E2 Elimination Question: Zaitsev or Hoffman?

[kevin2400]

Hey guys, I'm a bit confused on what exactly will lead to Hoffman instead of Zaitsev for E2. I've tried searching, but some people say in general, if it's a bulky base, it's going to be Hoffman, but is it possible for a bulky base to result in Zaitsev as well? What about a tertiary halide?

I just wanted to calrify for for the purposes on the MCAT, can we just assume would we just have to that bulky base will result in Hoffman?

Also, what exactly counts as a "bulky" base? cause I'm not really sure how to tell if a base is bulky or not, it seems kind of subjective.

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[LoLCareerGoals]

There is no linear metric here. Mixtures of products could form. Generally bulky base is used to discourage competition from SN2 reaction when the substrate is moderately substituted (2 subs).

What about tertiary halide? You mean when tertiary halide is used as a substrate? Generally when the substrate is 3 tertiary, there is no competition from SN2 reaction and there is no need to use a bulky base.

Problem is when a bulky base is used it can no longer consistently abstract the most subbed proton (to result in the most subbed and thus stable carbcation) and when it abstracts less subbed proton Hoffman product forms. All you can really say is the bulkier the base the more (as a %tage) Hoffman product will form.

 

[dlquan]

A good example I've seen often of a bulky base is tert-butoxide. Whereas, something like methoxide (OCH3) is a small base.

 

[orgohacks]

Yep. In some cases you can see LDA as a bulky base, but tert-butoxide is the industry standard.

 

[chemtopper]

A strong bulky base with tertiary substrate will give nearly 90% E2 mechanism with Hoffmann's elimination.Strong base will attack the proton very fast and if it is bulky natuarlly it will attack the least hindered proton.So this results in Hoffmann's product.

 

[BerkReviewTeach]

Hey guys, I'm a bit confused on what exactly will lead to Hoffman instead of Zaitsev for E2. I've tried searching, but some people say in general, if it's a bulky base, it's going to be Hoffman, but is it possible for a bulky base to result in Zaitsev as well? What about a tertiary halide?

I just wanted to calrify for for the purposes on the MCAT, can we just assume would we just have to that bulky base will result in Hoffman?

Also, what exactly counts as a "bulky" base? cause I'm not really sure how to tell if a base is bulky or not, it seems kind of subjective.

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 E2 elimination, 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.