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Role of alcoholic potassium hydroxide in an E2 reaction

Chemistry Asked by Shubham Sagar on November 11, 2021

Which of the following compound will give least substituted alkene as major product with alcoholic $ce{KOH}$.

The options are:

1.enter image description here
2.
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3.enter image description here
4.
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I am confused between options 1 and 3. Both seem to give more substituted alkene as the major product. Does alcoholic $ce{KOH}$ always mean it is an E2 elimination?

What is the mechanism here?

One Answer

Haloalkanes react with alc. $ce{KOH}$ mainly via $E_2$ mechanism. But, in some cases, where $beta-ce{H}$ isn't present, it proceeds via $E_1$, as in case of $(4)$ one. Let's analyze each option,

Option $(1)$:

Firstly, the $ce{-Cl}$ departs, and then the base picks up a $beta-ce{H}$ from the terminal methyl group (i.e., $ce{C-5}$), resulting in mono- substituted alkene (Hofmann product as major). Steric factors are significant in selectivity of $beta-ce{H}$. Therefore, Zaitsev product will be formed in very minor concentrations, bcoz $ce{-C(CH3)2Ph}$ is more bulkier substituent than $ce{^tBu}$.

Option $(3)$:

This one is similar to $(1)$ case, except for the fact that due to less steric hindrance, it will give di- substituted alkene (Zaitsev product).

Option $(2)$: It's clear that this molecule will form a tri- substituted alkene.

Option $(4)$: As hinted above, it will proceed through $E_1$, resulting in a tri- substituted alkene.


Different levels of substitution are shown below, (source: www.chemistrysteps.com)

substitution-in-alkene

Answered by Rahul Verma on November 11, 2021

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