Transcript for:
AQA A-Level Chemistry Exam Preparation

hello lovelies and welcome to the hold of AQA AEV chemistry in one shorsh video I'm am going to sign post the time stamps in the description down below which will help you jump between the bits that you need so if you're advising for a topic test or whether you're advising for paper one or whether you're advising for paper th you can get all the information down below because this is everything in one video and the only time you need to watch a whole video in one go is for Al paper 3 which is right at the end if you want just spits for paper one use time stamps if you just want bits for an as topic test use the timestamps down below to go with this video there is the free verion guide which you can get on my website which takes you through everything that the examiners expect you to know we're going to go over it briefly in this video to refresh your mind maybe the morning of the exam maybe before you're starting your revision and then if there anything you don't understand you can use the links in the free origion guide to take you through to the longer teaching video where I go over everything in much more detail with much longer examples to go with this also again all the links to these are in the description down below there are the predicted papers for this year's exams written based on loads and loads of experience and checked by experienced teachers and examiners there is also the free multiple choice questions the whole course over my website where there are thousands of questions to help you rise and there's going to be linked to some live revision workshops that we are running for you okay [Music] [Music] [Music] [Music] here is our basic atomic 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 + one neutrons are also in the nucleus have a mass of one but a charge of zero electrons are found in the outer shells their mass is very small it's 1 1,036 the mass of a proton and they have charge of minus one when we say the charge is + one Z and minus one this is the relative charge compared to other things the actual charge on this we can measure in and is very very small but it's is much easier to say plus one and minus one the mass of protons neutrons and electrons has been worked out based on carbon 12 as a reference standard and it is really important to remember that this drawing of an atom is not too scar it is mainly empty space the nucleus the diameter of the nucleus is basically 10 to Theus 15 M whereas the whole spherical diameter of the atom is 10^ - 10 m 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 Zed 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 a decimal is is a bit strange but it can be a 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 six 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 they different masses means they might have different physical properties here we have a crude diagram of a Time oflight Mass spectrometer your unknown sample is mixed with a polar solvent and inserted under high pressure ionization will occur by electrons or by spraying acceleration will occur by electric field with smaller ions going faster and here we can use the equation for kinetic energy half * Mass * velocity squared the speed will depend on the mass and the charge with heavy things going slower and the detector will see the irons creating a small ch charge this is what we will get as the results and they can be used to calculate relative mass the MZ is the mass charge ratio and all of these numbers are positive the mass number that we see on the periodic table is an average of all of the naturally occurring isotopes of an element we can use the mass spec to work this this out here we see that 20% of the naturally occurring boron has a mass of 10 and 80% of the natural occurring boron has a mass of 11 so we can do 20 * 10 + 80 the percentage * 11 the mass divided by 100 which is the total of the percentage this will give us 10.8 as a relative atomic mass for Boron and this is the number that you will see written on on the periodic table 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 sh shells and orbitals and we can draw them like this so for example if we look at C 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 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 that is the electronic structure of calcium but we can write it out a bit neater 1 S2 2 S2 2 P6 3 S2 3 P6 4 S2 here we have a shell this is a subshell 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 Orit tools can be looked at as spfd they can be spheres they can be dumbbells and all are present at once we start with 1 s and then two s and three 2 p orbitals each orbital can hold two electrons so the total number of electrons in the first shell is two in the second shell it is 2 + 6 giving us 8 the shorthand way of writing this literally looks at the shells and the orbitals and number of electrons so argon's 18 electrons are 1 S 2 2 S 2 2p 6 3 S2 3 P6 the short hand for that would be Aron in square brackets calcium has 20 electrons 1 S2 2 S2 2 P6 3 S2 3 p64 S2 but that's a bit of a mouthful so we can say argon which covers that first bit there 4 S2 because it's got the noble gas Arrangements ions are atoms that have lost or gained electrons for example here we see the electronic config ation for calcium 1 S2 2 S2 2 P6 3 S2 3 P6 4 S2 now you can either remember this which you can do for the group one and group two as plus one and + 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 variable oxidation state group seven will form minus1 ions and group six will form minus 2 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 ion losing another one will give us a plus two ion and we can change the way we write the electronic structure to reflect that 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 REM removal of one electron from each atom within one mole of atoms in a gaseous form to make one mole of gasius + one ions hydrogen and using our state symbols gas will be turned into hydrogen ion gas plus one electron sodium gaseous form it is really important to get these right will be turned into sodium Plus in the gaseous 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 + one ions in a gaseous form to make one mole of gasius plus 2 ions are equations and again it is important to get the state symbols correct here gasius helium plus will go to helium 2+ + 1 electron sodium plus will go to sodium 2+ in a gous 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 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 7 and electrons 8 there is a big jump in ionization energy it is always going to get harder to remove 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 burum to Boron and magnesium to aluminium this drop can be used as evidence for electron configuration if we look at burum and Boron the fifth electron is the first one in the 2p because a new subshell has been started the fifth electron is easier to remove another drop can be between groups five and six if we look at nitrogen and oxygen with seven and eight electrons you can see that e electron is the first one to be paired in the 2 p so the ionization energy is giving us evidence for electron configuration 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 small small 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 maass 0.02 can have a wide range of numbers if we're looking at upper and lower bounds you will fre L be asked to convert between units especially for temperature to go from Celsius to Kelvin you add 273 kelv CSI - 273 cm cubed decim cubed IDE 1000 cm cubed to M cubed / millium decim cubed to M cubed / th000 and the other way around decim cubed cm cubed * th000 m Cub CM Cub * million and then M Cub to decim Cub * th000 for concentration calculations if you want to change from moles per decim cubed to G per decim cubed you need to change it by the Mr of the substance these are a couple of important definitions you will see these phrases used a lot and 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 12th the mass of one atom of carbon the relative atomic mass is the average mass of one atom compared to 112th the average mass of one atom of carbon you can also see these referred to as a r and m a mole is the amount of a substance that contains the same number of particles as the number of carbon 12 atoms in 12 G avagadro's number is the number of particles in a mole 6.02 * 20 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 m r you might also seen this written as n = m/ m r mass is in G and Mr is in Gs per mole the number of particles is equal to the number of moles times avagadro's number so we could get a question such as this calculate the number number of particles in 7 g of gold we would need to do moles equals 7 that is the mass divided by the Mr of gold which you can look up on your periodic table so we have calculated the number of moles where we then need to take the number of moles and times it by avagadro's number to give us 2.14 * 10 22 particles and it is important we look at the number of sign significant figures here here I wrote down all of the significant figures on my calculator here I have gone to the same resolution as avagadro's number 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 don't explicitly ask for in the question it is expected so solid S gas g l for liquid AQ for aquous 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 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 oxygen because they're both only in one place on each side and then redo your your numbers so we have 10 hydrogens two iodines one sulfur and four oxygens we 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 going to 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 eight idin one sulfur four oxygens so now our hydrogens our sulfas and our oxygens are balanced we can just put two in front of the iodine 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 your working out always write the equation out in full studies clear to The Examiner exactly which bit they should be looking at 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 the equations that use moles we have the ideal gas law so PV equal NR T the concentration of solution so n for moles equals concentration time volume the equation for Mass where moles is mass / Mr and looking at the number of particles where it is moles times the avagadro 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 for the concentration of solutions we need to know that concentration equals mass over volume concentration is measured in G per decim cubed mass in gr and volumes in decimet Cube for Ionic Solutions you need to remember that there are two different ions in there for example calcium 2+ 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 1 mole of sulfuric acid we will end up with 2 moles of hydrogen ions the first requir practical you will do is when you will become very very familiar with it is making a standard solution and then doing a titration these are skills that come up over and over again when you're making 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 waybo and then I'm continuously washing the waybo to get all of the powder off the waybo and then it is in the beaker where I'm 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 into the flask we're going to keep washing so we need to fill this flask all the way up to 500 M or 250 Ms or whatever it is just wash everything out so now I'm washing the flask out and then we just need to swirl it a little bit and then make it up to the volume that we were looking for you can see I'm washing the neck in case any powder has got down the neck of the flask we going to invert it a little bit to make sure is properly mixed and wash the neck again the aim is to get all of the powder that you have weighed out into solution now you'll see the little line on there 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 stand ion if you go over you're going to 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 care 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 bu 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 were looking for so you can go fast for the first cup and then slow down we are looking for concordant results and when I say concordant results I me results that are in within 0.1 1 0 cm 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 Tighter and your three actual tighters will be concordant you can record all of your results to two decimal places when you see a hydration C calculation there are some very specific things I want you to do first of all first of all is highlight all of 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 tooll 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 alkaline 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 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 have pulled out earlier so moles equals concentration time volume we need to adjust the volume so that we are work in liters because this is in cm 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 AL L the question is asking for the answer in G per decim 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 * m we can get our answer in G per decim cubed 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 we use the ideal gas law a lot and you need to remember this equation PV equal nrt 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 but 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 me cubed the gas constant is 8.31 JW per mole per Kelvin temperature is measured in kelvin now the convergence for this is often where people go wrong determine the mass of a 500 cm cubed sample of hydrochloride gas when at 20° C under 150 kilopascal pressure for a the 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 worthy 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 in non-standard units and convert them into standard units before we start so pressure was given in kilop pascals and we need it in pass Pascal so 150 kilopascals is going to be 150,000 pascals volume is given to us in cenm cubed and we need it in me cubed to go from ctime cubed to met cubed we need to divide by 1 million and temperature was given to us in De C and we need it in kelvin and to do that we need to add on 273 giving us 293 K we can then use a 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 = pv/ RT and then plug in the numbers once we have the number of moles we can use Mass equals moles time Mr to give us the number of GRS to give us the that the question is looking for 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 decim cubed and we can use P1 V1 equal P2 V2 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 two 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 DM cubed volume the second pressure is 2 so 24 = 2 V2 we can then use algebra to move the two over to the left hand side giving us 12 as the new volume 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 MR of 284 and is made from 62 G of phosphorus and 80 g of oxygen calculate the empirical and molecular formula this is how I want you to set it out very strictly if you do it 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 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 a number here and then we need to divide it by the lowest number the lowest number at these two is two so we're going to divide both of these by 2 2 / 2 = 1 5 / 2 = 2.5 because 2.5 isn't a whole number we need to multiply it by two to get the ratio so P2 5 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 we worked out p25 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 Mr from the question 284 ided 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 p410 and you can quickly check that that does add up to the right amount 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 yield and multiplying it by 100 so we get a percentage it was expected the reaction would give 14 G instead it gave 5.2 G calculate the percentage yield 5.2 the actual yield divided by 14 the theoretical yield time 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 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 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 it's percentage we times it by a th000 atom economy can be improved in two different ways we can find an altern ative reaction pathway that has less waste or by finding uses for the waste products from a reaction you need to be able to work out if something is in excess or if one of the reagents is limiting the reaction as with everything we need to start off with a balanced equation and work out the masses of everything we can see from the balanced reaction that one mole of Circ FC acid reacts with 2 mol of sodium hydroxide which turns into 1 mole of sodium sulfate and 2 mol of water this is balanced however in practice it might not always be like that for example we have 11.76 G of sulfuric acid reaction with 2.4 G of sodium hydroxide work out which one is the limiting reagent and this is the equation that we are going to be using so for sulfuric acid we have the mass so 11 .6 / the Mr will give us12 M of sulfuric acid and .6 moles of sodium hydroxide if we divide by the lowest number to get the ratio we will have one mole of sodium hydroxide with 2 moles of sulfuric acid this doesn't match up with the equation with the moles we worked out from the equation so will tell us which one is the limiting reagent when we are looking at a hydrated salt we can talk about the water of crystallization for example here we're going to look at hydrated and anhydrous copper sulfate they are different colors and the formula can reflect whether it is hydrated or anhydrous the anhydrous form is white here it is as a white crystal you can then add water to it and it will go to the blue crystals that perhaps your bit more familiar with we can then turn it back to the anhydrous white crystal by heating it again the anhydrous salt will have the formula of C so4 whereas the hydrated copper sulfate is cuso4 dot this dot in here 5 H2O this is a reversible reaction because upon heating or adding water you can get the sort change forms you might be asked a calculation based on this in the exams here we have 7.5 G of hydrated aluminium 3 nitrate which was heated and 3.24 G of water was lost from the sample determine the formula the first thing we need to do is determine the formula of the aluminium 3 nitrate aluminium is a 3 plus ion nitrate is a minus ion that is going to give us aluminium Open brackets n O3 Clos brackets three then we can write in the dot X cuz that's bit we're trying to find out H2O trying to find out the formula of the hydrated salt we can work out the mass of the anhydr salt by taking the MH of the hydrated aluminium nitrate and removing the water giving us 4.26 G we can then go to an empirical formula where samees format we have what is in the question we have the numbers in the question or the masses we're then going to div divide that by the M now water I know this cuz it's 18 but aluminium nitrate I do have to do a calculation to work out when working our Mr always use the numbers given for mass on your periodic table with we divide the mass by the Mr that will give us the number of moles for each of these we can then work out the ratios so how many Waters were there for every aluminium nitrate that is just dividing it by the smallest number so if every one aluminium nitrate we had nine Waters giving us x = 9 and the formula Al Open brackets NO3 close brackets 3.9 H2O depending on the equipment that you use for something or the size of the equipment that you use for something there will be different levels of uncertainty involved for example here we have three measuring cylinders which can measure different volumes so are the first one the divisions are 4 cm cubed and the reading on it is 100 cm cubed the second one the divisions are 2 cm cubed and the reading is 70 cm cubed for the third one the divisions are. 2 cm cubed and the reading is 6 cm cubed the uncertainty involved in each is half a division so for the first one the reading could be plus or - 2 cm cubed plus or - 1 cm cubed plus -.1 cm 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 time 100 the total percentage un certainty in any experiment is the sum of all the individuals for ionic bonding we need metals and nonmetals 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 SC shell we can draw this using square brackets and having the charge on the outside of these square brackets this will give us the formula mgcl2 if we compare the electronic configurations of the atoms and the ions of both magnesium and chlorine you will see that magnesium has lost these 3 S2 so it is bringing it down to full shelf whereas chlorine has 3 P5 and it wants another one to make this 36 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 ion gained or lost electrons with all ions will feel this force not just single ones electrostatic attaction is stronger if ions have higher charges or if they are smaller 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 one 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 min-2 ions and ions of group seven elements are going to form minus1 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 of ionic compounds from knowing what the individual ions sodium carbonate sodium is going to have A+ one charge and carbonate has a min-2 charge since overall sodium carbonate has no charge we need the positive and the negative charges to balance each other out the three the Roman three in iron 3 means it has A+ three charge and sulfate has a min-2 charge now ion sofro over all has zero charge so we can look at this by doing 3 * two and just kind of swapping them over so because iron has a 3 plus charge we're going to need two of those to get to plus six and because sulfate has a minus 2 charge we're going to need three of those to get to minus 6 giving us the formula of fe2 Open brackets s so4 Clos brackets three often you'll be given a word equation and expected to construct the balanced equation from that and knowing your general sours equations is a really important part of being 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 aquous hydrogen ions plus aquous 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 iron so sodi 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 calent bonding occurs between nonmetals 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 calent 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 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 calent bond whereas the rest of them are traditional calent bonds giving us NH3 bcl3 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 the bonding the stronger bonding will happen with smaller ions more delocalized electrons and more positive nucleus when we are thinking about the properties oric structures it is really important that we think of these as a latis a large structure here we have sodium chloride and we write is NAC but it is massive it's not just one sodium and one chlorine each sodium plus ion 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 ltis wide not just to the ones that donated or received the electrons G ionic glates will have high melting points and high boiling points this is because the strong electrostatic attraction requires large amounts of energy to overcome that strong electrostatic attraction they will be soluble in water 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 giant metallic lates 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 Co valent macromolecular structures and we are going to use silicon dioxide as an example these giant coent structures will have high melting and boiling points because the strong intra molecular bonds require large amounts of energy to overcome them they're 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 calent structures they are both made from carbon in Diamond each carbon makes four carboncarbon bonds whereas in graphite each carbon makes three carboncarbon bonds diamond is very hard in a ltis structure graphite is soft as the atoms are arranged in layers that can slide over each other Diamond does not conduct Electric it 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 Cove valent ones there are some common examples that you should be familiar with water H2O ammonia NH3 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 weak inter molecular bonds they are insoluble in water they do not conduct there are two commonly confused because of the spelling types of bonding here we are going to look at Cove valent int molecular bonds which are the strong Cove valent bonds between atoms in a compound these are very very strong and then there are the weaker inter molecular bonds these ones are much easier to overcome and give rise to properties such as the low melting and boiling points 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 H CN b e F2 and these are linear that Bond angles are going to be 180° 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 mly 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° Bond angle for S O3 bcl3 and al3 these are trigonal planer they will have Bond angles of 120° and these have no lone pairs they are flat and we can see the bonding here for CH4 and nh4+ these are tetrahedral they have Bond angles of 19.5 and there are no lone pairs when we are drawing 3D structures a line like this shows it is in plain with 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 plain with paper going backwards and going forwards if we just have the dot and cross d diagram over here it is really hard to see how the bonds are arranged for NH3 and cl3 these are trigonal pyramidal that Bond angles are 107 so roughly 2.5 less than that of T hoodle 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 plain 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 line 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 that the W angle is going to be reduced due to the veence electron Theory the lone pairs can be seen here on the Doon cross diagram pcl5 is trigonal B pyramidal it has two different Bond angles of 120° and 90° there are no lone pairs the bond angles are really easy to see the difference in in the 3D structure using M but much harder to see it when you are drawing it out in the 3D way or even in the dot and cross diagram if this is something you struggle with remembering then either using M mods or flash cards cuz 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 SF six is octahedral all of the bond angles are 90° and there are no lone pairs we can see this in the 3D model we can see this when we actually holding the Molly mods in our hands that the bond angles are 90° whereas on here it is much harder to visualize these being 90° Bond angles this is another one with an expanded octet around the central atom the SE theory is a bit of a mouthful but then so is veilance 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 109.5° in our bond angle NH3 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 loone pairs the bond angle decreases a loone pair loone pair is more repulsive than a lone pair bonding pair is more repulsive than a bonding pair bonding pair electr negativity is a measure of how much an element will attract electrons we have Florine over here as the most electr negative elements as we move across a period the electro negativity will increase as the number of protons increases in a period the number of electron shells Remains the Same same so the atomic radius is decreasing as the electrons are pulled in electr negativity decreases as we move down groups the increasing number of shells increases the shielding around the nucleus calent 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 Cove valent bonding with the electron cloud shared evenly if we have different elements in a calent 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 with in the bond and this can all be due to difference in electro negativity if there is more than a two difference in electro negativity then we're going to be seeing ionic bonding sharing electrons is a continuous Spectrum with Co valent bonding at one end and ionic bonding at the other end and for example in hydrogen gas we have equal electr negativity between these two elements so they share electrons equally in HCL the chlorine is more electron negative 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 electr negativities 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 a tractions 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 we can see this as HDL 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 oldfashioned 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 typod Dio 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 electron Rons 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 isoma there are much fewer opportunities for contact meaning the strength is going to be reduced straight isomers will have more contact points meaning the inter molcular 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 Florine right over here in the very electr negative area of the periodic table between these two elements there is a large difference in electro negativity at the same time as hydrogen bonding you can have other forces occurring such as Vandal but hydrogen bonding is stronger than the other intermolecular forces and this leads to a few properties they will have anomalously high oiling points and we will see in water that ice is less dense and water has a very high specific heat capacity how we use words and definitions are important so here we're going to be looking at enl 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're looking at the eny 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 an endothermic reaction will get colder while an exothermic reaction gets hotter enthalpy change is measured in KJ per mole enthalpy profile diagrams can tell us a lots about what is happening in a reaction we have energy going up the side and the reaction going along the bottom the products 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 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 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 genuinely given under standard conditions and you will be told this the temperature is always considered to be room temperature so 25° C or 29 98° Kelvin the pressure is one atmosphere pressure which is atmospheric pressure this is also 100 kilop pascals when we are talking about the standard enthropy change of combustion this is the eny change that occurs when one mole is burnt completely in an excess of oxygen under standard condition we can write that in short hand here we have Delta for changing little C showing it a combustion H for enthalpy and under standard conditions these values are generally going to be exoic because it is combustion or if something doesn't burn then zero when we are talking about the standard eny change of formation this is the eny 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 short hand again Delta for change in F for formation H for enthalpy and under standard conditions this value can be positive or negative it can be endothermic or exothermic we can look at the enthropy change of a reaction using calorimetry notice the spelling calorimetry not colorimetry the equation for this is energy change equals mass time 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 the specific heat Capac so we can work out the energy change that has gone on for energy change we're going to be using Jewels for 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 measuring the enthalpy change of a reaction is a required picle 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 cuz 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 polyarene 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 will see through this we need to draw a line of best fit it's not going to look beautiful CU 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 all 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 time and it will even draw your graph for you errors might occur when we have energy loss to surroundings a lid or the colerin cup helps with that or the reaction could be incomplete thus you wouldn't get a true reading hess's Law is a 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 eny change of reactions that we can't measure by using data from reactions that we can me so for the eny change of combustion we have our reactants going to our products but as an alternative path we have our combustion products and I always draw boxes around these that's just the way that I was thought 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 always lay out by writing on the data so here we have a carbon and two lots of hydrogen gas combusting new notice these are negative values and the units are KJ 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 - 394 + 2 * - 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 cuz exothermic 776 K per mole eny change of formation looks very similar these are formed from the elements so that is what we have at the bottom except the arrows here go in the opposite dire Direction because this is 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 in opposite direction to Arrow so this one is the one we need to change the signs for different direction different signs same direction same signs so it is now + 75 and plus 242 - 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 Bond enthalpies because each of them will be slightly different so the value that we use is the mean averaged across the lot 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 be released or given out taken in during the reaction here 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 show 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 start very methodically to label this out so here is our bond 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've counted it to make sure you do not miss any out you can see here quickly going through the calculations that 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 were 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 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 dependent on the temperature the concentration the pressure or if a catalyst is involved we can use a Maxwell boltzman distribution curve to look at the energy that particles in a reaction have up the side here we have particles with that energy lots of different graphs will say number of particles fraction of particles percentage of particles it is just the amount horrible word of particles with that energy and energy along the bottom we will have some particles with low energy over here the peak is the most probable energy slightly shifted from the mean 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 pass 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 the 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 temperature has an effect on the rate of reaction here we have our Max volum 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 the activation energy roughly a 10% increase in temperature will double the rate of a reaction as well as more par having sufficient energy 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 energy of those collisions here we're going to use the example of sodium thy sulfate 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 bit of card with a cross on it at school and you can put your chicle flask on top of that mix the Solutions in the chicle 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 throw back to the really old way my videos used to look Way Way Back years ago so the sodium thyr sulfate and the hydrochloric acid will make sodium chloride sulfur dioxide and sulur the sulfur is the bit that you can see because it is a solid it will precipitate out of the reaction and it cause it to go cloudy the time taken for a cross dispar is a rough value you can put numbers on it time it but if we want to improve this we can use a colomer to give us proper more accurate quantitative data you can use a Data Logger with this as well and it's beautiful it'll 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° too much gas will be evolved and that will be poisonous and not good for you we can draw some graphs of one over time versus the mean temperature put on a beautiful line of best fit and the initial rate of reaction is proportional to one over time 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 or 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 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 lelia's principle and dynamic equilibrium is a great topic for questions Leia's principle says that if an external condition is altered the equilibrium will work to counter that change it is important to remember that a dynamic equilibrium 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 what 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 2 moles so an increase in pressure will favor the forward reaction the right hand side has fewer moles shifter in 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 a 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 homogenous system if we have a reaction here with the capital letters being our compounds and then the lowercase letters being their multiples we can work out the equation for KC by putting the product on top and then these bits the the compounds 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 units to KC will vary and the easiest most simplest way to work this 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 are going to have nitrogen and hydrogen three lots of them so that is our expression for KC if we're doing calculation we 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 going to write one two three out down here and then its algebra we can cancel them out and see what we have left giving us 1 over moles per decim cubed squared giving us moles to- 2 decim to 6 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 KC you will see an increase in KC for endothermic reactions and a decrease in KC for exothermic reactions 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 for oxidation 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 compounds will add up to zero the total oxidation States in an ion will add up to the charge on that ion group one elements will be+ one group two will be plus two hydrogen is + one except in metal hydrides when it is minus one oxygen is min-2 except in peroxides or when combined with Florine chlorine and bromine are minus one except in compounds with oxygen or Florine here we have sulfuric acid hydrogen we know is + one and we have two of those + 1 * 2 gives us + 2 in total oxygen we know is -2 we have four of those - 2 * 4 gives us - 8 in total this adds up to zero meaning sulfur must be + 6 for our carbon ion we have an overall charge of -2 o we know is -2 and there are three of them - 2 * 3 is -6 overall we need to get to -2 so what plus - 6 makes - 2 well that is + 4 for the carbon in a metal hydride we know sodium in group one will be + one this is an exception for hydrogen so hydrogen will be minus one giving us zero overall hydrogen peroxide H2O2 hydrogen is going to be + one and we have two of those so + 1 * 2 gives us + 2 this needs to be 0 overall which means the oxygen needs to in total add up to min-2 meaning there needs to be min-1 each -1 * 2 gives us - 2 we can look at the names and the Roman numerals and things to work out what they are copper 2 means it has a 2 plus iron nitrate 5 means the nitrogen is going to be + 5 and from copper 2 nitrate 5 we can work out the formula of this compound really useful if you have ion compounds we know the nitrate is going to be Min -1 oxygen is min -2 we have three of those that gives us minus 6 we know nitrate is + 5 giving us NO3 minus copper is + 2 so we need two nitrate ions to go with the copper when we are looking at Redux 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 1 metal be + 1 oxygen - 2 hydrogen + 1 these are from our rules sodium + 1 chlorine minus1 we can see chlorine started off with zero and went to minus1 and + 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 Redux 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 Oxygen 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 in a four Waters will balance out the oxygen 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 oxygen 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 there's 10 electrons and the top one by two so there are 10 electrons so I'm just going to write out here with this reaction being multiplied by two and with the bottom reac being multiplied by five 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 some 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] oh [Music] there is structure and order to the periodic table and understanding it will help you immensely it is arranged by increasing atomic number not by mass for example we have argon here and Krypton here argon has 18 protons and a mass of 40 whereas 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 with the in 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 in 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 and the right eight electrons in that shell 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 of Trends are seen in other periods for example period two just above it as we move across the group we're 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 aluminium 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 between phosphor and sulfur as pairing starts repulsion increases slightly here sodium magnesium and aluminium will have strong bonds leading to high melting and boiling points silicon also has strong calent bonds it is a dry structure it will have high melting and boiling points chlorine sulfur and phosphorus are simple calent so they will have a weak intermolecular bonds leading to low boiling points and argon is monoatomic very low boiling points for group two we are looking at the ones going down on the left hand side these are the alkaline earth metals and they will 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 two 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 can't be extracted just by carbon it is a batch process making it very expensive it also uses large amounts of electricity group two sulfates get less soluble As you move down the group berium sulfate is insoluble it is given to patients when they are needed to have an x-ray of soft internal organs the barium means that when you have an x-ray you can actually see the structure of these organs when we are testing for sulfate ions we need something we sulfate in hydrochloric acid and berium chloride acidified berium 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 group seven or group 17 sits over on the right hand side and these are the halogens Florine 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 iodine is a gray solid that's sublimes sublimes me 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 electr negativity decreases the increasing number of shells and 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 hens 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 are 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 haly ions are acting as a base and the sulfuric acid is reduced Florine and chloride are not strong enough reducing agents so here you will only see the acid base reactions occurring with sodium bromide it is a moderately strong reducing agent so we will see the acid base reaction as well as the Redux reaction taking place I on IES are a strong reducing agent so here is our overall equation but from this there are lots of other reactions going on not only acid base reactions but Redux reactions as well leading to a wide range of products not just the hydrogen iodine but iodine sulfur dioxide sulfur water and hydrogen sulfide with all of these reactions what you will get out is a white steamy fumes of the hydrogen haly gas which will turn blue litmus paper red when we are testing for haly irons you are going to need a solution of that haly ion 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 because telling 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 ion so the white color will disappear the same will happen to silver bromide on 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 an oxidation state of zero will have an oxidation state of minus1 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 + 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 disproportionation reaction hclo is chloric 1 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 CL minus is the chlorate 1 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 1 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 minus 2 and there are three of them giving us - 6 in total the whole thing is zero so chlorine must be + 5 so this is sodium chlorate 5 this is one big example where using the Roman numerals is important if you see them write them down and pay attention to them a favorite practical in Alo chemistry and a favorite exam question is being given a mystery solution or a mystery white powder 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 cat and anions there are also the flame tests we can add on to this as well when you 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 for sulfate ions we will add berium chloride and dilute hydrochloric acid a positive result 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 Lim water cloudy hydroxide ions can be added to ammonium chloride and it will start to be very smelly leading on to the inverse test for ammonium ions and we were 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 two ions we're going to be adding two different things sodium hydroxide and sulfuric acid both dilute and concentrated for 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 precipitation for calcium ions the addition of either dilute or concentrated Sulu 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 berium chloride so if you want to test both sulfate ions and halide ions in a single example in a single test tube if you do this one first you are adding in chloride ion and you will get a false positive for halide ions the order does matter [Music] [Music] organic chemistry lature 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 sceletal formula will not show any carbons or any hydrogens just skeletons and other functional groups 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 the 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 alkal group will be potentially a side chain with the formula CN h2n +1 an aliphatic compound we straight chains branched on nonaromatic rings based around hydrogen and carbon alicyclic compounds will have nonaromatic rings whereas aromatic compounds will will contain Benzene Rings a 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 e in there that is a double bond it's crude but it works 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 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 this is propan one o here is another example again picking out the carbons as the turning points in our sceletal formula drawing those points those dots on as our backbone and connecting them up we 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 of our skeletal formula drawing on the back bone and the functional group for the structural formula we have a ch3 ch2 ch2 and a CO giving us butanoic acid 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 your revision 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 die 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 we are 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 the prefix comes from the number of carbons so one is meth eth Pro four is but five five is pent six is hex 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 ear is going to have a but prefix here is a side branch there's only one of them this is a methy Side branch putting all together this will be methy but a the a tells us it has single Bonds in it here's something else a little bit more complicated start 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 a 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 a methy and an eth St 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 ethy and two methy we can now start to put the name together because they need to be in alphabetical order not in number order we have three ethy two methy 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 the longest carbon chain and here we can see it's not straight at all please feel free to use color pencils and 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 carbon in so they are all methes 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 way rounds we get it it's either going to be 3 4 4 5 now we have all that 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 methal groups say it's three four four five Dash Tetra methy heane one word which I did manage to fit on one line different functional groups will change the suffix of the word so we have a n e for alkanes an alken will have a double bond in it somewhere and it has an e ne but to here an alcohol will have an oh function group this one four carbons butan 2 O alkanes with hallogen attached to them in front so this is two chlorobutane aldhy will have this functional group on the end and this is butanal an ism of there ketones will have the functional group in the middle and this is but an own carboxilic acids have their functional group on the end and this is beamic acid esters will have their functional group in the middle and we have four on each side so this is butti butanoate there is a priority list of functional groups starting with carboxylic acids moving down to aldhy 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 alken we have a methy 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 1 2 3 4 5 6 so this is hex we have a methy 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 uh 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 branches going in alphabetical not numerical order so B becomes 4 C so we have five bromo four chloro five methy hex 2 another example that looks horrible but once you follow the rules is absolutely fine we are going to start to be drawing lots of reaction mechanisms and 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 Vision 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 ion and a minus ion here we can see the breaking of a CO valent bond with the arrows starting at the bond and then the electrons going somewhere else you always have to be very careful with where your arrows start and where they actually end up they start here and they will go here when you're writing on charges make sure they are 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 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 methy propane they both have a formula C4 h10 but as you can see from the mly 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 propenol but the alcohol group is in a different position we have propan one o and we have propan 2 o one of them will have the O group on the end whereas the other has the O 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 isoma it's not on the end it's in the middle this gives us a different functional group these are functional group isomers because buttin L is an alahh and buttin 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 but 2 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 isomer so this is e but 2 e and when they're on the same side it is the Zed ice masses gives us Zed but 2 as a point but 1 doesn't show EZ isomerism 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 iser or a zed iser your e isers will have priority group on different sides of the double bond your Zed icers will have priority groups on the same or the Z orful 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 buttin so giving us two three D chlorine 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 on the same Z side so this is the Zed Isa and this one is the E isoma 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 Zed iart 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 fractional distillation is a way of separating out crude oil and crude oil 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 Vandal 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 condens ing 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 not enough of them is produced from fractional distillation the long ones are not used rarts and lots of them are left over and 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 will have double bonds we can have catalytic cracking which uses a zealite catalyst it's done at 400 150° 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 branched and cyclic alanes and aromatics such as Benzene thermal cracking is done at higher temperatures 400 to 900° c a high pressure of 7,000 kilopascals and will give us lots of double bonded alenes 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 bunson 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 do water vapor as they're both greenhouse gases carbon monoxide is a toxic gas that can lead to death carbon is suit that can lead to global dimming and atmosphere pollutions leading to breathing difficulties sulfur dioxide can lead to acid rain and various nitrous oxides can be toxic gases and lead to acid rain unburnt 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 hydro carbons and react them with nitrogen oxides giving us safe for products the catalytic converter will have a honey cream structure it will have Platinum Palladium and radium as a catalyst the chlorination of ALC canes 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 substitute ution 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 vision 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 c H3 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 chloromethane and then give us back our chlorine radical this is why it's a propagation reaction because we start and end with the same thing the electron from the chlorine radicle will go in to this Bond here and the electron here will move over then this radical will then going to attack this Bond and the electron will move over giving us back another radical step three is termination 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 it is important that you learn nucleophilic substitution reaction mechanisms carefully and you can draw them accurately in an exam this is an example that can be applied to lots of other situations nucleophiles are electron pair donors and the ones you need to know about are hydroxide ammonia and Cyanide and for substitution we are swapping one thing for another thing here we have chloroethane and our nucleophile is important that you draw on your charges these are our partial charges and then the arrow will start at the electron it will start at the charge that is attacking the bond and it will end exactly where it is going to that positive Delta positive carbon and then the electrons from the bond will move over to the hogen then we will have a swapping a substitution as that nucleophile pops itself in there there are slightly more complicated examples if we have ammonia as the nucleophile again our Arrow needs to start where it starts and end where it actually ends accuracy is so important when you are drawing these otherwise you will not get the marks and this time we have an intermediate which is attacked by another nucleophile with the electrons from the right hand side moving over and then we will have two products in the end a chloride ion and an ammonium ion the rate of these reactions depends on how strong the bonds are the carbon Florine Bond will be the strongest whereas the chlorine iodine Bond will be the weakest meaning any reaction that is involving carbon Florine will be slow whereas any reaction involving carbon iodine will be fast elimination reactions can happen with hydroxide if we have a metal hydroxide and alcohol solution it will be in elimination reaction whereas a metal hydroxide in aquous solution we will get a substitution reaction here we have bromoethane and our hydroxide ion it is really important when we are drawing mechanisms that your arrows start and end in the correct place our Arrow starts from the negative charge and go to the hydrogen from the middle bond connecting the hydrogen and the carbon the electron will then move to the bond connecting the carbon and the carbon and an electron will move from the calent bond over to the bromine this will create a double bond and will release a bromide ion and water hydrogen is lost from the carbon adjacent to the carbon that has the hogen we can see this reaction happen when we have potassium hydroxide in alcohol the depletion of ozone has an important set of reactions in it chlorofluorocarbons cfc's are very very useful they useful as solvents in refrigeration and aerosols they were widely used but not disposed of properly or safely carbon we have carbon Florine bonds which which are short and very strong and we have carbon chlorine bonds which are longer and weaker the carbon floring bonds are not affected by UVS which is why scientists have now shifted towards using hydrofluorocarbons instead of chlorofluorocarbons as they are safer this is a free radical reaction we will start with initiation using UV as an essential this will give us our chlorine radicals the chlorine radicals can then go on to react with ozone O3 to give us another radical and oxygen gas these radicals then keep the reaction going reacting again with ozone to remake this propagation remake the chlorine radical and give us more oxygen gas the reaction will eventually end at some point with two radicals reacting with each other and the overall reaction for this is two lots of ozone will be turned into three lots of oxygen gas ozone in the upper atmosphere is important for protecting us from UV radiation breakdown of ozon by cfc's contributes to the hole in the ozone layer bonding and reactivity of alkenes is fascinating they are unsaturated they will have a double bond so an alen has that double e in there to help you remember the position of the double bond doesn't always stay the same here we have buttin but two different versions of buttin depending on where the carbon carbon double bond is around carbon in the double bond we will have a plain AR geometry whereas around other carbons in the compound we will have a tetrahedral geometry these two examples here give us but 1 in and but 2 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 mechanisms electrophilic addition reaction mechanisms of alkenes that you need to know and be able to draw properly the first one is the addition of hydrogen bromide hydrogen bromide is polar so the electron Rich double bond will be attracted to the Delta positive which will move the electron in the coent bond down to the bromine this will produce a carbocation intermediate and the negative bromide ion will be attracted to the positive carbon and the bromium will be added on electrophilic addition reactions with sulfuric acid will look very similar with the electron dens attacking the Delta positive hydrogen and the electron moving over to the oxygen giving us a carbocation as an intermediate the negative oxygen can then attack the positive carbon giving us yet another intermediate at the end of stage one at the end of electrophilic Edition stage two is a hydrolysis reaction with water whereas we will get a hydroxide iron added on and sulfuric acid remade as the Catalyst for stage one you need cold concentrated sulfuric acid and for stage two you need warm and you need to add in water if you wanted to add on bromine it will look very similar to our first reaction over here except you will get bromine in both positions you will see an orange to colorless change and this is the test for [Music] alkenes working out the major and minor products in a reaction is also known as mnov rol here we have an asymmetrical Aline which we're going to add hydrogen bromide 2 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 bromine we will get different compounds either one bromobutane or two bromobutane and if you're in 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 will get a carbo cat ion 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 electr fire will preferentially add on to this carbon making two bromobutane the 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 carbocation whereas this one is a primary carbocation addition polymers can occur with alkenes here we have chloroethene 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 then become poly chloroethane it is very long to draw out which is why we generally don't draw the whole thing but it will have multiple repeating units of chloroethene polychloro ethine is also known as PVC and it is incredibly useful it's waterproof it's an insulator and it's very unreactive the very strong intermolecular bonds prevent chains from moving over each other making it very very strong a plasticizer something to break those strong intermolecular bonds to get in the way can be added to help chains move over each other so that PVC can be used for clothing and wires unplasticized PVC UVC can be used for window frames something that needs a very strong plastic the naming of alcohols is very similar to Al Al canes here we have three different alcohols with the functional group in a different place this is butan 2 it is numbered so the functional group has the lowest possible number if we use the green numbers it would be butan 3 or which is not correct this is butan one or with the functional group on the end the last one has a two methy group and the functional group on the two carbon making it to methy butan 2 oil butan 1 o is a primary alcohol butan 2 O is a secondary alcohol and two methy butan tuol 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 O 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 bent geometry as in water around the O 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 inter molecular bonding leading to the properties of low volatility low boiling points and then being good solvents there are two methods for the production of alcohol the hydration of ethine here we have Ethan with the electron dense region around the double bond attacking the hydrogen ion giving us a carbocation water or steam the lone paay will then be attracted to that carbocation giving us an intermediate with a positive charge the electron from the calent bond will break that giving us an alcohol and regenerating the hydrogen ion the hydrogen ion from the phosphoric acid Catalyst 300° Centigrade so very hot and 70 atmospheres high pressure steam will also be needed to be added the advantages for this are that is fast it is a continuous process and we will get a very pure product at the end the disadvantages however is that the ethine comes from crude oil it is a nonrenewable resource it will run out and is in a high demand for other things due to the high temperatures and the high pressures this has a high energy consumption making it expensive limiting its use the second method is a ferment ation of sugars glucose will be fermented to ethanol and carbon dioxide we will need yeast for this and this is an anerobic respiration reaction it will need a moderately low temperature for the yeast to be able to work too hot and yeast just doesn't work the advantage is that sugar is a renewable resource it is a pretty low Tech solution so lots of developing countries can use it with low startup costs and low startup needs the disadvantages are is that is a batch process this makes it very slow and laborious it also gives an impure product and needs to be an additional step of distillation afterwards ethanol can be used as a biofuel the ethanol is produced via fermentation growing the sugar cane will remove carbon dioxide from the atmosph spere during photosynthesis the glucose from photosynthesis will then be fermented and the alcohol from fermentation will be burnt in combustion as a fuel six Caron oxides are removed from the atmosphere in photosynthesis two are released during fermentation and four during combustion making this process carbon neutral as no net carbon dioxide is added to the atmosphere in this set of equations however this doesn't take into account the energy used to grow the crops the irrigation the farm machinery and the distillation that's needed to purify it afterwards if the electricity the energy for this comes from a fossil fuel Power Station then this process actually does add carbon dioxide to the atmosphere there are also a few ethical issues associ iated with this the sugar cane that is used as a boil fuel could be used to feed hungry people a large amount of land is used for this this land could be used to grow other crops which could be used to feed people and the habitats the destruction of habitats of what was there before is a massive problem when we looking at oxidization of alcohols you need to be aware of the differences between primary secondary and tertiary here we have a prim 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 ethanal at the end this is going to give us an alahh this is after gentle heating the alahh can then be further oxidized to give us a carboxilic acid this requires more continuous vigorous Heating this happens under reflx the further oxidization to give us the carboxilic 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 di chromate we will start off with the chromate 6 ion 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 is reaction 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 into an alahh and a secondary alcohol into a ketone reflux is more vigorous continuous Heating and it will turn a primary alcohol into a carboxilic acid it will also turn any alahh into a carboxilic acid Tolen reagent can be used to a silver mirror test for an alahh tlins is going to be made from ammonia and silver nitrate giving us a complex ion we will then have alahh reacting with the silver ions producing silver this will give us the silver mirror on the side of a teste 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 irons from tap water will stop this working you need to heat this very gently to get it to work bailing solution can also be used to test for an alahh 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 carboxilic acid produced and the copper oxide side 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 co electrons in the calent bonds are going to move breaking that Bond giving us a carbocation and the electrons in one of the co valent bonds to a hydrogen will break giving us a double bond this dark purple example has given us but 2 in but is not always necessarily the same bond that breaks this example here the slightly lighter purple Bond could break and then we could get but one in out of this but 2 in will show EZ 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 guilty of forgetting it a lot from a reaction is one of your required practicals and there are lots of different examples reactions you could do for this you could prepare cyclohexene by dehydration of cyclohexanol distill off the cyclohexene and then test it for a double bond you could prepare ethanal by the oxidation of ethanol and then distill off the SNL and use toen silver mirror to test for the alahh hopefully you've managed to do this in the lab and are familiar with quick fit apparatus which we can see here in the lab I've done it and a slightly neater nicer drawing of what is going on you need to be aware of the equipment and the safety and where the water goes in and out that is a very important thing in the lab I have used an electronic heater because some of the chemicals involved in this are very volatile and if we use bunson burner for this then there is a risk of chemicals catching fly making it a bit more dangerous you want to heat this to the boiling point of the required product too much and you might get the products you're not looking for and we're going to be separating things by boiling points for required practical six we are going to be testing for alcohols aldhy alkenes and carboxilic acids when you are testing for an alcohol put ethanol into a dried test tube add solid sodium and a posit positive results are bubbles being given off we can use a failings test for an aldah A positive solution is the blue turning into a bright orange so a color change here when we are testing for an alken we can use orange bromine water and with a positive result it will go colorless not clear colorless when we are testing for a carboxilic acid we can add sodium hydrogen carbonate and a positive result be lots of bubbles of carbon dioxide gas being given off we can confirm that the gas is carbon dioxide by using lime water if we're going to be testing for halogens we need to add night tric acid silver nitrate and a positive test will be a color change for your halogens the white cream yellow are very hard to distinguish from each other so it's good to have reference samples so you don't get confused Mass spectrometry can be used to determine the molecular formula of a compound there are several steps involved we start with ionization where electrons are knocked off to give a positive ion they are then accelerated so they all have the same kinetic energy and then they are deflected by a magnetic field the level of deflection is based on mass and charge lighter ions are deflected more and more charged ions are deflected more we can use the computer data at the end to determine the formula of the compound 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 stage 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 ch3 ch2 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 low logical answer if it had different molecular paks if it didn't contain this for example methal 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 so some data from infrared spectroscopy you need to look for some character istic regions there are three you need to know different groups absorb infrared at different set frequencies you will get given a data sheet in exam so don't worry 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 in aldah and ketones you're looking to this char istic region here and for carboxilic acids which will have an 08 and a carbon oxygen double bonds it has kind of a double region one in the same place as the O 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 infrared radiation is closely linked to greenhouse effect the carbon oxygen double Bonds in carbon dioxide absorb infrared radiation thus preventing it from escaping the atmosphere the light that we get from the Sun is visible light or ultraviolet light this light is is not absorbed by carbon dioxide by the carbon oxygen double bond so this light manages to reach the earth once it is emitted from the earth it is at the lower frequency radiation the infrared radiation that gets absorbed by the carbon oxygen double bond preventing it escaping the atmosphere [Music] [Music] [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 is 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 enthropy change of formation this is the standard enthropy 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 one atmosphere of pressure the enthalpy of latis formation is a standard enthropy change when one mole of ionic Lattis is formed from its ions in gaseous form under standard conditions the enthalpy of latis dissociation is a standard enthalpy change when one mole of an ionic lce is dissociated into its ions in gaseous form the first ionization enthropy is an enthropy 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 enthalpy change when one mole of electrons is removed from one mole of + one ions to give one mole of 2 plus ions in a gaseous form the enthalpy of atomization is the enthalpy change when one mole of atom ATS in a gaseous form are formed from that elements in a standard State Bond enthalpy this is the enthalpy chained when one mole of a calent bond is broken homolytically in a gaseous state electron affinity is the enthropy change when one mole of atoms in a gaseous form gain one mole of electrons to form one mole of minus1 ions in a gaseous form the enthropy change of hydration is the eny change when one mole of gaseous ions becomes one mole of aquous ions it is really important to have accurate descriptions for these terms because these could easily come up as exam questions born Harbor Cycles once you get to grips with them are very very elegant but you need to display 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 same the answer will be the same irrespective of the root 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 ltis enthropy of sodium chloride the eny change of formation of sodium chloride the enthropy change of atomization or the enthropy change of sublimation the enthropy change of atomization for chlorine and the first ionization enthropy 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 lce enthropy of sodium chloride from start to end in the solid green line it is a combination of all of the other figures in the highlight lighted 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 sign so it is minus - 348 we are then going in the opposite direction of the first ionization enthalpy of sodium the opposite direction of the eny change of atomization of chlorine the opposite direction so it's minus for the enth change of atomization of sodium and in the same direction as a arrow for the en we change a 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 enthropy step and 2 moles of C minus must be made all of these numbers are based on real experimental data theoretical values can differ based on the calent character of the bonds entropy or Delta s is is a measure of disorder in the system the higher the entropy the higher the entropy value 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 are 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 if we're 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 Gibs free energy has a symbol G or Delta G for changing if a reaction happens or not is the feasibility of a reaction this is a balance between Delta H and Delta s so Delta G the Gibs free energy is equal to Delta H the change in enthalpy minus t e temperature Delta s change in entropy Delta G is in KJ per mole Delta H is in KJ per mole T temperature is in kelvin and entropy is in jewles 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 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 eny change and a positive entropy chain it will be spontaneous at all temperatures however if we have a positive enthropy chain 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 spontaneous Above This temperature 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 the order and B concentration and Y is the order the Little Numbers in our original equation are the stochiometric 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 units moles perm cubed per second for a reaction that is first order overall we can look at the rate equation rearrange it to give k equal rate over the concentration of a replace what we can with our units and then start cancelling and what we have left is seconds the minus one so the units for first order reaction first order overall the units of the rate constant are seconds to the minus one for a second order overall reaction and this doesn't matter whether it is um a squar or whether it's the rate constant and then 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 a subject with rate over a and over B replace what we can with the actual units and then start cancelling out again it is worth writing this out in full every time you see it just to ensure you don't make any mistakes so we reaction at this second order overall the units for K are moles -1 decim Cub seconds-1 K is for a set temperature and this will change this will increase as the temperature increases when you first see the arenus 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 arenus equation describes the link between the rate constant and that temperature temperature up the top there is T this is in kelvin R is the molar gas constant you will get given this value in the exam you do not need to learn it however it is 8.31 Jew per mole per Kelvin EA is the activation energy for the reaction and that is in jewles per mole that e there the lowercase e is the mathematical constant e the uppercase a is the orous constant which is reactant dependent this is more commonly rearranged in this way so Ln k equals L A minus ea/ RT if is going to be in graphical format we're going to have Ln K up one side and then along the bottom we can have 1 over T the gradient for this is minus the activation energy over R we can determine the rate equations and the re 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 reacting with nitrogen dioxide 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 ratees of reaction and this is the rate determining step we can measure the rate of reaction by initial rate method this is also known as the iodine clock and is one of your required practic 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 thy sulfate ions which is in reaction two any excess iodine is then reacted in a solution and will turn blue altering the concentration of iodide ions you can experiment and experimentally determine the order of reaction with respect to iodide ions here we're going to 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 that I've read through the method and already drawn my table out before the experiment start so here I have2 G of magnesium and 50 cm cubed of 1 M of hydrochloric acid I'm now going to add them together and use a gas rangee to measure what's collected so this is quite a complicated experiment to do because your hands need to be doing a lot of things at once you need to be adding the Magnesium to the GL at exactly same time need to bit bung off and you need to practice the same timer at the same time we've got a gas syringe here gas syringes are quite um expensive quite um delicate piece of equipment so you also need to make sure that this doesn't shoot out the end and smash so using a bit of skill and put that on there start the timer and going to keep an eye on the gas image now every 15 seconds I need to be recording the volume of gas produced you can see this gas syringe is moving filling up quite quickly sometimes there's a bit of lag at the beginning um as the gas wi gets stuck you can see this moving quite quickly along just need to be careful that your isn't too quick and that gas ring blows out the end you'll know the reaction is finished when the gas ring 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 liest FIT then what you need to do right down here at the beginning is to get your ruler and line your ruler up on the on the line and you going to need to work out the initial rate so what I've drawn here here is a tangent to the line so we are going to get the line and its steepest part and then going to work out the gradient of this line here draw my tangent worked out the gradient of tangent and I've worked out the gradient of the line when you do different concentrations you can work out the gradient line for each them and compare it 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 say 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 start we can assume that there are 100% of the moles of gas are reactant and we've got 0% no products for example in this situation we're going to have 2 moles of gas a and .5 mol of gas B then we're going to have 1.3 Mol 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. 2 / 2 we can then find the total pressure this might be given in the exam say it's 5 kilopascals the mole fraction of a which is .1 we can then do1 * 5 to find the partial pressure of a KP is going to vary with temperature but not be affected by Catalyst 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 to 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 + 1.10 Vols 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 the 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 a 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 of 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 1 mle decim cubed solution of the iron all standard Electro potentials are the difference between any given half cell and the standard hydrogen electrode one of your required practical is 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 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 the solution of the metal ions and this needs to be connected up to a voltmeter with wires we can then take the reading from the voltmeter we need to have a salt bridge here the ends are buned with a little bit of cotton wool and then fills 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 there are a number of commercial uses for electrochemical cells including fuel cells we can have an alkaline hydrogen oxygen fuel cell cell this is made up from a hydrogen half cell and an oxygen half cell the hydrogen half cell is where oxidation reaction will take place and the oxygen half cell will we have a reduction reaction taking place the overall reaction for this is half oxygen gas plus hydrogen gas gives us water since water is the only product it is more environmentally friendly than some of the other ways of producing energy it is highly efficient however there are disadvantages to this hydrogen is very flammable making storage difficult and dangerous the cells will have a limited lifespan and if we were to do a life cycle assessment of a fuel cell the production of it involves tox XC chemicals all of which needs to be taken into account a commercial use for electrochemical cells is rechargeable batteries before we get into this we need to look at a technical definition the thing that we call a battery a single AA or AAA battery is actually a cell for it to be considered a battery we need to have multiple cells so multiple cells are needed for it to actually be a battery battery non-rechargeable cells have an irreversible reaction happening in them rechargeable cells involve a reversible reaction lithium cells are the ones that are found in Mobile phones if we can charge our phones they have a rechargeable battery or a rechargeable cell in them we have lithium and cobal oxide at the positive electrode the cobal will be reduced in this reaction and we have lithium at the negative electrode we can write it out in a traditional way that we would recognize from our sales and see we still have assault bridge in there so the rechargeable batteries that you have in your mobile phones have this smaller version of the very traditional EMF cells that we're used to seeing before we look at bronzed lar acidbase 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 and nh4+ 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 will accept and the part that will donate the proton a strong acid is an Aid that will fully dissociate you should be able to recognize some strong acid for example hydrochloric acid hydr bromic acid Hydro iodic acid sulfuric acid nitric acid those are the common ones that you should be familiar with but just to expand it a little bit per chloric acid and chloric acid are also strong so because they fully dissociate whenever we're do these calculations we can assume that the concentration of acid is equal to the concentration of hydrogen ions for example if we have1 mole per decim 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 log1 and here I'm going to put into the calculator for you so you can see how to actually use it which buttons you actually need to press cuz this is an area people really fall down on so we can see the pH of nitric acid .1 mes per decim cubed is 1 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 die 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 hydrogen please pay attention to this in the question and check exactly what they're asking you for if we want to calculate the pH of a strong base we can use KW water would associate inch hydrogen ions and hydroxide ions this will weakly dissociate so some will dissociate 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 of hydrons 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 KD concentration of water and call that KW which is the ionic product of water and this varies weird temperature at 25° C KW is 1 * 10 -4 M DM -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 is fully dissociated so 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 of2 moles per decim cubed sodium hydroxide at a given temperature 25° C we can use KW we can write our KW constant rearrange it replace KW with 1 * 10-4 replace the concentration of hydroxide ions because we know that from the question 2 once we have the concentration of hydrogen ions we can do minus log concentration of hydrogen ions to find the pH in this case it would be 13.30 to two decimal places if we want to calculate the pH of a weak acid is a tiny little bit more complicated than calculating a pH of a strong acid but only a tinely little bit more complicated weak acids are one that partially dissociate in water there are a few that you should be familiar with methanoic acid ethanoic acid benzoic 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 ha being the hydron ion and the base and a being the base here water is in excess and because is in excess we can rewrite that as ha so the acid dissociates into the hydrogen ion and the base we can turn that into an equilibrium equation with a concentration of hydron and concentration of base on the top and the concentration of ha on the bottom Ka is the acid dissociation constant you might also see p Ka a 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 ha doesn't change because they are so weakly dissociated we can assume that the concentration of ha at equilibrium is the same as ha at the start meaning we can rewrite 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 ha 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 ion Square divided by the concentration of ethanolic acid at the start we know what Ka is because we were told in the question so we can replace that we know what concentration of ethanoic acid is at the start so we can replace that in the equation then we can rearrange the equation giving us 8.5 * 10- 7 is concentration of hydrogen ions squared little bit of algebra to get rid of that squared we need to square root the other side giving us 9.22 * menus 4 is a concentration of hydrogen ions and always keep your calculator values when you're doing this if you don't know how to keep your calculator 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 pH of 3.04 to two 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 are 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 phenol failin which works at a very high ph and 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 one of your required practicals is investigating how the pH of a solution changes using a pH probe we're going to be looking at the reaction between a weak acid and a strong base the first thing that you need to do is to calibrate your pH probe because they will never be exactly the same you need to have your standards so you know what they are and you need to calibrate them you will need to wash your probe at every single step to make sure there are no stray hydrogen ions or stray hydroxide ions left over from the last solution this needs to be done in distilled water to avoid any errors we can then use our Solutions which we know the value of we can then see what the value is on the pH probe and we can draw our calibration curve like this here this should be ph7 but the probe is coming up a 6.3 we can then do our titration slowly adding in the different amounts of base and then recording the pH as the tiny little bits of Base get added in and this is very very small increments depending on exactly what your experiment says you can then draw your graph your calibration curve the actual pH versus what the pH should have been according to the buffer and then adjust your pH readings so that you know what you've actually got a buffer solution is one that maintains a steady pH even after additions of small volumes of 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 sorts of that weak base one example from biology is that 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 ethanolic acid and sodium ethanolate 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 a 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 clear 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 * 10- 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] [Music] we're going to look at the reactions of period 3 elements with water sodium will react to give sodium hydroxide and hydrogen gas it is a very exothermic reaction that will take place so you might see flames you might see fizzing happening sodium hydroxide is going to be very very alkaline magnesium will react with water to give us magnesium hydroxide and Hy gas this is a much less impressive reaction it will react slowly with cold water but quickly with steam it will still be alkaline but less alkaline solution will be produced roughly ph10 as magnesium hydroxide is less soluble in water sodium or react gives sodium oxide magnesium will react to give magnesium oxide aluminium metal is generally always covered with a thin layer of aluminium oxide and to get it to react you need to rub it and it will react again silicon reacting with oxygen will give us silicon dioxide phosphorus and oxygen will give us p410 or p203 in limited oxygen sulfur reactant will give us sulfur dioxide and some sulfur trioxide and these are all Co valent compounds the ionic compounds will have high melting and boiling points whereas the other ones will have lower ones this is the reaction of period 3 oxides with water sodium oxide will react to give us sodium hydroxide this is a strongly alkaline solution this reaction is very exothermic magnesium oxide will react with water to give us magnesium hydroxide it is alkaline but slow less so at ph10 aluminium oxide is insoluble in water so there is no reaction silicon oxide is also insoluble in water these will give a pH of seven the pH of water P4 h10 will react with water to give us phosphoric acid sulfur dioxide will react to give a sulfurous acid it will have a pH of 2 to3 Tre and this is a weak acid sulfur trioxide will react to give us sulfuric acid for the structure of phosphoric acid for the ion we lose one of the hydrogens and it is replaced with negative charge for sulfurous acid and again for the ion one of those will lose the hydrogen and get a negative charge similar for sulfuric acid except we have two double bonded oxygens on there and one of those will lose the hydrogen period 3 oxides can also react with acids and bases basic oxides will react with acids to Produce Salt and watera with hydrochloric acid to give us sodium chloride and water this follows the very familiar reaction of base plus acid equals salt and water and works for both sodium oxide and magnesium oxide with whichever acid you would like to react it with here magnesium oxide is reacting with sulfuric acid to produce magnesium sulfate and water anic oxides can act as both acids and bases aluminum oxide can react with hydrochloric acid and is acting as a base to give us aluminium chloride and water the rest of period 3 form acidic oxides silicon oxide will react with very concentrated sodium hydroxide and only very concentrated as it is insoluble in a weak base or in water the rest will act as acids as follows sulfur dioxide when released into the environment is a polluting gas that can cause acid rain this reaction is one way of removing polluting gases from Industrial Waste transition metals are fascinating things that sit here right in the middle of the periodic table and the ones you need to know about are titanium through to Copper they will form complexes they have a range of beautiful colored ions they are veryable 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 subshell 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 I metal because it has a full 3d subshell neither is scum the 4S is a lower energy subshell so it is removed first so for cobal 2+ 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 ligans ligans will bond in a dative CO valent Way by donating both of the electrons in the bond a few new terms we're going to be using as well as Li and complex I the coordination number is the number of bonds to the central ion a monodentate ligant 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 iron in the middle square brackets and surrounded by the liens when you are writing it out we have square brackets the transition metal iron rounded brackets with a liant in the middle the subscript number would indicate the number of lians square brackets and then the charge on the outside all monodentate six coordinate complexes have an octahedral geometry there are two by dentate liant need to know about ethane one two diamine and each of them will make two bonds to the central ion or Ethan diano and again each will make two bonds to the central ion both of these have an octahedral geometry we can also have multi- dentate lians here we have EDT PA and there six places that it forms bonds are highlighted here heem in hemoglobin is another multidentate liand 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 we can have substitution reac actions where ligans exchange there are three monodentate ligans 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's been a change in coordination number from 6 to 4 also there will be a change in Geometry as it will now be tetrahedral this can happen with cobal copper or iron we can also have liant exchange by multi- dentate or b dentate lians f by dentate lians we have ethane one2 diamine and Ethan diate you need to learn these however you do not need to learn the structure of your multident liant ad 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 ethan1 two diamines so the exact same coordination number six but because the B dentate Lian makes two bonds we only need three by dentate lians or if we're replacing it with a multi- denate ligan Eda which has a six coordination number will Mak six bonds we replace six Waters with one Eda and this this is the keyate effect bidentate and multi- dentate ligans will replace monodentate lians 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 we need to know the different shapes that complex ions make because they have a 3D shape hopefully you'll be familiar with wedges and dashes from earlier in the course for a maning take Li with the six coordination number we're going to have an octahedral shape these are for small ligans for your larger lians 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 square planer complexes or we can have linear complexes for example in tollins complex ions can show Cy trans isomerism here we have a platinum Central ion and we can have CIS platin or we can have transplatin cisplatin is a very useful anti-cancer drug transplatin is not a very use useful anticancer drug for octahedral complexes where not all of the lians are the same they can also show Cy trans isomerism or if we have a complex ion that has two bidentate ligans and two monodentate ligans these can show CIS trans isomerism it is well worth spending some time time practicing drawing these out in a logical manner complex ions can also show Optical isomerism for example with a hexadentate ligan like Eda if we have two monodentate lians and two bidentate lians or if we have three bidentate lians these two are Optical isens as are these two you'll notice they are mirror images of each other it is possible for something to show Optical isomerism and C Str isomerism complex ions are beautifully colored it's one of the things that makes some 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 liant or the coordination number is changed we see color when some of the wave length invisible light absorbed and removed and then the rest of the wav lengths are transmitted or reflected and this is all in the D electrons we have the average or the ground state of D electrons and when they absorb a light they are excited from the ground state the difference between the ground state and the excited state can be given by Delta eal h v or HC over Lambda Delta is our Chang in E is energy which is meas in Jewels H is planks constant which is 6 33 * 10 - 34w per second we have the frequency of light measured in hertz C is the speed of light which is three * 10^ 8 m/s and Lambda is the wavelength to measured in met if we want to look at the color or the color change we can use spectroscopy or we can use a Colorimeter transition metals have variable oxidation States for example vadium will lose two 4S electrons to become vadium 2+ they will lose the 4S electrons both or they lose the 3D electrons as they have a lower energy vadium with an oxidation state of five is yellow vadium with an oxidation state of four is blue three plus vadium is green whereas vadium with an oxidation state of two is a lovely violet color so we can see lots of color changes with badium and we can see them all in the following reaction starting off with vadium with a plus five oxidation state at yellow it will then go through all of the different colors and oxidation States in this reaction to end up with vadium 2+ we can have silver in a complex ion and when it meets that functional group that alide the silver will come out of the complex iron and you will see a silver mirror the silver will then form and the outside of the test tube this is the silver mirror test alahh and the complex iron with silver in is more commonly known to you as toins the Redux potential of any transition metal ion is influenced by the ph and by the ligant [Music] 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 pottassium manganate in your buet and then your known volume of the solution that we analyzing goes into your chronical flask this needs to be done in an excess of acid indicated for this as potassium per manganate decolorizes as it react 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 manganate is in excess because of the gorgeous deep purple color of the potassium of Mangan 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 tighter again with this question we are going to work through sorting out the numbers and then we are going to do the calculation so un but it was made up in 250 cm Cub solution and from this 25 cm cubed was titrated against 0.02 moles CU potassium manganate 7 the Titan needed to oxidize the iron sulfate was 24.2 CM Cub 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 exam or you might have to work these out for yourself we have our manganates being reduced to manganese 2 and remember that only thing we can add to these is hydron electrons and water to balance out the four oxens 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 balanc Ion is going to be oxidized so it's going to start as Ion 2 and then get oxidized to ion 3 and for this we just need to add on our electrons now we need to balance these to make an over all reaction which just means we need to times the bottom 1 by five when you're balancing combining 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 and you don't make mistakes then we can go back and cross things off afterwards we're going to work out the moles of potassium manganate yeast from that the moles of iron sulfate from that the moles of iron sulfate in 250 cm Cube and then we're going to use that to work out the original mass for our mol of pottassium manganate we can do concentration time will give us 4.8 * 10-4 moles we can look at the ratio in the equation and see that for every one mole of manganite ions we have five of ion giving us 2.4 * 10-3 Mo now we need to work out how much we had in 250 cm Cub to work out the original Mass which is what the question is asking us mass is moles * m which will give us 3648 G transition metals can act as homogeneous and heterogeneous catalysts homogeneous means they're in the same phase hetrogeneous means they're in a different phase 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 a higher rate of reaction homogeneous catalysts where the reaction will happen via an intermediate and the intermediate will generally have a different oxidation state to the reactants or the products the contact process uses a hetrogeneous catalyst badium oxide in the production of sulfuric acid step one we will sulfur dioxide reacting with vadium in the plus five oxidation state and vadium in the products will be in the plus4 oxidation state the products from the first step the vadium 204 with vadium in plus 4 oxidation state will then react with oxygen and go back to being vadium 5 oxide we can cancel out some of the vadium in the two different steps to give us overall equation of sufur dioxide reacting with oxygen to give us sulfur trioxide and it is a sulfur trioxide is actually used in the production of sulfuric acid a harra process using a hetrogeneous iron Catalyst we have nitrogen mixing with hydrogen to produce ammonia and this is one instance where a catalyst can be seen to be poisoned by impurities found in the reactants poisoning of a catalyst can reduce the efficiency of the Catalyst reducing the yield of the reaction and increase the cost of the reaction as you will need to replace the Catalyst this is the reaction between iodide ions and persulfate ions this is a homogeneous catalyst the overall reaction for this is the reaction between pulite ions and iodite 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 persulfate ions reacting with positive ion ions in step two the ion 3+ ions from the products of Step One react with the iodide ions to remake the two plus ion ions which were reactant in step one and to give us iodide ION three 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 mang ions and Ethan diate this uses a homogeneous catalyst and is an example of autocatalysis where one of the products acts as a catalyst so the reaction will start off slow and then will get faster as more products are produced the overall reaction has the manganate ions which are negative reacting with negative Ethan diate ions this is a slow reaction as two negative ions will repel each other and there is a high activation energy in the catalyze reaction manganese 2+ will react with the manganate ions and we will have manganese 3+ as a product the manganese 3+ can then go on to further act as a catalyst reacting with the Ethan diare ions to produce manganate 2+ the activation energy of this is lowered we can then use algebra to cancel out the mean 2 plus and the magnate 3+ ions on either side come up with the overall reaction for our metal Aqua ions the acidity of a complex ion which has a central transition metal with a three plus charge is going to be greater than one with a two plus charge as the three plus Hons have a higher charge density they will attract water more strongly weakening the bonds and more easily releasing hydrogen ions there are lots of metal 2+ reactions that you need to be aware of you need to know the reactions of copper and hydroxide ions all of these follow a very similar pattern so if you learned the example you should be able to apply it in lots of different situations you also need to know copper and ion reacting with carbonate ions and be able to work out the colors of the precipitates the metal Aqua ions with 3+ are a few reactions you need to know you need to know Iron 3+ and aluminium 3+ reacting with both hydroxide and ammonia again these react in very similar ways so if you learn the example and and practice it applying you should be able to do it in an exam and the metal Aqua 3+ ions reacting with carbonate ions these have slightly different products so you can learn these examples these will have carbon dioxide and water as additional products aluminium hydroxide is amphoteric which means it can act as an acid or can act as a base again you should learn these two examples here and then be able to apply them in an exam we can use practical methods to determine what is in our complex ions and this is one of the key practicals we are going to be looking at iron three nitrate copper 2 chloride an ammonium Ion 2 sulfate if you are using the AQA practical instructions these are q r and s ion 3 nitrate will start off as a yellow solution copper 2 chloride starts off as a light blue solution and ammonium Ion 2 sulfate starts off as a pale green solution test number one is going to be the addition of sodium hydroxide we can do this slowly drop by drop until it is in excess here you can see my three experiments Ion 2 nitrate will go from a yellow solution to forming an orange brown precipitate copper 2 chloride will go from a light blue solution to have a deeper blue precipitate an ammonium Ion 2 sulfate will go from a pale green solution to a kind of gray green precipitate you can see the precipitate cuz it goes cloudy in test number two we start with 10 drops of sodium carbonate in the test tube and we slowly add in our test solution so ion un nitrate will give us an orange brown precipitate copper 2 chloride will give us a blue green precipitate and ammonium Ion 2 sulfate will give us a gray green precipitate for test number three we are adding silver nitrate iron 3 nitrate we will not see any visible change happening here no change will be observed for copper 2 chloride we will see a white precipitate and unfortunately I don't have a video for the third experiment here but what you will see is that there will be a light round precipitate forms if I had a video of this which I don't and I'm truly exed [Music] [Music] Optical isomerism is also known as chyal 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 mods is much easier to understand these are drawn flat the same way but you see they are mirror images of each other they cannot be superimposed on top of of each other these are different en antias and these will affect light differently Optical isas in antias have similar physical and chemical properties but rotate polarized light differently if the polarized L 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 produce a mixture a 5050 mixture of each an antia this is a racemic mixture because 50% of the light is being rotated clockwise and 50% is being rotated anticlockwise there is no over all rotation of light the difference between the enanas is important when we looking at drugs and enzymes some of them will only work with one an antima there are a few famous examples of this one being thalidomide where one enantia will cause birth defects and the other nanti won't and the other one is iprof where one is useful and helpful and the other an antima isn't so is important that drug companies have a way of separating out the enanas we're going to look at alides and ketones together because these carbonal compounds are very similar alhida carbon is double bonded to an oxygen and a hydrogen at the end of a group whereas a ketone the functional group will be in the middle a carbon double bonded to an oxygen this alahh you can see here in the Molly mods is buttin Al the AL bit tells us it's an alahh and the someone at the bottom is butan own own bit tells us it's a ketone they are soluble in water due to the ability to form hydrogen bonds there are two tests to tell the difference between alides and ketones we have a test with Tolen reagent and we have a test with failings reagent I covered both of these in more detail earlier in this video both will only give a positive result with an alahh so toins 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 alahh and this aldah can be further oxidized to a carboxilic 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 Tetra hydroborate in aquous solution this will give us hydrogen minus ions we can have ethanal with its partial charges within the molecule being attacked by the hydrogen nucleophile this will give us an inti mediate 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+ ions 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 conditions 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 alahh 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+ ion to give us our end product and now lots of students forget that s 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 three carbon chain here making this two hydroxy propan nitr here we've gone from an alahh to a hydroxy nital we have a similar reaction with our Ketone we have our Ketone here known 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 we will 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 two methy propan nitr the methy is a group The hydroxy is a group but the nitr is the higher priority group so is the main stem of the name the hydroxy nitrol that we got from the reduction of an alahh has a chyro carbon there in the middle so the reduction of alahh highes and asymmetric ketones will produce an antias when we are drawing and naming car oxyc 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 carboxilic acid this is propan OIC acid the OIC acid bit 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 salts of carboxilic acids here if the sodium positive ion is attracted to the negative charge we can get a sodium propanoate salt we can identify carboxilic 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 Esters are made when we react carboxilic acid with an alcohol here we have ethanolic acid and propan one all and then we have our Esther coming out of it at the end the name of the Esther comes from the alcohol and the carboxilic acid so ethanolic acid and propile one will give us propile ethan8 we will also get water out at the end of it and the highlight atoms in here are the ones that go on to make water for the condensation reaction it's 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 it is reflux with dilute hydrochloric acid or in alkaline conditions it is reflux with sodium hydroxide in alkaline conditions we will then num up with a sodium salt of carb carboxylic acid esters 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 animal fats and vegetable oils are made up from propan 123 trial which is also known as glycerol and longchain carboxilic acids which are also known as fatty acids each faty acid will react with an alcohol group in a condensation reaction to give us a very large Esther and water oils and fats can undergo hydrolysis reactions in Alkali conditions and this will give us glycerol and salts o of our carboxilic acid and these can be used as soap biodiesel is made up from methy Esters of carboxilic acids here we have our Esther we can react it with methanol to get glycerol which is remember propan 123 trial and our methy Esters acid and hydrides have a rather complicated looking functional group but it's really not that scary this is ethanoic anhydride and this is safer to use or acid and hydrides are safer to use than oo chlorides in the manufacturer of aspid or other industrial processes AO chlorides will give off hydrochloric acid as a toxic byproducts whereas acid and hydrides will give off carboxilic acids acid and hydrides will react with water and I've drawn the acid and hydride the two different halves in different colors here because when the acid and hydride reacts with water it will give of carboxilic acids they don't have to be the same carboxilic acids here is giving off two different carboxilic acids for this reaction you just need water and it will occur at room temperature acid and hydrides will react with alcohol to give us an Esther and a carb oxyc acid for this reaction you need alcohol and it will occur at room temperature acien hydrides will react with ammonia to give us a primary amide and a carboxilic acid if there is excess ammonia this carboxilic acid can be turned into a salt for this you need ammonia and the reaction would occur at room temperature carboxilic acids will react with Prime primary amines to give us secondary amides and a carboxylic acid again if we have excess then we can get a salt from the carboxilic acid for this reaction to occur we need the primary amine and it will occur at room temperature ASO chlorides have a similar structure to carboxilic acids or alides with a carbon double bonded to and oxy but the other thing bonded to the carbon is a chloride this is much more reactive than carboxilic 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 the erors start where the ARs go to a Sol chlorid 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 iron and then at the end we will get carboxilic acid and hydrochloric acid for this you need water and and it will occur at room temperature ASO chlorides will react with alcohols this is a very very similar reaction you can see I've just replaced one of the hydrogens with a ch3 group if you learn one of these mechanisms you should be in a really good place to do all of the mechanisms the oxygen the nucleophile will attack the carbon there with partial positive charge we will get an intermediate set up and our products at the end will be an Esther and hydrochloric acid for this you need alcohol and the reaction will occur at room temperature asoc chlorid will also react 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 giving us an intermediate at the end we will get a primary a and hydrochloric acid for this reaction you need ammonia and it will occur at room temperature ASO chlorides will react with primary amines to give us secondary amides and hydrochloric acid this reaction needs a primary amine and will occur at room temperature we're going to be testing the purity of this solid here I have melting point tube which is a very very fine tube which has an opening on each end and what I need to do is just fill this with the organic solid to about 5 mm in depth now you don't want to get stuff on the end you just want it in the center of the tube here so this is our ancient melting Pointes these are melting point tubes with a tiny bit of our solid that we just made in there and then what we need to do in the top here is place in the thermometer three samples and Tes two points and we can see through this little hole here we can see our three samples and what we need to be watching is for when they melt and they should all melt at a certain point it shouldn't be over a range and that is a pure you can see here that the sample has melted and it is way below the temperature than it should be cuz it wented about 78° and we wanted it to between 100 120 so a bit of fail for us there but you can see how the melting point apparatus works so here we're going to be using a round bottle flask and obviously these um are round so to hold that study I'm just going to put it in a large speaker to hold it as a secure um base we are going to making an organic solvent so I have 50 cm Cub of ethanol 50 cm CU of glacial ethic acid this stuff is quite nasty um we are using some quite nasty things have doing all of this in a fume hood and give that a bit a swell to mix it and then really slowly I'm going to add some concentrated of furic acid you can see if you look closely um the concentrated s furic acid changing the um the bated then my finish so after mix all those things I've set it up to reflex here I have my mantle um you could use a budson burner or water bath if you don't have one of these available here's my round bottom flask this is just sitting ever so slightly above the basket here so it's not sitting on it it's sitting so slightly above it got my connector and I've got my reflux condenser here the important thing to 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 the top so we can see this is bubbling away now and if we look really really carefully in here you can see if I just wiggle you around a little bit you can see that it's evaporating and it's condensing so you can see that the cold water is causing um anything that's evaporated to condense and then drop back down into the flask where it can be reboiled so reflux just let the same things be reboiled over and over again has been refluxing for about 10 minutes we just need to change all the CR glass around so it's now going to distill Now we' set this up for brief um distillation we've got exactly the same um bottom flass bubbling away here we've got um a corner and stop on here and we've got our condenser and then all everything that's distilling off is collecting in this little V here if I show you the drops so it evaporates it condenses and then the the gas sent along here it condenses travels all the way down here and drops into the end so now we've um distilled off 2/3 of our solution we need to add some uh sodium carbonate to it 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 stop off the top pour in my sodium carbite pour in um my just atill here and then what you need to do is pop the lead back on this we're going to be inverting it so we need to make sure that it's quite secure you need to have Qui firm holding this iners it several times to give it a mix and what you need to do is open the Gap open the um tap to release any gas that comes off I would do this maybe two three times invert it until until you stop hearing the little um fiz the little release that tells you a gas been produced and then turn it back up the right way and allow it to separate out into two different layers you can already see here that we've got um 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 clamp if you want to I'm just going to hold it for a second so I've just open this tab ever so slightly if I want to open it Fuller it will go faster but I want to have control of how this works so I'm just opening it drop by drop and we are 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 get get discarded with my lower layer that's absolutely fine if you wanted to be keeping the lower layer and discarding the top layer then I would stop so there's a little bit of the lower layer still in here in here but since we're going to be discarding the lower layer stop that let it settle down for a bit since we're going to be discarding the lower layer I'm going to go over so slightly past um the line so I lose a little bit my top layer but I make sure that it's all pure so I'm slow that down so I don't know bubbles you can see the layer is just here stop that drop a couple more fuses so I can still see the layer it's just about here now now in here I just have my top layer and that's what I'm going to use for the next parts so now I've done that once I need to do that again with 20 cm cubed of saturated chaosium chloride so my ster is closed and this is the top layer from before so in that goes again I need to mix it then make sure secure grip mix it a couple of times open the tap to really release the gas mix it a couple more times open the tap to release any gas and then keep mixing it until you can't get here any more gas released and let it separate out so again we can see two layers separating out 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 to relas spit the pressure the top start running that down there okay so it goes quite quickly once you take the top off I'm just going to let it settle 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 my top layer I'm just going to let that go through into there taking the top off gas now I'm just going to add some um solid um an hydrous calcium chloride doesn't say how much so I'm going to add that much give it a bit SW and then what I need to do is to transfer the liquid into a round bottom glass and set up a distillation again now we're going to distill off again but what we've added into the apparatus is a thermometer just here so what we can do is collect of the different fractions which come off at different temperatures because we only want a certain uh fraction we are waiting for something to distill off between 74 and 79° so everything that is coming out at the moment we can just discard 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 hexagons 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 cyclohex 135 Trine with double Bonds in there alternating between double and single bonds however there's some evidence that supports the ring structure with Deliz electrons over the double single Bond alization it has a PL art structure it is flat if it was alterating 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 cyclohex 13 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 these are the important bits you need to know for the exam but I don't think it hurts to tell you that 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 h23 plus ion and a hso4 minus ion which I'm just going to rub out cu we don't need that for the moment the h23 plus ion will then go on to give water and n2+ for step two we have our Benzene and it is going to react with our electrophile the no2+ we are going to get an intermediate our hso4 minus ion 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 at 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 aluminium chloride and our asol 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 al4 is going to drag off that hydrogen giving us our product reforming the Catalyst and hydrochloric acid as a byproduct we can to 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 propile Amin is a primary amine but if your amine is in the middle if it has one hydrogen attached to it not two it is a secondary amine so this one would be n methy which is that group there ethal 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 methy ethy propile Etc if we have ammonia and hogen or alcane we will get a 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 where 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 nit trial to a primary amine step one would be the hogen or alane plus cyanide ion come from something like pottassium cyanide to give us our Nitro set up here and then in step two we can take our nit trial reduce it to give us our Primary aine in this situation the hydrogen for reduction comes from l i a l H4 when we have aromatic aiming we're going to take our nitrobenzene and add in a reducing agent that's hydrogen square brackets which is going to turn the N2 group into an nh2 group and water this will give us pheny amine we can use tin or iron as a catalyst and we need to heat it with hydrochloric acid pheny amine is used in the manufacture of dyes amines can act as bases as they can accept a proton starting with the weakest we have aromatic amines here we have pheny aine 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 we 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 accept the making it a better base there are lots of reactions of amines these can act as nucleophiles halogeno Aline plus ammonia will give us a primary amine this primary amine plus a halogeno alane will give us a secondary amine this secondary amine plus halogeno alcane will give us a tertiary amine this tertiary amine you can see where I'm going with this can't plus a hogen will give us aary 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 hogen alcane and ammonia the ammonia the lone pair is going to be attracted to the carbon and then the broing is going to get shifted out the way we're going to have an intermediate which ammonia is going to be be attracted to again and then we are going to have our primary aiming we're going to start from the same place the same hogen alane except this time instead of ammonia we've got our primary aine it's still exactly the same with the Lan pair on the nitrogen being involved our intermediate this time is slightly more complicated but only slightly more complicated and my still going to do the same thing in the same place and our product at the end is going to be a secondary aain these may look really horrible and complicated but once you get used to drawing it it's fine starting with the same hogen or alkan again we are then going to be using our secondary aine with nitrogen acting exactly the same way our intermediate is looking a little bit more complicated but amonu is going to be doing exactly the same thing and we're going to end up with a tertiary aing at the end back again to our halogeno alane and this time we have our tertiary amine the nitrogen that being involved and this time we go to aary ammonium salt and these are used as cationic scts condensation polymerization occurs between dicarboxylic acids and diales dicarboxylic acids and diamines or between amino acids dicarboxylic acids and dioles will give us polyesters because that's the linkage dicarboxylic acids and diamines will give us polyamides because that is the linkage and amino acids will polymerize to give us proteins here we have our dicarboxylic acid with a carboxilic 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 will be an Esther linkage set up please note carefully how I'm drawing this with 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 diester is benzene for dicarboxylic acid reacting with ethane one2 dial and this will give us terene 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 hex dioic acid this is reacting with hexan 16 diamine to give us Nyon 66 you can have different numbers in the middle of nylon you can have nylon 46 depending on the number of carbons in the middle both polyesters and polyamides nylon terene 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 14 dicarboxylic acid reacting with Benzene 14 diamine and the common name for the product is kevlar this is used in blit proof vests here we have a polymer of 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 three strong bonds between strands of polymers biodegradable polymers are a brilliant alternative to sending polymers 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 biod degradable 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 finite raw materials lots of polyalkenes are made from crude oil and this is finite and has 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 are 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 common place thing to do and we have the systems in place for 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's also the issue of pollution visual pollution air pollution from the degradation of things and mixing of products poos 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 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 replaceable with any different things to make the different amino acids but it would always have the same basic structure and amino group on one end and a carboxilic acid group on the other end the r groups are changeable and that will lead to all of the different properties this carbon in the middle means it is chyal in nearly all circumstances apart from when the R Group is H and then it is not chyal 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've 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 zit 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 wheat buffer there are different levels of structure within proteins here we have two amino acids they going to have different R groups so I drawn them as R and R Dash we are going to have a condensation reaction between the amino group and the carboxilic acid group this is going to give us a peptide linkage in in the middle our monomer here is an amino acid and two of them joined together is a DI peptide this is a condensation reaction so it will lose water you can take your dipeptide or your protein chain and it will undergo hydrolysis reaction to remake the constituent amino acids the long chain of different amino acids is the primary structure this will fold up in two different ways to make the secondary structure it will either be an alpha helix or a beta plet sheet the next level of folding will give us the tertiary structure within which you can see helices and ple sheets there is then the complex quary structure which is more than one polypeptide chain the structure of the alpha Helix and the beta plet sheet is held together for intermolecular bonding these can be hydrogen bonds as you can see here disulfide Bridges when your R Group has a sulfur in it or ionic bonds all of these help to hold together the secondary and the tertiary and the quaternary structure enzymes are very complicated proteins they generally have a quaternary structure the active site of an enzyme where the reaction actually happens is very specific for the substrate so it might only work with one n antima most amino acids are all the L or negative n antimo this orange representation here this enzyme is required for the maturation of the Corona virus if we know the shape of the active site of the protein then chemists can develop Inhibitors which will fit in the middle here drugs that can block the active site and can be effective medicines there is a lot of computer AED design that can be used in the development of drugs if we know what the structures look like and we know what the active site looks like we can play around trying to fit things in you need to know the structure of DNA and you might be aware now that I love Molly mods so here is my Molly mod building of DNA the structures are given to you on the data sheet so you do not need to learn these but you do need to be familiar with them you get given the phosphate iron and the two dark ribos and you get given the structure of all of the bases there are four DNA bases that you need to know about and another one that's just in RNA but that's a bit more biology the purines are guanine and adenine and the pyramides are cytosine thyine and only in RNA which is a bit beyond this uracil C and G will make three hydrogen bonds to each other and a and t will make two hydrogen bonds to each other a nucleotide is a phosphate ion plus a two dioxy ribos plus a base a strand of DNA is a polymer of nucleotides and your overall structure of DNA is a double helix of two complementary strands CIS platin is a complex ion it has platinum in the middle it has two chloride lians and two ammonium lians and it is CIS platin that we are interested in as an active anti-cancer drug transplatin is not active as an anti-cancer drug because of the different structure here you can see CIS platin in the DNA the chloride ions in the structure are replaced by nitrogen ing guanine this is a ligan substitution reaction the Helix sheep is deformed and DNA replication is prevented so it is effective at stopping cells replicating and effective anti-cancer drugs but when you are using it there needs to be a balance between killing off the cancerous cells and killing off the healthy living cells because there will be negative side effects from killing off healthy living cells 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 where I 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 to pause it copy stuff down and then use that to answer questions [Music] [Music] [Music] [Music] [Music] [Music] n [Music] n [Music] [Music] n [Music] [Applause] [Music] proton NMR 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 shows the number of different environments we're looking at the chemical Shi shift will shows 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 the spin coupling pattern will show us the adjacent hydrogens when we looking at chemical shift TMS is going to be the standard this is given the value of zero and everything is compared to this this is tetramethyl siline and is used because it is symmetrical so all the protons are in the same environment the solvent we're looking at could be deuterated chloroform and deuterium doesn't absorb so it won't give a peak other ones could be deuterated D methyl sulfoxide 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 chance are it's parts of a carboxilic acid this is all going to be shown on your data sheet 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 proton is attached to a carbon that has three hydrogens attached to it whereas if you look at this second one it's got a slightly different proton environment set up there are three different types of environment 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 group adjacent to a ch3 group and there are two protons in this environment the Orin ones can be described 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 it is going to have integration number of three or 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 integration number will be two the orange ones are adjacent to a ch2 group and there are three 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+ 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 0 + 1 gives us 1 which means we'll have a single Peak which is called a singlet if it's adjacent to a c h 1 + 1 is two means we will get two peaks or it will look like a doublet if it's add to a ch2 2 + 1 is three we will get Three Peaks or a triplet if it's adjacent to a ch3 3 + 1 is four means we'll get a quartet with four pigs 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 going to be a ch2 group and a ch2 group we can have a triplet and a doublet meaning we're going to have a CH group and a ch2 group we could have quartet and a triplet showing a ch2 group and 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+1 rule seven paks is going to mean six hydrogens and most common example of this is going to be two ch3 groups or two methal groups attached to the same carbon now when we are looking at Peaks you have to remember that it is next to so the ch2 group is responsible for the triplet 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 generally going to be a carbon dou oxygen the chemical environment of a particular carbon 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 to look at a couple of [Music] examples all have the same formula but a different arrangement in space we going to look at the different carbon environments but one has four different carbon environments whereas as we move through the compounds we can start to see that the methal groups become interchangeable and they are in the same carbon environments if something has four carbon environments it will have Four Peaks on a carbon NMR if something has two carbon environments there 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 moved by the spots 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're never going to be beautifully neat this can be used to separate amino acids and you can visualize spots using Lin hydren or UV this is based on the separation being a balance between the solubility in the mobile phase and retention in the stationary phase column chomatography 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 chromy 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 now we're going be looking at TLC so I've got my tablet of aspirine which I've crushed with a pecal and mot here I've Got My TLC plates and particular plates that we use are shiny on one side and matte on the other side is the matte side that we want to be using and the first thing we need to do and you should really be doing this with the ruler is about a cenm up you need to draw a pencil line so now I've solved my aspirin of ethanol and I have my um dots drawn a pencil at the bottom of my TLC paper I have in my hand here a little bit of capillar tubing this is very very fine open tubing and what I'm going to do is just pick up some of the aspirin in the tube the aspirin solution in the tube it's capillary tubing it should run up the tubing a little bit and then I'm going to dot it onto spot number one now you'll need to do this quite a few times allowing it to dry in between each time the reason we allow it to dry and do it quite a few times is because if you just do a lot in one go you'll get a big spludge whereas we want a tiny concentrated spludge so do a little bit let it dry do a little bit let it dry do a little bit let it dry and then you get the TR concentrated spge that'll give us the best results so now I have my TLC plate in my chamber I'm just going to PO a lid on top of this to make sure that um none of my solvent evaporates now I just want to point out that um here I actually have a little bit of a gap in my lid if you can find because when you don't get a gap that'll work much better now you can see the Sol 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 about 1 cm on the top and then we can stop okay so after you bu it you're going to get something that look like this I will admit this isn't the plate I just showed you because ruv light is broken at the moment but you'll get big blobs look like this are probably slightly elongated and in a regular shape and you want to measure from the middle of this blob down to here we can use this value to calculate the RF value ouch this is why in some videos I explain scratches [Music]