2H quartet in a 1H NMR spectrum, from TBR OCHEM

[arc5005]

First I want to say that this question was part of a 50 question quiz, and was one of my best scores so far on these practice quizzes in the TBR books. I got a 44/50, which put me in the 130-132 range for this section, so that felt great. However, I still need some clarification on this topic, please!

Which compound shows a 2H quartet in its 1H NMR spectrum?

A) 2-chloropentane
B) 3-chloropentane
C) 2,2-dichloropentane
D) 3,3-dichloropentane



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Answer:

D) 3,3-dichloropentane

A 2H quartet results from a CH2 group adjacent to a CH3 group on one side and a carbon with no hydrogens on the other. In 2-chloropentane, there is no isolated CH2 groups, so choice A is eliminated. In 3-chloropentane, there are two CH2 groups adjacent to CH3 groups, but each is adjacent to a secondary carbon bonded to an H. Choice B is eliminated. In 2,2-dichloropentane, there is an isolated CH3 group, but there is no isolated CH2 group, so choice C is eliminated. In 3,3-dichloropentane, there are two CH2 groups adjacent to CH3 groups, and each is adjacent to a CCl2 group (which has no hydrogens). The result is that the CH2 groups are isolated quartets, making chocie D the correct answer.

random <- I drew each molecule, click link (not spam).


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Clarification:

1. So I wasn't sure if the answer for this was option A or option D. I ended up choosing A (2-chloropentane) because I thought maybe the CH2 on carbon 3 would show a quartet since it is adjacent to a CH2 group & a CHCl group. But I knew that this might also show a multiplet or another signal, because the molecules it is coupling with are not the same, so I kind of knew option A might be incorrect. When this situation occurs, what would the signal look like, and what is it called?

2. The reason I didn't choose choice D even though it did have an obvious CH2 adjacent to an alone CH3 group was because there are two of them, and it's a symmetric molecule. So this is where I need some clarification please. I thought maybe because there was 2 identical CH2 groups that it wouldn't appear as a 2H quartet, and it would instead appear as a 4H quartet. Could I get clarification on this please?

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[Doogie.Howser]

For 1, it is most likely that the coupling constants between the the CHCl proton and CH2 proton nearby aren't equal and would result in a multiplet such as a doublet of doublets (I don't believe it's a doublet of doublets but it is one of those scenarios). So be careful in assuming a quartet unless the protons involved in coupling on both sides of the proton in question are in equal environments.

For 2, the lowest possible whole numbers are used in discerning the integration. You're correct that there are four protons hidden under that quartet, but due to symmetry the 4 is reduced to 2 and the 6 is reduced to 3.

If you need further clarification, let me know!

 

[arc5005]

For 1, it is most likely that the coupling constants between the the CHCl proton and CH2 proton nearby aren't equal and would result in a multiplet such as a doublet of doublets (I don't believe it's a doublet of doublets but it is one of those scenarios). So be careful in assuming a quartet unless the protons involved in coupling on both sides of the proton in question are in equal environments.

For 2, the lowest possible whole numbers are used in discerning the integration. You're correct that there are four protons hidden under that quartet, but due to symmetry the 4 is reduced to 2 and the 6 is reduced to 3.

If you need further clarification, let me know!

doublet of doublets refers specifically to the hydrogens of a disubstituted benzene with with para-substitution when they are different molecules. thank you

 

[Doogie.Howser]

I think that's too specific of a scenario for how often I've seen a doublet of doublets on a proton NMR! Lol. I thought the extended splitting patterns came up any time a proton is coupled with 2+ different groups that give it different coupling constants?

 

[arc5005]

I think that's too specific of a scenario for how often I've seen a doublet of doublets on a proton NMR! Lol. I thought the extended splitting patterns came up any time a proton is coupled with 2+ different groups that give it different coupling constants?

oh maybe i'm just thinking for just the benzene range, then.

 

[AdaptPrep]

When this situation occurs, what would the signal look like, and what is it called?

If you want to know what the signal for the hydrogens on carbon #3 in 2-chloropentane would look like, remember the "n+1" rule and that splitting of splitting is multiplicative and not additive.

Carbon #3 is adjacent to two different carbons, carbon #2 and carbon #4.

Carbon #2 has one hydrogen. Therefore, one hydrogen on carbon #2 would split the hydrogens on carbon #3 into a doublet IF THEY WERE THE ONLY NEIGHBORS (n + 1 = 1 + 1 = 2).

Carbon #4 has two hydrogens. Therefore, the two hydrogens on carbon #4 would split the hydrogens on carbon #3 into a triplet IF THEY WERE THE ONLY NEIGHBORS (n + 1 = 2 + 1 = 3).

However, since carbon #3 has two types of neighboring hydrogens, their respective splitting pattern is multiplied: 2 x 3 = 6 (think of it as each of the two peaks of the doublet being split into a triplet). The splitting pattern of the hydrogens on carbon #3 can now have up to 6 peaks (of course, that is the maximum, it could have less based on the j-coupling values and also the resolution of the H-NMR machine). Since it can have up to 6 peaks on carbon #3, we can safely assume that is not the answer they are looking for. Of course, you would have to use the same method on all the different types of hydrogens to rule out the whole compound. But I think you were asking for the splitting pattern of only the hydrogens on carbon #3.

If you want to check yourself on the other hydrogens in compound A:
hydrogens on C-1 would be a doublet (1 + 1)
hydrogen on C-2 could have up to 12 peaks--a multiplet ( [3 + 1] * [2 + 1] )
hydrogens on C-3 could have up to 6 peaks (as shown above)
hydrogens on C-4 could have up to 12 peaks--a mutliplet ( [2 +1] * [3 +1] )
hydrogens on C-5 would be a triplet (2 + 1)

I hope that all makes sense. Let me know if you need clarification.

 

[aldol16]

1. So I wasn't sure if the answer for this was option A or option D. I ended up choosing A (2-chloropentane) because I thought maybe the CH2 on carbon 3 would show a quartet since it is adjacent to a CH2 group & a CHCl group. But I knew that this might also show a multiplet or another signal, because the molecules it is coupling with are not the same, so I kind of knew option A might be incorrect. When this situation occurs, what would the signal look like, and what is it called?

This signal would be a doublet of triplets or triplet of doublets depending on the relative coupling constants. The C3 protons would be split by the methylene protons on C4 and the methine proton on C2.

2. The reason I didn't choose choice D even though it did have an obvious CH2 adjacent to an alone CH3 group was because there are two of them, and it's a symmetric molecule. So this is where I need some clarification please. I thought maybe because there was 2 identical CH2 groups that it wouldn't appear as a 2H quartet, and it would instead appear as a 4H quartet. Could I get clarification on this please?

In NMR, the ratios of protons are what you are looking for. When we integrate a spectrum, we choose a peak to correspond to unity and the rest of the spectrum is integrated by the program relative to the area of that reference peak. So in other words, you will get the correct ratio of protons always but not necessarily the correct absolute integrations. In molecules of symmetry like 3,3 dichloropentane, the two methylene signals will be the exact same as well as the two methyl signals. Therefore, you will see two signals in the NMR spectrum with relative integrations of 2:3, where the absolute numbers are actually 4:6.

 

[aldol16]

For 1, it is most likely that the coupling constants between the the CHCl proton and CH2 proton nearby aren't equal and would result in a multiplet such as a doublet of doublets (I don't believe it's a doublet of doublets but it is one of those scenarios). So be careful in assuming a quartet unless the protons involved in coupling on both sides of the proton in question are in equal environments.

Splitting by methylene protons yields triplets and splitting by methine protons yields doublets. So you couldn't possibly get anything resembling a quartet or doublet of doublets in this case.