[Music] this video is going to be looking at topic 10 deep for the international es level chemistry from edit cell and we're going to be focusing on mass spectrometry and and threaded so we're going to be looking at how we interpret the data from mass spectra in order to suggest possible structures for organic compounds so this is using our master charge ratio and also a fragmentation patterns we're also going to be using em Fredette spectra in order to determine the functional groups that are present or to be able to predict the EM Fredette absorptions or give wave number data depending on what functional drips are present so let's start off by looking at the mass spectrometry of organic molecules so we've met mass spectrometry back and topic two when we were looking at simple diatomic molecules such as chlorine or bromine now we can actually use mass spectrometry to interpret much more complex structures we're gonna focus on some very simple organic structures and we can figure out not only the molecular mass but also the actual structure itself and then the mass spectrum of any compound is going to contain lots of definite Peaks and this is due to fragmentation and what that means as we're getting the compound breaking down into specific ions that have a positive charge and the mass spectrometry will figured out what mass they have and people come out on the spectrum so the peak with the highest mass to charge ratio or MZ value is known as the molecular ion peak now this peak represents the relative molecular mass of the organic molecule so it basically means that if you have a molecule that has a mass of 56 you're going to have a peak at 50 secs and it should be the highest MZ value there should be nothing else after that please note that this is different from the tallest peak the tallest peak and the spectra is known as the base peak so as important that we do distinguish between these two things especially when you are answering a question ever asks what one represents the molecular mass you need to be seeing the highest mass-to-charge ratio not the tallest so fragmentation patterns are commonly used to determine the molecular structure of an organic molecule and what actually happens with these fragments as they are formed when we get carbon to carbon bonds and the molecule being broken and we form a positive ion and a neutral species which is usually like a free radical so if we look at the example here we've got ch3 ch3 we're gonna break this bond here and we're gonna be left with a ch3 iron and a ch3 free radical if we have a longer chain you can see that we can get different fragments we can get upon possibly a ch3 plus or a ch2 ch3 plus and each of these fragments is going to have a different mass and the mass will correspond to the MZ peaks and the spectrum remember that Z is your charge and in this case the charge is 1 so it would be the mass divided by 1 and that's where we would see our peak so let's have a look at an example if we have a mass spectra of butane if you can't remember butan as a four-membered hydrocarbon as an au kin completely saturated with hydrogen's all around the carbons so we have a c4h10 now what happens and here as we've said we get this fragmentation pattern and that causes us to get some breakage between our carbon to carbon bonds and when that happens then we make a couple of different Peaks now you can see that we do have some that are not present but we haven't labeled them we're only going to be labeling the most common Peaks so they tend to be the ones that are the highest so we've got a peak at 15 a peak at 29 a peak at 43 and a peak at 58 this one here because it has the highest mass to charge ratio this as our molecular ion peak that's tells us that the molecule has a mass a molecular mass of 58 that's one here at 43 this as our base peak and it just means that this is the most abundant the most common fragment pattern so one of the most common fragments that we'll see as with a mass of 15 and that as a ch3 our methyl fragment we can then have an ethyl at 29 or propel at 43 and of course the butane with this positive charge is going to be our molecular ion peak so you can see that these Lane up what the main piece here so when we're analyzing a mass spectrum there are a number of different key Peaks that we're going to come across the more you practice this the more you will start to memorize some of these Peaks but I do recommend that you try to commit this to memory because it would become a lot quicker for you and you will see that it's the same Peaks that come over and over so we can have Peaks like our methyl at 15 we can have a carbon oxygen bond at 28 we can have aldehyde group at 29 or carboxylic acids at 45 it just depends on the molecule itself so let's have a look at an example so compounds a and B have the molecular formula of C 3 each sex or their mass spectra are both shown below and we want to be able to deduce the structures now when we have c3h6 oh hopefully you realize that can't be obviously a hydrocarbon it also cannot be an alcohol because we don't have enough hydrogen's so we're gonna have two possible answers here we're either going to have opener which has the aldehyde or we're going to have propanone which has the ketone and we want to determine from these structures which one is going to be watch now both mass spectra show a peak at 58 now this is our molecular ion peak and because both of them are they're showing that same P that doesn't actually help us they also both have a peak at 15 which means that we must have a ch3 fragment and now again because this appears on both of them it does not help to distinguish the two structures so what we actually need to look at is we need to look at the peak that is left now and here we have a peak at 29 whereas NB we have a peak at 43 now as we have a peak at 29 or if you go back to our table we said that 29 was either going to be a ch3 ch2 or a CH oh okay so based on that well we know actually that we have both of those Peaks because this must be proper now because proper now is going to be drawn like this and what we see here is we have this bond breaking and we end up with this fragment here and this fragment here both of which are 29 so that must mean that a is proper now now we can confirm this by looking at the piece for Part B so for structure B we've got a peak at 43 and if we go back we said that that could either be our peak that we see in a ketone or that could be any of the other ones as well and there because these two here are not going to be produced by this particular molecule this must be our ch3co structure so just to confirm propanol and it's drawn like this so we are seeing a fragment happening here and the speak at 43 is going to be that part of the structure here so Part B must be propanone so looking at the answer you can see that it's telling us what we just seen there that the major peak at 29 must be our CH or positive or our ch3 ch2 so this as the aldehyde proper now whereas our peak at 43 is going to be our CH 3 CH or because it cannot be either one of these because they would also give in order to get that you would have a radical being formed that doesn't show on our spectrum so this must be propanone now sometimes you may be asked to write the equations well what happens is we've got our ch3 ch2 CH over here so our aldehyde and then fragmenting to form the ether fragment and also fragmenting to form our CH both and for our ketone we have our CH d co and our CH defragment being formed so let's look at another example this time we're going to be looking at C and D and we're looking at these compounds that have the molecular formula C 3 H 8 or now that tells us that it must fall into our alcohol so we've got two possible structures we can either have propound one o which is our primary alcohol or we could have propane 2 O which is our secondary alcohol then we're going to use the mass spec to tell us what 1s watch even though of course both of them have this peak at sexted which is our molecular ion peak but again that doesn't help us with the structure they also both have a peak at 15 which we know as for a ch3 peak so then we have to look at what is left over so for C we have a peak at 31 whereas for d we have a peak at 45 so if we start with D well if we ever peak at 45 and I'll pick it 15 is because we've had a breakage here and we better than left over with a fragment that must add up to 45 and that is we can see from an oxygen two carbons and five hydrogens so that must mean that D has propun to oh let's just check to see FC as actually propound one all and here we can confirm it well we look at this peak at 31 and let's pick at 31 is caused when we take a fragment here between these two carbons and we're left with one carbon one oxygen and 50 hydrogens and that gives us a peak at 31 so that tells us that C must be pull pan 1 oh so when we're trying to find out the structures we want to look at what possible fragments should happen from our structure so the best thing to do is to draw it and then start to break the carbon-carbon bonds then you simply just add the amount of mass for each of the fragments and see if it matches up with the peaks that we have and our structure so you can see that we've got the answers here still confirming that and see we have this 31 which tells us it was proper 1 all whereas D has the peak at 45 and it tells us that it is full power - oh that's it for mass spec if you want to practice that more there are some questions that you can find on the internet or you can check past papers to see if there are any specific questions you may wish to check unit 2 and unit FLE so let's move on to infrared spectroscopy or infrared is a type of radiation that has a frequency just below that a red light in the visible spectrum so if our visible spectrum is here this section here is prime credit now when our molecules absorb infrared radiation there are two possible effects that weak have we can either have the bond stretching like this so going back and forth or we can have the bonds bending and moving like that now an international ear level we only can set at the stretching vibration so you don't need to worry about the bending the amount of radiation that is going to be absorbed will depend on three things they can depend on the length of the bond the strength of the bond and the mass of the atom involved in the bond now because we've got these specific factors that do have an effect on the amount of radiation that is absorbed we can then start to look at characteristic absorptions so specific functional groups because of the bonds that they have with their specific lens strengths and atoms they are going to have a very specific absorption and we can use that to an advantage and less type of spectroscopy the absorption of infrared radiation is linked to changes and platitudes so any small nonpolar molecules are not going to absorb infrared radiation so just please be aware of that you do need to have a level of polarity so IR spectra have a vertical axis which you can see on the diagram here and there is known as transmittance and that is just another way of seeing how much radiation has been absorbed the horizontal axis is known as wave number and it has a frequency with the unit centimeters to the mainland Sun now what you must notice is that it goes and the opposite direction from what you would expect it starts at 4000 and it goes down to 500 and you can also see that we have a change and scale here I don't want it too much about that you're not going to be asked to explain why we do it like that it's just to make you aware that there has a slight difference on the scale so you need to take that into account when you're checking your piece there each of these Peaks that I'm circling here we'll all be characteristic to the vibration of a specific group so when the molecule is absorbing that infrared radiation each of its functional groups will absorb a specific amount and it will cause a stretching vibration and we will then see a peak on the spectra compared to that so when we're looking at alert and fed spectrum we are interested and the intensity of each peak because it shows how much radiation has been absorbed if we have a weak intensity that refers to a high transmittance value and high intensities refer to a low transmittance value so that just tells us how high or low the peak is going to be we also of course need to know the wave number of values that is one of the most important things because that tells us what functional group is and we can also wait to see if it's going to be a sharp peak or a broad peak for example if you look at the alcohols the alcohols have a peak of wave numbers from 3 75023 to zero as you don't know that is actually a very broad peak and it's very wide it can be very very overpowering and an infinite spectrum if we compare that to something like an aldehyde that is got as peak between 1740 and 1720 that would be a very sharp peak because it's for very small difference and we've numbers so we should be able to do three things with infrared spectroscopy we should be able to protect the absorptions that we would see in the spectrum based on the functional groups that are present we should be able to deduce the functional groups that are present if we're given a less than wave numbers and if we're given the wave number and the molecular formula we can deduce structure so it all just depends on what you're given and the question and that will also depend on how many marks the question is worth there is a section known as the fingerprint region which is between 1,500 and 500 wave numbers and these are results from bending vibrations so we're not going to consider those and less course you just need to be aware that they will exist but we're only looking at any Peaks there are above 1500 wave numbers so this is a page taken from your data book of course you all have a copy on your data focal if you don't you can find the copy over on the in Excel website and you will have less and you exam so you are not expected to memorize any of the wave numbers so you'll be given this table and your data book and you can use it to determine what the Peaks are notice it does have the bending vibrations and for your CH but as we said these are going to be in the fingerprint region so we're not actually interested in those you'll be focusing on the stretching vibrations between all the colors so let's look at a few watch examples if we have the infrared spectrum of proper now what would we expect to see well prop allow is going to be an aldehyde so it's going to have this particular structure so we should be seeing CH bonds um absorptions which will give us in the range of 2900 to 28 20 and from 2775 to 2700 so again you can check that on here there's your aldehydes and we will also have that very sharp carbonyl bond Chun transmittance peak at between 1740 and 1720 so this is the way that we would be predicting the functional groups if we had the for what example two if we're given the infrared spectrum of this particular molecule what what would we be seeing well again we can draw it if we're not sure from the spectrum sorry from the structure so this is going to be profound - oh and the most common one that we're gonna see of course yes we will see the CH bonds but the most common one is going to be this voltage and because as an alcohol we're going to protect a very broad absorption between those wave numbers so it's important that we're distinguishing that for the aldehyde as we've said this is going to be a sharp absorption whereas for the alcohol that is going to be a broad absorption white example who three gives us the wave numbers and then asks us to figure out what functional groups it in tins so we're gonna have absorptions at 3675 2,870 and 1735 so we would match each of these up to the data book and we would see that the 3675 would be our alcohol the chili 7-0 would be our CH bonds and are 175 will be our how to height this is simply just using your data book so you must be comfortable with being able to flip through the data book and find the vibrations that are characteristic for each of these functional groups so please make sure that the first thing you see the data break is not when you go into your exam make sure that you're using are all driven your past papers and you are familiar with square each watch each page s and what each page has I feel like it worked example 4 so we have an organic compound that has absorptions and the wave number ranges of 3500 to 3300 and three five zero zero two two five zero zero so we want to again these are data book to help us with that and this is going to be an H direct but this time it is not for an alcohol as a no H for a carboxylic acid so you will see and the data book if we go back that our or witch's efforts and alcohol our phenyl is going to be different to a carboxylic acid so just make sure that you are packing the cadet wave number range and our absorptions between 3300 and two five zero zero are going to be for our amine group and that could be for an amino acid but don't worry too much about knowing what amino acids are because that is you haven't covered that yet that is covered in year 13 but if that was going to be the answer they would give you enough information probably and the question would be at the start saying this is the structure of an amino acids showing that it has a carboxylic acid and an amine group so again it's making sure that we read the question as closely as we can if we look at what example five we're less time looking to deduce the possible structure so we're given with number absorptions and we're also given c2h4 Auto so we've got looking at these particular structures we know these particular wave number sorry we know that we have to have an aldehyde functional group and we have to have an alcohol functional group not that we cannot have it as ethanoic acid being our carboxylic acid because our range for ROH is out west back so we've got three four five zero and it's at what the range so it can't possibly be a carboxylic acids so our structure could only be an aldehyde group with a carbon that must have an H group attached to it now you've probably never seen this molecule before but that does not matter you can still work out its structure from all of the information that you're a given wide example sex is looking at an organic compound with the structure of C 3 H sex altered and you can see that we're given the infrared spectrum the same as opposed to given the wave numbers and we're asked to work out the possible structure so we've got a couple of different things that we have to look at here so we're going to have a peak here at 1700 and what we do have a peak here that is approximately 3000 and notice that it is not a particularly broad peak so this is going to be for our or inch and because it's a standard peak then it must be a carboxylic acid and this is going to be the carbonyl group of a carboxylic acid so we go back up to our structure so our molecular formula c3h6 4 - well we're using our products like a cigarette then we've got 3 carbons the only thing that it can be is propanoic acid hopefully these worked examples are making sense but let's have a look at past paper question to give us some information of its what type of question that we're going to see so this is from the June 2018 paper and we are given organic compounds that have been analyzed using physical methods like mass spectrometry and and Fredette now there are other parts of those questions that are a bit chemical tests and quantitative measurements but we're only going to focus on the spectroscopy so we're given best mass spectrum and we are also given that IR and they've given some infrared data and the table so you don't necessarily have to use your data book here they are giving your hand to figure it out which one you need so we want to know what can be deduced about ethanol from the presence of the peak at the mass-to-charge ratio of 46 so let's just go back here and let's see let's the speak here is 46 so we already know that this has the molecular ion peak and what does that tell us hopefully you can remember it tells us the molecular mass so it means that ethanol has an M R of 46 M R being your molar mass or your molecular mass we want to identify the species as responsible for the peak at 31 and the mass spectrum of ethanol and we want to state how its formed so just to make this a little bit easier let's draw our ethanol we have our two carbons are all H group and we're surrounded by our hydrogen's so we need to figure out a peak at 31 where we can only have one fragmentation of our CC bonds so that must be best linkage here and if we just go back we can see that this is a very abundant peak at 31 so we can check that this side is going to be both 12 13 14 15 if we fragment it here whereas here we're going to have 12 13 14 15 plus 16 is giving us a peak of 31 so that peak has to be our ch2 or itch and always remember to put your plus sign we will not accept any other answer it cannot be written as ch3 or because that is not the way that bin marshal is matchless meetup is the actual and then for the second market is how that is formed well it's that we are losing a ch3 and that is due to the carbon-carbon bond breaking you cannot simply just save it as fragmentation you have to see what is happening and the fragmentation so we get this carbon-carbon bond breaking and we're losing that when we fail and that leaves us with the ch2oh plus and then part three is asking us to identify one feature of the infrared spectrum which confirms the functional group and ethanol or our functional group that we're going to be looking at is going to be our or H so if we go back you can see that we have this lovely broad peak here and we can check that it falls within our range of thirty thousand two hundred to three thousand seven hundred and twenty notice that it can't possibly be the carboxylic acid of course because that's not an ethanol but just to check it does not fit with them the wave number range here the carboxylic acid would fit somewhere and two there so we want to make sure that we're including the appropriate wave numbers so it's going to be our H clip and that gives us wave numbers between three seven five zero two three two zero zero centimeters to the minus one well hopefully that's made spectroscopy a little bit cleaner for you again if you do want to try out some more examples there are plenty of our equations that are available and pass papers or you can train someone link or even check the textbook there are what examples there are also the checkpoint questions and the chapter questions if you have any questions or anything that you're not sure about in this video please leave a comment below and I hope to see you back on the channel and future to check it what other videos we have [Music] you