Transcript for:
A Level Chemistry Summary

hello lovelies in this video we're going to go over all the content that you need for your OC a level chemistry and this is everything all in one video we're going to start from module two and then work our way through to module six if you Rising for a topic test or if you're Rising for one of the exams towards the end and you just want to revise a specific section then the time stamps in the description down below is where you're going to find the bits that you need also in the time stamps down below you'll find the link to the free revision guide which you can get from on my website now in this it's the same audit as the video it has everything the examiners think that you know everything you're expected to know before you walk into the exam and you can take off the bits that you do know and if there's anything you don't know you can click on the link in there and it will take you through to my longer teaching video which will go over everything in a lot more detail for you also in the description you'll find the links to the predictive papers that are written specifically for this year and having been check by experienced examiners there is also the free course on my website which has thousands of multiple choice questions to help your advise and links to the life revision workshops that we're running [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 plus one neutrons are also in the nucleus have a mass of one but a charge of zero electrons are found in the outer shells their mass is very small it's 1,36 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 atch or charge on this we can measure in kums and is very very small but it's is much easier to say + 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 scare it is mainly empty space the nucleus the diameter of the nucleus is basically 10^ the -50 M whereas the whole spherical diameter of the atom is 10 to the minus 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 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 decimal it 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 elements 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 8 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 that different masses means they might have different physical properties ions are atoms that have lost or gained electrons for example here we see the electronic configuration 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+ one and plus two or you can work it out group one will form plus one ions group two will form plus two ions the transition metals have a variable oxidation state group seven will form minus1 ions and group six will form minus 2 ions but they are all aiming for noble gas configuration so here we have calcium 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+ 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 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 112th the mass of one atom of carbon the relative atomic mass is the average mass of one atom compared to one 12th the average mass of one atom of carbon you can also see these referred to as AR and M the mass number that we see on the periodic table is an average of all of the naturally occurring isotopes of an elements we can use the mass spec to work this out here we see that 20% the naturally occur in boron has a mass of 10 and 80% of the natural luaran 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 the periodic table it is 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 minus 2 ions and ions of group seven elements are going to form minus one ions however the transition metals have variable oxidation state and you also need to know the formula and the charges on some of the complex ions we can to look at a couple of examples of forming the formulas of ion compounds from knowing what the individual ions are sodium carbonate sodium is going to have a plus one charge and carbonate has a min-2 charge since overall sodium carbonate has no charge need the positive and the negative charges to balance each other out the three the Roman three in ION three means it has a + three charge and sulfate has a min-2 charge now ion sulr over all has zero charge so we can look at this by doing 3 * 2 and just kind of swapping them over so because iron has a three plus charge we're going to need two of those to get to + 6 and because sulfate has a minus 2 charge we're going to need three of those to get to Min -6 giving us the formula of fe2 Open brackets s o4 close brackets 3 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 it 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 left hand side we have three hydrogens one iodine one sulfur and four oxygens over on the right hand side we have four hydrogens two iodines one sulfur and one oxygen so we can see straight away that there is a difference here the easiest thing to do is to start with increasing the number of oxygens because they're both only in one place on each side and then redo your numbers so we have 10 hydrogens two iodines one sulfur and four oxygens we've fixed the oxygens we can now move over to the left hand side and look at something else now we could increase the iodines next but that is just 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 iodines one sulfur four oxygens so now our hydrogens our sulfur 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 series clear to The Examiner exactly which bit they should be looking at 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 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 the smallest number of significant figures so this is what we need to mirror in the answer otherwise we are changing the resolution of the answer we cannot give that to that accuracy because this number is to not that accuracy if we think back to our GCSE math 0.02 can have a wide range of numbers if we're looking at upper and lower bounds you will frequently be asked to convert between units especially for temperature to go from Celsius to Kelvin you add 273 Kelvin to Celsius minus 273 cm cubed decim cubed / th000 cm cubed to M cubed IDE million decimet cubed to me cubed IDE th and the other way around decim cubed cm cubed * th000 m cubed cm cubed time million and then me cubed to decim cubed times by, for concentration calculation if you want to change from moles per decim cubed to G per decim cubed you need to change it by the M of the substance a mole is the amount of a substance that contains the same number of particles as the number of carbon 12 atoms in 12 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 seeing this written as n = m m mass is in G and Mr is in G per mole the number of particles is equal to the number of moles times avagadro's number so we could get a questions such as this calculate the 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 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 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 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 DM 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 to two atmospheres at room temperature because we know the pressure was doubled P2 is two atmospheres which means P1 is high 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 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 ided by m 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 Mr there is a difference between the molecular and the empirical formula the molecular form for 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 it equals divided by the lowest number and this is going to give us the ratio if we follow this format here we should be fine so the elements in the question phosphorus oxygen find the numbers in the question it doesn't matter whether it's grams or percentages or anything find the numbers in the question and write them down in the appropriate place we then need to divide that by the AR and you can get the number from your periodic table and whatever your periodic table says use the number that it gives on there don't use the whole number or any other number use the number it gives on your periodic table a number here and then we need to divide it by the lowest number the lowest number at these 2 is 2 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 p25 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 five 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 when we are looking at 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 you're a 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 to do 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 to determine the formula of the aluminium 3 nitrate aluminium is a three plus ion nitrate is a minus ion that is going to give us aluminium Open brackets n O3 close 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 mash 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 divide that by the M the 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 if 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 ratio iOS so how many Waters were there for every aluminium nitrate that is just dividing it by the smallest number so for every one aluminum nitrate we had nine Waters giving us xal 9 and the formula a Open brackets N3 close brackets 3.9 H2O before we jump into going over the details on the Practical activity group I just want to explain a few things to you there are lots of different practical activities so practical activity group one has three different experiments in it they are all very similar because they are teaching you their skills I'm going to go over all of these in detail but if there's one that you have or haven't done in class don't worry don't think your teachers missed something out but also don't think you can skip it because you might have done something you might have something different but in the end all of the practicals in the Practical activity group are going to help you in the exam so it is worth going over all of them even if they're not the exact one that you have done in class the practicals you come across in the exam will be similar using similar skills to what you have done but in a new context and the Practical activity groups pop up all all through the course so something that came up really early on in the course might also come up later on in the course there was a real mix of topic areas for practical activity group 1.1 you might have done this or you might have done something different or something similar this specific activity is looking at the determination of the composition of copper 2 basic carbonates the term copper two basic carbonate might be a little bit confusing but let's just focus on the formula because you don't actually need to know the basic and you will understand what that dot means when we get around hydrated salts but here we have copper that is been left out and it's gone kind of that greeny color that you might see on Old roofs and it is reacted with sulfuric acid out of this we're going to get copper sulfate water and carbon dioxide and it is the carbon dioxide that we're interested in here because we can collect this gas and when we can collect it we can measure the volume of gas that's been released and then use Mass use equation to work backwards determine the composition of copper two basic carbonate because your practical activity groups are skills based you might have done this in a number of different ways you might have done it with an inverted measuring cylinder you might have used a gas arrange from the equation we know that copper two basic carbonate and carbon dioxide andal 1: one mol ratio and if we know the volume of carbon dioxide we can work out the moles of carbon dioxide and then we can work out the mass of carbonate moles of carbon dioxide is equal to the volume of carbon dioxide over 24,000 which is the volume of gas the mass of copper carbonate is going to be equal to the moles of copper carbonate Times by the m r of copper carbonate copper has a mass of 63.5 carbon has a mass of 12 oxygen has a mass of 16 and there are three of those so it's 16 * 3 giving us an MR copper carbonate of 123.5 once we know the mass of copper carbonate having worked it out from the previous reaction we can take that mass of copper carbonate that was reacted in the experiment divide it by the mass of copper to basic carbonate that we used in the experiment times it by 100 to work out the composition practical activity group 1.2 sounds very different but it has the same skills and a very similar calculation so even if you didn't do this exact one in class it is really worth going over it in this video we are going to be experimentally determining the relative atomic mass of magnesium not just looking up on the periodic table so if we reacts magnesium with sulfuric acid we'll get magnesium sulfate and hydrogen gas this hydrogen gas is being released which means we can collect the volume of it and once we know the volume of it we can then use that in our calculations here is a slightly different practical to the illustration I showed you before where we have the inverted measuring cylinder and we are collecting the gas that has been revolved from within the chical flas and then using a measuring cylinder to collect the volume of gas that is displaced in the measuring cylinder again looking at our equation we have a 1: one ratio of gas hydrogen involved to magnesium so we can do moles of hydrogen gas is equal to the volume of hydrogen gas collected divided by 24,000 very similar to the previous experiment we can see the Mr of magnesium is equal to the mass of magnesium that we used in the experiment divided by the moles of magnesium that were reacted and we know that there a one: one ratio here so the moles of magnesium are going to be the same as moles of hydrogen practical activity group 1.3 is determination of the formula of magnesium oxide this has always slightly irritated me because the formula of magnesium oxide is quite easy to work out so we're going to be experimentally determining the formula of magnesium oxide because magnesium is a two plus ion and oxygen is a minus two ion so the formula of magnesium oxide is simply mg o I would expect my GCC students to be able to work that out but for a level you need to show how you can work it out we are going to be heating magnesium in a crucible and you need to know the mass accurately at every single point so you need to know the mass of the empty Crucible with a lid the mass of magnesium plus The Crucible and then after you've heated it the mass of magnesium oxide plus The Crucible we can take the mass of magnesium plus The Crucible Al together and minus the mass of the crucible that would give us the exact mass of magnesium that was used then we can take the value after we've heated it so the mass of magnesium oxide plus the crucial minus the mass of the crule that would give us the mass of the magnesium oxide so now we know the mass of magnesium we worked it out in blue what we need to do now is to work out the mass of the oxygen so for this we can take the mass of magnesium oxide minus the mass of magnesium and that will give us the mass of oxygen once we have the mass of magnesium and the mass of oxygen we can just follow the process for an empirical formula calculation to work out the formula for magnesium oxide 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 G and volumes in decim 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 Chlor IDE 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 we use the ideal gas law a lot and you need to remember this equation PV equal NR T P stands for pressure V is the volume n is the number of moles R is the gas constant you need to remember it's gas constant 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 juw per mole per Kelvin temperature is measured in kelvin now the conversion for this is often where people go wrong determine the mass of a 500 cm cubed sample of hydrochloric gas when at 20° C under 150 kilopascal pressure for any calculation question the first thing I like to do is to highlight all the numbers and then pull them out of the question and write them down separately some questions can be very 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 kilopascals and we need it in pascals so 150 kilopascals is going to be 150,000 pascals volume is given to us in centime cubed and we need it in me cubed to go from centim Cub to me 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 the equation and the first thing we're going to find out is the number of moles so we can rearrange the equation to give us n = pv/ RT and then plug in in the numbers once we have the number of moles we can use Mass equals moles * Mr to give us the number of grams to give us the mass that the question is looking for often in a reaction there is a difference between how much we think we're going to make and how much you actually get this is called the percentage yield we can calculate percentage yields by dividing the actual yields by the theoretical yields and multiplying it by 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 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 it could be hard to collect it safely it could be difficult to separate it from the unreacted reactants there could be side reactions occurring that were not predicted at the beginning giving you products but not the product that you want atom economy when you do a reaction you will end up with a certain mass of stuff at the end the mass of the products but not all of that is actually useful some of it is going to be waste and some of it is going to be useful so we can calculate percentage atom economy by looking at the mass of useful products divided by the mass of all reactants and because it's percentage we times it by a th000 atom economy can be improved in two different ways we can find an alternative reaction pathway that has less waste or by finding uses for the waste products it is going to make your life a lot easier if you learn all of these formulas within acids hydrochloric acid is HCl this will give us a H+ ion and a CL minus ion sulfuric acid is h2so4 this will give us hydrogen ions and sulfate ions and hydrogen sulfate ions nitric acid is hno3 this will give us hydrogen ions and nitrate ions ethanolic acid is a carboxilic acid it has that functional group on the end it will partially dissociate so that we will have a negative charge spread over the oxygens and a hydrogen for the alkaly sodium hydroxide is n AOH that will give us positive sodium ion and negative hydroxide potassium hydroxide is Koh giving us a positive potassium and a negative hydroxide and then we have ammonia which is NH when we talk about strong and weak acids it is important we don't get them used because there is a difference between the strength of an acid and the concentration of an acid strong acids will fully dissociate they will have a very low PH weak acids will partially dissociate strong acids will generally have a pH less than three these include a hydrochloric acid or hydronic acid and sulfuric acid weak acids will still have a pH less than seven but maybe not all the way down to zero examples include ethanolic acid this is going to represent water here we have our acid and the rest of the compound whatever that might be if we have a strong acid at a low concentration the acid will be fully disso iated from the rest of the compound they will be completely separate the hydrogen ions if we have a strong acid at a high concentration we're going to have the same number of water molecules in here but more acid and they are still all fully dissociated there's just more of them because they're at a higher concentration if we have a weak acid at a low concentration you will see they are partially dissociated some will be dissociated and some will not be a weak acid at a high concentration will have more acid in and again some of them will be dissociated and some of them won't be dissociated so the concentration refers to how much there is of whatever it is we're looking at and the strength of the acid is the ability to dissociate hydrogen ions the neutralization equation is a core one in chemistry and it is important that you remember that a hydrogen ion and a hydroxide ion in a reversible reaction will form water this equation comes up a lot and forms the basis for lots of the chemistry we look at often you'll be given a word equation and expected to construct the balanced equation from that and knowing your general sords equations is a really important part of being able to do that a metal plus an acid will give us salt plus hydrogen a metal oxide plus acid will give us salt plus water a metal hydroxide plus acid will give us salt plus water a metal carbonate plus added will give us salt plus water plus carbon dioxide using Hydrochloric acid will give you chloride salts using sulfuric acid will give you metal sulfate salts and using nitric acid will give you metal nitrate salts at the heart of this is the neutralization equation which comes up a lot 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 ions so sodium hydroxide plus hydrochloric acid the sodium ions and the chloride ions will give you the salt sodium chloride and then the water will be formed from the hydroxide ions and the hydrogen ions practical activity groups 2.1 2.2 and 2.3 are all very very similar they going to vary slightly in the details but the experiment that you'll get in the exam will vary slightly in the details as well for all three of these you need to make a standard solution and then carry out an acidbase titration so it does not matter if you making a standard solution for sodium hydrogen carbonates or for any of the other things that are in 2.2 and 2.3 or a completely different one in an exam where you might be given the method and then asked to follow it through the main thing to focus on when you're making a standard solution is that all of the powder you have weighed out is in the solution and that the volume is accurate so all of the powder that was in the waybo needs to be washed into the beaker generally three times as a minimum you need to be making this up to whatever volume of your flask is so 250 or 500 so we can wash this several times we've now poured it into the flask and we need to wash wash all of the powder out of bacer again going around the edges three times and then because you've poured it into the funnel there is going to be some powder in the funnel wash the funnel we can swirl it to mix it up and then just add loads running the liquid down the side of the flour so we get anything that might have been left on there when you get close to the end go slowly there is a line that you need to go up to and it is up to not over the line go drop by drop at the end now if you make a mistake and go over the line do not remove any liquid you going to have to start again because if you remove stuff you are changing the concentration of the solution for the titration the method is the same for all the Practical activity groups it doesn't matter which one you're doing because this is a basic method for the titration the things that might change are the indicators that you're going to be using depending on the PH range you are looking for so you could e use phenol filen or you can use methy orange here phenol philen is my absolute favorite now what you might want to do what I suggest you start with for a titration is a rough titration so you know roughly where you are aiming for you want to get concordant results which are very very close to each other so you know that what you're measuring is accurate go slowly for this you are going to need to go drop by drop at the end because you do not want to miss the end point and then we are need to go get the mean of the concordant results and that is the bit that we use in our titration calculations when you see a titration calculation there are some very specific things I want you to do first of all first of all is highlight all 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 Alkali and the concentration of The Alkali concentration of The Alkali is X that's what we're going to put in there and the rest of it we're going to grab from the question and write it down so it's easy for us to find later with all of the information in one place we don't have to dive back into the complicated text to find out what we need when we need it we're also just going to note down the equations that we are using to help us so that we don't get confused trying to think of things trying to remember when we are deep in answering the question the first thing you need to do for this is to write a balanced equation so we have sulfuric acid and sodium hydroxide making our sodium sulfate salt and water step one in any tiren 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 working in lers CU this is in cm cubed giving us 025 mol of hydrogen ions we then need to find the ratio of hydrogen ions to hydroxide ions and because of the ratio in the balanced equation it's doubled we can then use moles concentration volume we have the number of moles we have the volume so we can find the concentration of The Alkali the question is asking for the answer in grams per decimeter cubed so we need to convert our units third thing we need to do is to find find out the mass of sodium hydroxide using moles * m we can get our answer in gam per decimet cubes whenever you have anything reoccurring 1.6 reoccurring for example always use your calculat value otherwise you're going to be introducing rounding errors into your answers 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 your 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+ two hydrogen is+ one except in metal hydrides when it is minus1 oxygen is min-2 except in peroxides or when combined with Florine chlorine and bromine are minus one excepting 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 0er meaning sulfur must be+ 6 for our carbon ion we have an overall charge of -2 oxygen we know is -2 and there are three of them -2 * 3 is - 6 overall we need to get to -2 so what's plus - 6 makes -2 well that is plus four 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 minus1 giving us zero overall hydrogen peroxide H22 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 -2 meaning there need to be -1 each -1 * 2 2 gives us minus 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 iron compounds we know the nitrate is going to be Min - one oxygen is min -2 we have three of those that gives us - 6 we know Nitro is + 5 giving us N3 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 one metal be + one oxygen - 2 hydrogen + 1 these are from our rules sodium + 1 chlorine minus1 we can see chlorine started off with zero and went to minus one 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 oxygens are balanced with water hydrogens are balanced with hydrogen ions and any charges that are uneven are balanced with electrons here we have a reaction that we know occurs but it is not balanced adding in 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 charge 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 it out here with this reaction being multiplied by two and with the bottom reaction 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 sum of the hydrogens we can then write this as a complete overall reaction this may seem very complic ated but with a bit of practice you'll get the hang of it no problem 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 calcium it has 20 electrons so we are going to start filling from the bottom each needs to be filled singly each electron and within each box they must have opposite spins two electrons 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 2 s and three 2 p orbitals each orbital can hold two electrons so the total number of electrons in first shell is 2 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 S2 2 S2 2 p 6 3 S2 3 p six the short hand for that would be argon in square brackets calcium has 20 electrons 1 S2 2 S2 2 P6 3 S2 3 P6 4s2 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 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 it the outer shell and chlorine has seven electrons in the outer shell magnesium will transfer one electron from its outer shell to each chlorine outer 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 electron 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 3p5 and it wants another one to make this 3 P6 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 when we are thinking about the properties of an ex structures it is really important that we think of these as a ltis a large structure here we have sodium chloride and we write it as na 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 one that donated or received the electrons gion ionic lates 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 Cove valent bonding occurs between non-metals it is the sharing of electrons between two non-metals when we are drawing it a single bond is one pair of electrons being shared a double bond is two pairs of electrons being shared so four in total a triple bond is three pairs of electrons being shared six electrons in total when we are drawing things at a level circles are optional and I will draw them from now on without circles so here we have oxygen X is from one O's from the other we have four electrons or two pairs in the middle making our double bond for nitrogen we will have six electrons in the middle or three pairs in dative 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 Calum bonds giving us NH3 bcl3 you need to know the shapes of lots of different molecules or at very least be able to work them out so here are some examples we have carbon dioxide hcn b e F2 and these are linear the 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 Molly mods and actually sit there building these and playing with them you can see how the shape is linear how the bonding sets up and then we can draw the dotts and cross diagram down here and relate it to the the stick diagram with the 180° Bond angle for S SO3 B3 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 ch 4 and nh4 plus these are tetrahedral they have Bond angles of 109.5 and there are no lone pairs when we are drawing 3D structures aign like this shows it is in plain with the paper here we have backwards and then forwards so we can have two Bonds in line with the paper one coming forwards out of the paper and one going backwards into the paper and it is much easier to see if you actually have this in your hands and you can hold it and you can align it so that you have the bonds in playing with paper going backwards and going forwards if we just have the do and cross 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 tet 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 is 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 pyramid or and and it has two lone pairs again shown here in pink it really helps to remind you if you add in the lone pairs compared to water without the lone pairs that the bond angle is going to be reduced due to the veilance electron Theory the lone pairs can be seen here on the dots and cross diagram pcl5 is trigonal by 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 mod 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 sf6 is octahedral all of the bond angles are 90° and there are no lone hairs we can see this in the 3D model we can see this when we are actually holding the mly mods in our hand 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 veon 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 oun 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 NH H3 will have 107 and H2O has 104.5 this is because lone pairs are more repulsive than bonding pairs when you increase the number of Lone pairs is the bond angle decreases a lone pair lone pair is more repulsive than a lone pair bonding pair is more repulsive than a bonding pair bonding pair electro negativity is a measure of how much an element will attract electrons we have Florine over here as the most Electro negative elements as we move across period the electr negativity will increase as the number of protons increases in a period the number of electron shells Remains the Same so the atomic radius is decreasing as the electrons are pulled in electr negativity decreases as we move down groups the increasing number of shells increases the shielding around the nule coent 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 Co 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 W one side and as the polarity increases the the shift of the electron cloud is going to increase as well until we get to the point where the electrons have been transferred and we have ionic bonding for lots of these partial dipoles will be set up within the bond and this can all be due to 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 calent bonding at one end and ionic bonding at the other end and for example in hydrogen gas we have equal electro negativity between these two elements so they share electrons equally in HCL the chlorine is more electr 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 HC L 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 htl molecules they will all have these dipoles and this will result in attractions between the Delta negative and the Delta positive Parts this will result in a higher melting and boiling point as the intermolecular bonding is stronger we can see this as HCL is asymmetrical for a symmetrical molecule the forces are going to balance each other out and it will not be polar when we are looking at induced droles they can be refer 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 ster 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 electrons the more electrons involved the stronger it will be the shape of the molecule the more surface area that allows more contact the stronger the forces will be for example between these two linear compound there are lots of opportunities for contact whereas in the branch isomar there are much fewer opportunities for contact meaning the strength is going to be reduced straight isomers will have more contact points meaning the intermolecular forces will be stronger and they will have a higher boiling point hydrogen bonding is an area where we see a lot of of crossover with Biology it occurs when hydrogen is bonded to either nitrogen oxygen or floring right over here in the very electr negative area of the periodic table between these two elements there is a large difference in electr negativity at the same time as hydrogen bonding you can have other forces occurring such vaval but hydrogen bonding is stronger than the other intermolecular forces and this leads to a few properties they will have enormously high boiling points and we will see in water that ice is less dense and water has a very high specific heat capacity when we are talking about simple molecular structures coent ones there are some common examples that you should be familiar with with water H2O ammonia NH3 nitrogen gas N2 carbon dioxide CO2 and oxygen gas O2 obviously there are 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 coent intra mular bonds which are the strong Cove valent bonds between atoms in a compounds these are very very strong and then there are the weaker in tur molecular bombs these ones are much easier to overcome and give rise to properties such as the low melting and boiling points [Music] [Music] [Music] there is structure and order to 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 pottassium which comes after it in the periodic table has 19 protons and a mass of 39 a lower Mass which might make it seem like it's in the wrong order but it is arranged by atomic number everything within a group will have similar chemical properties they have the same number of electrons on the outer shell periods will go across the periodic table they will show a repeating Trend this is periodicity they will have the same number of electron shells but an increasing number of electrons on the outer shell the structure can also help you remember how many electrons go 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 there are 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 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 outmost 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 starts filling up P orbitals there is also another drop between GRE groups five and six through 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 coent bonds it is a dry structure it will have high melting and boiling points chlorine sulfur and phosphorus are simple coent so they will have a weak intermolecular bonds leading to low boiling points and argon is mono Atomic very low boiling points these are some really important definitions to learn and to make sure you understand properly the first ionization energy is the energy that is required for the removal of one electron from each atom within one mole of atoms in a gaseous form to make one mole of gaseous + 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 gas form plus one electron the second ionization energy is the energy that is required for the removal of one electron from each ion within one mole of plus one ions in a gaseous form to make one mole of gasius plus 2 ions are equations and again it is important to get the state symbols correct here gaseous helium plus will go to helium 2++ one electron sodium plus will go to sodium 2+ in a gaseous form plus one electron there are a number of B 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 its 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 s and electrons 8 there is a big jump in ionization energy it is always going to get harder to remove electrons and as electrons are removed the repulsion between the remaining ones is less when we are looking at Trends in ionization energies we will see that there is an increase across periods there is a sharp drop in the first ionization energy between the end of one period and the beginning of the next and these two bits of data give us the evidence for shells ionization energy can provide evidence for electron structure if we look at our graph here we have increasing atomic number here are the groups and this is the ionization energy there is a small drop in ionization energy between in groups 2 and three for example brillium 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 seen between groups five and six if we look at nitrogen and oxygen with seven and eight electrons you can see that eighth electron is the first one to be paired in the 2p so the ionization energy is giving us evidence for electron configuration 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 a more positive nucleus giant metallic luses have a few particular 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 Cove 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 graphi are special examples of giant calent structures they are both made from carbon in Diamond each carbon makes four carbon carbon bonds whereas in graphite each carbon makes three carbon carbon bonds diamond is very hard in a ltis structure graphite is soft as the atoms are arranged in layers that can slide over each other Diamond does not conduct electricity all the electrons are involved in bonding and there are no free electrons graphite does conduct electricity there is a free electron that is not involved in bonding meaning it can move around and conduct 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 electon 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 outmost 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 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 sublimes sublimes when something goes a solid straight to a gas bypassing liquid phase to give us a purple gas as you move down the group the electro 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 inre increasing number of electrons increases the strength of the intermolecular forces we can talk about displacement reactions involving halogens more reactive halogens can act as stronger oxidizing agents and the stronger oxidizing agents will replace weaker oxidizing agents in a reaction when this happens you are likely to see a color change for example if we have chlorine added with bromide ions we're going to get chloride ions out and bromine we are going to go from a pale yellow solution to a solution of bromine which is an orange brown color and chloride ions the hallogen sit over to the right hand side of the periodic table in group seven or in group 17 which means on the outside we will find seven electrons they will have two in the S and five in the P for Florine this is 2 S2 2 P5 for chlorine this is 3 S2 3 P5 all of the hens make minus one ions so we will then change to have two S2 2 P6 or for chlorine 3 S2 3 3 P6 and this pattern follows throughout all of the halogens and the halides they all gain an electron when they turn from an atom into an ion if we think back to oil rig reduction is the gain of electrons so they're all being reduced we can see that when we look at the half equation and we look at the oxidation numbers Romine as BR has an oxidation number of zero whereas a bromide ion br minus we'll have a minus one oxidation state this has decreased as the reduction we have gained electrons they can act as oxidizing agents since the other parts of the reaction this whatever is reacting with the br2 has been oxidized 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 an oxygenation 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 C 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 chlorite however sodium here has an oxidation state of one we know oxygen is min-2 and there are three of them giving us minus 6 in total the whole thing is zero so chloro 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 when we are testing for halight irons you are going to need a solution of that halight Lon we are going to add to it silver nitrate and 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 iron this is a colorless complex iron so the white color will disappear the same will happen to silver bromide upon the addition of concentrated ammonia where silver iodide does not react with ammonia when we are testing for sulfate ions we need something we sulfate in hydrochloric acid and barium chloride acidified berium chloride with hydrochloric acid can be used 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 and the Order of tests is important the chloride ions in barium chloride will give you a false positive on a test for halide ions so the order of these reactions is important a favorite practical in 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 cats and anions there are also the flame tests we can add onto this as well when you testing the Halo ions we add nitric acid and silver nitrate fluoride ions will give us a white precipitate bromide ions will give us 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 will be a white precipitate for carbonate ions we will add hydrochloric acid we will see a gas given off this will be carbon dioxide gas to confirm its carbon dioxide gas we need to see the gas turning Lim water cloudy hydroxide ions can be added to ammonium chloride and it will start to be very smelly leading onto the inverse test for ammonium ions where we can add sodium hydroxide and we will get smelly ammonia released to confirm that it is ammonia what we we will be looking for is turning damp red litmus paper blue the order in which you do these reactions is important if you were given a unknown sample and you had to do a series of reactions on the same sample the order in which you do them is important the first thing you need to do is the test for carbonates then you need to do a test for sulfates and then you need to do the test for halodes when we are testing for sulfate we adding barium ion berium sulfate and berium carbonate both give precipitates but if you do the sulfate test before the carbonate test you will get a false positive the berium carbonate for the haly iron test you are adding silver ions silver carbonate and silver sulfate will give you false positives in this test leading you to perhaps think there are just halide ions in there as your anions and then not going on to test sulfate ions or carbonate irons so the order you do this in is important how we use words and definitions are important so here we're going to be looking at enthalpy change during a reaction heat energy can be taken in or given out this is the enthalpy change it is given the symbol Delta H Delta whenever you see this this triangle just means change in h isaly so Delta H is change in eny this shows us we 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 action will get colder while an exothermic reaction gets hotter enthropy change is measured in KJ per mole 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 298° kelvin the pressure is one atmosphere pressure which is atmospheric pressure this is also 100 kilop pascals enthalpy profile diagrams can tell us a lot about what is happening in a reaction we have energy going up 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 when we are talking about the standard enthalpy change of combustion this is the eny change that occurs when one mole is burnt completely in an excess of oxygen under standard conditions we can write that in shorthand here we have Delta for changing little C showing its a combustion H for enthalpy and under standard conditions these values are generally going to be exothermic because it is combustion or if something doesn't burn then zero when we are talking about the standard 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 changeing f for formation H for enthalpy and under standard conditions this value can be positive or negative it can be endothermic or exothermic the standard eny change of neutralization is the eny change when one Mo of water is formed by neutralization here is our neutralization equation again it is really really important one Delta neutralization en Delta is our changing H is enthalpy and N is neutralization 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 times specific heat capacity times the change in temperature and the change in temperature is the bit where we can actually get handson and measure this in a lab we can also measure the mass we generally know this specific heat capacity so we can work out the energy change that has gone on for energy change we're going to be using 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 when you are testing the enthropy change of combustion we can use something called calorimetry which has alcohol Spirit burning in and you might have a number of different alcohol Spirit burners methanol ethanol for you to test and you need to find the energy given out by combustion we can do this by measuring the temperature change of a known volume of water you need to know the change in temperature and you should record the start temperature and the end temperature not just the change in temperature as a control variable you need to have a constant distance between the two different parts and you need to measure the change in mass of the spirit burner before and after you burn it we can then use our equation for specific heat capacity so energy change be the mass of water Times by the specific heat capacity of water Times by the temperature change of the water that will give us the energy change but we are looking for the enal change here so we also need to know the moles of alcohol that we used up in creating this energy change we can do this by dividing the change in mass of alcohol so the change in mass of the spirit burner by the Mr of alcohol calculated energy change here might be in Jews so we need to find the enthropy change in combustion in kles per mole measuring the enthalpy change of a reaction is practical for your a level there is a more detailed video that you can go at 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 lots of recording at times you are going to need to record 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 into our calculations the beaker is to make sure it's steady the pooring 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 cuz temperature drops in intervals and then we need to go backwards find zero and extrapolate back what the maximum temperature reached was then we can put that into our calculations if you've got two of them you can combine these and follow the instructions this will allow you to find the maximum temperature reached or the temperature at zero which is difficult to do as you're doing it difficult to find the temperature at zero because that you are currently busy adding things in this experiment can be improved by using a data loger 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 polyserine cup helps with that or the reaction could be incomplete thus you wouldn't get a true reading 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 bond 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 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 you 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 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 for you this follows the first law of thermodynamics that 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 enthropy change of reactions that we can't measure by using data from reactions that we can measure 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 reactor ANS 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 it out by writing on the data so here we have a carbon and two lots of hydrogen gas combusting 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 it's 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 direction because this is a formation from elements to the reactants and from elements to the products here is another reaction we can't directly measure but we can look at the formation from the elements down here now remember this needs to be balanced so we have three lots of hydrogen and half oxygen gas draw in your arrows add on the data draw with color pen or highlighter the arrow that you are going to be following notice this one here is now in the opposite direction we're going opposite direction to Arrow so this one is the one we need to change the signs for different direction different signs same direction same same signs so it is now + 75 and + 242 -0 same signs different signs always remember to put the positive or negative on here and add in your 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 depending on the temperature the concentration the pressure or if a catalyst is involved 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 increas in concentration is more particles in the same volume an increasing pressure is the same number of particles in a smaller volume thus the frequency of collisions will increase we can use the gradients on the graph to visually see a change in the ratees of reaction or or to actually calculate the rate of reaction so here is a beautiful graph with a line of best fit going through you'll see it starts at zero and it is a smooth line remember this is chemistry not not now I'm doing this very roughly so please forgive me for not completely feeling in my Axis for rate of reaction we need to be comparing two things so here we have cm cubed of gas released per second so what are we measuring and how long are we measuring it for it could be volume it could be math it could be hours or seconds in an exam you might be asked to compare the rates of reaction at 10 seconds and at 3 minutes so I'm roughly going to Mark these on my graph 10 seconds and 180 seconds and what we're going to be measuring is the changing up divided by the change in AC cross so change in cm cubed divided by the change in seconds always always always draw your construction lines on a graph so the examiner can see what you are doing so they know you know what you're talking about and you haven't just randomly pulled numbers out of the air you need to put your ruler close to the graph so you are measuring the exact gradient at the point you are looking at and draw your line we're going to be drawing a triangle please try and draw the biggest triangle that you can fit and then we going to need to change it up divided by the changeing across again at 180 seconds we need to draw a tangent that touches the graph just at 180 seconds so you can see the differences in my two lines there and again we need to do a triangle please the biggest triangle you can and then for each of them we need to divide the changing up divided by the change in across and this will give us Radiance we can see that steeper things gim going to go faster and shallower things shallower gradients are going to be slower introduce a catalyst will have an effect on the rate of reaction they increase the rate of reaction by providing an alternative pathway with a lower activation energy causing there to be an increase in the number of particles that are able to react and an increase in the number of successful collisions if we look at our reaction profile this is an uncatalyzed reaction and the activation energy that it takes for the reaction to happen however a catalyzed reaction will have a lower activation energy transition metals can act as homogeneous and heterogeneous catalysts homogeneous means they're in the same phase heterogeneous means they're in a different phase and a catalyst is something that increases the rate of reaction by providing 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 really need to have a large surface area such as a honeycomb structure as this is where the reaction actually takes place the reactants are absorbed and the active site on the Catalyst and this can weaken the bonds or hold the reactants in a more reactive configuration when you combine this with a higher concentration of reactants at the Catalyst you will get higher rate of reaction homogeneous 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 all of the experiments in Practical activity group N have something in common we are following the rate of reaction continuously and there are lots of different ways you can do this you can do this by collecting a gas you can do this by the loss of mass you can do this by following a change in PH or change in color or change in how opaque something is a way to improve any of these would be to use a Data Logger it is more accurate because it's computer reading it and it can generally draw your graph as you are going along so the different ways you can do this you can have an inverted measuring cylinder or a buet collected collection tube where your reaction is taking place with these you need to be wary of any gas that might have been trapped up here before you started same with a measuring cylinder or a buet this could happen very quickly you can measure the loss of mass so a gas would be released we have cotton wool bung at the top here to allow the gas to get out but not to allow anything to fall in that would affect our readings of mass or you could use a gas syringe to collect gas release 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 reduced 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 hydrochloric acid I'm going to add them together and use a gas ringe 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 we need to be adding the Magnesium to the glass at exactly same time we need to bit bangle and you need to practi the same timer at the same time we've got a gas syringe here gas syringes are quite um expensive um delicate piece of equipment so you also need to make sure that this doesn't shoot out the end and smash so using AIT of skill put that on there start the timer and going to keep an eye the gas 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 wind gets stuck you can see this moving quite quickly long you need to be careful that your reaction isn't too quick and that gasoline blows at the end you know the reaction is finished when the gasoline stops collectic 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 L es bit 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 are going to need to work out the initial right so what I've drawn 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 of them and compare it 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 grph 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 par the activation energy these are the ones that will react if we increase the temperature we will shift the curve now small 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 the 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 particles 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 the energy of those collisions lelia's principle and dynamic equilibrium is a great to topic for questions Leia's principle says if an external condition is altered the equilibrium will work to counter that change it is important to remember that a dynamic 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 balanced 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 homogeneous 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 of 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 it's algebra we can cancel them out and see what we have left giving us 1 over moles per decim cubed squared giving us M to- 2 decim to six there are some things that will have no effect on KC that is concentration pressure or the presence of a catalyst however an increase in temperature will affect KC you will see an increase in KC for endothermic reactions and a decrease in KC for exothermic reactions [Music] [Music] [Music] [Music] [Music] organic chemistry lature is very important because it tells you what an exam question it's asking you for the empirical formula is the simplest whole number ratio of each elements within a compound the molecular formula is the real number of atoms of each individual element in a compound the general formula will be the formula that covers a homologous series the structural formula is the minimum level of detail you need to draw something so ch3 ch2 ch3 the displayed formula will show all of the bonds whereas the skeletal formula will not show any carbons or any hydrogens just skeletons and other functional groups when drawing you can draw the displayed the skeletal or the structural formula if one of these is mentioned in a question please give the examiner what they are looking for the displayed formula will show all of the bonds here we can see each of the carbons and these will become points on the scill little 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 as the turning points in our sceletal formula drawing those points those dots on as our backbone and connecting them up you have a ch3 a CH and a ch2 and this is propane slightly more complicated one now again we are going to start by using our carbons as our points as our backbone of our skeletal formula drawing on the backbone and the functional group for the structural formula we have a ch3 ch2 ch2 and a c 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 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 carbon so one is meth eth Pro four is but 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 for it 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 it all together this will be methal but a the a tells us it has single Bond 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 ethy 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 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 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 so is 3 4 4 5 Das 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 in here an alcohol will have an oh functional group this one 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 iser 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 butal butanoate there is a priority list of functional groups starting with carboxilic 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 in another example that looks horrible but once you follow the rules is absolutely fine 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 br branched or nonaromatic rings based around hydrogen and carbon alicyclic compounds will have non-aromatic rings whereas aromatic compounds 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 structural isers 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 Molly mods and as you can see from the displayed formula they are different Arrangements these are chain isomers and they will have different physical properties for example branching will give different boiling points here we have 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 out holes 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 Al is an alahh and buttin O is a Ketone the same formula but the different position in this place of double bonded oxygen 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 El 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 Cove valent bond with the arrow starting at the bond and then the electrons going somewhere else you always have to be very careful with where your 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 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 carbon and hydrogen in and they only have single bonds a really stupid way for you to remember the difference between alkanes and alkenes is that alkanes has one e in it whereas an alken has two e in it and has a double bond these are Sigma bonds here and they have free rotation around the bonds so even though I have drawn this in a linear fashion it is not a linear fion around each of the carbons we have a tetrahedral arrangement of other atoms the bond angle will be 109.5 around each and every carbon so it is not a straight line at all it is all bendy this is the arrangement that means the electrons can be as far apart from each other in space this is the electron repulsion Theory when you have long chains of hydrocarbons straight chains can make lots of bonds lots of London lots of forces between each chain the longer the chains the more bonds they can make meaning is going to have a higher boiling point any branching that we can see in longchain hydrocarbons is going to reduce the number of points where forces can be made which means things like branching will decrease the boiling 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 oxide and carbon there is a lot of pollution from combustion carbon dioxide contribute to climate change as does water vapor as they're both greenhouse gases carbon monoxide is a toxic gas that can lead to death carbon is soit that can lead to global dimming and atmosphere pollutions leading to breathing difficulties sulfur dioxide can lead to acid rain and then 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 as bad and they can take the unburnt hydrocarbons and react them with nitrogen oxides giving us saf products the catalytic inverter will have a honey cream structure it will have platinum palladian 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 substitution Step One is initiation step where with UV light chlorine cl2 will produce two chlorine radicals with this free unpaired electron this is the process of homolytic fision 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 onto 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 of chlorine radical this is why it's a propagation reaction because we start and end with the same thing the electron from the chlorine radical will go in to this Bond here and the electron here will move over then this radical will then go and 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 radical 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 State and our reactants are used in the initiation and in propagation bonding and reactivity of alkenes is fascinating they are unsaturated they will have a double bond so an alken 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 but but two different versions of buttin depending on where the carbon carbon double bond is around carbons in the double bond we will have a planer 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 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 a lots of electrons all in one place 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 ch 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 iass this is e but 2 in and when they're on the same side it is the Zed ice mases gives us Z but2 in as a point but one in doesn't show EO 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 we the same these are the priority rules for deciding whether something is an e Isa or a zed Isa your e isers will have priority groups on different sides of the double bond your Zed icers will have priority gops on the same or the Z oh I know I'm sorry side of the double bond priority is determined by the atomic number here is some lovely examples we have two chlorines on a buttin so giving us two three D chloro buttin 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 same side so this is the Zed isoma and this one is the E isomer isomers can differ in polarity because the E isoma is symmetrical it has polar bits on both sides so it cancels each other out overall and is not polar whereas here The Zed iite 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 there are a few electrophilic addition reactions that need to know you need to know the reactants you need to know the products and you need to know any catalysts or conditions so for addition of a hydrogen you might know this is hydrogenation with a nickel Catalyst this is going to be a solid catalyst so the alken is going to be absorbed onto the surface of the Catalyst that is where the reaction is going to take place the two hydrogens in H2 will each get added to the spaces across double bond we can have the addition of hens and for this one we have a reaction mechanism the electrons in the electron dense region will induce a dipole in brine we will get an intermediate carbocation giving us a negative bromide ion which will then go on to attack the carbocation this is the test for an alen decolorizing of bromine water this will give us a dih Halo alkane at the end or to be more specific one two di bromoethane we can have the addition of hydrogen halides in a very similar reaction mechanism so if you learn one learn it well you should be able to apply it to everything else the electron dense region will attack the Delta negative Shifting the electrons to the bromine again we will get a carboca and a negative bromide ion the elect will move towards the positive carbocation and we will end up with hydrogen and bromine being added across the double bonds giving us a Halo alane if at any point in this video I say halogeno iane that just means I've been teaching this for a really long time it's the same thing we can have the addition of steam in the presence of an acid Catalyst and this is generally going to be phosphoric acid catalist the electron dense region will interact with the hydrogen on one of the arms of phosphoric acid giving us an intermediate carbocation as it has taken one of the hydrogens and this will leave the phosphoric acid with a negative charge phosphoric acid is the Catalyst for this so we need some water which will get added onto the carbocation giving us a second intermediate this carbocation we'll see the breaking of one of the bonds between hydrogen and the oxygen and then this hydrogen will go on to reform the phosphoric acid Catalyst and we will end up with an alcohol and our Catalyst at the end working out the major and minor products in a reaction is also known as mnol 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 bromo butane 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 hydren 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 caroa i the negative bromide ion will be attracted to the positive charge on the carbo cation 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 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 the square brackets put on square brackets and put an n on there to show it's a repeating unit this will then become polychloro ethane 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 in there polychloro Ethan is also known as p DC and it is incredibly useful 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 biodegradable they cannot be broken down by holding hsis 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 were recycling trying to make something new things can be sent to landfill and I will admit I did struggle to find an advantage for sending things to landfill but it is a very common place thing to do and we have the systems in place for it already it's pretty easy however we are running out of space to put stuff our landfills are filling up rapidly things that are already in there take years and years to break down there is also the issue of pollution visual pollution air pollution from the degradation of things and mixing of products polymers can be used as a fuel which can be used to generate electricity and energy the disadvantage of this is the toxic air pollution that comes from this the naming of alcohols is very similar to alkanes here we have three different alcohols with the functional group in a different place this is butan 2 all it is numbered so that 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 two methy butan 2 o butan 1 o is a primary alcohol butan 2 O is a secondary alcohol and two methyl 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 tetral 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 but between alcohols we are going to get inter molecular bonding leading to that properties of low volatility low boiling points and then being good solvents when we look at oxidization of alcohols you need to be aware of the differences between primary secondary and tertiary here we have primary alcohol and we are going to be using an oxidizing agent which we show by square brackets with an o in it this is going to give us 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 reflux 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 a chromate 6 iron being orange and after the oxidation reaction has happened the cremate 3 Iron is going to be green so we will see a color change in here however this is a reaction you've probably seen in real life and you Noti 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 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 to 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 1 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 an alcohol group can be substituted with a haly iron to give a Halo Alan we will have an alcohol plus a hydrogen halide to give us a Halo alane but everything that we need to put in there is a bit more complicated ated we need alcohol a sodium halide and sulfuric acid and this needs to be heated under reflux the first thing that will happen is sodium bromide in this example will react with the sulfuric acid and that will give us the hydrogen haly that we are looking for this hydrogen haly will then go on to react with the alcohol here we have ethanol plus hydrogen bromide which which will give us the Halo alane Plus water when we're looking at the hydrolysis of Halo alanes we are looking at nucleophilic substitution reactions here we have our bromo alane and here we have our negative hydroxide ion bromine is more electronegative so we're going to get a partial charge on here Brom is going to be slightly negative carbon slightly positive so the negative charge on the hydroxide could go in and be attracted to the slightly positive carbon this will give us a substitution reaction as the electrons move over towards the bromine that will separate off and we will end up with an alcohol and a bromide ion this can be done at room temperature but very slowly it's best done when heated under reflux with sodium hydroxide we can also have hydrolysis by water with silver nitrates and ethanol the reaction will look very similar but at the end we will not only get bromide ions we will get hydrogen ions as well silver nitrate you may remember is part of identification of Halid ions so that broide will then go on and react with the silver ions from the silver nitrate to give us a precipitate and this precipitate is visible and the color will tell us that it's silver bromide and this is a test for Halo ions we can look at the trend in the rates of hydrolysis to give us information about the strength of the carbon Halo bonds and the test that we do this is also the test for haly so you're probably familiar with it the rate at which the precipitate forms can tell us all about the strength of the carbon halide Bond because for the precipitate to actually be made the carbon Halo Bond needs to be broken floro alkanes are very unreactive you are not going to break that bond by hydrolysis iodo alanes react very quickly that bond is going to be broken by hsis very quickly and you'll see a quick formation of the precipitates and then in between those we have bromo alanes and chloro alanes going down we see the speed of hydrolysis increasing this is also the speed at which the bond is being broken the speed of which the precipitate is formed this is opposite to the strength of the bond the weaker bonds will break faster toal six we are going for alcohols alahh highes alkenes and carboxilic acids when you are testing for an alcohol put ethanol into a dried test tube add solid sodium and a positive results are bub being given off we can use a failings test for an alide A positive solution is the blue turning into a bright orange so a color change here when we are testing for an alkine 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 with 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 nitric 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 now we looking at TLC so I've got my tabing which I've crushed with a peel and motar here I've Got My TLC plates and the 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 centimeter up you need to draw a pencil line so now I've soled my aspirine 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 capillary 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 capillar with cubing it should run up the tube 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 you 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 tiny concentrated spge that'll give us the best results so now I have my TLC plate in my chainb 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 where 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 youve about it you're going to get something that looks like this I will Adit this isn't the plate I just showed you because our light is broken at the moment but you'll get big blobs look like this or 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 the depletion of ozone has an important set of reactions in it chlorofluorocarbons cfc's are very very useful they're 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 are short and very strong and we have carbon chlorine bonds which are longer and weaker the carbon Florine bonds are not affected by UVS which is why scientists have now shifted towards using hydrocarbons 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 so here we're going to be using a round Bott flask and obviously these um are round so to hold that study I'm just going to put it in a large beer to hold it as a secure um base we are going to making an organic solvent so I have a 50 c of ethanol 50 cm Cub of glacial ethic acid this stuff is quite nasty um we have been using some quite nasty things have all of this in a food give that a bit of a SW to mix it and then really slowly I'm going to add some concentration of puric acid you can see if you look closely um the concentrated suric acid changing the um the liqu it because is concentrated and I finish so after mixing all those things I set it up to reflex here I have my mantle um you could use a bubs and 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 reflx condenser here the important thing to remember with reflux is that the water needs to go in at bottom the cold water comes in here and then the hot water or the water that's been warmed up a little bit needs to go out at the top so you can see this is bubbling away now and if we look really really carefully in here you can see if I just wgg you around a little bit you can see that it's evaporating and it's condensing you can see that the cold water is causing um anything that's evaporated to condense and then drop back down into the FL FL where it can be reboiled so reflux just let the same things be reboiled over and over again after it's been refluxing for about 10 minutes we just need to change all the CIT glass around so it's now going to distill now we've set this up for re 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 big here show you 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 the stop off the top pour in my sodium carbonate pour in um my just a steel here and then what you need to do is pop the Le 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 five holding thist 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 do this been for two three times invert it until you stop hearing the little um fiz the little release that tells you a gas has 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 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 top layer so it doesn't matter if a little bit of my top layer goes out um and gets discarded with my lower layer that's absolutely fine if you 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 ever 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 just SL that down so my bubbles you can see the layer is just here stop that drop a couple more hes 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 part so now that I've done that once I need to do that again with 20 cm cubed of saturated calcium chloride so my socer is closed and this is the top layer from before so in that goes again I need to mix it B there make sure we secure GP mix it a couple of times open the tap to release the gas mix it a couple more times open the tap to release any guas and then keep mixing it until you can't get here any more gas released and then 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 release bit the fres the top start running that down there okay so it goes quite quickly once you take the top off I'm just going to it settle out for a little bit and then go very slowly drop by drop watch 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 of I guess now I'm just going to add some um solid um an hydrous calcium chloride do say how much so I'm going to add that much give it bit SWR and then what I need to do is to transfer the liquid into a round botom flass 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 off the different fractions which come off at different temperatures because we only want a certain uh fraction we are waiting for uh some you distill off between 74 and 79° so everything that is coming out at the moment we can just discard 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 of this part and the identity of this part now if we know the compound definitely contains this and has an overall mass of 58 then butane is The Logical answer if it had different molecular Peaks if it didn't contain this for example methyl propane will have the same mass and the same formula but it will give different fragmentation paks it will not give this fragmentation Peak here when you have a question SE some data from infrared spectroscopy you need to look for some characteristic regions there are three you need to know different groups absorb infrared at different set frequencies you will get given a data sheet in exam so don't worry you don't need to learn these but you do need to be familiar with the data sheet and what it looks like here are some example graphs for an alcohol you were looking for this characteristic region here if we have a carbon oxygen double bond for example in aldhy and ketones you looking at this characteristic region here and for carboxilic acids which will have an 08 and a carbon oxygen double bond 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 ground graphs and be able to refer to them in the exam and pick them out of data given to you in an exam [Music] [Music] 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 styom metric 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 equals 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 a first order reaction first order overall the units of the rate constant are seconds 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 the concentration of a and the concentration of B because we still have two things there the overall order of both of those is still second order again we need to rearrange it so we've got K as 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 a moles to theus1 decim Cub seconds theus1 K is for a set temperature and this will change this will increase as the temperature increases we can measure the rate of reaction by initial rate method this is also known as the iodine clock the reaction equation for this is hydrogen peroxide plus hydrogen plus iodide ions will give us iodine the color and water when all of the iodine produced in a reaction has reacted with all of the available thy sulfate ions which is in reaction two any excess iodine is then unreacted in a solution and will turn blue altering the concentration of iodid ions you can experiment and experimentally determine the order of reaction with respect to iodide ions 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 floss 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 thy sulfate and the hydrochloric acid will make sodium chloride sulfur dioxide and sulfur the sulfur is the bit that you can see because it is a solid it will precipitate out of the reaction and cause it to go cloudy the time taken for a cross disp 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 grass 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 we can use time concentration graphs to work out the half lives and the Order of reactions for a zero order graph we are going to get a straight line going down for a first order graph it is a curved line going down a second order graph it is still a curved line going down and you'll notice that it stand steeper and then shallower for a zero order reaction the gradient is equal to K for first order reactions they're going to have a constant half life so we go down to half on the graph go across and then down half of a half is a quarter so at quarter we go across and then down half of a quarter is 1/8 so at 1/8 we go across and then down and we have identified three half lives here here is the first half life the second half life I'm just going to call T2 and the third half life T3 they are all the same although I will confess this was a very rough graft or so the fact they look all the same is actually pure luck when you do this in the lab hopefully there will be very very close values because the half life we have worked out here is a constant half life this is the time that it takes so the bits for T1 T2 and T3 are going to be the halflife and they are going to be the same we can call this T half to work out the weight constant for this is ln2 over the half line here that's shown as T half we can determine the rate equations and the reaction mechanisms because unsurprisingly reactions are more complicated than the overall equation lets on here we have what looks like a very simple reaction however that is not what happens it goes through a series of different steps in step one we've got nitrogen dioxide 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 and what is left over will give us our overall equation for the reaction the rate equation for this is rate the constant and nitrogen dioxide is second order changing the concentration of a reactant will affect the rate of the slow step and not the rate of the fast step because it is second order with respect to nitrogen dioxide this is the slow step the one with nitrogen dioxide in it is zero order with respect to hydrogen making this step step two the one with hydrogen in the fast step the slowest step will be the one that determines the overall rate of reaction and this is the rate determining step when you first see the 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 a orous constant which is reactant dependent this is more commonly rearranged in this way so Ln k equals N A minus EA over RT if is going to be in graphical format we're going to have Ln k at one side and then along the bottom we can have 1 over T the gradient for this is minus activation energy over R the equilibrium constant KP for reactions involving gas we're 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 and5 moles of gas B then we're to have 1.3 moles of gas C we can add them all together so our total number of moles are 2 moles of gas the mole fraction for gas a is2 ided 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 do .1 * 5 to find the partial pressure of a KP is going to vary with temperature but not to be affected by a catalyst before we look at bronstad lar acid base equilibriums we need to look at a few definitions an acid is a proton donor a base is a proton acceptor and a proton is a hydrogen ion so hydrochlor Chic 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 plus can donate a proton so in this situation NH3 is the base and nh4 is the acid these will make a conjugate acid base pair the part that will accept and the part that will donate the proton the relationship between Ka and PKA is pretty simple but can look pretty scary so I'm going to show you on calculator Ka is acid dissociation constant but because these can be quite complicated numbers written in standard form with 10-4 10us 5 PKA is another value that is linked and just looks a lot nicer so it's easier to work with to change from PKA to Ka we need to use logs for a strong acid we are going to have a large value for KA and a small value for PKA for a weak acid we are going to have a small Ka and a large PKA so for ethanoic acid Ka is equal = to 1.8 * 10- 5 show you this on the calculator how exactly you would put it in I've done it here Open brackets 1.8 * 10 to the minus 5 and make sure you put your bracket in the right place if you're not familiar with your calculator please practice doing that and this gives us the value for PKA for ethanoic acid for hydrofluoric acid the pka is equal to 3.17 so we can now work out what the KA a is and we need to do 10 to the minus 3.17 10 to the power minus 3.17 = 6.76 * 10- 4 you can see the difference in these two numbers how k a is very licated number whereas PKA is a much nicer number to work with a strong acid is an acid that will fully dissociate you should be able to recognize some strong acid for example hydrochloric acid hydrobromic acid hydroiodic acid sulfuric acid nitric acid those are 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 doing these calculations we can assume that the concentration of acid is equal to the concentration of hydrogen ions for example if we have .1 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 con ation of acid so that is minus log1 and here I'm going to put it 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 s uric acid is a d 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 I 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 will dissociate 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 the top concentration of hydroxide ions on the top and concentration of water on the bottom if we rearrange that we can then take the KC concentration of water and call that KW which is the ionic product of water and this varies with temperature at 25° C KW is 1 * 10 -14 m s decm to - 6 if we have a strong base for example sodium hydroxide that would 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 mes per decm 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 Point 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 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 ions and the base we can turn that into an equilibrium equation with a concentration of hydrogen 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 PKA which is minus log of Ka so you'll need your calculator to work that one out when 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 ion Square divided by the concentration of ethanolic acid at the start we know what Ka is because we were told it in the question so we can replace that we know what concentration of ethanolic 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 square do we need to square root the other side giving us 9.22 * menus 4 as 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 a pH of 3 04 to two decimal places a buffer solution is one that maintains a steady pH even after additions of small volumes of acid or Alkali an acidic buffer solution is made from a weak acid and the salt of that weak acid a basic buffer solution is made from a weak base and the salt of that weak base one example from biology is that 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 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 * 10us 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 hyd 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 2 decimal places in one of your practicals you might have done some pH probe work some titration work and come up with some pH curves there are four different ones you need to be able to recognize starting with a strong acid and a strong base because they're strong they're going to start low and end High a strong acid in 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 concentration 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 phalin which works at a very high ph and methy 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 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 buffer 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 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 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 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 or one atmosphere of pressure the enthalpy of latis formation is a standard enthalpy chained when one mole of ionic Gattis 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 lattice is dissociated into its ions in gaseous form the first ionization enthalpy is an eny change when one mole of electrons is removed from one mole of atoms in a gaseous form to give one mole of + 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 one mole of 2 plus ions in a gaseous form the enthalpy of atomization is the enthalpy chained when one mole of atoms in a gaseous form are formed from that elements in a standard State Bond enthalpy this is the enl be chained when one mole of a calent bond is broken homolytically in a gaseous state electron affinity is the eny change when one mole of atoms in a gaseous form gain one mole of electrons to form one mole of minus one ions in a gaseous form the enthropy change of hydration is the enthropy 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 har 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 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 latis 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 ltis 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 highlighted green line starting off with the electron affinity of chlorine we're starting in the same place but because we are going in the opposite direction to the arrow it needs to change signs so is min - - 348 we are then going in the opposite direction of the first 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 eny change of atomization of sodium and in the same direction as a arrow for the eny change of 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 cleal minus must be made all of these numbers are based on real experimental data theoretical values can differ based on the co valent character of the bonds entropy or Delta s 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 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 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 these 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 temperature Delta s change in entropy Delta G is in K per mole Delta H is in K 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 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 eny change it will be spontaneous at all temperatures however if we have a positive eny change and a negative entropy change then it will not happen at any temperature if both enthropy and entropy are positive when Delta G is zero then the temperature will be the enthalpy divided by the entropy spontaneous Above This temperature this has a self indicating color change and when we see the color change we can answer the question how much has been oxidized the important thing to remember when you're working out equations is that the manganate ions are going to be reduced you're going to have your potassium manganate in your buet and then your known volume of the solution that we are analyzing goes into your chical flask this needs to be done in an excess of acid indicated for this as potassium manganate colorizes 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 Mangan it's really hard to see the bottom of the meniscus so for these titrations we read it from the top of of the meniscus if you do this consistency the difference between the top and the top then you will still get the actual Titus again with this question we are going to work through sorting out the numbers and then we are going to do the calculation so an sulfate was made up in 250 cm Cube solution and from this 25 cm cubed was titrated against 0.02 moles VM cubed pottassium 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 in exam or you might have to work these out for yourself we have our manganates being reduced to manganese 2 and remember the only thing we can add to these hydron electrons and water to balance out the four oxygen on the left hand side we need to add four 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 that charges balance 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 add on our electrons now we need to balance these to make an overall reaction which just means we need to times the bottom one 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 so 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 V sulfate in 250 cm cubed and then we're going to use that to work out the original mass for our MTH of pottassium manganate we can do concent rtion * 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 r which will give 3648 G iodine and sodium thiosulfate Redux hydrations are a touch more complicated because they have an extra step an extra reaction in there your thy sufate ions are going to be oxidized where your iodine ions are going to be reduced while looking at the two individual equations is useful it is more likely that you will see the all equation for the reaction given at once for your iodine and sodium thos hydration you going to have your sodium thiosulfate in your PX in your chronical FL you going to have a known volume of what we are analyzing we are also going to need to have in there an excess of potassium iodide in the chical flask you going to have re reaction between what we're analyzing the oxidizing agent and the iodide ions from potassium iodide this is going to start to produce iodine this is going to give the solution its distinctive yellow color as the titration goes on the iodine is going to get back reduced back to iodide and the color is going to fade this makes the end point rather hard to see as it's rather subtle color change this is where we can start to use starch as an indicator this means the solution will be um I kind of like it nice blue black color and as we get closer to the end point of the titration the blue black color will start to disappear and when the blue black color has gone that will be the end point of our titration 2.8 G of copper Alo is reacted with acid neutralized and made up to 250 cm cubed in water from this 25 cm cubed solution is reacted with excess acous potassium iodide and two products acous iodine and copper 1 iodide the mixture is then titrated against 29.8 cm cubed of1 m per decim Cub 30um thyro sulate calculate the percentage copper in the original alloy and again the first thing I'm doing doing is just pulling all of the numbers that I'm going to need out of the question so that I can use them straight away so the first thing I'm going to do is to use the information in the question to work out my equations so the first one is a reaction that happens during the titration now the equation we need to work out are copper ions are going to react with our iodide ions and as it tells us in the question we are going to get copper 1 iodide and iodine now the hard part for these is working out the ratios so we're going to go from the copper the iodine the th sulfate two moles of th sufate react with one mole of the iodine and two moles of the copper ions will give us one Mo of iodine so here our copper and our th sulfate are actually in a 1: one ratio for our moles of our th sulfate volume * concentration over th will give us 2.98 * 10-3 Mo moles of the copper because it's in a 1: one ratio are exactly the same we use 25 cm cued in our titration so to get it back up to 250 what we need to do is just times that by 10 to find the mass of the copper in the original alloy we take the number of moles and then we times it by the mass of copper from the periodic table giving us 1.89 G now we need that as a percentage of copper in the middle original alloy now use the number from your calculator for this even though I've written down 1.89 that is not full calculator value and if you use that you're likely to introduce a rounding error in your Sun so 1.89 / 2.80 * 100 gives us 67.6% we can use cells to work out Electro potentials a simple cell is a metal electrode in a solution containing that metal for example here we have a zinc electrode in a zinc salt solution and a copper electrode in a copper salt solution a salt bridge is used to connect them together to allow electrons to flow and this will create the voltaic Cell at each side we're going to have oxidation or reduction reactions happening for example here zinc is being oxidized while copper is being reduced and we can add those together to give us the overall reaction from the voltmeter we can get the E cell for this reaction and for this example it is + 1.10 Vol 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 ion 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 29 Kelvin 100 kilopascals and a 1 mole decim cubed solution of the ion all standard electrode potentials are the difference between any given half cell and the standard hydrogen electrode measuring the voltage in an EMF cell this is a lovely practical and I hope you've had the chance to do it we need to start off with some very clean electrodes so you can rub them down with a bit of sandpaper and then you can clean them so they're free from Grease with some propanone if your electrod 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 your reading from the voltmeter we need to have a salt bridge here the ends are buned with a little bit of cotton wall 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 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 we call a battery a single a A 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 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 cells 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 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 variable oxidation states which makes them so useful and they can act as catalysts the reason for all of these properties the reason they are transition metals is because they have an incomplete D subshell when they're atoms or ions an interesting thing to point out here is cromium and copper where the 3D is filled before the 4S because it is more stable to have a half full or a full the D shell than it is to have a full or half full 4S shell zinc is not really a transition metal because it has a full 3d subshell neither is 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 ligant and complex ion the coordination number is the number of bonds to the central ion a monodentate ligant will form one Bond whereas a bidentate li 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 ligans when you are writing it out we have square brackets the transition metal ion 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 B dentate liant you need to know about ethane one2 diing 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 octed geometry we can also have multi dentate lians here we have EDTA and there six places that it forms bonds 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 need to know the different shapes that complex irons 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 ligans 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 plar complexes or we can have linear complexes for example in toins complex ions can show CIS trans isomerism here we have a platinum Central ion and we can have CIS platin or we can have trans platin CIS platin is a useful anti-cancer drug transplatin is not a very useful anti-cancer 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 B dentate lians and two monodentate lians these can show CIS trans isomerism it is well worth spending some time practicing drawing these out in a logical manner complex ions can also show Optical isomerism for example with a hexadentate ligan like Eda if we have two monodentate ligans and two b dentate ligans or if we have three bidentate ligans these two are Optical ements as on these two you'll notice they are mirror images of each other it is possible for something to show Optical isomerism and CIS trans isomerism there are a few ligan substitution reactions you need to be familiar with you need to know the complex how the complex is written what is substituted the geometry the coordination number the shape and the color for each for this we're going to be looking at monodentate lians so water ammonia and chloride ions water and ammonia are a similar size and they're both uncharged so here we have copper in a complex with six Waters and it has a two plus charge this is going to have a partial substitution with four ammonia which gives us a complex ion that has four ammonia and two waters in it and there has been no change in the charge this is still 2 plus we go from a pale blue solution to a dark blue solution and with this there has been no change in Geometry there's been no change in charge and there's been no change in coordination number the chloride iron is larger and it is charged so if we start off with the same original complex iron the hex 2 and then we add in four chloride ions we are going to get copper in a complex with four chloride ions this has changed its charge from plus two to minus 2 CU we've added in four negative charges so we have seen a change in the geometry we've seen a change in the charge and we've seen a change in the coordination number the changing color will again be from pale blue to Yellow this time chromium in a complex with six water so hex Aqua chromium 4 will be fully substituted with six ammonia in this reaction this is going to go from a violet color to a purple color you will see there's been no change in Geometry no change in coordination number and no change in charge one of the things I love about transition metal chemistry is the beautiful color precipitates that they form these examples are all going to be with aquous sodium hydroxide so copper 2 plus ions with two hydroxide ions will go on and form a solid precipitate of copper hydroxide which is a beautiful blue color Ion 2 will go on with hydroxide to form a green precipitate ion 3 will go on with hydroxide to form a brown orange precipitate not the most beautiful one then manganese hydroxide is a brown precipitate and chromium 3 gives us a green precipitate now if we add in excess hydroxide copper ion manganese are going to be insoluble in excess aquous hydroxide whereas chromium will go on to form a complex Ion with more hydroxide in there we will get hexa Aqua 6 promium 3 complex iron which is dark green no longer a precipitate this is a solution the precipitation reactions with aquous ammonia are similar to the ones with hydroxide they look very similar but the chemistry is slightly different because ammonia is acting as a base so writing an equation for the precipitation reaction with ammonia doesn't really work because it it just doesn't work out like that what actually happens is the copper is going to be form copper hydroxide again remember this is aquous ammonia ammonia is acting as a base when we have excess ammonia we can actually see the equation properly and we will see it forming complex Ion with four amonia in there and two water in there with limited ammonia we get a blue precipitate and then with excess ammonia we get a deep blue precipitate chromium with ammonia will go on to form ammonium hydroxide precipitate exactly the same as before again the ammonia is acting as base here and when we add in excess ammonia we will see that the complex ion that is formed will have six ammonia acting as ligans the precipitate will no longer will have a purple solution at the end ion and manganese are exactly the same reaction as we had with aquous ammonia they will form hydroxide precipitates you need to be aware of the reduxx reactions that happen with transition metals and then be able to apply this to a number of different situations so ion 3 can be converted to Ion 2 and then back again manganese and hydrogen ions can be used to change ion Tu to ion 3 the Ion 2 is being oxidized ion 3 can be reduced in a similar reaction with iodide ions and we're going to go from orange brown to pale three chromium 3 can be converted to chromium 2 chromium 3 goes to chromium two with hydrogen peroxide and an alkal where it is being oxidized we can see this when we look at the oxidation numbers of everything chromium 3 plus ion is chromium 3 in chromium on the right hand side we have four oxygen that's min-2 so Min - 8 overall we need to get to min-2 for the whole compound so the oxidation state of chromium has to be plus six chromium can also be reduced and this might be a more familiar color change where we go from that bright orange to the green if we look at the oxidation numbers again the overall is -2 oxygen is-2 there are seven of those that gives us -4 we have two chromium so we need 2 x + -4 = -2 so X is going to have to equal + 6 and chromium on the right hand side is+ three copper ions can be reduced from copper 2 ions to copper 1 one ions in Copper iodide where we're going to go from a pale blue solution to a solution with a white precipitating copper oxide can react with sulfuric acid to give us a number of different products solid copper which is going to be the coppery brown color copper sulfate which hopefully you're familiar with is a lovely blue solution and water if we look at the oxidation numbers copper on the left hand side in Copper oxide we know that oxygen is min-2 we have two coppers so to get to zero they have to be plus one each over on the right hand side copper as a solid is zero we know that A sulfate ion is min-2 making copper plus two so we can see copper has changed oxidation state in two different ways it's gone from + one to zero where it's been reduced and it's gone from + one to plus2 where it has been oxidized and this type of reaction is one of my favorite words or the name of this reaction is one of my favorite words this is a disproportionation reaction where the same thing has been oxidized and reduced in one reaction [Music] [Music] [Music] there is an old joke in organic chemistry that you spend 10% of your time learning chemistry and 90% of the time learning how to draw perfect hexagon to help me draw perfect hexagons I've got this paper which you can download completely free off 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 sixc 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 cyclohexyl bonds oranization it has a plar structure it is flat if it was alternating double and single bonds you'd expect there to be a change in geometry the bond length in Benzene is an intermediate Bond length between single bonds and double bonds Benzene is more stable than cyclohexaamylose 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 niter 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 HS s so4 minus ion which I'm just going to rub out cuz we don't need that for the moment the h23 plus ion will then go on to give us 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 nitri Benzene for this we need concentrated nitric acid and our catalyst is concentrated sulfuric acid and it needs to be reflexed at a moderately high temperature you need to be able to draw some isoli reactions the first step is the form 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 Bening 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 when we have our familiar shape with Benzene but we have hydroxide group and it this is phenol phenol is a weak acid and we know it's a weak acid because of its reactions with two different things it will react with a strong base for example sodium hydroxide but it does not react with weak bases for example any carbonates it can undergo a number of electrophilic substitution reactions for example if we add bromine in then bromine will get added onto three different positions in the phenol and the hydrogen that was there in the first place will come out in the other product as hydrogen bromide the end product is 246 Tri bromen we can have another reaction with nitric acid where we get an N2 group being put on at the bottom or the NO2 group can go onto one of the sides giving us either four nitrophenol or two nitren these reactions are much easier the ease of electrophilic substitution of phenol compared to the same reactions with benzene is because one of the lone pairs on oxygen is actually helping the electron system in the middle it's donated to the P system so increasing electron density thus increasing the ease of reactions phenol can undergo electrophilic substitution so the hydroxide group can be replaced with an nh2 group or an NO2 group giving that's either phenol amine or nitrobenzene phenol phenol amine are two for directing and nitrobenzene is three directed if we number these starting from the phenol group number it in the opposite way around so you can see two is in a different place but four is in the same place o and nh2 are both electron donating groups whereas NO2 is an electron withdrawing group the electron donating groups can have multiple substitution reactions happening so that's why we are going to get bromine being added on in three different locations both the twos and the fours whereas NO2 is electron with drawing group so we're only going to get a single substitution happening this knowledge the ability to add things on in a certain place is important when you are designing organic synthesis pathway we're going to look at alahh and ketones together because these carbonal compounds are very similar alahh highes will have a functional group where the carbon is double bonded to an oxygen and a hydrogen at the end of a group whereas in Ketone the functional group will be in the middle a carbon double bonded to an oxygen this alahh you can see here in the mly mods is buttin Al the AL bit tells us it's an alahh and the cell one at the bottom is buttin 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 test to tell the difference between alides and ketones we have a test with Tolen reagent and we have a test with failings reagent I have covered both of these in more detail earlier in this video both will only give a positive result with an alahh so tolins 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 alahh can be further oxidized to a carboxilic acid a second 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 Edition 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 in 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 slight safer to handle in a laboratory and is used in preference we have our alahh with our paral 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 now lots of students forget that cyanide is actually carbon and nitrogen so our longest carbon chain in here actually goes into the cyanide the CN that is a carbon in there so we now have a three carbon chain here making this two hydroxy propan nitrile here we've gone from an alahh to a hydroxy nitr we have a similar reaction with our Ketone we have our Ketone here propanone and it has partial charges in there the Delta positive carbon in there will be attached 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 nitrile 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 chyal carbon there in the middle so the reduction of aldah and asymmetric ketones will produce an antias Tolen reagent can be used to a silver mirror test for an alahh to is going to be made from ammonia and silver nitrate giving us a complex ion we will then have aldah reacting with the silver ions producing silver this will give us the silver mirror on the side of a testud 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 one thing we can use to test for carbonal compounds is two for dinitrophenol hydrazine or two for dnp which is a much easier thing to say this is going to give us a positive test for either an aldah or a ketone you can see in the video of it here we are going to get a lovely precipitate form the positive result is going to be this orange precipitate that will form whenever you have an aldah or a ketone you can then go on to do further test if you want to differentiate these two when we are drawing and naming carboxylic acids it is this group here at the end that we are looking at that's the functional group that tells us it's a carboxilic acid this is propen OIC acid the OIC acid bit at the end tells us from the name that it's a carboxilic 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 salts 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 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 highlighted 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 L up with a sodium salt of carboxilic acid if we want to make an ASO chloride we need to start with the carboxilic acid we need to add in s O2 so the O group will be replaced with a c and we will end up with HCL and SO2 coming out as well as our ASO chloride ASO chlorides have a similar structure to carboxilic acids or aldhy with a carbon double bonded to an oxygen 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 errors start where the erors go to AO chloro 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 set up will give us an intermediate ion and then at the end we will get carboxylic acid and hydrochloric acid for this you need water and it will occur at room temperature AO 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 a partial positive charge we will get an intermediate setup and our products at the end will be an Esther and hydrochloric acid for this you need alcohol and a reaction will occur at room temperature asoc chlorides will also eract 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 amide 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 prim Amin and will occur at room temperature amines can act as a base now I'm going to use ammonia as the example here but the chemistry sits for any aing here the nitrogen has a lone pair on it when there is a proton around it can accept that proton and we will get a dative coent bond being formed amines can react with acids acting as a base in exactly the same way we will get a salt out at the end we're going to start this by looking at the naming of aliphatic aines this has three carbons in it so it's going to be propile aine similar to alcohols we have primary secondary and tertiary amines propile amine is a primary amine but if your a 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 methal which is that group there ethal which is the other alcohol group one aain and the N tells us it's a secondary if it has two NS 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 halogeno alane we will get to primary aine 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 aiming as a product this this is a nucleophilic substitution reactions we can also go from a nit trial to a primary aine step one would be the hogen or Alan plus cyanide ion come from something like pottassium cyanide to give us our n trial set up here and then in step two we can take our nit trial reduce it to give us our Primary aing in this situation the hydrogen reduction comes from l i a l H4 when we have aromatic aines we're going to take our nitrobenzene and add in a reducing agent that's hydrogen square brackets which is going to turn the NO2 group into an nh2 group and water this will give us fenile aiming we can use tin or iron as a catalyst and we need to heat it with hydrochloric acid pheny amine is used in the manufactur of dyes there are lots of reactions of amines these can act as nucleophiles halogeno Alan 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 hogen oane will give us a tertiary amine this tertiary amine you can see where I'm going with this can't you 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 orcane and ammonia the ammonia the lone pair is going to be attracted to the carbon and then the bra is going to get shifted out the way we're going to have an intermediate which ammonia is going to be attracted to again and then we are going to have our primary aiming we're going to start from the same place 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 complic ated and my is 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 aain with nitrogen acting exactly the same way our intermediate is looking a little bit more complicated but a m is going to be doing exactly the same thing and we're going to end up with a tertiary aiming at the end back again to our halogeno alane and this time we have our tertiary Amin the nitrogen that being involved and this time we go to aary ammonium salt and these are used as cic sulfa amino acids might feel like a bology 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 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 will always have the same basic structure an 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 chyro 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 my 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 I've drawn these the other way around ease of showing you the reaction carboxylic acid group on an amino acid can react with a hydroxide here I've used sodium hydroxide to give us a salt or it can react with an alcohol to give us an Esther the amino group can react with an acid and again we will get a salt at the end you can have three different types of amides primary secondary and tertiary because this carbon over here can be any long alkohol group it could be ch3 it could be ch2 ch3 could be ch2 ch2 ch3 I'm just going to draw it as a carbon because it can be anything this is a primary amide because there is one carbon nitrogen bond this is a secondary amide and again this and this is any long alcohol group here we have two carbon nitren bonds and we can show this in the naming by putting an N at the front of it tertiary amides and again 1 2 three are just any alcohol grp groups have three carbon nitrogen bonds and we can show this in the naming by putting NN at the front before we then go on to list all the alal groups 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 each other 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 poly amides because that is the L 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 rackets large square bracket surrounding everything and the EST 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 benzine 14 dicarboxylic acid reacting with ethane 12 dial and this will give us terene which is a fabric here is another dicarboxylic acid acid reacting with the di aiming 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 hexan dioic acid this is reacting with hexan 16 diamine to give us Nylon 66 you can have different numbers in the middle of nylon you can 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 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 Bullit proof vests here we have a polymer of Kevlar and you can see within lots of the bonds here there are partial dipoles so that when we get two polymers lining up next to each other we can get bonding between the polymers happening we can get hydrogen bonding occurring between the oxygens in the carbon oxygen double bond and the hydrogens in the nitrogen hydrogen bond giving very strong bonds between strands of polymers when we are looking at organic senses Pathways when we like looking at designing things we need to be able to extend the carbon chain length we can do this by adding on a nitrol group here we have halyo alane and potassium cyanide the nitrol group is going to replace the hallogen group in a nucleophilic substitution reaction and we're going to get a longer chain L here in our halo Alan we have two carbons whereas over on right hand side after the reaction we have three carbons this can also happen with Al hides or ketones with the addition of hydrogen cyanide we are going to go from an oxygen double bonded to a carbon to a hydroxide group and a nital group here we have introduced a branch but this could also be these are intermediates in organic synthesis Pathways because then we can take the nitrs that have been formed this one we are adding hydrogen and the Catalyst and that nitrol group on the end we're going to be breaking the triple bonds and we're going to be adding in hydrogens an alternative thing that you can do is to add water and hydrochloric acid heast up a little bit and turn your nitr into a carboxilic acid you have the choice of a reduction reaction to produce an aing if that is what you looking for in your organic synthesis Pathways or a hydrolysis reaction to produce a carboxilic acid if that is what you need in your organic synthesis the next two slides are going to be summary slides of all of the organic reactions that we've come across in the course so far it is going to go pretty quickly because I've gone over these in detail in other places in this video what you can do is pause this slide and copy down the sections that you need and practice going from place to place if you want a more detailed video why 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 pause it copy stuff down and then use that to answer questions [Music] [Music] [Music] [Music] [Music] n [Music] n [Music] [Music] n [Music] pron 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 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 else 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 due to ated chlorop form and deuterium doesn't absorb so it won't give a peak other ones could be deated 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 parts of a Benzene ring and if it's all the way over by 11 then chance are its 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 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 adjacent to a ch3 group and there are two protons in this environment the orange 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 paks 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 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 pigs 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 one 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 doublets if it's stent to a ch2 2 plus one is three we will get Three Peaks or a triplet if it's adjacent to a ch3 3 + 1 is 4 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 direct next to each other we can have two doublets which means there's going to be a CH group and a CH group we can have two triplets which means there's going to be a ch2 group and a ch2 group we can have a triplet and a dublet meaning we're going to have a CH group and a ch2 group we could have quartet and triplets showing a ch2 group and a ch3 group a quartet and a with a CH group and a ch3 group or a multipet which this one has seven paks in it and thinking about our n+ one rule seven PS is going to mean six hydrogens and the most common example of this is going to be two ch3 groups or two methy groups attached to the same carbon now when we are looking at Peaks you have to remember that it is next to so this ch2 group is is responsible for the triplet and the ch3 group is responsible for the quartet carbon andmr 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 the far left is defitely going to be a carbon a Wonder 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're can to look at a couple of examples all have the same formula but a different arrangement in space we going look at the different carbon environments but one all has four different carbon environments whereas as we move through the compounds we can start to see that the methy 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 or two peaks and three carbon environments Three Peaks chromatography can be used to separate out different things in a mixture thin lay 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 there 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 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 fit each part to leave the column can be recorded and each fragment can be identified even more sophisticated is gas chromatography which works on very similar principles but gives us much more detailed results this can be used to separate volatile liquids or gases again you will have a column packed with solid and gas will pass over it at high temperature and high pressure the retention time can be used for ident ification of samples ouch this is why in some videos I explain scratches [Music]