Re: Explanation for 119

I understand all of the explanation except the initial determination of peaks. Why are there 3 peaks? I thought there would be 5 peaks. What happens to the hydrogens attached to C1 and C4?
-----------
Sorry that I didn't respond to this earlier -- I didn't realize that the question had been put as a comment on the previous one.

The hydrogens on C1 are part of the two methyl groups bound to C-2. While, in nomenclature, we would distinguish this carbon from the carbon in the methyl group on C-2, you can see that they're actually equivalent.

Please check out my unparalleled MS Paint skills that show the three groups of protons by color:

Question #119 on AAMC Practice Test #9

This is asking what the H NMR of the Compound II would look like (I tried to recreate Compound II but it didn't show up well).


The explanation provided was rather brief. Can you please provide a bit more detail? Thanks.

-----------------

This question asks us what the proton NMR of 2-bromo-2,3-dimethylbutane would look like. Just to walk through the nomenclature real quick, our longest carbon chain is four carbons. Carbon "2" would be the one that has the bromine on it, to give it the lowest possible number. This carbon also has a methyl group attached. Carbon "3" has just a methyl group.

With proton NMR, there are four things to consider:

(1) groups of equivalent hydrogens (corresponds to the number of peaks)

(2) number of hydrogens within each of these groups (corresponds to the "integration," or height of the peaks)

(3) electron-withdrawing groups and hybridization of the carbons (corresponds to "left-shifting" or "de-shielding" of the protons)

(4) splitting of peaks (corresponds to how many hydrogens are connected to the adjacent carbon).

For this question, since it's only asking us how many peaks there are, and what the splitting is, we can focus only on #1 and #4 above.

First step: how many groups of equivalent hydrogens are there? To answer this, I usually think about how I would describe the hydrogens in the molecule. Here, there are three types:

-Those that are in the methyl groups connected to carbon 2

-Those that are in the methyl groups connected to carbon 3

-The one hydrogen by itself bound to carbon 3

Automatically, this means there are three peaks. Now, let's talk about splitting.

For the first group (methyl group hydrogens on carbon 2), there will be no splitting. This is because there are no hydrogens on carbon 2. Remember that splitting will correspond to n+1, where n is the number of hydrogens on the adjacent carbon. For this peak, we end up with a singlet.

For the second group (methyl groups on carbon 3), there will be splitting caused by the lone hydrogen on carbon 3. This will split this peak into n+1 = 1+1 = 2 "subpeaks," or a doublet.

Finally, for the last carbon (the one by itself on carbon 3), it will be split by the six carbons found on the methyl groups attached to carbon 3. This will split the peak into n+1 = 6+1 = 7 "subpeaks," or a septet.

Thus, we end up with a singlet, a doublet, and a septet.