Predict the multiplicity (the number of peaks as a result of spli...

Question

Predict the multiplicity (the number of peaks as a result of splitting) and the chemical shift for each shaded proton in the following compounds. (a) CH3-CH2-CCl2-CH3 (b)

Structures of two organic compounds with bromine and methoxy groups for NMR analysis.

Chemical structures of two compounds for predicting NMR multiplicity and shifts.

Accepted Answer

Hey friends. Let's do this problem. It says, predict the chemical shift and the multiplicity for each type of proton in the compound shown below. I have attached this table to determine the chemical shift for different types of protons. So first we're going to identify how many signals our compound will produce. And we do that by determining the number of unique non equivalent protons. Right? Then we'll determine the multiplicity of each signal. And that depends on the number of neighboring protons to the protons in question. Right. So if we're looking at protons a then we want to look at how many protons or neighboring proton A add one to that number. And that will give us our multiplicity. Then we will determine the chemical shift of that signal. And we can use our table and the characteristics of the structure to help us determine that chemical shift. So, starting with structure one, I'm going to draw all the carbons and hydrogen. We have four carbons Starting from left to right. Ch 3, 2, CBR, BR and another CH three. So, how many types of unique non equivalent protons do we have? We have these three hydrogen zon this method group. Those are equivalent to each other but non equivalent to the remaining hydra jin's. So, those will produce one signal. These protons on the next carbon are equivalent to each other. Right? There is no other carbon carbon. They are Ananta topic. So they will produce one signal and they are different than the other carbons. Right? There's no symmetry in our compound or anything. So our last methyl group will also produce its own signal with those three protons. So that gives us three signals proton A. proton B. proton C. So now will determine their multiplicity. So how many protons are on the carbon neighboring the carbon that has protons A. Well the neighboring carbon has two protons two plus one is three. So that will give us a triplet to represent carbon. Or sorry, protons A. Than protons be. They have a neighboring carbon with three protons, three plus one is four. So that will give us a quartet for those protons B. And proton. See they have no hydrogen on the neighboring carbon zero plus one will give us a single it. Alright. So lastly, let's determine our chemical shift for each proton signal. So carbons A. They are three single bonds. Ch ch. C. H. That carbon is single bonded to methylene carbon. So there's nothing really fancy or electron withdrawing. So that chemical shift will be at around one PPm. Alright. Our methylene protons sit same situation, right? They are single bonds ch ch bonds next to the methyl group. But they are next to this carbon that has bromine atoms. Two bromine atoms. So they are still going to be pretty upfield. But a little bit more downfield than the methyl protons. So that quartet will be at about 2.2 PPm. And lastly our single it for our methyl group over here. Well these protons are attached to the carbon next to the carbon. With those electro negative bromine atoms. So again they're going to be a little bit more downfield than the other methyl group. So that signal it will be at about 1.7 ppm. Alright, that's it for our first structure. Moving on to our second structure, we have three carbon chain 123. And those end carbons are metals. So we have one hydrogen on our central carbon and an O. Methyl group on the central carbon as well. So the number of signals for this structure. Well, these methyl groups are equivalent. Right? The structure is symmetrical because of the O. Methyl group in the middle. Both ends on the outsides are equivalent. So those will give the same structure signal. Those methyl protons, the middle hydrogen will have its own signal. And the hydrogen on the O. Methyl group will also give its own signal. So that will have three signals as well. The multiplicity of each signal. Alright, well, for protons A. There is one neighboring carbon and that carbon has one proton, one plus one is two. So that will give us a double it proton B has two neighboring carbons and each carbon has three protons. So that's six plus one is seven. That will give us a septet and R. O. Methyl protons have no neighboring carbon. Right? It's attached to an oxygen. So there are no neighboring hydrogen ins and that will give us a single it. Alright. Looking at the chemical shift, our methyl protons are just a C. H bond, right? No electro negative groups directly nearby. So that will be at about 1.2 PPm. Our septet for this hydrogen. Well, that's going to be a little bit more downfield, right? Because we have a hydrogen attached to a carbon attached to an oxygen. So in our table we see that when we have that relationship, the signal is going to be around 2-4 ppm. So our septet will be at four ppm. In similarly, our hydrogen On the O. Methyl group are right next to that oxygen atom. So that will be also more downfield at around 3. ppm. But there are three hydrogen is here. Right? So they will be a little bit more shielded than this singular proton. Right? That will have more D shielding happening on it because there's only one proton. So that is it for structure number two. Which gives C. To be the correct answer for this problem. That's it for this one. I hope this video is helpful and thanks for watching

Predict the multiplicity (the number of peaks as a result of splitting) and the chemical shift for each shaded proton in the following compounds. (a) CH3-CH2-CCl2-CH3 (b)

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