in this example we're given the molecular formula of a compound and our goal is to use this nmr spectrum to determine the exact chemical structure of this compound how can we do this now because we have only three carbon atoms it might be wise to draw the different ways in which we can represent this molecule so let's draw all the constitutional isomers of c3h7br so we have three carbon atoms now there's only two locations in which we could put the bromine atom we could put it on carbon one or we can put it on carbon two so it's either one bromopropane or two bromopropane but now let's try these molecules a different way so we could draw one bromopropane like this ch3ch2 ch2br now 2-bromopropane will look like this is basically an isopropyl group attached to a bromine atom so looking at these two structures which one corresponds to the nmr spectrum that we see here now the simplest way to get the answer is to count the number of signals how many signals should we get for this molecule or how many different types of protons do you see this is proton a proton b proton c these three protons in the methyl group will show up as one signal but they're different from these two protons because they exist in a different chemical environment these protons are closer to the bromine atom than the ones in the methyl group so they will show up as a different signal so therefore this molecule cannot represent the structure that we see here because it has a total of three signals whereas we need two signals now let's analyze the second molecule we know by default this has to be the answer but let's understand why these two methyl groups are identical to each other they exist in the exact same chemical environment so therefore they will show up as one signal which we can refer to it as signal a signal b has to be the ch group because that's different from the methyl groups so right then and there we can see that we have the two signals now which peak corresponds to signal a and which one corresponds to signal b what would you say looking at the chemical shift the chemical shift for a ch next to a bromine atom is usually around three to four so therefore this has to be signal b and this will be signal a a method group typically is around one ish but if it's one carbon away from a carbon with a bromine atom it's going to be a little bit more downfield or towards the left so it's going to be a little bit more than one let's look at the split in pattern the ch group is adjacent to six hydrogen atoms so using the n plus one rule we're going to get basically a signal with seven peaks and so this is going to be a septem now analyzing the splitting pattern of the method group because these two are identical to each other you only need to focus on one at a time so focusing on this methyl group it's only adjacent to one hydrogen atom and so using the n plus one rule one plus one is two and so we can see the doubling that we have here around 1.4 1.5 and so it's very clear that this structure represents two bromo propane or we can call it isopropyl bromide now let's move on to our next example c4h9br now recommend you pause the video and try this example so based on the last example use what you've learned to draw the chemical structure that corresponds to this particular nmr spectrum and then once you have your solution you know play the video to see if the answer is correct so let's begin let's draw the different constitutional isomers for c4h9br and let's do so using line structures initially so the first thing we can do is we can put all four carbons in a straight chain now we can add a bromine atom on carbon one so this is going to be one bromobutane or we can add a bromine atom on carbon two and this is going to be two bromobutane now instead of having a straight structure like a straight line structure this could be a branch structure so we could have four carbons like that and then we could put the bromine atom on the primary carbon or we could put the bromine atom on the tertiary carbon so those are the four possibilities that we can get for c4h9br now let's convert the line structure into a condensed structure so let's start with one bromobutane now this particular carbon atom is attached to one other carbon atom so this is going to be a ch3 you always have a methyl group at the end of a carbon chain this carbon atom has two hydrogen atoms and it's attached to two of the carbon atoms so keep in mind carbon can only form four bonds at most so next we have a ch2 and this two also has two hydrogen atoms so that's going to be a ch2 this last carbon has two hydrogen atoms and a bromine so that's going to be ch2 br that's the condensed structure for one bromobutane now let's do the same for two bromobutane so we always have a method group at the end so this is going to be ch3 next we have a methylene group of ch2 this carbon is connected to two carbon atoms in a bromine atom so there's only one hydrogen left since carbon has four b onds so this is going to be a ch attached to a bromine atom and here we have a methyl group so that's two bromo butane now here we have one bromo two methyl propane so let's convert that into a condensed structure so let's focus on this tertiary carbon first that carbon is attached to three other carbon atoms so the fourth bond must be a bond to a hydrogen atom so i'm going to draw that first that's ch and then we have two methyl groups so we have a ch3 on top and a ch3 on the bottom and then this is a ch2 there's two hydrogens on it attached to a bromine atom so we have ch2 br now finally let's convert this one into a condensed structure so we have a carbon here and there's no hydrogens on this carbon because that carbon is quaternary it already has four bonds but we do have three methyl groups so we could simply just write the three methyl groups so that's tert-butyl bromide and we can get rid of this so those are the four condensed structures that we can draw for c4h9br now what do you recommend that we should do next in order to identify the correct chemical structure let's use the process of elimination so looking at our proton nmr spectrum we have four signals one two three four so let's eliminate the structures that do not contain four signals so for this structure this is going to be signal a b c d we have four different types of hydrogen atoms so we can't eliminate that one for this one too this we can call this a b c and d so that too has four different structures I mean four different signals now for this structure these two I mean these three methyl groups they will appear as one signal so we can definitely eliminate that structure because it doesn't have four different types of hydrogen atoms so this is not the answer that we're looking for now looking at the next example these two methyl groups are identical so they will appear as one signal which we can call signal a and then we have a ch group that's b and then the ch2 br group so there's three different types of hydrogen atoms in that chemical structure we don't have four so this molecule will give us three signals not four so now we have two options we need to determine if it's going to be we'll call this structure number one or structure number two that is is it going to be two 2-bromobutane or 1-bromobutane so let's find out so what we have here is a triplet this is a doublet and here this is a quintet and we have a sextet or six signals so let's identify the splitting pattern for the hydrogen atoms in this structure so proton a what's going to be the splitting pattern for the methyl group here well if we look at the adjacent carbon it has two hydrogen atoms using the n plus one rule two plus one is three so this will appear as a triplet now for protons b there's five adjacent hydrogen atoms so using the n plus one rule five plus one is six so this will appear as a sex ten now for proton c this is two plus two which is four and then plus one that's five so this will appear as a quintet now for signal d or the protons that correspond to signal d there's only two hydrogen atoms adjacent to it or on the adjacent carbon two plus one is three so we're going to get triplet so note that we do not get a doublet with this particular structure so that tells us that we could eliminate this one which means two bromobutane must be our answer but let's confirm it so let's call this proton a or signal a b c and d so this methyl group is adjacent to two hydrogen atoms two plus one is three so that's going to appear as a triplet for signal b the ch2 is adjacent to three plus one or four hydrogen atoms so plus one that's going to be five we're going to get a quintet and for signal c we have two plus three or five adjacent hydrogen atoms plus one so that's going to be a sextet and for the last one the ch3 is adjacent to one hydrogen atom one plus one is two so we get a doubling thus this structure has all the signals that we need so we have a sextet here which is between three to four or it has a chemical shift between three to four due to the electronegative bromine atom attached to ch group so that makes sense this ch stream we can see a doublet just above one and for signal b we can see it's a quartet between one and two and then the last one the ch3 at the end that one is further up field and it's a triplet it's further away from the carbon with the bromine atom so it's more shielded thus it has the lowest chemical shift so that's it for this particular example so that's one way in which you could draw the chemical structure given the proton nmr spectrum