TBR confusion: Oxygens and Nitrogens in H-NMR

[Episome]

In Organic Chemistry I was taught that hydrogens bound to oxygens or nitrogens "ignore" neighboring hydrogens in H-NMR. The Berkeley Review seems to suggest otherwise.

TBR O-Chem Part II, Page 224, #34.:

"After neutralization, the major organic product from the reaction of H3CNH2 and H3CCl would show which of the following proton NMR peak patterns?"

a. Singlet (3H), singlet (3H), septet (1H)
b. Singlet (3H), singlet (3H), quartet (1H)
c. Singlet (6H), sextet (1H)
d. Doublet (6H), septet (1H)

According to The Berkeley Review, the correct answer is d.

First off, I would be inclined to believe that the secondary amine formed after one reaction would react again, forming a tertiary amine, but the answer claims that "the final organic product is (H3C)2NH," which is a secondary amine. Even if this is true, shouldn't the answer be Singlet (6H), singlet (1H) instead of (d.)?

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[reising1]

In Organic Chemistry I was taught that hydrogens bound to oxygens or nitrogens "ignore" neighboring hydrogens in H-NMR. The Berkeley Review seems to suggest otherwise.

TBR O-Chem Part II, Page 224, #34.:

"After neutralization, the major organic product from the reaction of H3CNH2 and H3CCl would show which of the following proton NMR peak patterns?"

a. Singlet (3H), singlet (3H), septet (1H)
b. Singlet (3H), singlet (3H), quartet (1H)
c. Singlet (6H), sextet (1H)
d. Doublet (6H), septet (1H)

According to The Berkeley Review, the correct answer is d.

First off, I would be inclined to believe that the secondary amine formed after one reaction would react again, forming a tertiary amine, but the answer claims that "the final organic product is (H3C)2NH," which is a secondary amine. Even if this is true, shouldn't the answer be Singlet (6H), singlet (1H) instead of (d.)?

If we are talking about HNMR for (CH3)2NH, then the correct answer is D. You have to remember that splitting happens on non-identical neighboring H atoms. The 6 Hs on the 2 CH3 groups will contribute to a doublet (6H) as a result of the H on the N atom. The H on one CH3 group will not be split by an H on the other CH3 group, because the two groups are chemically identical. And then the H on the N atom will contribute to a septet (1H) because it is a single H that has 6 H atoms on its neighboring atoms (namely, the two methyl groups).

 

[Episome]

"And then the H on the N atom will contribute to a septet (1H) because it is a single H that has 6 H atoms on its neighboring atoms (namely, the two methyl groups)."

I understand the rest of the solution. What I don't understand is why the hydrogen on the nitrogen contributes a septet instead of a singlet. I thought that similar to oxygen, no coupling is observed between the hydrogen on a nitrogen and hydrogens on the carbon atom to which the nitrogen is attached.

See these links for references to what I'm talking about:
Oxygen Coupling
Nitrogen Coupling

 

[reising1]

"And then the H on the N atom will contribute to a septet (1H) because it is a single H that has 6 H atoms on its neighboring atoms (namely, the two methyl groups)."

I understand the rest of the solution. What I don't understand is why the hydrogen on the nitrogen contributes a septet instead of a singlet. I thought that similar to oxygen, no coupling is observed between the hydrogen on a nitrogen and hydrogens on the carbon atom to which the nitrogen is attached.

See these links for references to what I'm talking about:
Oxygen Coupling
Nitrogen Coupling

I'm actually not sure about that. I tend to count them as contributing.

 

[Synapsis]

This is confusing. Amines alkylate really easily (someone else also just asked a question on this). Why wouldn't the product be the quaternary ammonium? The thing is, with each successive methyl added, the nitrogen actually becomes a better nucleophile, and since the product is more reactive than the starting material, you should fully alkylate this thing. Am I missing something?

 

[reising1]

I'm also confused now because I've been doing some research about an H bonded to a N and it seems that it does count but it counts in that it makes a singlet. Take a look at the structure of morpholine online. This has 3 peaks (two triplets and a singlet).

 

[Deadlifts]

This is confusing. Amines alkylate really easily (someone else also just asked a question on this). Why wouldn't the product be the quaternary ammonium? The thing is, with each successive methyl added, the nitrogen actually becomes a better nucleophile, and since the product is more reactive than the starting material, you should fully alkylate this thing. Am I missing something?

For exhaustive alkylation to take place and produce an ammonium salt, you need to have enough moles of the alkyl halide AND a base.

The paragraph mentions no base (like K2CO3) so we can only add one methyl group.