Czarcasm

Hakuna matata, no worries.
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Jun 22, 2013
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I just want to make sure if I'm understanding this correctly:

The non-Zaitsev (Hoffman) product typically forms under one of two conditions:

- Using a bulky base (ex. t-Butoxide). The bulky base abstracts the less hindered proton to form the less substituted alkene.
- Using a bulky LG on a substrate. Only instance I've really seen this is for quaternary amines that undergo elimination to form an alkene. I suppose since the LG itself is bulky (Amine with 3 alkyl groups attached), it also causes the base (I suppose any strong base in this scenario: bulky or not) to abstract the less hindered proton, also forming the less substituted alkene.

It's really this second instance I'm not too sure about. The first one I recall learning, but the second one I'm not too familiar with. "Hoffman Elimination" is on the AAMC Outline, so I figured this was important to know.

If anyone is familiar with this and can shed some insight, that would be greatly appreciated.

Thanks!
 
OP
Czarcasm

Czarcasm

Hakuna matata, no worries.
5+ Year Member
Jun 22, 2013
964
407
Crypts of Lieberkühn
Status
Medical Student (Accepted)
Nevermind, found my answer thanks to one of @BerkReviewTeach's older Q&A responses:

Hofmann Elimination is a very specific reaction that results in the formation of the non-Zaitsev alkene. The MCAT test writers are free to word things any way they wish, so perhaps they will use the term Hofmann product in lieu of saying the non-Zaitsev product.

The thing to note about the Hofmann Elimination is that rather than emphasizing a bulky base (which is a necessaity in E2 reactions in order to reduce the amount of substitution reaction that competes), they use a bulky leaving group. An amine is exhaustively methylated to form trimethyl ammonium cation (which is the equivalent of a positively charged version of a t-butyl group). This leaving group is so large that the conformer it forms must have the bulky alkyl chain anti to it, so when undergoing an E2elimination, the more hindered alkene (Zaitsev product) cannot be formed. Typically you use a base that has a cation that complexes with the counteranion of the amonium cation (usually iodide) and you don't worry as much about the bulkiness of the base. Silver oxide (Ag2O) is a typical base used for the elimination reaction.