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. enter image description here

3.enter image description here 4. enter image description here

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?

  • 2
    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – andselisk Jul 22 at 11:59

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)


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