metal and water we will get a group two metal hydroxide and hydrogen a group two metal plus oxygen will give us a metal oxide group 2 hydroxides become more soluble as we go down the group magnesium hydroxide is very sparingly soluble almost insoluble and it's used to treat stomach acid calcium hydroxide is used to neutralize acidic soils calcium oxide and calcium carbonate is used to remove sulfur dioxide from flu gases magnesium can be used to extract titanium titanium needs to be very pure so it can't be extracted just by carbon it is a batch process making it very expensive it also uses large amounts of electricity group 2 sulfates get less soluble As you move down the group barium sulfate is insoluble it is given to patients when they are needed to have an x-ray of soft internal organs the barrier means that when you have an x-ray you can actually see the structure of these organs flame tests are one of my favorite things to do they are really good fun but you need to learn the colors for them lithium is Crimson potassium lilac Airmen green sodium yellow calcium red and a few other rubidium red cesium orange red strontium Red Copper Blue grain to differentiate between these you would then have to go on to do further tests when were you looking at the thermal decomposition of Group 1 and group 2 carbonate and nitrites there was some specific examples you need to learn four group one carbonates lithium carbonate will decompose lithium oxide and carbon dioxide and this is similar Trend throughout all of group one as we go down you'll see an increase in stability of the carbonate and thus decomposition temperature we increase down the group lithium nitrate will decompose lithium oxide NO2 and O2 but the Lays just lithium nitrate the others in the group won't decompose as much the group two we have a similar reaction to group one which will give us a group two oxide and carbon dioxide there is also a similar pattern in the nitrates the increase in stability is due to the larger ions having a lower charge density ratio group 7 or group 17 sits over on the right hand side and these are the halogens fluorine gas is a highly reactive pale yellow gas chlorine gas is a poisonous pale green gas bromine liquid will give off poisonous fumes it is also used to test for alkenes ID is a gray solid that sublimes sublimes when something goes from a solid straight to a gas bypassing the liquid phase to give us a purple gas As you move down the group the electronegativity decreases the increasing number of shells and the increases the atomic radius reducing the ability of the nucleus to attract the electrons As you move down the group The Melting and the boiling points also increase the increasing number of electrons increases the strength of the intermolecular forces we can talk about displacement reactions involving halogens more reactive halogens can act as stronger oxidizing agents and the stronger oxidizing agents will replace weaker oxidizing agents in a reaction when this happens you are likely to see a color change for example if we have chlorine added with bromide ions we're going to get chloride ions out and bromine we're going to go from a pale yellow solution to a solution of bromine which is an orange brown color and chloride ions when we are talking about the halogens we can see a lot of Trends in their reactions they have an increasing ability as reducing agents as we move down the group sulfuric acid added to a sodium halide will generally give us a hydrogen halide gas because here the halide ions are acting as a base and the sulfuric acid is reduced fluorine and chloride are not strong enough reducing agents so here you will only see that acid-base reactions occurring with sodium bromide it is a moderately strong reducing agent so you will see the acid-base reaction as well as the redox reaction taking place iodines are a strong reducing agent so he is our overall equation but from this there are lots of other reactions going on not only acid-base reactions but redox reactions as well leading to a wide range of products not just the hydrogen iodide but iodine sulfur dioxide sulfur water and hydrogen sulfide with all of these reactions what you will get out is a white steamy fumes over the hydrogen halide gas which will turn blue litmus paper red when we are testing for halide ions you are going to need a solution of that halide line we are going to add to it silver nitrate and nitric acid we will then see a faintly colored precipitate a solid coming out silver chloride will give us a white precipitate silver bromide will give us a cream precipitate and silver iodide will give us a yellow precipitate if you look at the video you will see that the colors are very very close to each other if you are doing this in a lab as a practical it is a really good idea to have a standard set you can refer to your consider the difference between white and cream when you've got no reference is really really hard the nitric acid is there to remove any carbonate ions they will react and turn into carbon dioxide since the silver carbonate will give a false positive here if we add dilute ammonia to silver chloride then we will get a complex ion this is a colorless complex iron so the white color will disappear the same will happen to silver bromide upon the addition of concentrated ammonia where silver iodide does not react with ammonia we can start by looking at the reaction of chlorine with water chlorine in a reversible reaction with water will give us hydrochloric acid and chloric acid chlorine with the oxidation state of zero will have an oxidation state of -1 in hydrochloric acid this is a decrease in the oxidation state it has gained electrons and been reduced we can also see chlorine having oxidation of zero and going to plus one this is an increase in oxidation state it has lost electrons and been oxidized where the same thing has been reduced and oxidized in a reaction this is called a disproportonation reaction hclo is chloric one acid this is what we use in swimming pools to kill bacteria it is also used to treat drinking water be dangerous and this leads to a balance and potentially a controversy however on the balance of things the benefits of killing bacteria to provide Safe Drinking Water for people outweigh any minor potential tiny um risk of there being too much chlorine in the water chlorine can also react with cold dilute alkali a reaction with sodium hydroxide will give us sodium chloride and chloric acid and water again this is the disproportionation reaction clo minus is the chlorate one ion sodium chlorate is bleach we are just going to take a tiny segue here to talk about naming things and iron this is sodium chlorate sodium one chlorate because we know in here the oxidation state of chlorine is one but this is also sodium chlorate however sodium here has an oxidation state of one we know oxygen is -2 and there are three of them giving us minus six in total the whole thing is zero so chlorine must be plus five so this is sodium chlorate five this is one big example where using the Roman numerals is important if you see them write them down and pay attention to them if we want to test for carbonate ions or for hydrogen carbonate ions we can add an acid to our polder and a positive result will be giving off the evolution of carbon dioxide gas here we have our powder in our tube going this way we're going to add in a little bit of acid and then we're going to collect the gas down here the gas will move into lime water and you can see it has gone cloudy giving us a positive result for carbon dioxide when we are testing for sulfate ions we need something we sulfate in hydrochloric acid and barium chloride acidified barium chloride with hydrochloric acid can be used to test for sulfate ions it will give a very nice positive result of going white a white precipitate will be formed if you are doing a single test tube set of reactions in the order of tests is important the chloride ions in barium chloride will give you a false positive on a test for halide ions so the order of these reactions is important a favorite practical in allele chemistry and a favorite exam question is being given a mystery solution or a mystery white powder and being asked to use the knowledge of tests to work out what it is so here is a summary of all the test tube tests for cations and anions there are also the flame tests we can add on to this as well when you are testing the halide ions we add nitric acid and silver nitrate chloride ions will give us a white precipitate bromide ions will give us a cream precipitate iodide ions will give us a yellow precipitate they can go colorless if we add dilute or concentrated ammonia to test the sulfate ions we will add barium chloride and dilute hydrochloric acid a positive result will be a white precipitate for carbonate ions we will add hydrochloric acid we will see a gas given off this will be carbon dioxide gas to confirm its carbon dioxide gas we need to see the gas turning lime water cloudy hydroxide ions can be added to ammonium chloride and it will start to be very smelly leading on to the invest test for ammonium ions and we will get smelly ammonia released to confirm that it is ammonia what we will be looking for is turning damp red litmus paper blue to test for group 2 ions we're going to be adding two different things sodium hydroxide and sulfuric acid both dilute and concentrated barium either the dilute or concentrated sodium hydroxide will give us a colorless solution either dilute or concentrated sulfuric acid will give us a white precipitate for calcium ions the addition of either dilute or concentrated sulfuric acid or sodium hydroxide will give us a slight white precipitate magnesium ions with dilute sodium hydroxide will give us a slight white precipitate whereas concentrated sodium hydroxide will give us a white precipitate dilute sulfuric acid will give a slight white precipitate and concentrated sulfuric acid will give us a colorless solution strontium and either dilute or concentrated sodium hydroxide will give a slight white precipitate whereas the addition of dilute or concentrated sulfuric acid will give a white precipitate if you are asked to do all of these in a single test tube then the order of reactions does matter for example here we are adding barium chloride so if you want to test for sulfate ions and halide ions in a single sample in a single test tube if you do this one first you are adding in chloride ion and you'll get a false positive for halide ions the order does matter foreign [Music] [Music] there are a few random bits you need to know in chemistry that will pop up all over the place but don't fit into any particular topic when a question mentions significant figures give your answer to the same number of significant figures that is the smallest number of significant figures in the question any change in that will affect the resolution here we have two examples this one and this one are two different numbers of significant figures however this answer here is not the correct answer because it is not to the right number of significant figures this has a smallest number of significant figures so this is what we need to mirror in the answer otherwise we are changing the resolution of the answer we cannot give that to that accuracy because this number is to not that accuracy if we think back to our GCSE math 0.02 can have a wide range of numbers if we're looking at upper and lower bounds you will frequently be asked to convert between units especially for temperature to go from Celsius to Kelvin you add 273 Kelvin to Celsius minus 273. centimeters cubed decimeters cubed divided by a thousand a centimeters cubed to meters cubed divide by a million decimeters cubed to meters cubed divided by a thousand and the other way around decimeters cubed centimeters cubed times five thousand meters cubed centimeters cubed times by a million and then meters cubed to decimeters cubed times by a thousand for concentration calculations if you want to change from moles to decimeters cubed to grams per decimeters cubed you need to change it by the Mr of the substance a mole is the amount of a substance that contains the same number of particles as the number of carbon 12 atoms in 12 grams Avogadro's number is the number of particles in a mole 6.02 times 20 to 23 quite a lot fortunately you will get given this value in the exam you don't need to remember it you do however need to remember some equations which we can use moles in moles is equal to mass over Mr you might also seeing this written as n equals m over m r mass is in grams and Mr is in grams per mole the number of particles is equal to the number of moles times Avogadro's number so we could get a question such as this calculate the number of particles in seven grams of gold we would need to do moles equals seven that is the mass divided by the m r of gold which you can look up on your periodic table so we have calculated the number of moles we would then need to take the number of moles and times it by Avogadro's number to give us 2.14 times 10 to the 22 particles and it is important we look at the number of significant figures here here I wrote down all of the significant figures on my calculator here I have gone to the same resolution as Avogadro's number there are lots of equations where moles come up and they can be used to switch from one equation to another equation as an intermediary so this is just a summary slide of all of the equations that use moles we have the ideal gas law so PV equals n r t the concentration of solution so n for moles equals concentration times volume the equation for Mass whether moles is mass divided by m r and looking at the number of particles where it is moles times the Avogadro number the constant so you can go from one place to another place using the ratio of moles so we can go from the number of particles to the volume this way or we can go from volume over here to m r there is a difference between the molecular and the empirical formula the molecular formula will give us the exact number and identity of each element in the compound the empirical formula will give us the simplest whole number ratio of the elements within that compound in an exam question we might see something like this a compound of phosphorus 5 oxide has an mr284 and is made from 62 grams of phosphorus and 80 grams of oxygen calculate the empirical and molecular formula this is how I would need to set it out very strictly if you do like this we shouldn't come up across any problems we're going to start by making a table with your element the number in the question divided by the AR equals the whatever number it equals divided by the lowest number and this is going to give us the ratio if we follow this format here we should be fine so the elements in the question phosphorus oxygen find the numbers in the question it doesn't matter whether it's grams or percentages or anything find the numbers in the question and write them down in the appropriate place we then need to divide that by the AR and you can get the number from your periodic table and whatever your periodic table says use the number that it gives on there don't use the whole number or any other number use the number it gives on your periodic table there's a number here and then we need to divide it by the lowest number the lowest number out of these two is two so we're going to divide both of these by two D divided by two equals one five divided by 2 equals 2.5 because 2.5 isn't a whole number we need to multiply it by 2 to get the ratio so p205 this is the empirical formula now we know the Mr of this is 284. so we need to start by working out the Mr of the empirical formula that worked out p2o5 so this is our Mass from the periodic table and these are the numbers that we have in the empirical formula this gives us 142. we take the m r from the question 284 divided by 142 it means there are going to be two lots of the empirical formula in the molecular formula thus we are going to have p4010 and you can quickly check that that does add up to the white amount balancing equations is an incredibly important skill you will use it in nearly every single lesson it is definitely worth taking the time to practice this at a level you have to include state symbols in your equations even if they didn't explicitly ask for it in the question it is expected so solid S gas g l for liquid AQ for atheists you have to include these so your first step is just going to be drawing circles around the equations we cannot change anything in these bubbles but we can change the number of Bubbles and then list what you've got in the same order on both sides it just makes things easier on the left hand side we have three hydrogens one iodine one sulfur and four oxygens over on the right hand side we have four hydrogens two iodines one sulfur and one oxygen so we can see straight away that there is a difference here the easiest thing to do is to start with increasing the number of oxygens because they're both only in one place on each side and then redo your numbers so we have 10 hydrogens two iodines one sulfur and four oxygens we've fixed the oxygens we can now move over to the left hand side and look at something else now we could increase the iodines next but that is just gonna cause US problems later on so we're going to look at the hydrogens we have two hydrogens in sulfuric acid and eight in the hydrogen iodide giving us 10 in total aidenes one sulfur four oxygens so now our hydrogens our sulfurs and our oxygens are balanced we can just put two influencer ID and balance that as well this is a vitally important skill you need to practice this so you can do it quickly after you've done it you're working out always write the equation out in full so this clear to The Examiner exactly which bit they should be looking at when we are looking at the volume of gases it is important to remember that one mole of gas under room temperature and room pressure will occupy 24 decimeters cubed and we can use P1 V1 equals p2v2 pressure and volume at a constant temperature to determine volumes so determine the volume of one mole of gas would occupy if the pressure was doubled to T atmospheres at room temperature because we know the pressure was doubled P2 is two atmospheres which means P1 is half of two atmospheres giving us one atmosphere we know from laws that this is 24 decimeters cubed volume the second pressure is two so 24 equals 2v2 we can then use algebra to move the 2 over to the left hand side giving us 12 as the new volume we use the ideal gas law a lot and you need to remember this equation PV equals n r t p stands for pressure V is the volume n is the number of moles R is the gas constant you need to remember it's gas constant B you don't need to remember the value for it you get given that in the exam and T is the temperature pressure is measured in pascals volume is measured in meters cubed the gas constant is 8.31 joules per mole per Kelvin temperature is measured in kelvins now the conversions for this is often where people go wrong determine the mass of a 500 centimeter Cube sample of hydrochloric gas when at 20 degrees C under 150 kilopascals pressure for any calculation question the first thing I like to do is to highlight all the numbers and then pull them out of the question and write them down separately some questions can be very wordy and having the numbers that you need right there in front of you can make things a lot easier we can then see which ones of these are a non-standard units and convert them into standard units before we start so pressure was given in kilopascals and we need it in pascals so 150 kilopascals is going to be 150 000 pascals volume was given to us in centimeters cubed and we need it in meters cubed to go from centimeters cubed to meters cubed we need to divide by 1 million and temperature was given to us in degrees Celsius and we need it in to kill them and to do that we need to add on 273 giving us two nine three Kelvin we can then use the equation and the first thing we're going to find out is the number of moles so we can rearrange the equation to give us n equals PV over RT and then plug in the numbers once we have the number of moles we can use Mass equals moles times Mr to give us the number of grams to give us the mass that the question is looking for the first of your core practicals is to measure the molar volume of a gas here we are adding a known mass of calcium carbonate to an acid and recording the volume of carbon dioxide is given off that we are collecting in these measuring tubes you're going to need to do this with several different volumes of calcium carbonate once you've done that for several different masses of calcium carbonate and recorded the volume of gas release you can draw a graph you can then work out how much carbon dioxide will be released from the equation as one mole of calcium carbonate will release one mole of carbon dioxide if you do know I have that exact value on the graph you can then work it out for the concentration of solutions we need to know that concentration equals mass over volume concentration is measured in grams per decimeter cubed mass in grams and volumes in decimeters Cube for Ionic Solutions you need to remember that there are two different ions in there for example calcium two plus and two chlorine ions here so from one mole of calcium chloride we will get one mole of calcium ions but two moles of chloride ions the same is true for acidic Solutions and this is important for titration calculations sulfuric acid for every one mole of sulfuric acid we will end up with two moles of hydrogen ions making a standard solution and then doing a titration these are skills that come up over and over again when you're making a standard solution the thing that is most important is that you want to be very very accurate so you can see in this video here I've weighed out my powder into the Weber and then I'm continuously washing the weibo to get all of the powder off the weibo and then it is in the beaker while mixing it and what I'm going to do is to wash the sides of the beaker to get all of the powder that might be on the sides of the beaker actually in to the flask we're going to keep washing because we need to fill this flask all the way up to 500 mL or 250 mils or whatever it is just wash everything out so now I'm washing the flask out and then we should swirl it a little bit and then make it up to the volume that we're looking for you can see I'm washing the neck in case any powder has got down the neck of the flask we're going to invert it a little bit to make sure it is properly mixed and wash the neck again the aim is to get all of the powder that you have weighed out in to solution now you'll see the little line on that you want to go up to that line but do not go over it so at the end we can go drop by drop the Temptation if you go over it is to remove some do not remove any from your standard solution if you go over you're gonna have to start again if you go over and then remove some you are removing some of the powder that you have weighed out so you are changing the concentration of the solution so if you go over do not remove any that would be one source of error K is key when you are doing a titration you need to take care on every single step because there are lots of different possible sources of error you might have a bit of confusion reading the numbers on the buret or you might have a bubble in your tap you might go over the end point by a couple of drops you need to do this super super slowly a good thing to do is to always start with a rough tighter so you know roughly the number that you're looking for so you can go fast for the first part and then slow down we are looking for concordant results and when I say concordant results I mean results that are in within 0.10 centimeters cubed of each other you would be aiming for three concordant results and when you get practiced at this you can do a rough Tighter and then three actual titles and your three actual titers will be concordant you can record all of your results to two decimal places when you see a titration calculation there are some very specific things I want you to do first of all first of all is highlight all the information in the text and then we're going to pull out that information so it is all in one place when you need it and you don't have to Tool through all of the information again when you're trying to answer the question so we need the volume of acid the concentration of acid the volume of The Alkali and the concentration of The Alkali concentration of The Alkali is X that's what we're going to put in there and the rest of it we're going to grab from the question and write it down so it's easy for us to find later with all of the information in one place we don't have to dive back into the complicated text to find out what we need when we need it we're also just going to note down the equations that we are using to help us so that we don't get confused trying to think of things trying to remember when we are deep in answering the question the first thing you need to do for this is to write a balanced equation so we have sulfuric acid and sodium hydroxide making our sodium sulfate salt and water step one in any titration calculation is to find your moles of known here we know the acid so that's what we're going to be finding the moles of and we're going to be using the numbers that we've pulled out earlier so moles equals concentration times volume we need to adjust the volume so that we are working in liters because this is in centimeters cubed giving us 0.0025 moles of hydrogen ions we then need to find the ratio of hydrogen ions to hydroxide ions and because of the ratio in the balanced equation it's doubled we can then use moles concentration volume we have the number of moles we have the volume so we can find the concentration of The Alkali the question is asking for the answer in grams per decimeter cubed so we need to convert our units first thing we need to do is to find out the mass of sodium hydroxide using moles times m r we can get our answer in grams for decimated cubes whenever you have anything reoccurring 1.6 reoccurring for example always use your calculator value otherwise you're going to be introducing rounding errors into your answers depending on the equipment that you use for something or the size of the equipment that you use for something there'll be different levels of uncertainty involved for example here we have three measuring cylinders which can measure different volumes so the first one the divisions are four centimeters cubed and the reading on it is a hundred centimeters cubed the second one the divisions are two centimeters cubed and the reading is 70 centimeters cubed for the third one the divisions are 0.2 centimeters cubed and the reading is six centimeters cubed the uncertainty involved in each is half a division so for the first one the reading could be plus of minus two centimeters cubed plus minus one centimeter cubed plus minus 0.1 centimeters cubed however any experiment is more than likely to involve several stages several measurings thus several levels of uncertainty to work out the percentage uncertainty it is plus or minus the uncertainty divided by the measurement times 100. the total percentage uncertainty in any experiment is the sum of all the individuals often in a reaction there is a difference between how much we think we're going to make and how much you actually get this is called the percentage yield we can calculate percentage yields by dividing the actual yields by the theoretical yields and multiplying it by a hundred so we get a percentage it was expected the reaction would give 14 grams instead it gave 5.2 grams calculate the percentage yield 5.2 the actual yield divided by 14 the theoretical yield times 100 gives us 37 as the percentage yield there are a number of reasons for a lower than expected percentage yield reactions do not always go to completion some of the reactants just don't react and are left over at the end there could be a loss of product it could be difficult to collect it could be hard to collect it safely it could be difficult to separate it from the unreacted reactants there could be side reactions occurring that were not predicted at the beginning giving you products but not the product that you want atom economy when you do a reaction you will end up with a certain mass of stuff at the end the mass of the products but not all of that is actually useful some of it is going to be waste and some of it is going to be useful so we can calculate percentage atom economy by looking at the mass of useful products divided by the mass of all reactants and because this percentage we times it by a thousand atom economy can be improved in two different ways we can find an alternative reaction pathway that has less waste or by finding uses for the waste products often you'll be given a word equation and expected to construct the balanced equation from that and knowing your general sort equations is a really important part of being able to do that a metal plus an acid will give us salt plus hydrogen a metal oxide plus acid will give us salt plus water a metal hydroxide plus acid will give us salt plus water a metal carbonate plus added will give us salt plus water plus carbon dioxide using Hydrochloric acid will give you chloride salts using sulfuric acid will give you metal sulfate salts and using nitric acid will give you metal nitrate salts at the heart of this is the neutralization equation which comes up a lot aqueous hydrogen ions plus aqueous hydroxide ions in a reversible reaction with water if you struggle to work out the products of a reaction it helps to break it down into ions so sodium hydroxide plus hydrochloric acid the sodium ions and the chloride ions will give you the salt sodium chloride and then the water will be formed from the hydroxide ions and the hydrogen ions it is expected that you know the formula of ionic compounds or at the very least can work them out from the periodic table ions of Group 1 elements are going to form plus one ions ions of group two elements are going to form plus two ions ions of group six elements are going to form minus two ions and ions of group seven elements are going to form minus one ions however the transition metals have variable oxidation States and you also need to know the formula and the charges on some of the complex ions we can look at a couple of examples of forming the formulas Violet compounds from knowing what the individual ions are sodium carbonate sodium is going to have a plus one charge and carbonate has a minus two charge since overall sodium carbonate has no charge we need the positive of the negative charges to balance each other else the three the Roman 3 in Iron three means it has a plus three charge and sulfate has a minus two charge now iron safe rate over all has zero charge so we can look at this by doing three times two and just kind of swapping them over so because iron has a three plus charge we're going to need two of these to get to plus six and because sulfate has a minus two charge we're going to need three of those to get to minus six giving us the formula of fe2 Open brackets so4 close brackets three [Music] [Music] organic chemistry nomenclature is very important because it tells you what an exam question is asking you for the empirical formula is the simplest whole number ratio of each element within a compound the molecular formula is the real number of atoms of each individual element in a compound the general formula will be the formula that covers a homologous series the structural formula is the minimum level of detail you need to draw something so ch3 ch2 ch3 the displayed formula will show all of the bonds whereas the skeletal formula will not show any carbons or any hydrogens just skeletons and other functional groups when drawing you can draw the displayed the skeletal or the structural formula if one of these is mentioned in a question please give the examiner what they are looking for the displayed formula will show all of the bonds here we can see each of the carbons and these will become points on the skeletal formula and we are going to draw in the backbone between these carbons no hydrogens and then we need to draw on the functional group as well for our structural formula we take it bit by bit here we have a ch3 here is a ch2 here is another ch2 and then an o h this is propan one all here is another example again picking out the carbons as the turning points in our skeletal formula drawing those points those dots on as our backbone and connecting them up you have a ch3 a CH and a ch2 and this is propane slightly more complicated one now again we are going to start by using our carbons as our points as our backbone over our skeletal formula drawing on the backbone and the functional group or the structural formula we have a ch3 ch2 ch2 and a c o o h giving us butanoic acid lots of the words we use in organic chemistry might be new to you so it's worth spending a bit of time going over them a homologous series is a group of compounds that has these same functional group but each one will have a different length carbon chain a functional group is the group of atoms within a compound that give the compound its properties an alcohol group will be potentially a side chain with the formula cnh2m plus one an aliphatic compound we straight change branched on non-aromatic rings based around hydrogen and carbon alicyclic compounds will have non-ameriatic rings whereas aromatic compounds will contain Benzene rings saturated compound will only have single bonds between carbons whereas unsaturated compounds will have double bond between carbons an easy way to remember this is that when we have saturated things they are alkanes and there is one e in there so they are single bonds and unsaturated will be alkenes there are two is in there that is a double bond it's crude but it works when we are naming things in organic chemistry we use the IUPAC rules I have a large number of very long videos going over lots and lots and lots of examples of naming things in organic chemistry this remember is just a brief summary for Eurovision if you're confused by any of these bits go and watch one of the longer videos these are the rules we follow whenever we are naming something in organic chemistry find the longest carbon chain this is not always a straight chain this is the backbone for the naming identify all of the side branches Circle and identify all of the functional groups number the chain so the branch with the highest priority functional group has the lowest number we're going to use di try for more than one of a branch of the same type branches always go in alphabetical order there are commas in between numbers hyphens separate numbers from letters and there are no gaps between names of things different functional groups will change the suffix of the word so we have a n e for alkanes and alkene will have a double bond in it somewhere and it has an e n e Bute two in here and alcohol will have an o h functional group this one for carbons butane two o alkanes with halogen attached to them in front so this is two chlorobutane aldehydes will have this functional group on the end and this is butanal an isomer of theirs ketones will have the functional group in the middle and this is Butte and own carboxylic acids have their functional group on the end and this is butanamic acid Esters will have their functional group in the middle and we have four on each side so this is butyl butanoate there is a priority list of functional groups starting with carboxylic acids moving down to aldehydes ketones alcohols alkenes and halogens so in a compound that has more than one functional group this is the order that we need to go through when we are naming things here we have an alkene we have a methyl and we have some halogens using our rules exactly the same way that we have before the first thing we need to do is to find the longest carbon-carbon chain one two three four five six so this is hex we have a methyl group a chloro group a bromo group and trial and error numbering to see which ones give the highest priority group the lowest numbers do it from both sides and we can see that the blue numbering will give the double bond the highest priority group the lowest numbering so this is what we need to go with this blue numbering here now the chances of me fitting this name on one line are slim but remember this is all one long name putting it all together we have branch is going in alphabetical not numerical order so B becomes 4C so we have five bromo four chloro 5 methyl hex 2 in another example that looks horrible but once you follow the rules is absolutely fine structural isomers will have the same formula the same number of each element of each atom but will have a different structure here we have some examples here is four carbons in a straight chain and here is four carbons but with a branch so in a straight chain we have butane but when it is Branched we have methyl propane they both have a formula c4h10 but as you can see from the Molly mods and as you can see from the displayed formula they are different Arrangements these are chain isomers and they will have different physical properties for example branching will give different boiling points here we have propanol but the alcohol group is in a different position we have profound one oil and we have propan 2 all one of them will have the oh group on the end whereas the other has the oh group in the middle these are primary and secondary alcohols these are position isomers they will have the same functional group but in a different place here we have an oxygen added in but we can see that here it is on the end whereas in the isomer it's not on the end it's in the middle this gives us a different functional group these are functional group isomers because butane L is an aldehyde and butane own is a ketone the same formula but the different position in this place of double bonded oxygen stereo isomers are also called EZ isomers here we have Bute two in with the double bond in the middle but you can see they look different these are not exactly the same even though the displayed formula would indicate to you that they are we can see these are going up and down and these are going up they have a different arrangement in space this is because there is no free rotation around the double bond and this only works if there are different groups here they are drawn out a little bit clearer so you can see here the ch3 group are on opposite sides and here the ch3 group are on the same side where they are on opposite sides this gives us the E isomers this is e but two in and when they're on the same side it is the Z ice masses gives us Z but two in as a point but one name doesn't show e z or some reason because on one of the carbons on the double bond these two groups are the same so it doesn't matter which way up or round it goes it will be the same these are the priority rules for deciding whether something is an e-i smart or a z isomer your e-isomers would have priority groups on different sides of the double bond your Z isomers would have priority groups on the same or the same oh I know I'm sorry side of the double bond priority is determined by the atomic number here are some lovely examples we have two chlorines on a butene so giving us two three dye chlorobutane chlorine at 17 has a higher atomic number than carbon so it has the higher priority here are the chlorines highlighted and you can see in this one they are when the same same side so this is the Z isomer and this one is the e isomer isomers can differ in polarity because the E isomer is symmetrical it has polar bits on both sides so it cancels each other out overall and is not polar whereas here the Z isomite is polar because the polar bits are both on the same side giving it a polarity they can also differ in their boiling points due to the differences in the intermolecular forces made you need to know your prefixes up to 10 this may seem like a lot but once you've got the walls down it's not a problem at all one is me two is E3 is pro four is Butte five is Pence six is hex seven is hept eight is Ox nine is nine and ten is dead we're going to use these rules to name things starting with the basics alkanes you're going to see me use this template a lot it is super helpful for organic chemistry and you can download it for free from my website names have different parts to them first names surnames prefixes and suffixes so here is something that I've drawn the first thing we do is to identify the longest carbon chain highlighted here in green and then counts the number of carbons in that chain so four is going to have a Butte prefix here is a side branch there's only one of them this is a methyl Side branch putting it all together this will be methyl butane the Ain tells us it has single Bonds in it here's something else a little bit more complicated starts in the same way and find the longest carbon chain count things so we know that we have six carbons giving us hex as the prefix and aim because it's an alkane here are our two branches one of them has one carbon in and one of them has two carbons in giving us the methyl and an ethyl we need to start numbering from over here because that is where we are going to get the lowest numbers possible so this gives us three ethyl and two methyl we can now start to put the name together because they need to be in alphabetical order not in number order we have three ethyl two methyl hexane now with practice all of these skills will come very naturally to you so we can start to look at horrible looking things and name them the first thing we need to do as always is highlight quite the longest carbon chain and here we can see it's not straight at all please feel free to use color pencils add highlighters in this sort of question seven carbons in the chain gives us heptane now we need to find all those branches there are four different branches on this one they all have one carbons in so they are all methyls now we need to look at the numbering so we end up with the lowest possible numbering I'm just going to draw it on from both directions trial and error to see which way will give me the lowest possible numbers and we can see this is an example where it actually doesn't matter which three rounds we get it these are going to be three four four five now we have all our information we can start to build the name up now this is one word even if I can't fit it all on one line we have four methyl groups stage three four four five Dash Tetra methyl heptane one word which I did manage to fit on one line fraxum distillation is a way of separating out crude oil and crude Oiler is a mixture of different length hydrocarbons and a hydrocarbon is something that only has hydrogen and carbon in it the length will influence the properties so longer chains will have an increasing number of Van developer forces holding them together and the more intermolecular forces the higher the energy needed to separate them thus the higher the boiling point very briefly the oil is heated it goes into the column and it separates out a different boiling or condensing points thus separating them by chain length with short ones at the top and long ones at the bottom crude oil is a mixture of different length hydrocarbon chains the short ones are very useful and are used a lot but are not enough of them is produced from fractional distillation the long ones are not used very much and lots of them are left over at the end cracking is a way of turning long chain hydrocarbons into shorter chain hydrocarbons alkanes and alkenes when we are talking about alkanes and alkenes alkanes are saturated because they have all single bonds and alkenes are unsaturated because they would have double bonds we can have catalytic cracking which uses a zeolite catalyst it's done at 450 degrees C and a moderate pressure just above one atmosphere this is more efficient as it uses less energy lower temperatures and lower pressures it will give us branch and cyclic alkanes and aromatics such as benzene thermal cracking is done at higher temperatures 400 to 900 Degrees c a higher pressure of 7 000 kilopascals and will give us lots of double bonded alkenes carbon and hydrogen in and they only have single bonds a really stupid way for you to remember the difference between alkanes and alkenes is that alkanes has one e in it whereas an alkene have two e's in it and has a double bond these are Sigma bonds here and they have free rotation around the bonds so even though I have drawn this in a linear fashion it is not a linear fashion around each of the carbons we have a tetrahedral arrangement of other atoms the bond angle will be 109.5 around age and every carbon so it is not a straight line at all it is all bendy this is the arrangement that means the electrons can be as far apart from each other in space this is the electron function Theory when you have long chains of hydrocarbons straight chains can make lots of bonds lots of London lots of forces between each chain the longer the chains the more bonds they can make meaning is going to have a higher boiling any branching that we can see in long chain hydrocarbons is going to reduce the number of points where forces can be made which means things like branching will decrease the boiling points when we are combusting alkanes we are using them as a fuel and they have a wide range of uses as a fuel gas for heating an example in your Bunsen burners Petrol in vehicles kerosene for example in airplanes even through to the simple wax you have in candles complete combustion is done with lots and excess of oxygen the alkane plus oxygen will give us carbon dioxide and water incomplete combustion is in a limited supply of oxygen and the range of products come out of this are much wider and variable depending on the conditions again we will get carbon dioxide and water in addition we will get carbon monoxide and carbon there is a lot of pollution from combustion carbon dioxide contribute to climate change as does water vapor as they are both greenhouse gases carbon monoxide is a toxic gas that can lead to death carbon is soot that can lead to global dimming and atmosphere pollution leading to breathing difficulties sulfur dioxide can lead to acid rain and there is nitrous oxides can be toxic gases and lead to acid rain unburned hydrocarbons are another pollutant in Industry sulfur dioxide can be removed from flu gases by reacting it with calcium oxide which will neutralize it in vehicles can remove carbon monoxide and nitrogen oxides turning them into nitrogen gas and carbon dioxide still pollutants but not as bad and they can take the unburnt hydrocarbons and react them with nitrogen oxides giving us safer products the calculating converter will have a honeychrome structure it will have platinum palladian and rhodium as a catalyst we are going to start to be drawing lots of reaction mechanisms it is important you understand what all of the different things mean if you want to get full marks on an exam question you need to have really careful drawing of arrows a DOT is an unpaired electron a fish hook Arrow a half Arrow shows the movement of one electron a double-headed Arrow will show the movement of two electrons for example in homolytic fission we will get electrons moving from the middle one to each chlorine giving us two chlorine radicals in heterolytic vision and remember homo is the same so both the products at the end will be the same hetero is different so the products at the end will be different here we have both electrons going in One Direction so we will get a plus iron and a minus ion here we can see the breaking of a covalent bond with the arrow starting at the bond and then the electrons going somewhere else you always have to be very careful with where your arrow starts and where they actually end up they start here and they will go here when you're writing on charges make sure they're next to the thing they are actually on and when you're doing formational Bond the curly Arrow starts at the electron that will be in the bond the chlorineation of alkanes can be looked at by adding chlorine to methane for this UV light is needed however Beyond this reaction here it is actually much more complicated than it seems we're going to look at the process of free radical substitution Step One is initiation step where with UV light chlorine cl2 will produce two chlorine radicals with this free unpaired electron this is the process of homolytic fission where one electron from the bond will go evenly to each chlorine step two is a propagation reaction methane will react with those chlorine radicals to give a ch3 radical which will then go on to the next step of the reaction notice here how I've drawn the radical on the carbon because that's where the electron actually is this ch3 radical will react with cl2 to give us chlorine methane and then give us back our chlorine radical this is why it's a propagation reaction because we start at n with the same thing the electron from the chlorine radical will go in to this Bond here and the electron here will move over then this radical we're then going to attack this Bond and the electron move other giving us back another radical step three is determination of this where we will get two radicals reacting together to form something that is not a radical these can be a range of different termination steps following on from a range of different propagation steps this one here is the one that is more likely to be shown this can then go back to the beginning in a chain reaction whereas this is a minor waste product if we look at our original equation the actual products are the ones that are produced in the propagation step and our reactants are used in the initiation and in propagation bonding and reactivity of alkenes is fascinating they are unsaturated they will have a double bond so an alkene has that double e in there to help you remember the position of double bond doesn't always stay the same here we have Bute in but two different versions of butane depending on where the carbon carbon double bond is around Karma's in the double bond we would have a planar geometry whereas around other carbons in the compound we will have a tetrahedral geometry these two examples here give us Bute one in and Bute two in this double bond in the middle is an area of high electron density electrophilic addition reactions are going to happen in this type of reaction because of the area of high electron density in a double bond we have two types of bond we have Sigma bonds and we have Pi bonds lots of electrons all in one place there are a few electrophilic addition reactions that you need to know you need to know the reactants you need to know the products you need to know any catalysts or conditions so for addition of a hydrogen you might know this is hydrogenation with a nickel catalyst this is going to be a solid catalyst so the alkene is going to be absorbed onto the surface of the Catalyst that is where the reaction is going to take place the two hydrogens in H2 we each get added to the spaces across the double bond we can have the addition of halogens and for this one we have a reaction mechanism the electrons in the electron dense region will induce a dipole in bromine we will get an intermediate carbocation giving us a negative bromide ion which will then go on to attack the carbocation this is the test for an alkene decolorizing of bromine water this will give us a DI Halo alkane at the end or to be more specific one two dibromoethane we can have the addition of hydrogen halides in a very similar reaction mechanism so if you learn one learn it well you should be able to apply it to everything else the electron dense region will attack the Delta negative Shifting the electrons to the bromine again we will get a carbocation and a negative bromide ion the electrons will move towards the positive carbocation and we will end up with hydrogen and bromine being added across the double bond giving us a Halo alkane we can have the addition of steam in the presence of an acid Catalyst and this is generally going to be phosphoric acid catalyst the electron dense region will interact with the hydrogen on one of the arms of phosphoric acid giving us an intermediate carbocation as it has taken one of the hydrogens and this relieves the phosphoric acid with a negative charge phosphoric acid is the Catalyst for this so we need some water which will get added onto the carbocation giving us a second intermediate this carbocation will see the breaking of one of the bonds between hydrogen and the oxygen and then this hydrogen will go into reform the phosphoric acid Catalyst and we would end up with an alcohol and our Catalyst at the end we're going to react alkenes with potassium manganate in acidic conditions we would see the breaking of that double bond there and then we would get a diode formed as oxygen as hydroxide group is added across that double bond we don't have to write out the whole thing because the potassium manganate is acting as an oxidizing agent and we can just write that as an o in square brackets working out the major and minor products in a reaction is also known as my coffin cough syrol here we have an asymmetrical alkene which we're going to add hydrogen bromide to now where the hydrogen goes and where the bromine goes we don't yet know it can go onto either carbon depending on the position of the braving we will get to different compounds either one bromobutane or two bromobutane and if you're an industry and you want to make a particular thing then this is important here we have our hydrogen bromide which is a polar molecule the electron dense region of the double bond will be attracted to the Delta positive hydrogen and the electrons in the bond will move down to the bromine a hydrogen will be added on and we'll get a carbocation the negative bromide ion will be attracted to the positive charge on the carbocation added to the carbon with the fewest hydrogens on it so this will be the minor product this carbon here has fewer carbons than this carbon here so the electrophile will preferentially add on to this carbon making two bromobutane a major product the major products will come from the more stable carbocation in terms of stability primary are the weakest whereas tertiary carbocations are the most stable here we have a secondary Karma cation whereas this one is a primary carbocation addition polymers can occur with alkenes here we have chloroethine and to draw it as a polymer we need to extend the bonds outside of square brackets put on square brackets and put an n on there to show is a repeating unit this will then become polychloroethane it is very long to draw out which is why we generally don't draw the whole thing they will have multiple repeating units of chloroethine in there polychloroethine is also known as PVC and it is incredibly useful biodegradable polymers are a brilliant alternative to sending polymath to landfill polyesters and polyamides are made by condensation reactions thus they can be broken down by hydrolysis reactions this makes them biodegradable polyalkenes from addition polymerization are not biodegradable they cannot be broken down by hydrolysis reactions these must be disposed of in one of the following ways with recycling the advantage is that it reduces the need for the finite raw materials lots of polyalkenes are made from crude oil and this finite and there's lots of other uses when they're recycled they can be melted down and made into other things and reshaped however the disadvantages is that they need to be collected and sorted out since each type has to be melted down has to be recycled with its own type of plastic you cannot mix different types of plastic when you were recycling trying to make something new things can be sent to landfill and I will admit I did struggle to find an advantage for sending things to landfill but it is a very commonplace thing to do and we have the systems in place for it already it's pretty easy however we are running out of space to put stuff our landfills are filling up rapidly things that are already in there take years and years to break down there is also the issue of pollution visual pollution air pollution from the degradation of things and mixing of products polymers can be used as a fuel which can be used to generate electricity and energy the disadvantage of this is the toxic air pollution that comes from this initial difference between the different halogenic alkanes we have a primary and the carbon that is attached to the halogen is also attached to other hydrogens and this massive complicated compound which you should be able to name has both secondary and tertiary halogens in here is the secondary one because the carbon that it is attached to is attached to only one hydrogen the fluorine here in pink is a tertiary because the carbonate is attached to is attached to no hydrogens the primary secondary and tertiary is very similar to alcohols if we're looking at the prefixes we can go bromo chloro or fluoro during hydrolysis reactions which is one of your core practicals primary halogenic alkanes will react slowly to form that precipitate whereas tertiary will react quickly to form precipitate when we're looking at the hydrolysis of haloalkanes we are looking at nucleophilic substitution reactions here we have our bromo alkane and here we have our negative hydroxide ion bromine is more electronegative so we're going to get a partial charge on here where we mean it's going to be slightly negative carbon slightly positive so the negative charge on the hydroxide could go in and be attracted to the slightly positive carbon this will give us a substitution reaction as the electrons move over towards the bromine that will separate off and we will end up with an alcohol and a bromide iron this can be done at room temperature but very slowly it's best done when heated under reflux with sodium hydroxide we can also have hydrolysis by water with silver nitrates and ethanol the reaction will look very similar but at the end we will not only get bromide ions we will get hydrogen ions as well silver nitrate you may remember is part of identification of halide so that bromide down would then go on and react with the silver ions from the silver nitrate to give us a precipitate and this precipitate is visible and the color will tell us to its silver by my and this is a test for haloid ions so we want to increase the chain length of a halogeno arcane we can react it with the potassium cyanide and where we start off with halogen or alkane with two carbons in we can see here we have added a third carbon onto this in a similar looking reaction we can react it with ammonia but instead of being CN added on we get NH3 giving us either a primary amine or a nitrile we can look at the trend in the rates of hydrolysis to give information about the strength of the carbon halide bonds and the test that we do this is also the test for halide ions so you're probably familiar with it the rate at which the precipitate forms can tell us all about the strength of the carbon halide Bond because for the precipitate to actually be made the carbon Halo Bond needs to be broken fluoroalkanes are very unreactive you are not going to break that bond by hydrolysis iodine alkanes react very quickly that bond is going to be broken by highly dorises very quickly and you'll see a quick formation of the precipitate and then in between those we have bromoalkanes and chloride alkanes going down we see the speed of hydrolysis increasing this is also the speed at which the bond is being broken the speed at which the precipitate is formed this is opposite to the strength of the bond the weaker bonds will break faster the naming of alcohols is very similar to alkanes here we have three different alcohols with the functional group in a different place this is butane 2 all it is numbered so the functional group has the lowest possible number if we use the green numbers it will be butan 3 or which is not correct this is be Time Warner with the functional group on the end the last one has a two methyl group and the functional group on the two carbon making it two methyl butane two all butane oneol is a primary alcohol butane 2 are what is a secondary alcohol and two methyl butane two wall is a tertiary alcohol it is not the names that determines whether it's primary secondary or tertiary but what is attached to the carbon that the functional group that the oh group is attached to how many carbons and how many hydrogens it's attached to if we have more than one alcohol functional group then we can call it a trial or a dial there is a tetrahedral geometry around the carbons but there is a bench geometry as in water around the oh functional group remember this oxygen is going to have lone pairs on it and the oxygen in the hydrogen in this are going to have a dipole established this means between alcohols we are going to get intermolecular bonding leading to their properties of low volatility low boiling points and then being good solvents when we look at oxidization of alcohols do you need to be aware of the differences between primary secondary and tertiary here we have primary alcohol and we are going to be using an oxidizing agent which we show by square brackets with an o in it this is going to give us ethanol at the end this is going to give us an aldehyde this is after a gentle Heating the aldehyde can then be further oxidized to give us a carboxylic acid this requires more continuous vigorous Heating this happens under reflux the further oxidization to give us the carboxylic acid secondary alcohols can be oxidized to give us ketones which have a similar but different position of the functional group tertiary alcohols cannot be oxidized it is important to remember the differences in what can and can't be oxidized and how far they can be oxidized the oxidizing agent that is used is acidified potassium dichromate we will start off with a chromate 6 iron being orange and after the oxidation reaction has happened the cremate 3 Iron is going to be green so we will see a color change in here however this Resurrection you've probably seen in real life and you'll notice these are not nice colors it's a pretty dirty green we get at the end distillation will turn a primary alcohol in to an aldehyde and a secondary alcohol into a ketone reflux is more vigorous continuous Heating and it will turn a primary alcohol into a carboxylic acid it will also turn any aldehyde into a carboxylic acid elimination reactions of alcohol are also known as dehydration reactions because we are going to be losing water we are going to need a concentrated acid for this and it needs to be done under reflux here we have our alcohol the lone pairs on oxygen are going to be attracted to the positive hydrogen ion giving us a positively charged intermediate the electrons in the covalent bonds are going to move breaking that Bond giving this a carbocation in the electrons in one of the covalent bonds to a hydrogen will break giving us a double bond this dark purple example has given this Bute two in but is not always necessarily the same bond that breaks this example here the slightly lighter purple ones could break and then we get put one in out of this but each two in will show e z isomerism so we could get a wide range of products out of this turning alcohols into alkenes will allow us to produce polymers without crude oil the alcohol could come from sugar making this a renewable process and do not forget to write down that the water is lost because it is important and I am guilty of forgetting it a lot an alcohol group can be substituted with a halide iron Halo Alka we will have an alcohol plus a hydrogen halide to give us a Halo alkane but everything that we need to put in there is a bit more complicated we need alcohol a sodium halide and sulfuric acid this needs to be heated under reflux the first thing that will happen is the sodium bromide in this example will react with the sulfuric acid and that will give us the hydrogen halide that we are looking for this hydrogen halide will then go on to react with the alcohol here we have ethanol plus hydrogen bromide which will give us the haloalkane plus water when we are looking at the reactions for alcohols they were going to go combustion and oxygen to give us water and carbon dioxide and they will give us hydration with pcl5 to give us a Halo arcane they can also undergo how donation if we had red phosphorus which is P4 combining with iodine these two will react to give us P i3 and this will give us an idoalkane or we can undergo oxidation reactions by reacting it with acidified potassium dichranate either under distillation or reflux to give us either an aldehyde a ketone or a carboxylic acid depending on exactly what you're looking for we can have an elimination reaction or we can react the alcohol with phosphoric acid to give us alkenes and there are more details about each of these reactions on individual slides so here we're going to be using a round of flask and obviously these um are round so to hold that study I'm just going to put it in a large Beaker to hang out in Secure um base we are going to make an organic silver so I have 50 centimeters cubes of ethanol 50 centimeters cubed of glacial ethyleric acid this stuff is quite nasty and we are raising some quite nasty things happening all of this in the PM head and give that every other smell to mix it and then it really slowly I'm going to add some concentrated sulfuric acid thank you you know you can see if you go closer [Music] um that concentrations I'm clearing that's it changing I think ability so after explore things I've said after reflux here I have my mental um you could use a Bunsen burner oh Lord's wife if you don't have one of these for you well here's my brown bottom flask this is just sitting ever so slightly above the basket here so it's not sitting on it it's just not trying to find out for my connector and I've got my reflux condenser here the important things remember with reflux is that the water needs to go in at the bottom the cold water comes in here and then the hot water or the water that's been warmed up a little bit needs to go out at the top so you can see this is bubbling away now and if we look really really carefully in here you can see if I just move you around a little bit you'll see that it's evaporating and it's condensing so you can see the cold water is causing um anything that's evaporated you can bend so then drop back down into the flask where it can be reboiled so reflux just like the same things being reboiled over and over again after it's been reflection for about 10 minutes we just need to change all equip it also around so it's going to steal now we've set this up for brief um distillation we've got exactly the same and bottom plus bubbling away here we've got um a corner and stuff on here and we've got our condenser and then all everything that's distilling off is connecting in this little vehicle here if I show you the drops so it evaporates it condenses and then the the gas is sent along here it dependencies travel all the way down here and drops into the end so now we've um distilled off two thirds of our solution we need to add some uh sodium carbonate to there and put it in a separating funnel so you need to make sure that it's closed when it is in line it's open when it's not in line it's closed so I'm going to take it off at the top pour in my sodium carburetes causing my practically just distilled here and then what you need to do pops then lead back on you're gonna be in 13 we need to make sure that it's quite secure you need to have quite where I'm holding this is several times to give it a mix and we need to do is open the Gap open the um tap to release any guess that comes off what I would do is maybe two three times without it until you stop hearing that it'll um fit that it will release that tells you a gas has been produced and then turn it back out the right way and allow it to separate out into two different nodes you can already see here that we've got a lower layer and top layer separating out I'm just going to leave that for a bit and then I'm going to take it over here and pull off the lower layer now you can put this in a can if you want to I'm just going to hold it for a second so I've just opened this tab ever so slightly if I was open it full a little bit faster but I want to have control like how this works so I'm just opening it drop by drop and we're going to be discarding the lower layer in this case and we're going to be keeping the top layer so it doesn't matter if a little bit of my top layer goes out um and gets discarded with my lower layer that's absolutely fine if you really wanted to be keeping the lower layer and discarding the top layer then I would stop so there's a little bit of a lower layer still here in here but since we're going to be discarding the lower layer just stop that let it settle down for a bit since we're going to be discarding the lower layer I'm going to go ever so slightly past the line so I lose a little bit more top layer but I make sure that it's all pure bubbles you can see the layer is just here foreign calcium chloride starting with my stock price closed and this is the top layer from full so in that goes again I need to mix it down there make sure it's cured with mix a couple of times open the tap to release the gas mix it a couple more times open attach release and guess and then keep mixing it until you can't get hearing more gas released and let it separate out so again we can see two layers separating out of separating out here and again I want to discard the lower layer so I'm just going to run this off until the layer has gone all the way through so I don't really see it fresh at the top we start writing that down there okay so it goes quite quickly once you take the top off I'm just going to set it out for a little bit and then go very slowly drop by drop watching the um part between the layers separate out through and then here I just have my top layer left so now that I've just got my top layer I'm just gonna let that go through into there taking the time to release any gas now I'm just going to add some um so I'll need um and I just uh calcium chloride don't say how much so I'm going to add that much put it slow and then what I need to do is to transfer the liquid into a round muscle glass and instead of a distillation again now we're going to steal off again but what we've added into the apparatus is a thermometer just here so what we can do is collect off the different fractions which come off at different temperatures because we only want a certain fraction we are waiting for a thing which did this deal off between 74 and 79 degrees so everything in the back is coming out at the moment we can just discard [Music] foreign [Music] [Music] thank you we can determine the identity of organic compounds from a mass spec for example here we're going to look at butane here is a sample Mass Spec that we might get from butane now in the ionization state it is going to be broken down the ionization safe breaks it into different parts the biggest Peak will be your molecular iron Peak and the rest of them will be fragments and from the fragments we can work out the identity the peak that has a mass of 29 could be ch3 connected to a ch2 gradually working out the little bit starting with the ch3 and then adding on the ch2 and working out the mass the 43 Peak well we know it's already bigger than the ch3ch2 because we've just worked out to be 29. so if we add on another ch2 just the carbon is 41 adding on two more hydrogens will take us up to 43 T of this part and the identity of this part now if we know the compound definitely contains this and has an overall mass of 58 then butane is The Logical answer if it had different molecular Peaks if it didn't contain this for example methyl propane will have the same mass and the same formula but it will give different fragmentation Peaks it will not give this fragmentation Peak here when you have a question sending some data from infrared spectroscopy you need to look for some characteristic regions there are three you need to know different groups absorb infrared at different set frequencies you will get given a data seasoning exam so don't worry you don't need to learn these but you do need to be familiar with the data sheet and what it looks like here are some example graphs for an alcohol you were looking for this characteristic region here if we have a carbon oxygen double bond for example an aldehydes and ketones you're looking for this characteristic region here and for carboxylic acids which will have an O8 and a carbon oxygen double bond it has kind of a double region one in the same place as the O8 and one in the same place as the carbon oxygen double bond you need to be familiar with these characteristic regions on the graphs and be able to refer to them in the exam and pick them out of data given to you in an exam [Music] you will see chemists talk about standard conditions a lot and it is important that you know what they are we use standard conditions because lots of reactions will change with a change in temperature or pressure for example the reaction could get faster at higher temperatures so whenever you see values given in a calculation these calculations these values will change as well depending on the temperature and the pressure so they are generally given under standard conditions and you will be told this the temperature is always considered to be room temperature so 25 degrees Celsius or 298 degrees Kelvin the pressure is one atmosphere pressure which is atmospheric pressure this is also a hundred Kila pascals enthalpy profile diagrams can tell us a lot about what is happening in a reaction we have energy going outside and the reaction alone along the bottom the product and reactants are labeled with their different amounts of energy if the reactants have more energy than the products the products will have less energy than reactants energy is released this is an exothermic reaction or we can say that Delta H is negative on the other side the products will have more energy than the reactants energy is absorbed this is an endothermic reaction Delta H will be positive many reactions will have an activation energy the hump of energy that is required for a reaction to start here it is in green on the diagram how we use words and definitions are important so here we're going to be looking at enthalpy change during a reaction heat energy can be taken in or given out this is the enthalpy change it is given the symbol Delta H Delta whenever you see this this triangle just means changing H is enthalpy so Delta H is change in enthalpy this shows us we are looking at the enthalpy change under standard conditions if we have a negative Delta H we are going to have an exothermic reaction heat energy is given out a positive Delta H is an endothermic reaction heat energy is taken in endothermic reaction will get colder while an exothermic reaction gets hotter enthalpy change is measured in kilojoules per mole when we are talking about the standard enthalpy change of combustion this is the N3 change that occurs when one mole is burnt completely in an excess of oxygen under standard conditions we can write that in shorthand here we have Delta for changing little C showing it's a combustion H for enthalpy and under standard conditions these values are generally going to be exothermic because it is combustion or if something doesn't burn then zero when we are talking about the standard entry change of formation this is the enthal we change when one mole of a substance is formed from its elements under standard conditions for this everything needs to be in their standard States here is our shorthand again Delta for changing f for formation H for enthalpy and under standard conditions this value can be positive or negative it can be endothermic or exothermic the standard enthalpy change of neutralization is the length that we change when one mole of water is formed by neutralization here is our neutralization equation again is really really important one Delta neutralization enthalpy Delta is our change in H is enthalpy and Newt is neutralization or we can look at the enthalpy change of a reaction using calorimetry notice the spelling calorimetry not colorimetry the equation for this is energy change equals mass times specific heat capacity times the change in temperature and the change in temperature is a bit where we can actually get Hands-On and measure this in a lab we can also measure the mass we generally know this specific heat capacity so we can work out the energy change that has gone on for energy change we're going to be using joules the mass we're going to be using grams for temperature we're going to be using Kelvin your data sheet should be able to help you with this if Delta H is negative it gets hotter if you have a positive Delta H it will get colder when you are testing the enthalpy change of combustion we can use something called calorimetry which has alcohol Spirit burners in and you might have a number of different alcohol Spirit burners methanol ethanol for you to test and you need to find the energy given else by combustion we can do this by measuring the temperature change of a known volume of water you need to know the change in temperature and you should record this start temperature and the end temperature not just the change in temperature as a control variable you need to have a constant distance between the two different parts and you need to measure the change in mass of the spirit Burnet before and after you burn it we can then use our equation for specific heat capacity so energy change with a mass of water Times by the specific heat capacity of water Times by the temperature change of the water that will give us the energy change what we are looking for the enthalpy change here so we also need to know the moles of alcohol that we used up in creating this energy change we can do this by dividing the change in mass of alcohol so the changing mass of the spirit burner by the Mr of alcohol calculated energy change here might be in joules so we need to find the N3 change in combustion in kilojoules per mole measuring the enthalpy change of a reaction is practical for your a level there is a more detailed video that you can go and watch of this but very briefly here it is you are going to need to make some nice tables because you're going to be doing a lot of recording at times you are going to need to record at which point you add in the powder to the solution or mix your two solutions you are going to need to know the mass that you are adding so we can add this in to our calculations the beaker is to make sure it's steady the polystyrene cup is for insulation and we need a thermometer so we can actually record the temperature you can measure the temperature change of reaction over a period of time here is when we added in the powder to the solution so we couldn't take a measurement there but we can draw a graph of time skipping the one where we added it in and you'll see see through this we need to draw a line of best fit it's not going to look beautiful because temperature drops in intervals and then we need to go backwards find zero and extrapolate back what the maximum temperature reached was then we can put that into our calculations if you've got two of them you can combine these and follow the instructions this will allow you to find the maximum temperature reached or the temperature at zero which is difficult to do as you're doing it difficult to find the temperature at zero because you are currently busy adding things in this experiment can be improved by using a data log up which will continuously monitor the temperature over that time and it will even draw your graph for you errors might occur when we have energy lost to surroundings a lid or the polystyrene cup helps with that or the reaction could be incomplete thus you wouldn't get a true reading hess's law is an subject I really enjoy I think it's really elegant and I have done lots and lots of other videos full of examples to help you work it out if this summary slide doesn't make it clear for you this follows the first law of thermodynamics the energy is always conserved so if we're going from the start to the end it will always take the same amount of energy to get there it doesn't matter which path you take the enthalpy change for any reaction depends on the initial and the final points and is independent of the reaction pathway we can use this to find the enthalpy change of reactions that we can't measure by using data from reactions that we can measure so for the enthalpy change of combustion we have our reactants going to our products that is an alternative path we have our combustion products and I always draw boxes around these that's just the way that I was taught to do it if we burn our reactants we will get the combustion products and if we burn our products we will get the combustion products so if we want to find information going from reactants to products we can use the combustion products as a bypass we can't directly measure carbon and hydrogen going to methane but we can know the data for the combustion products the combustion products being carbon dioxide and water I would lay out by writing on the data so here we have a carbon and a two lots of hydrogen gas combusting notice these are negative values and the units are kilojoules per mole now this is the path that we are going to be taking I always encourage my students to draw this on in a highlighter or a colored pencil so you can see it now for the start we are going in the same direction as the arrow so the signs are the same minus 394 plus 2 times minus 286 for the other parts we are going in the opposite direction so the sign is opposite plus 890 gives us an overall value of minus because it's exothermic 776 kilojoules per mole and to be change of formation looks very similar these are formed from the elements so that is what we have at the bottom except that arrows here go in the opposite direction because this is a formation from elements to the reactants and from elements to the products here is another reaction we can't directly measure but we can look at the formation from the elements down here now remember this needs to be balanced so we have three lots of hydrogen and half oxygen gas draw in your arrows add on the data draw with color pen or highlighter the arrow that you are going to be following notice this one here is now in the opposite direction we're going the opposite direction to Arrow see this one is the one we need to change the signs for different direction different signs same direction same signs so it is now plus 75 and plus 242 minus 110 same signs different signs always remember to put the positive or negative on here and add in your units when we talk about Bond enthalpies we mean mean abundant enthalm is because each of them will be slightly different so the value that we use is the mean averaged across the law Bond making is exothermic bond breaking is endothermic to work out Delta H it is the energy to break the bonds in the reactants minus the energy to make the bonds in the products that is energy that we released or given out taken in during the reaction he is our example and I always encourage my students to draw out what they can see it makes you take note of the actual bonds that are involved and is less likely to cause mistakes later on especially when we get to something like water because there are actually four bonds involved and it is much easier to see those four bonds if you draw it out fully and then it starts very methodically to label this out so here is our bonds here is how many we have of them and then you can pop in the data that you are given for the question either number or tick off Each Bond once you counted it to make sure you do not miss any out you can see here quickly going through the calculations so I am laying this out in a very clear and methodical way making it very easy for the examiner to see what I've done to see potentially if there are any mistakes in writing any of these numbers down so if there was a mistake an error carried forward could be given always please always lay your calculations out as clearly as possible and give the examiner instructions as to what is actually happening we can then do the final calculation and work out the answer this is an exothermic reaction because it is negative always with this type of calculation where you can have negative and positive numbers add in the sign and the units [Music] [Music] Collision theory is simply that reactions will happen when particles with sufficient energy collide here we have our reaction profile with our reactants up here and our products down here it could also be that the products could be up here for a reaction this energy here is the activation energy The energy needed to get a reaction started for a reaction to take place it could be either making bonds or breaking bonds or a combination of these depending on what reaction it is the rate of a reaction will change depending on the temperature the concentration the pressure or if a catalyst is involved we can use the gradient on the graph to visually see a change in the rate of rational or to actually calculate the rate of reaction so here is a beautiful graph with a line of best fit going through you'll see it starts at zero and it's a smooth line remember this is chemistry not up no I'm doing this very roughly so please forgive me for not completely filling in my axes for rate of reaction we need to be comparing two things so here we have centimeters cubed of gas released per second so what are we measuring and how long are we measuring it for it could be volume it could be math it could be hours or seconds in an exam you might be asked to compare the rates of reaction at 10 seconds and at three minutes so I'm roughly going to Mark these on my graph 10 seconds and 180 seconds and what we're going to be measuring is the changing up divided by the change in a cross so change in centimeters cubed divided by the change in seconds always always always draw your construction lines on a graph so the examiner can see what you are doing so they know you know what you're talking about and you haven't just randomly pulled numbers out of the air you need to put your ruler close to the graph so you are measuring the exact gradient at the point you are looking at and draw your line we're going to be drawing a triangle please triangle the biggest triangle that you can fit and then we're going to need to change it up divided by the change in across again at 180 seconds we need to draw a tangent that that touches the graph just at 180 seconds so you can see the differences in my two lines there and again we need to draw a triangle please the biggest triangle you can and then for each of them we need to divide the change in up divided by the change in across and this will give us gradients we can see that steeper things generally going to go faster and shallower things shallower gradients are going to be slower concentration and pressure also affect the rate of reaction and we're going to do them together because they are very very similar an increase in concentration on pressure will lead to an increase in the rate of reaction an increase in concentration is more particles in the same volume and increasing pressure is the same number of particles in a smaller volume thus the frequency of collisions will increase temperature has an effect on the rate of reaction here we have our Maxwell Watson distribution curve with our original temperature T1 T2 is an increase in temperature if the activation energy lies here we can see that shifting temperature to T2 means more particles now have enough energy to overcome that activation energy roughly a 10 increase in temperature will double the rate of a reaction as well as more particles having sufficient energy and they will have more kinetic energy they are moving around more so they are more likely to collide leading to an increase in the number of collisions and an increase in the energy of those collisions we can use a Maxwell boltzmann distribution curve to look at the energy that particles in a reaction have the side here we have particles with that energy lots of different graphs will say number of particles fractional particles percentage of particles it is just the amount of word of particles with that energy and energy along the bottom we will have summer particles with low energy over here the peak is the most probable energy a slightly shifted from the main energy the area under the curve is the total number of particles and here we have the activation energy there is no maximum energy that a particle can have and it will go through the origin since there are no particles that have no energy the particles in this bit here that have passed the activation energy these are the ones that will react if we increase the temperature we will shift the Curve now more particles have passed that activation energy increasing the numbers that are available to react a catalyst will shift the position of the activation energy by providing an alternative route introducing a catalyst will have an effect on the rate of reaction they increase the rate of reaction by providing an alternative pathway with a lower activation energy causing there to be an increase in the number of particles that are able to react and an increase in the number of successful collisions if we look at our reaction profile this is an uncatalyzed reaction and the activation energy that it takes for the reaction to happen however a catalyzed reaction will have a lower activation energy transition metals can act as homogeneous and heterogeneous catalysts homogeneous means they're in the same phase heterogeneous means they're in a different ways and a catalyst is something that increases the rate of reaction by providing an alternative pathway with a lower activation energy and that is the key thing here for a heterogeneous Catalyst generally the catalyst is solid and the reactants will generally be gases or liquids a catalyst we need to have a large surface area such as a honeycomb structure as this is where the reaction actually takes place the reactants are adsorbed at the active site on the Catalyst and this can weaken the bonds or hold the reactants in a more reactive configuration when you combine this with a higher concentration of reactants at the catalyst you will get higher rate of reaction homogeneous calculus where the reaction will happen via an intermediate and the intermediate would generally have a different oxidation state to the reactants or the products [Music] [Music] lucitellier's principle and dynamic equilibrium is a great topic for questions Leisure talia's principle says if an external condition is altered the equilibrium will work to counter that change it is important to remember that a dynamic equaniblium is not static both the forward and the backwards reactions are occurring at the same time but not always at the same rate the forward reaction is exothermic meaning if we increase the temperature the equilibrium will shift to the left hand side to counteract this the reverse reaction is endothermic and this will increase to lower the temperature this will give us a lower yield of ammonia conversely water to what you think increasing the temperature gives a lower yield of ammonia but the industry conditions have to balance rates of reaction which will increase with the yield the left hand side has four moles whereas the right hand side has two moles so an increasing pressure will favor the forward reaction the right hand side has fewer moles Shifting the reaction this way will reduce the pressure high pressure in Industry might increase the yield but maintaining that high pressure is expensive and can potentially be dangerous if we change the concentrations for example increasing reactant it would shift the reaction this way to help counter it also removing a product would shift it this way to counter it but adding in a catalyst will have no effect since both the forward and the reverse reaction will be sped up by this KC is the equilibrium constant for a homogeneous system if we have a reaction here with the capital letters being our compounds and then the lowercase letters being that multiples we can work out the equation for KC and by putting the product on top and then these bits the the compounds and substances go in Brackets and then their multiples go outside of the brackets the reactants go on the bottom we use square brackets to show concentration the unit to KC will vary and the easiest most simplest way to work these out without getting confused is just by writing it all out not by trying to skip guess and do it in your head just take a little bit of extra time to write it out here is our equation for ammonia so we are going to have ammonia on top with two outside down below the reactants we're going to have nitrogen and hydrogen three lots of them so that is our expression for KC if we're doing calculation we'll just take the numbers from the question and pop them in here to work out the units we just need to put the units in not forgetting our multiples up here and then to help you not get confused write it out fully so here we had two of them so I've written out two of them this one goes here here I've got three of them so I'm gonna write one two three out down here and then it's algebra we can cancel them out and see what we have left giving us one over moles per decimeter cubed squared giving us most amounts two decimeters to six there are some things that will have no effect on KC that is concentration pressure or the presence of a catalyst however an increase in temperature will affect casein you will see an increase in KC for endothermic reactions and a decrease in KC for exothermic reactions the equilibrium constant KP for reactions involving gas we are going to be using partial pressure instead of concentration we need to determine the mole fraction which is the number of moles of a gas divided by the total number of moles of all gases in the reaction the partial pressure is the mole fraction multiplied by the total pressure so for a reaction we can write an equation so KP the equilibrium constant will be equal to the partial pressure of C to its coefficient over A over B for this example reaction at the starts we can assume that there are 100 of moles of gas are reactant and we've got zero percent no products for example in this situation we're going to have 0.2 moles of gas a and 0.5 moles of gas B then we're going to have 1.3 moles of gas C we can add them all together so our total number of moles are two moles of gas the mole fraction for gas a is 0.2 divided by 2. we can then find the total pressure this might be given in the exam say it's five kilopascals the mole fraction of a which is 0.1 we can then do 0.1 times 5 to find the partial pressure of a KP is going to vary with temperature but not to be affected by a catalyst foreign [Music] thank you [Music] before we look at Bronsted-Lowry acid-base equilibriums we need to look at a few definitions an acid is a proton donor a base is a proton acceptor and a proton is a hydrogen ion so hydrochloric acid can dissociate into hydrogen ions protons and chloride ions hydrochloric acid is the acid and the chloride ions are the base because HCL hydrochloric acid can donate a proton and chloride ions the base can accept a proton NH3 can accept a proton an nh4 plus can donate a proton so in this situation NH3 is the base and nh4 is the acid these will make a conjugate acid base pair the part that we'll accept and the part that would donate the proton the relationship between Ka and PKA is pretty simple but can look pretty scary so I'm going to show you on calculator Ka is the acid dissociation constant but because these can be quite complicated numbers written in standard form with 10 to the minus 4 10 to the minus 5 PKA is another value that is length and just looks a lot nicer so it's easier to work with to change from PKA to Ka we need to use logs for a strong acid we are going to have a large value for KA and a small value for PKA for a weak acid we are going to have a small Ka and a large PKA so for ethanoic acid Ka is 1.8 times 10 to the minus 5. show you this on the calculator how exactly you would put it in I've done it here Open brackets one 0.8 times 10 to the minus five and make sure you put your bracket in the right place if you're not familiar with your calculator please practice doing that and this gives us the value for PKA for ethanoic acid for hydrofluoric acid the pka is equal to 3.17 so we can now work out what the k a is and we need to do 10 to the minus 3.17 10. to the power minus 3.1 7 equals 6.76 times 10 to the minus 4. you can see the difference in these two numbers how Ka is very complicated number whereas PK is a much nicer number to work with a strong acid is an acid that will fully dissociate you should be able to recognize some strong acid for example hydrochloric acid hydrobromic acid hydroiodoic acid sulfuric acid nitric acid those are the common ones that you should be familiar with but just to expand it a little bit perchloric acid and chloric acid are also strong so because they fully dissociate whenever we're doing these calculations we can assume that the concentration of acid is equal to the concentration of hydrogen ions for example if we have 0.1 mole per decimeter cubed of nitric acid the pH of this is going to be minus log of the concentration of hydrogen ions we are assuming the concentration of hydrogen ions equals the concentration of acid so that is minus log 0.1 and here I'm going to put into the calculator for you so you can see how to actually use it which ones you actually need to press because this is an area people really fall down on so we can see the pH of nitric acid at 0.1 moles per decimeter cubed is one and please do this to two decimal places nitric acid is a monobasic acid this is what most of your questions will be about but I just want to make you aware that sulfuric acid is a Dye basic acid it will have two hydrogen ions dissociate so watch out for this in questions you might have sulfuric acid giving off two hydrogen ions or it might go to one hydrogenide please pay attention to this in the question and check exactly what they're asking you for when we talk about strong and weak acids it is important we don't get confused because there is a difference between the strength of an acid and the concentration of an acid strong acids will fully dissociate they will have a very low ph weak acids will partially dissociate strong acids will generally have a pH less than three these include a hydrochloric acid or hydrobromic acid and sulfuric acid weak acids will still have a pH less than seven but maybe not all the way down to zero examples include ethanoic acid this is going to represent water here we have our acid and the rest of the compound whatever that might be if we have a strong acid at a low concentration the acid will be fully dissociated from the rest of the compound they will be completely separate for hydrogen ions if we have a strong acid at a high concentration we can have the same number of water molecules in here but more acid and they are still all fully dissociated there's just more for them because they're at a higher concentration if we have a weak acid at a low concentration you will see they are partially dissociated some will be Associated and some will not be a weak acid at a high concentration will have more acid in and again some of them will be dissociated and some of them won't be dissociated so the concentration refers to how much there is of whatever it is we're looking at and the strength of the acid is the ability to dissociate the hydrogen ions if we want to calculate the pH of a weak acid is a tiny little bit more complicated than calculating a pH with strong acid but only it's highly little bit more complicated weak acids are one that partially dissociate in water there are a few that you should be familiar with methanolic acid ethanoic acid benzonoic acid hydrofluoric acid which is always a surprise to me and then expanding it a little bit further nitrous acid sulfurous acid and phosphoric acid when an acid partially Associates we can assume an equilibrium is set up with h a being the hydrogen ion and the base and a being the base here water is in excess and because it's in excess we can rewrite that as h a so the acid dissociates into the hydrogen ions and the base we can turn that into an equilibrium equation with a concentration of hydrogen ions and concentration of base on the top and the concentration of h a on the bottom Ka is the acid dissociation constant you might also see PKA which is minus log of Ka so you'll need your calculator to work that one out when we are doing these calculations we can make two assumptions the first assumption is that the concentration of hydrogen ions is equal to the concentration of Base ions at equilibrium and the second assumption is that h a doesn't change because they are so weakly dissociated we can assume the concentration of H A at equilibrium is the same as h a at the start meaning we can re-write our equation as the concentration of hydrogen ions squared because concentration of hydrogen ions equals concentration of Base at equilibrium divided by the concentration of H A at the start which will generally get given in the question so we're going to put this into practice first thing you need to do is to always write down your equation with state symbols and ensure that it is balanced work out your equilibrium and here we're going to be doing concentration of hydrogen ions squared divided by the concentration of ethanoic acid at the start we know what Ka is because we were told it in the question so we can replace that we know what concentration of ethanoic acid is at start so we can replace that in the equation then we can rearrange the equation you're giving us 8.5 times 10 to the minus 7 is concentration of hydrogen ions squared a little bit of algebra to get rid of that squared we need to square root the other side giving us 9.22 times mint minus 4 as a concentration of hydrogen ions and always keep your calculated values when you're doing this if you don't know how to keep your calculated values use the answer button or the memory buttons and practice with this before you go into the exam because it is vitally important to avoid rounding errors pH is a minus log concentration of hydrogen ions giving us a pH of 3.04 to two decimal places if we want to calculate the pH of a strong base we can use KW water would associate inched hydrogen ions and hydroxide ions this will weakly dissociate so some would associate and some won't if we want to write this as an equilibrium then we can do it the same as we do the others KC with concentration hydroxide on top concentration of hydroxide ions on the top and concentration of water on the bottom if we rearrange that we can then take the case D concentration of water and call that KW which is the ionic product of water and this varies wear temperature at 25 degrees c k w is 1 times 10 to the minus 14 mole squared decimeters to minus 6. if we have a strong base for example sodium hydroxide that will give us sodium ions and hydroxide ions because it is a strong base we can assume it is fully dissociated to the concentration the initial concentration of the base is going to be equal to the concentration of hydroxide ions if we want to find the pH of 0.2 moles per decimeter cubed sodium hydroxide at a given temperature 25 degrees C we can use KW we can write our KW constant rearrange it we'll replace KW with 1 times 10 to the minus 14. we'll place the concentration of hydroxide ions because we know that from the question 0.2 once we have the concentration of hydrogen ions we can do minus log concentration of hydronized to find the pH in this case it would be 13.30 to 2 decimal places in one of your practicals you might have done some pH probe work some titration work and come up with some pH curves there are four different ones you need to be able to recognize starting with a strong acid and a strong base because they're strong they're going to start low and end High a strong acid and a weak base is going to look different weak acid and a strong base and a weak acid and a weak base the straight up part of the graph in the middle might look a little bit odd this is the equivalence point this is where the concentrations are similar or the same and neither the acid or the base is in excess around this point the pH will change very quickly when you were doing a titration you need to pick an indicator not all indicators are suitable for all reactions two that you may be familiar with are phenolphthalein which works at a very high pH and methyl Orange which works much lower down the pH scale so when you are picking an indicator you need to make sure that it is one that will pick up the equivalence point and not be outside of it a pH probe will work at any value a buffer solution is one that maintains a steady pH even after editions of small volumes of acid or alkali an acidic buffer solution is made from a weak acid and the salts of that weak acid a basic buffer solution is made from a weak base and the salt of that weak base one example from biology is that the blood is a buffer it will maintain a constant pH of roughly 7.4 and hydrogen carbonate ions are used as the buffer one example is ethanoic acid and sodium ethanoate as a buffer when acid is added the minus ions will pick that up when Alkali is added hydroxide ions from water will pick up the hydrogen ions producing more water and Shifting the equilibrium to ensure that the equilibrium is maintained we need to know how to calculate the pH of the buffer solution as with all long calculations in chemistry the very first thing that I want you to do is to highlight the key bits of information in different colors if that will help you and then pull all of the information out write it down by the side so you don't need to keep dipping into the complicated question every time you need to get a little bit of information we've sorted it all out we've laid it out clearly in one place so this is all the information we are going to need for this calculation because we've got different volumes of our salt and our acid we can work in moles to make them all into the same volume so the first thing I'm doing is working out our moles of salt and our moles of acid we can have our equilibrium equation and we can rearrange that so we are finding out the concentration of hydrogen ions we can then replace it with all the numbers that we know this is the same Ka so 1.7 times 10 to the minus 5 and because we've put the concentrations into the same volume we can use moles here instead of concentration we have our concentration of hydrogen ions we can then do minus log concentration of hydrogen ions to find the pH and always use your calculator value do not write something down and then use the rounded value that you've written down you will introduce rounding errors use your calculator value and you should get a pH of 4.37 to two decimal places [Music] [Music] there are some areas where the language that you use is very important and thermodynamics is one of those areas so we're going to go over some key terms it's important that you learn them well and you can use and apply them properly in an exam so take your time with this slide write down the answers copy down the key terms and learn them the enthalpy change of formation this is the standard enthalpy change of formation for a compound equal to the energy that is transferred when one mole of the compound is formed from its elements when they are under standard conditions and in their standard States standard conditions is another thing you need to learn they are 298 Kelvin or one atmosphere of pressure the enthalpy of lattice formation is a standard enthalpy change when one mole of ionic lattice is formed from its ions in gaseous form under standard conditions the enthalpy of lattice dissociation is a standard enthalpy change when one mole of an ionic lattice is dissociated into its ions in gaseous form the first ionization enthalpy is the enthalpy change when one mole of electrons is removed from one mole of atoms in a gaseous form to give one mole of plus one ions in a gaseous form the second ionization enthalpy is the enthal we change when one mole of electrons is removed from one mole of plus one ions to give one mole of two plus ions in a gaseous form the enthalpy of atomization is the enthalpy change when one mole of atoms in a gaseous form are formed from that elements in a standard state Bond enthalpy this is the enthalpy change when one mole of a covalent bond is broken homolytically in a gaseous state electron affinity is the enthalpy change when one mole of atoms in a gaseous form gain one mole of electrons to form one mole of minus one ions in a gaseous form the enthalpy change of hydration is the enthalpy change when one mole of gaseous ions becomes one mole of aqueous ions it is really important to have accurate descriptions for these terms because these could easily come up as exam questions born Hard Cycles once you get to grips with them are very very elegant but you need to spell your working clearly so we don't get confused we can use them to calculate data that we can't directly measure from bits of data that we can directly measure similar to hess's law the data will be the same the answer will be the same irrespective of the route that we take so here we have sodium chloride as a solid and we're going to go all the way up to sodium ions and chlorine gas with lots of different steps in between we have our ions and that is the electron affinity of chlorine down to the lattice enthalpy of sodium chloride the enthalpy change the formation of sodium chloride the enthalpy change of atomization or the entropy change of sublimation the end to be change of atomization for chlorine and the first ionization enthalpy of sodium this drawing this structure can look very confusing but if you take it carefully and you take it logically it's no problem at all this is where I like to use highlighters in the exam so you know where you start and you know where you finish and you know which route you are taking so we can make it very clear which way we're going and which ones need to change sign so if we want to work out the lattice enthalpy of sodium chloride from start to end in the solid green line it is a combination of all of the other figures in the highlighted green line starting off with the electron affinity of chlorine we're starting in the same place but because we are going in the opposite direction to the arrow it needs to change signs so it is minus minus three four eight we are then going the opposite direction of the first ionization enthalpy of sodium the opposite direction of the entropy change of atomization of chlorine the opposite direction so it's minus for the N3 change of atomization of sodium and in the same direction as an arrow for the enthal we change the formation of sodium chloride so it's a positive we don't change the sign on that one once you have clearly laid out all of your data and please clearly lay it out so the examiner can see where everything's coming from and if you make a mistake we can just do the math and get the answer at the end an exam question might start and end in different places you follow exactly the same method to find different data a few other things like magnesium chloride magnesium will undergo a second ionization entry step and two moles of clear minus must be made all of these numbers are based on real experimental data theoretical values can differ based on the covalent character of the bonds entropy or Delta s is a measure of disorder in the system the higher the entropy the higher the entropy valued the more disorder there is thus the more stable the system is because there are more ways of rearranging the particles a reaction can happen spontaneously without any external influence if it's an exothermic reaction if it has products that are lower in potential energy and the more thermodynamically stable but there are some endothermic reactions which are also spontaneous a solid will have low entropy whereas a gas will have high entropy simple compounds will have low entropy whereas complex ones will have high entropy pure substances will have low entropy whereas a mixture will have high entropy we can see that entropy Delta s is the sum of the entropy of the products minus the entropy of the reactants if entropy Delta s is positive there is an increase in entropy an increase in disorder and this will happen when we're moving from a solid to a gas or over increasing the number of moles Delta s is negative we have a decrease in entropy if there is an increase in entropy it is a likely reaction to happen spontaneously however if there is a decrease in entropy it is unlikely to happen Gibbs free energy has a symbol G or Delta G for change in the reaction happens or not is the feasibility of a reaction this is a balance between Delta H and Delta s so Delta G the Gibbs free energy it would Delta H the change in enthalpy minus t temperature Delta s change in entropy does g is in kilojoules per mole Delta H is in kilojoules per mole T temperature is in Kelvin an entropy is in joules per Kelvin per mole because we have temperature in the equation Delta G will vary with temperature if your free energy is negative the reaction will happen however this is nothing to do with rate so it may happen very very slowly if your free energy is positive then the reaction will not happen it is not a feasible reaction if we have a negative enthalpy change and a positive entropy change it will be spontaneous at all temperatures however if we have a positive enthalpy change and a negative entropy change then it will not happen at any temperature if both enthalpy and entropy are positive when Delta G is zero then the temperature will be the enthalpy divided by the entropy the spontaneous Above This temperature [Music] we can use cells to work out electrode potentials a simple cell is a metal electrode in a solution containing that metal for example here we have a zinc electrode in a zinc salt solution and a copper electrode in a copper salt solution a salt bridge is used to connect them together to allow electrons to flow and this will create the voltaic cell at each side we're going to have oxidation or reduction reactions happening for example here zinc is being oxidized while copper is being reduced and we can add those together to give us the overall reaction from the voltmeter we can get the E cell for this reaction and for this example it is plus 1.10 volts if we are going to be drawing or writing our cell our electrode there is a way that we do that ZN solid line and then the iron double solid line then we need our second iron solid line again and then the metal the double solid line in the middle is representative of the salt bridge this single solid line will differentiate between the two states and the most positive one is on the right hand side electrons move from negative to the more positive and we can predict if your reaction will happen based on the values that we know for E cell values are calculated against a reference sample and this is our standard hydrogen electrode here we have our standard hydrogen electrode we are going to get bubbles of hydrogen coming out in small gaps we will have a platinum electrode and because it is the standard reference half cell it needs to be done under standard conditions so this is 298 Kelvin 100 kilopascals and a one mole decimal cubed solution of the iron all standard electrode potentials are the difference between any given half cell and the standard hydrogen electrode measuring the voltage in an EMF cell this is a lovely practical and I hope you've had the chance to do it we need to start off with some very clean electrodes so you can rub them down with a bit of sandpaper and then you can clean them so they're free from Grease with some propanone if your electrodes are not clean then this could be a source of error in measuring the EMF the electrodes the metal electrodes need to be placed in a solution of the metal ions and this needs to be connected up to a volt meter with wires we can then take the reading from the voltmeter we need to have a salt bridge here the anger bungs with a little bit of cotton wool and then filled with salt solution in the middle you can see as the salt bridge goes in and out of the two solutions a voltage is able to be read on the voltmeter this is a very old voltmeter which I have to manually take the reading from myself you might be able to connect this up to a more sophisticated one a digital one to give you a better reading this has a self-indicating color change and when we see the color change we can answer the question how much has been oxidized the important thing to remember when you're working out equations is that the manganate ions are going to be reduced you're going to have your potassium manganate in your buret and then your known volume of the solution that we are analyzing goes into your conical flask this needs to be done in an excess of acid indicated for this as potassium permanganate decolorizes as it reacts we know we've reached the end point of the titration when we see the first permanent pink color this will happen when the potassium from angulate is in excess because of the gorgeous deep purple color of the potassium for manganate it's really hard to see the bottom of the meniscus so for these titrations we read it from the top of the meniscus if you do this consistency the difference between the top and the top then you will still get the actual Titus again with this question we are going to work through sorting out the numbers and then we are going to do the calculation so and sulfate was made up in 250 centimeter Cube solution and from this 25 centimeters cubed was titrated against 0.02 monstrousness Cube potassium manganate seven potato needed to oxidize the iron sulfate was 24.2 centimeters cubed calculate the original mass of the iron sulfate first thing we need to do is work out our equations now these may be given to you an example you might have to work these out for yourself we have our manganate being reduced to manganese two and remember the only thing we can add to these hydrogen ions electrons and water to balance out the four oxygens on the left hand side we need to add four waters on the right hand side to balance out the four waters on the right hand side we need to add eight hydrogen ions on the left hand side and five electrons to make sure the charge is balanced iron is going to be oxidized so it's going to start as iron 2 you and then get oxidized to iron three and for this we just need to add on our electrons now we need to balance these to make an oval all reaction which just means we need to times the bottom one by five when you're balancing combine the equations you need to look at the number of electrons my preference is for you to always write out everything in full so you don't forget things so you don't make mistakes then we can go back and cross things off afterwards if you're going to work out the moles of potassium permanganate yeast from that the moles of iron sulfate from that the moles by itself 8 in 250 centimeters cubed and then we're going to use that to work out the original Mass from most of potassium manganate we can do concentration times will give us 4.8 times 10 to the minus 4 moles we can look at the ratio in the equation and see that for every one mole of manganate ions we have five of iron giving us 2.4 times 10 to the minus three well now we need to work out how much we had in 250 centimeters cubed to work out the original Mass which is what the question is asking us mass is moles times Mr which will give us 3.648 grams iodine and sodium thiosulfite redox hydrations are attached more complicated because they have an extra step an extra reaction in there your thigh sofa ions are going to be oxidized whereas your iodine ions are going to be reduced while looking at the two individual equations is useful it is more likely that you will see the overall equation or the reaction given at once for your iodine and Valium thiosulfate titration you're going to have your sodium thiosulfate in your buret in your Carnival flask you are going to have a known volume of what we are analyzing we are also going to need to have in there an excess of potassium iodide in the conical flask you are going to have a reaction between what we're analyzing the oxidizing agent and the iodide ions from potassium iodide this is going to start to produce iodine this is going to give the solution its distinctive yellow color as the titration goes on the iodine is going to get back reduced back to iodide and the color is going to fade this makes the end point rather hard to see as it's rather subtle color change this is where we can start to start as an indicator this means the solution will be um I kind of like it I have a nice blue black color and as we get closer to the end point of the titration the blue black color will start to did this appear and when the blue black color has gone that will be the end point of our titration 2.8 grams of copper alloy is react with acid neutralized and made up to 250 centimeter cubed in water from this 25 centimeter Cube solution is reacted with excess aqueous potassium iodide and two products aqueous iodine and copper one iodide the mixture is then titrated against 29.8 centimeters cubed of 0.1 moles per decimeter cubed sodium thiosulfate calculate the percentage copper in the original alloy and again the first thing I'm doing is just pulling all of the numbers I'm going to need out of the question so that I can use them straight away so the first thing I'm going to do is to use the information of the question to work out my equations the first one is a reaction that happens during the titration now the equation we need to work out are copper ions are going to react with our iodide ions and as it tells us in the question we are going to get copper one iodide and iodine now the hard part for these is working out the ratios so we're going to go from a copper to the iodine to the thiosulfate two moles of thy sulfate react with one mole of the iodine and two moles of the copper ions will give us one mole of iodine so here our copper and our thiosulfate are actually in a one to one ratio for a moles or thi sulfate volume times concentration over a thousand will give us 2.98 times 10 to the minus 3 mole most of the copper because he's in a one-to-one ratio are exactly the same we use 25 centimeters cubed in our titration so to get it back up to 250 what we need to do is just times that by 10. to find the mass of the copper in the original alloy we take the number of moles and then we times it by the mass of copper from the periodic table giving us 1.89 grams now we need that as a percentage of copper in the middle original now use the number from your calculator for this even though I've written down 1.89 that is not the full calculated value and if you use that you're likely to introduce a rounding error in your sums so 1.89 divided by two point eight zero times 100 gives us 67.6 percent [Music] [Music] [Music] transition metals are fascinating things that sit here right in the middle of the periodic table and then once you need to know about my titanium through to Copper they will form complexes they have a range of beautiful colored ions they are variable oxidation states which makes them so useful and they can act as catalysts the reason for all of these properties the reason they are transition metals is because they have an incomplete d sub shelf when they're atoms or ions an interesting thing to point out here is chromium and copper where the 3D is filled before the 4S because it is more stable to have a half full or a full three D shell than it is to have a full or half full 4S shell zinc is not really a transition metal because it has a full 3d subshell neither is scandium the 4S is a lower energy subshell so it is removed first so for Cobalt 2 plus it will lose everything from the 4S before it loses anything from the 3D transition metals can form complex ions these are made up from a central transition metal ion surrounded by ligands ligands will bond in a dative covalent Way by donating both of the electrons in the bond a few new terms we're going to be using as well as ligand and complex eye the coordination number is the number of bonds to the central ion a monodentate ligand will form one Bond whereas a bidentate ligand will form two bonds there is a very particular way of drawing these we will have our transition metal ion in the middle square brackets and surrounded by the ligands when you are writing it out we have square brackets the transition metal ion rounded brackets with a ligand in the middle the subscript number will indicate the number of ligand square brackets and then the charge on the outside all monidentate six coordinate complexes have an octahedral geometry there are two bidentate ligands you need to know about ethane one two diamine and each of them will make two bonds to the central ion or Ethan diaanoate and again each will make two bonds to the central ion both of these have an octahedral geometry we can also have multi-dentate ligands here we have EDTA and there's six places that it forms bonds are highlighted here heme in hemoglobin is another multi-dentate ligand the hemoglobin will normally Bond oxygen but carbon monoxide will form a stronger bond with the complex unfortunately carbon monoxide is toxic to humans and can result in death transition metals have variable or station States for example Vanadium will lose two four S electrons to become Vanadium two plus they will lose the 4S electrons of all they lose the 3D electron as they over lower energy Vanadium with an oxalation instead of 5 is yellow Vanadium with an oxidization state of four is blue three plus Vanadium is green whereas Vanadium with an oxidation state of 2 is a lovely violet color so we can see lots of color changes with Vanadium and we can see them all in the following reaction starting off with Vanadium with a plus 5 oxidation state at yellow it will then go through all of the different colors and oxidative States in this reaction to end up with Vanadium two plus we can have silver in a complex ion and when it meets that functional group that aldehyde the silver will come out of the complex ion and you will see a silver mirror the silver will then form on the outside of the test tube this is the silver mirror test the aldehyde and the complex iron with silver in is more commonly known to you as tolerance the redox potential of any transition metal ion is influenced by the ph and by the ligand complex ions are beautifully colored important things that makes them my favorite topic and the change in color of the solution tells us that something is going on this could be a change in oxidation state the ligands or the coordination numbers changed we see color when some of the wave events invisible light absorbed and removed and then the rest of the wavelengths are transmitted or reflected and this is all in the D electrons we have the average or the ground state of the D electrons and when they absorb light they are excited from the ground state the difference between the ground state and the excited state can be given by Delta E equals HV or HC over Lambda Delta is our change in E is energy which is measured in joules H is Planck's constant which is 6.33 times 10 to the minus 34 joules per second we have the frequency of light measured in hertz C is the speed of light which is 3 times 10 to 8 meters per second and Lambda is the wavelength to measured in meters if we want to look at the color or the color change we can use spectroscopy or we can use a Colorimeter [Music] we need to know the different shapes that complex ions make because they have a 3D shape hopefully you'll be familiar with the wedges and dashes from earlier in the course foreign with the sixth coordination number we are going to have an octahedral shape these are the small ligands for your larger ligands we're going to have a tetrahedral shape because the charged chlorides don't really want to be anywhere near each other we can have a square planar complexes or we can have linear complexes for example in tolerance complex ions can show CIS trans isomerism here we have a platinum Central iron and we can have cisplatin or we can have transplantin cisplatin is a very useful anti-cancer drug trans Fatin is not a very useful anti-cancer drug for octahedral complexes where not all of the ligands are the same they can also show CIS trans isomerism or if we have a complex ion that has two bidentate ligands and two monothetic ligands these can show CIS transisomerism it is well worth spending some time practicing drawing these out in a logical manner complex ions can also show Optical isomerism for example with a hexadentate ligand like EDTA if we have two mod identity ligands and due by identity ligands or if you have three bidentate ligands these two are Optical isomers as are these two you'll notice they are mirror images of each other it is possible for something to show Optical isomism and CIS trans isomerism you need to be aware of the redox reactions that happen with transition metals and then be able to apply this to a number of different situations so iron three can be converted to iron2 and then back again manganese and hydrogen ions can be used to change int to ion3 the iron2 is being oxidized ion3 can be reduced in a similar reaction with iodide ions and we're going to go from orange brown to pale green chromium free can be converted to chromium II chromium 3 goes to chromium II with hydrogen peroxide and an alkali the way it is being oxidized we can see this when we look at the oxidation numbers of everything chromium three plus ions chromium three in chromium on the right hand side we have four oxygen that's minus two so minus eight overall we need to get to minus two for the whole compound so the oxidation state of chromium has to be plus six chromium can also be reduced and this might be a more familiar color change where we go from that bright orange to the green if we look at the oxidation numbers again the overall is minus two oxygen is minus two there are seven of those that gives us minus 14. we have two chromiums so we need two X plus minus 14 equals minus two so X is going to have to equal plus six and chromium on the right hand side is plus three copper ions can be reduced from copper two ions to a couple one ions in a copper iodide where we're going to go from a probably solution to a solution with a white precipitating copper oxide can react with sulfuric acid to give us a number of different products solid copper which is going to be the coppery brown color copper sulfate which hopefully you're familiar with is a lovely blue solution and water if we look at the oxidation numbers copper on the left hand side in Copper oxide we know that oxygen is -2 we have two coppers so to get to zero they have to be plus one each over on the right hand side copper as a solid is zero we know that A sulfate ion is minus two making copper plus two so we can see compost chains is always going to stay in two different ways it's gone from plus one to zero where it's been reduced and it's gone from plus one to plus two where it is being oxidized and this type of reaction is one of my favorite words or the name of this reaction is one of my favorite words this is the disproportionation reaction where the same thing has been oxidized and reduced in one reaction we can have substitution Reactions where ligands Exchange there are three more identity ligands you need to be aware of water and ammonia are similar in size and are uncharged the ammonia will replace water and we will see a color change when this happens we could also have an incomplete substitution where only four of the waters are replaced again we will see a color change here but there is not a change in coordination number or in geometry chloride ions are larger and they are charged so we will see some more changes not only will we see a color change but there has been a change in coordination number from six to four also there'll be a change in Geometry as it will now be tetrahedral this can happen with Cobalts copper or iron we can also have ligand exchange by multi-dente or by dentate ligands built by identity ligands we have ethane12 diamine and the Ethan dio8 we need to learn these however you do not need to learn the structure of your multi-dental ligand EDTA the equations for these are very long and look horrific but they follow exactly the same structure as all of the other one there is nothing in here that you can't do we will be replacing six Waters with three ethan12 diamines so the exact same coordination number six but because the bidentate ligand makes two bonds we only need three identity ligands or if you're replacing it with a multi-dentite leg in EDTA which has a six coordination number will make six bonds we'll Place six Waters with one EDTA and this is the chelate effect button take a multi-dentite ligands will replace mono dentate ligands there is an increase in entropy as there are more moles on the right hand side with the six Waters in these circumstances being released there is little effect on enthalpy due to similar Bond Energies meaning this reaction is likely to happen the contact process uses a heterogeneous Catalyst vanadium oxide in the production of sulfuric acid step one we will have sulfur dioxide reacting with the medium in the plus five oxidation state and Vanadium in the products will be in the Plus or oxidation state the products from the first step the Vanadium 204 with Vanadium in the plus four oxidative state will then react with oxygen and go back to being Vanadium 5 oxide we can cancel and some of the Vanadium in the two different steps to give us overall equation of sulfur dioxide reacting with oxygen to give us sulfur trioxide and it is a sulfur trioxide that is actually used in the production sulfuric acid this is the reaction between iodide ions and the sulfate ions this is a homogeneous catalyst the overall reaction for this is the reaction between the sulfite ions and iodide ions but this is a reaction between two negative ions and negative ions repel each other so the activation energy for this reaction is very high and the reaction is slow the catalyzed reaction is faster so in step one we have our negative per sulfite ion reacting with positive iron ions in step two the iron three plus ions from the products of Step One react with the iodide ions to remake the two plus ion ions which were reacting in step one and to give us iodide iron 3 can also be used to catalyze this reaction as step one and step two are not dependent on each other it doesn't matter which ones happens first because the catalyzed reaction has a positive ion and a negative ion as reactants it has a lower activation energy meaning the reaction happens faster this is the reaction between mangana Airlines and Ethan dio8 this uses a homogeneous catalyst and is an example of Auto catalysis where one of the products acts as a catalyst so the reaction will start off slow and then we'll get faster as more products are produced the overall reaction has the manganate ions which are negative reacting with negative Ethan dioaway ions this is a slow reaction as two negative ions will repel each other and there is a high activation energy in the catalyzed reaction manganese two plus will react with the manganate ions and we will have manganese three plus as a product the manganese three plus can then go onto further act as a catalyst reacting with the Ethan BioWare ions to produce manganate two plus the activation energy of this is lowered we can then use algebra to cancel out Omega Knight 2 plus and the magnate 3 plus ions on either side come up with the overall reaction [Music] foreign we can work out the mechanism for a reaction and the Order of a reaction from the data there is a link between the concentration of a reactant and the rate of that reaction if we have our equation we can take this and we can write rate equation where rate is K which is our constant so concentration of a X is order and B concentration Y is the order the Little Numbers in our original equation are the stoichiometric coefficients that's for the reaction the superscript numbers in our rate equation are the reaction orders they are different we can have a zero order reactant where the concentration of this reactant has no effect on the rate of reaction we can have a first order reactant where the rate of reaction is directly proportional to the concentration or it can be second order where the rate of reaction is proportional to the concentration squared the overall order is the individual orders summed please recognize the shape of these graphs in an exam we can determine the units for the rate constant from all the other units the rate of reaction is using the unit's moles per estimated cubed per second for a reaction that is first order overall we can look at the rate equation rearrange it to give k equals rate over the concentration of a replace what we can with our units and then start canceling and what we have left is seconds to minus one so the units for a first order reaction first order over all the units of the rate constant are second to the minus one for a second order overall reaction and this doesn't matter whether it is um a squared or whether it's the rate constant and then the concentration of a and the concentration of B because we still have two things there the overall order of both of those is still second order again we need to rearrange it so we've got K as the subject with rate over a and over B replace what we can with the actual units and then start canceling out again it is worth writing this out in full every time you see it just to ensure you don't make any mistakes super reaction is second order over all the units for K are moles to minus one decimeters cubed seconds to the minus one K is for a set temperature and this will change this will increase as the temperature increases or we can measure the rate of reaction by initial rate method this is also known as the iodine clock the reaction equation for this is hydrogen peroxide plus hydrogen plus iodide ions will give us iodine the color and water when all of the iodine produced in a reaction has reacted with all of the available thosulfate ions which is in reaction to any excess iodine is then unreacted in a solution and will turn blue altering the concentration of iodide ions you can experiment an experimentally determine the order of reaction with respect to iodide ions here we're going to use the example of sodium thiosulfate and hydrochloric acid for an investigation as to how the rate of reaction is affected by temperature you may well be given a bit of paper or better card with a cross on it at school and you can put your conical flask on top of that mix the Solutions in the conical flask and look at the solution changing over time it will gradually go cloudy so you can record the time that it takes for you to stop being able to see the crosses anymore and I just want to have a little throwback to the really old way my videos used to look Way Way Back years ago so sodium thosulfate in the hydrochloric acid will make sodium chloride sulfur dioxide and sulfur the sulfur is the bit that you can see because it is a solid it will precipitate out of the reaction and cause it to go cloudy the time taken for a cross GSP is a rough value you can put numbers on it climb it but if you want to improve this we can use a Colorimeter to give us proper more accurate quantitative data you can use a Data Logger with this as well and it's beautiful it will just draw the grass for you as you go along there are some risks with this if we make it too hot if you do it over 60 degrees too much gas will be evolved and that will be poisonous and not good for you we can draw some grass of one over time versus the mean at temperature put on a beautiful line of best fit and the initial rate of reaction is proportional to one over time here we've been measuring the rate of reaction by continuous monitoring this is between hydrochloric acid and magnesium chloride and what we're going to get is hydrogen produced here you can see I've read through the method and already drawn my table out before the external as well so here I have 0.2 grams of magnesium and 50 centimeters cubed of hormonal hydrochloric acid and I'm going to add them together and use a gas syringe to measure what's collected so this is quite a complicated experiment to do because your hands needs to be doing a lot of things at once because you're adding the Magnesium at exactly the same time and you need to practice the same amount of timer at the same time we've got a gas syringe here gas syringes are quite um expensive or it's um delicate piece of equipment so you also needs to make sure that this is Italian and smash so using now every 15 seconds I need to be recording the volume of gas produced you can see this gas range is moving feeling up quite quickly sometimes there's a bit of lag at the beginning so I suggest swing to get stuck because they didn't even quite quickly along you need to be careful that your reaction isn't too quick and that the ash swing throws out the end you'll know the reactions finished when the gas Moon stops collecting gas and when the Magnesium has disappeared from the hydrochloric acid so once we finish the reaction I need to draw My Graph here and my line of s bits then what you need to do right down here at the beginning is to get your ruler and align your ruler up on the on the line and you're going to need to work out the initial rate so what I've drawn it here is a tangent to the line so we are going to get the line at its steepest part when they're going to work out the gradient of this line here so my uh tangent I've worked on radio tangent I'm going to work out the gradient of the line when you do different concentrations you can work out the gradient line for each of them and compare it we can use time concentrating graphs to work out the half-lives and the Order of reactions for a zero autograph we are going to get a straight line going down for a first autograph it is a curved line going down my second autograph is still a curve line going down and you'll notice that it goes down steeper and then shallower for a zero order reaction the gradient is equal to K the first order reactions they're going to have a constant Half-Life so we go down to half on the graph go across and then down half of a half is a quarter so at a quarter we go across and then down half of a quarter is 1 8 so at 1 8 we go across and then down and we have identified three half-lives here here is the first Half-Life the second Half-Life I'm just going to call tier two and the third Half-Life T the weight they are all the same although I will confess this was a very rough graph drawing so the fact they look all the same is actually pure luck when you do this in the lab hopefully they will vary very close values because the half-life we have worked out here is a constant Half-Life this is the time that it takes so the bits for T1 T2 and T3 are going to be the half-life and they are going to be the same we can call this T half to work out the weight constant for this is and then 2 over the half line here that's shown as T ha we can determine the rate equations and the reaction mechanisms because unsurprisingly reactions are more complicated than the overall equation lets on here we have what looks like a very simple reaction however that is not what happens it goes through a series of different steps in step one we've got nitrogen dioxide we actually nitrogener oxide to make nitrogen trioxide and nitrogen oxide then in step two the nitrogen trioxide will react with hydrogen to give us more nitrogen dioxide and water we can then treat this a little bit like algebra and cross off things that are on both sides of the equation and what is left over will give us our overall equation for the reaction the rate equation for this is rate the constant and nitrogen dioxide is second order changing the concentration of a reactant will affect the rate of the slow step and not the rate of the fast step because it is second order with respect to nitrogen dioxide this is the slow step the one with nitrogen dioxide in it is zero order with respect to hydrogen making this step step two the one with hydrogen in the fast step the slowest step will be the one that determines the overall rate of reaction and this is the rate determining step when you first see the armenius equation it can look intimidating but it is actually very beautiful and elegant once you get used to it it is important to remember that the rate constant K is for a given temperature the Arrhenius equation describes the link between the rate constant and that temperature temperature up the top there is tea this is in Kelvins R is the molar gas constant you'll get given this value in the exam you do not need to learn it however it is 8.31 joules per mole per Kelvin E A is the activation energy for the reaction and that is in joules per mole that e there the lowercase e is the mathematical constant e the uppercase a is the Arrhenius constant which is reactant dependent this is more commonly rearranged in this way so lnk equals N A minus E A over RT if is going to be in graphical format we're going to have lnk up one side and then along the bottom we can have one over t the gradient for this is minus the activation energy over r foreign [Music] [Music] octal isomerism is also known as chiral isomerism here we have a very simple representation of a compound with a central carbon and it has bonds to four different groups and this is the key bit here because there are four different groups it has a different shape in space the two representations are mirror images of each other it can be hard to believe it when it's flat but when you see it in 3D with the Molly model is much easier to understand these are drawn flat the same way so you see they are mirror images of each other they cannot be superimposed on top of each other these are different enantiomers and these will affect light differently Optical isomers enantiomers have similar physical and chemical properties but rotate polarized light differently if the polarized light is rotated clockwise it is the positive or the D in antima if it is rotated anticlockwise then it is the negative or l in antima if you have come across D or L enzymes in biology this is where it comes from the majority of reactions will reproduce a mixture of 50 50 mixture of each enantiomer this is a racemic mixture because 50 of the light is being rotated clockwise and 50 has been rotated anti-clockwise there is no overall rotation of light the difference between the enantiomers is important when we are looking at drugs and enzymes some of them will only work with one enantioma there are a few famous examples of this one being thalidomide where one in antima will cause birth defects and the other enantiom won't and the other one is ibuprofen where one is useful and helpful and the other in antima isn't so it is important that drug companies have a way of separating out the enantiomers we can look at aldehydes and ketones together because these carbonyl compounds are very similar aldehydes will have a functional group where the carbon is double bonded to an oxygen and a hydrogen at the end of a group whereas in ketone the functional group will be in the middle a carbon double bonded to an oxygen this aldehyde you can see here in the Molly mods is butane Al that Albert tells us it's an aldehyde and someone at the bottom is butan own bit tells us it's a ketone they aren't soluble in water due to the ability to form hydrogen bonds there are two tests to tell the difference between aldehydes and ketones we have a test with tolerance reagent and we have a test with failings reagent I have covered both of these in more detail earlier in this video both will only give a positive results with an aldehyde so tolerance will give us that very distinctive silver mirror on the inside of the test tube whereas failings will go from Blue to red or brick red a brick red precipitate a primary alcohol can be oxidized to an aldehyde and this aldehyde can be further oxidized to a carboxylic acid a secondary alcohol can be oxidized to a ketone and just as a quick reminder that tertiary alcohols cannot be oxidized you will see the distinctive orange color being reduced to a dirty green color of your chromate as we can have oxidation we can also have reduction which is going in the opposite direction this is a nucleophilic addition reaction and it will need sodium tetrahydroborate in aqueous solution this will give us hydrogen minus ions ion we can have ethanal with its partial charges within the molecule being attacked by the hydrogen nucleophile this will give us an interim immediate ion which will then interact with hydrogen plus ions to give us a primary alcohol at the end for our Ketone it is a similar reaction we have partial charges within the molecule and they are attacked by the nucleophile hydrogen minus again we will get an intermediate which will react with h plus ions and this will give us a secondary alcohol they can also be reduced using potassium cyanide this needs to be done in acidic conditions and it will give us the cyanide the CN minus ion the acidic additions will give us the hydrogen ions hydrogen cyanide is a highly poisonous gas potassium cyanide is also highly poisonous but it is a solid making it slightly safer to handle in a laboratory and is used in preference we have our aldehyde with our parcel charges within it that is going to be attacked by the negative charge on the cyanide and it's going to give us an intermediate that is going to react with the h plus ion to give us our end product and now lots of students they get that cyanide is actually carbon and nitrogen so our longest carbon chain in here actually goes into the cyanide the CN that is a carbon in there so we now have a the Wii carbon chain here making this two hydroxy propane nitrile here we've gone from an aldehyde to a hydroxy nitrile we have a similar reaction with our Ketone we have our Ketone here propanone and it has partial charges in there the Delta positive carbon in there will be attacked by the nucleophile we will get our intermediate and then when we get our final product at the end now naming this one is going to be a little tricky because we need to look at our higher priority functional groups so we have two hydroxy 2 methyl propane nitrile the methyl is a group of hydroxy is a group but the nitrile is the higher priority group so is the main stem of the name the hydroxy nitrile that we got from the reduction of an aldehyde has a chiral carbon there in the middle so the reduction of aldehydes and asymmetric ketones will produce enantiomers tones reagent can be used to a silver mirror test for an aldehyde tolerance is going to be made from ammonia and silver nitrate to giving us a complex iron we will then have aldehyde reacting with the silver ions producing silver this will give us the silver mirror on the side of a test tube this is an incredibly temperamental reaction so if you didn't manage to get this to work in the lab do not worry lots of interfering ions from tap water will stop this working you need to heat this very gently to get it to work failing solution can also be used to test for an aldehyde we will get Blue Copper ions precipitating a red precipitate again not always the cleanest of looking reactions here giving us copper oxide at the end we will have a carboxylic acid produced and the copper oxide one thing you can use to test for carbonyl compounds is 2-4 dinotrophenyl hydrazine or two forward dnp which is a much easier thing to say this is going to give us a positive test for either an aldehyde or a ketone you can see in the video of it here we are going to get a lovely precipitate forming the positive result is going to be this orange precipitate and that will form whenever you have an aldehyde or a ketone you can then go into further tests if you want to differentiate these two when we are drawing and naming carboxylic acids it is this group here at the end that we are looking at that's the functional group that tells us it's a carboxylic acid this is propanoic acid the OIC acid bits at the end tells us from the name that it's a carboxylic acid these are weak acids the hydrogen ion will dissociate and there will be a delocalized negative charge over the two oxygens we can get sorts of carboxylic acids here if the sodium positive eye is attracted to the negative charge we can get a sodium propanamide salt we can identify carboxylic acids as they will react with carbonates to give off carbon dioxide gas you can confirm the identity of carbon dioxide gas using the lime water test a carboxylic acid can undergo reduction by reaction with lithium tetrahydro diluminum and you can write that more simply as a reducing agent this will give us a primary alcohol ICS an act is an assault reaction as an acid and base to give us salt and water or you can react it in another familiar action with an alcohol to give us an ester Esters are made when we react carboxylic acid with an alcohol here we have ethanoic acid and propan oneong and then we have our Ester coming out of it at the end the name of the Ester comes from the alcohol and the carboxylic acid so ethanoic acid and propane one will give us propyl ethanoate we will also get water out at the end of it and the highlighted atoms in here are the ones that go on to make water for the condensation reaction is going to be need to be reflux and concentrated sulfuric acid for the hydrolysis reaction to take place there are two different ways this can happen with acid is reflux with dilute hydrochloric acid or in alkaline conditions is reflux with sodium hydroxide in alkaline conditions we will then numb melt with a sodium sort of carboxylic acid Estes can be used in a wide range of things they can be used as plasticizers in polymers they are very sweet selling so they can be used as flavors or perfumes and they can be used as solvents for Polar Organic Solutions or substances if we want to make an acyl chloride we need to start with the carboxylic acid we need to add in socl2 so the o h group will replace with scl and we will end up with HCL and SO2 coming out as well as our acyl chloride isochlorides have a similar structured carboxylic acids or aldehydes with a carbon double bonded to an oxygen but the other thing a bonded to the carbon is a chloride this is much more reactive than carboxylic acids because it has this internal dipole it can be attacked by a nucleophile in a slight mouthful of a nucleophilic addition elimination reactions for the following reactions you not only need to know the products but you need to know the reaction mechanisms as well how to draw the reaction mechanisms where they are a start where the arrows go to a cell chlorides can react with water with water acting as our nucleophile it will attack the carbon atom there in the Middle where we have a partial dipole setup it will give us an intermediate ion and then at the end we will get carboxylic acid and hydrochloric acid for this you need water and it will occur at room temperature acyl chloride will react with alcohols this is a very very similar reaction you can see I've just replaced one of the hydrogens for this ch3 group if you learn one of these mechanisms you should be in a really good place to do all of the mechanisms oxygen the nucleophile will attack the carbon there with a partial positive charge we will get an intermediate setup and our products at the end will be an ester and hydrochloric acid for this you need alcohol and a reaction will occur at room temperature Acer chlorides will also erect with ammonia in a very similar reaction the nitrogen the lone pairs on the nitrogen this time will Target the carbon with a partial positive charge and giving us an intermediate at the end we will get a primary amide and hydrochloric acid for this reaction you need ammonia and it will occur at room temperature acyl chlorides will react with primary amines to give us secondary amide and hydrochloric acid this reaction needs a primary amine and will occur at room temperature condensation polymerization occurs between dicarboxylic acids and dials dicarboxylic acids and diamines or between amino acids dicarboxylic acids and dials will give us polyesters because that's the linkage dicarboxylic acids and diamines will give us poly amides because that is the linkage and amino acids will polymerize to give us proteins here we have our dicarboxylic acid with the carboxylic acid group on either end and our dial with an alcohol group on either end there will be a condensation reaction between the two and there'll be an ester linkage set up please note carefully how I'm drawing this within a bond extending outside of the square brackets large square brackets surrounding everything and the ester linkage in the middle a condensation reaction is one where we lose a small molecule most of the time this is water as we have lost here but you can also lose hydrochloric acid our diaste is Benzene one for dicarboxylic acid reacting with ethane12 dial and this will give us terylene which is a fabric here is another dicarboxylic acid reacting with the diamine we're again going to have a condensation reaction and lose water our dicarboxylic acid has six carbons in four here in the middle five six giving us hexane bioic acid this is reacting with hexane 1 6 diamine to give us Nylon 66 you can have different numbers in the middle of nylon you could have nylon 4 6 depending on the number of carbons in the middle both polyesters and polyamides nylon terylene are used as Fabrics here is another example of a dicarboxylic acid reacting with the diamine we will lose water again this time we have benzene one four dicarboxylic acid reacting with Benzene one for diamine and the common name for the product is kevlar this is used in bulletproof vests here we have a polymer of Kevlar and you can see within lots of the bonds here there are partial dipoles so that when we get two polymers lining up next to each other we can get bonding between the polymers happening we can get hydrogen bonding occurring between the oxygens in the carbon oxygen double bond and the hydrogens in the nitrogen hydrogen bond giving very strong bonds between strands of polymers [Music] there is an old joke in organic chemistry that you spend 10 of your time learning chemistry and 90 of the time learning how to draw perfect hexagon to help me draw perfect hexagons I've got this paper which you can download completely free of my website Benzene is a hexagon with a circle in the middle it is made up of six carbons and six hydrogens it is an aromatic six carbon ring with six delocalized electrons in the middle that's what the circle is showing us when the structure of this was being determined it was hypothesized this is actually cyclohexa135 train with double Bonds in there alternating between double and single bonds however there's some evidence that supports the ring structure with delocalized electrons over the double single bonds organization it has a planar structure it is flat it feels alternating double and single bonds you'd expect there to be a change in Geometry the bond length in Benzene is an intermediate Bond length between single bonds and double bonds Benzene is more stable than cyclohexa13 train should be if we look at the enthalpy of hydrogenation the actual experimental value that we get is different to the theoretical value showing it is more stable Benzene also doesn't react with bromine water again suggesting there are no double bonds there is a very specific mechanism you need to learn for the nitration of benzene step one we are generating the electrophile for this we need sulfuric acid and nitric acid this will give us h2no3 plus iron and hso4 minus ion which I'm just going to rub out because we don't need that for the moment the h2no3 plus iron will then go on to give us water and NO2 Plus for step two we have our Benzene and it is going to react with our electrophile the NO2 Plus we're going to get an intermediate our hso4 minus iron will come and take that hydrogen ion and the Catalyst will be regenerated we will then get nitrobenzene for this we need concentrated nitric acid and our catalyst is concentrated sulfuric acid and it needs to be refluxed in a moderately high temperature you need to be able to draw some isolation reactions the first step is the formation of the electrophile we have aluminum chloride and our acyl chloride which is going to give us the electrophile step two is the actual electrophilic substitution reaction the negative Benzene ring is going to be attracted to that positive charge on that carbon we're going to get an intermediate the negative alcl4 is going to drag off that hydrogen giving us our product reforming the Catalyst and hydrochloric acid as a byproduct when we have a familiar shape with Benzene but we have a hydroxide group in it this is phenol phenol is a weak acid and we know it's weak acid because of its reactions to two different things it will react with a strong base for example sodium hydroxide but it does not react with weak bases for example any carbonates it can undergo a number of electrophilic substitution reactions for example if we add bromine in then bromine will get added onto three different positions in the phenol and the hydrogen that was there in the first place will come out in the other product as hydrogen bromide the end product is 246 tribromophenol we can have another reaction with nitric acid where we get an NO2 group being put on at the bottom or the NO2 group can go onto one of the sides giving us either four nitrophenol or two nitrophenol these reactions are much easier the ease of electrically substitution of phenol compared to the same reactions with benzene is because one of the lone pairs on oxygen is actually helping the electron system in the middle it's donated to the pi system so increasing the electron density thus increasing the ease of reactions phenol can undergo electrophilic substitution so the hydroxide group can be replaced with an nh2 group or an NO2 group giving us either phenylamine or nitrobenzene phenol phenyl amine are two for directing and nitrobenzene is three-directed if we number these starting from the phenyl group number it in the opposite way around so you can see two is in a different place but four is in the same place oh and nh2 are both electron donating groups whereas NO2 is an electron withdrawing group the electron donating groups can have multiple substitution reactions happening so that's why we are going to get bromine being added on in three different locations both the twos and the fours whereas NO2 is electron withdrawing group so we're only going to get a single substitution happening this knowledge the ability to add things on it in a certain place is important when you are designing organic synthesis pathways there are lots of reactions of amines these can act as nucleophiles a halogenoalkane plus ammonia will give us a primary amine this primary amine plus a halogenoalkane will give us a secondary amine this secondary amine Plus halogenoicane will give us a tertiary aiming this tertiary aiming you can see where I'm going with this can't you plus a holiday in a while okay will give us a quaternary ammonium salt you do need to know the mechanisms for these however every single step is very very similar so if you learn the first one properly and carefully you should be fine here we have a hydrogener okay and ammonia the ammonia the lone pair is going to be attracted to the carbon and then the bromine is going to get shifted out of the way we're going to have an intermediate which a minor is going to be attracted to again and then we are going to have our primary aiming we're going to start from the same place this same halogenic alkane is that this time instead of ammonia we've got our primary amine it's still exactly the same with a lame pair on the nitrogen being involved I intermediate this time is slightly more complicated but only slightly more complicated ammonia is still going to do the same thing in the same place and our product at the end is going to be a secondary Amine these may look really horrible and complicated but once you get used to drawing it it's fine starting with the same halogen or alkane again we are then going to be using our secondary aiming with nitrogen actually exactly the same way our intermediate is looking a little bit more complicated but a mainly it's going to be doing exactly the same thing and we're going to end up with a tertiary amine at the end back again to our halogenoalkane and this one we have our tertiary amine the nitrogen that being involved and this time we go to a quaternary ammonium salt and these are used as cationic surfactants amines can act as bases as they can accept a proton starting with the weakest we have aromatic amines here we have phenyl amine for this to be able to accept a proton it needs to be added on to the nitrogen however the lone pair of electrons on the nitrogen has joined the delocalized electrons in the middle of the ring so it is not available to accept the proton ammonia comes in the middle so you have NH3 plus H2O gives us ammonium and hydroxide and then of the three of them primary amines are the strongest bases this is because the electrons move towards the nitrogen increasing the electron density so it can more readily accept the hydrogen ion it can more readily except the proton making it a better base we can start this by looking at the naming of aliphatic amines this has three carbons in it so it's going to be propile Amine similar to alcohols we have primary secondary and tertiary amines profile amine is a primary Amine but if your amine is in the middle if it has one hydrogen attached to it not to it is a secondary amine so this one would be n methyl which is that group there Ethan which is the other alcohol group one Amine and the N tells us it's a secondary if it has two ends it will be a tertiary and you could look at the name to build up the groups around it methyl ethyl profile Etc if we have ammonia and Hal General alkane we'll get to primary Amine here we have our primary amine and the ammonia the lone pairs on the nitrogen on ammonia are going to go in and get that carbon we're going to have a positive intermediate when more ammonia is involved and this will give us our primary amine as a product this is a nucleophilic substitution reactions we can also go from a nitrile to a primary Amine step one would be how General alkane plus cyanide come from something like potassium cyanide to give us our nitrile setup here and then in step two we can take our nitrile reduce it to give us our primary aiming in this situation the hydrogen for reduction comes from l i a l H4 when we have aromatic amines we're going to take our nitrobenzene and add in a reducing agent that's hydrogen square brackets which is going to turn the NO2 group into an nh2 group and water this will give us being our aiming we can use tin or iron as a catalyst and we need to heat it with hydrochloric acid phenyl amine is used in the manufacture of dyes amino acids might feel like a biology topic but there is lots of chemistry involved here you can see with the Molly mods I'm going to make all of the different amino acids and you can see in the background here we have the general structure with this pink bit being the general R Group you need to know the general structure of amino acid we have a carbon in the middle and this R is the r group that can be replaced with any different things to make the different amino acids but it will always have the same basic structure an amino group on one end and a carboxylic acid group on the other end that R groups are changeable and that will lead to all of the different properties this carbon in the middle means it is chiral in nearly all circumstances apart from when the R Group is H and then it is not chiral you do not need to remember all of these common names that are being displayed up here not even the biologists have to do that but it is expected that you'll be able to apply your skills in chemistry to naming a few of them I have gone over all of them all the ones you'll be expected to sensibly name in a whole separate video an amino acid is a zvita ion meaning it can have a positive charge on one half and a negative charge on the other half in low PH we are going to lose the negative charge and in high pH we're going to lose the positive charge this property of amino acids is an example of them acting as a weak buffer the next two slides are going to be summary slides of all of the organic reactions that we've come across in the course so far it is going to go pretty quickly because I've gone over these in detail in other places in this video what you can do is pause this slide and copy down the sections that you need and practice going from place to place if you want a more detailed video why go through lots and lots of examples of going from one place to another place that video is already made for you it's already waiting for you but this is going to go pretty quickly we're going to put some nice music over it I don't expect you to keep up I expect you to pause it copy stuff down and then use that to answer questions foreign [Music] [Music] [Music] foreign [Music] [Music] thank you [Music] thank you foreign [Music] [Music] [Music] foreign [Music] [Music] [Music] [Music] proton anymore is very similar to carbon NMR but the big difference is the amount of information we get given in the Spectra and how we read it the number of different sets of Peaks that we have show us a number of different environments we're looking at the chemical shift will show us the type of environment that we're looking at the integration numbers will show us the relative number or the ratio of hydrogens that are in that environment and then this spin coupling pattern will show us the adjacent hydrogens when we look at chemical shift TMS is going to be the standard this is given the value of zero and everything else is compared to this this is tetramethylsylene and is used because it is symmetrical so all the protons are in the same environment the solvent we're looking at can be deuterated chloroform and deuterium doesn't absorb so it won't give a peak other ones can be deuterated down methyl sulfox8 when are we looking at Spectra we are going to have a line across the bottom that's going to show us how far shifted along things are and this is going to tell us the type of environment that the protons we're looking at are in for example if it's between seven and eight then chances are it's part of a Benzene ring and if it's all the way over by 11 then chances are it's parts of a carboxylic acid this is organically shown on your data sheets so here we have a very simple hydrocarbon and if we look at this hydrogen if we look at this proton it is in an environment we can say that protons detached to your carbon that has three hydrogens attached to it whereas if we look at this second one it's got slightly different proton environment set up there are three different types of environments the pink protons are all the same we can describe each of these pink protons as being attached to a carbon that is adjacent to a carbon that has no hydrogens on it and there are three protons that we can describe in this way so there are three protons that are in an identical environment the purple protons can be described as being attached to a carbon which is attached to a ch3 group adjacent to a ch3 group and there are two protons in this environment the orange ones can we describe as being attached to a carbon that is adjacent to a ch2 group and there are three protons in this environment so for these pink protons if on a Spectra it is adjacent to no hydrogens then it becomes as a single Peak and because there are three protons in this environment these can have an integration number of three I'll be three High the purple ones are adjacent to a ch3 group which means it will show up as Four Peaks and because there are two protons in this environment the integration number will be two the orange ones are adjacent to a ch2 group and there are three protons in this environment and then where they are on the Spectra is all to do with the chemical shift the type of environment that they are in if we want to work out what the split pattern is saying we need to think about the N plus one rule the number of peaks in the splitting pattern will be one more than the number of hydrogens attached to the adjacent carbon for example if there are no hydrogens attached to the adjacent carbon zero plus one gives us one which means we'll have a single Peak which is called a singlet if it's adjacent to a CH one plus one is two means we will get two peaks or it'll look like a doublet if vegetation drew ch2 two plus one is three we will get Three Peaks or a triplet it was adjacent to a CH three three plus one is four means we'll get a quartet with Four Peaks when you do a lot of these you will start to recognize some patterns and some pairs of patterns that come up frequently now these pairs of patterns may not be directly next to each other we can have two doublets which means there's going to be a CH group and a CH group we can have two triplets which means there's going to be a ch2 group and a ch2 group we could have a tricklet and a doublet meaning we're going to have a CH group and a ch2 group we could have quartet and triplets showing a ch2 group in a ch3 group a quartet and a doublet with a CH group and a ch3 group or a multiplex which this one has Seven Peaks in it now thinking about our M plus one rule Seven Peaks is going to mean it's six hydrogens and the most common example of this is going to be two ch3 groups or two methyl groups attached to the same carbon now when we are looking at Peaks you have to remember that it is next 2. so the ch2 group is responsible for the triplets and the ch3 group is responsible for the quartet carbon NMR can tell us quite a lot of information about compound the number of Peaks tells us the number of different environments and the chemical shift tells us the type of chemical environments chemical shifts are going to be given on the data sheet but you should be familiar with reading them things over on the far left is generally going to be a carbon double engine oxygen the chemical environment of a particular colon atom is determined by its location within a compound it depends on what it is bonded to and what it is next to here we can look at a couple of examples all have the same formula but a different arrangement in space we're going to look at the different carbon environments if you want all has four different carbon environments whereas as we move through the compounds we can start to see that the methyl groups become interchangeable and they are in the same carbon environments if something has full common environments it will have Four Peaks on a carbon NMR if something has two carbon environments or two peaks and three carbon environments Three Peaks chromatography can be used to separate out different things in a mixture thin layer chromatography has plates that are coated in solid you will need to very carefully with a capillary tube dot on your sample the plates are coated with stationary phase this is the solid whereas the mobile phase the bit that will actually be moving up is the liquid the solvent once your plates are dry at the end you can work out the RF value by doing the distance move by the spot divided the distance moved by the solvent and it is really important that you are consistent when you're doing this and you pick the center of the spots they are never going to be beautifully neat this can be used to separate amino acids and you can visualize the spots using line hydrogen or UV this is based on the separation being a balance between solubility in the mobile phase and retention in the stationary phase column chromatography is a bit more complicated we have a column filled with powder or beads to give it our large surface area this is the stationary phase the mixture that you want to separate out will be dissolved in solvent and the mixture will separate out in the column the time for each part to leave the column can be recorded and each fragment can be identified even more sophisticated is gas chromatography which works on very similar principles but gives us much more detailed results this can be used to separate volatile liquids or gases again you will have a column packed with solid and gas will pass over it at high temperature and high pressure the retention time can be used for identification of samples so I've got more a tablet of aspirin which I've crushed with a person with a motor here I've Got My TLC plates and particularly so wings are shiny on one side and matte on the other side is the matte side that will be using and the first thing we need to do and you should really be doing this is the ruler is about a centimeter up you need to draw a pencil line so now I have to solve my aspirin of ethanol and I have my um dots drawn and pencil at the desktop or TLC paper I have in my hand here earlier about the clean retrieving this is very very fine and contributed and what I'm going to do is just pick up some of the aspirin in the tube yes we're in solution the tubes it's computer achieving it should run up the tube a little bit and then I'm going to dot it in front of the spot level one now you'll need to do this quite a few times and then it's dry in between each time the reason why now it's dry and do it quite a few times is because if you just do a lot in one go you'll get exploded whereas we want a tiny concentrated sponge so do a little bit let it dry do a little bit let it dried a little bit let it dry and then you've got to try to concentrated sponge that'll give us the best results so now I have my TLC plate in my chamber I'm just going to pop a lid on top of there to make sure that's um none of myself when it evaporates now I just want to point out that um here I actually have a look at the Gap in my that'll work much better now you can see saltment is just moving past my start line there and as it moves past the start line it's going to start to take some of the dissolved samples um up with it we need to wait until it gets out one centimeter on the top and then we can stop okay so after you've about it you're gonna hear something that looks like this I will look at this event the plate that I just showed you because I UV light is broken at the moment but you'll get big blobs that look like this so probably like the elongated in a regular shade so you want to measure from the middle of this blob down to here and you can use this value you need to calculate the RF value ouch this is when somebody is I haven't explained scratches [Music]