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The question asks for the rate of reaction in $\beta$-elimination using alcoholic $\ce{KOH}$.

Arrange the following alkyl halides in decreasing order of rate of $\beta$-elimination reaction with alcoholic $\ce{KOH}$:

(A) $\ce{CH3-CH(CH3)-CH2Br}$

(B) $\ce{CH3-CH2Br}$

(C) $\ce{CH3-CH2-CH2Br}$

The answer I got was:

B > C > A

The answer given is:

A > C > B

In many sources it is given that the order depends on the stability of the double bond formed, that the more substituted alkene is more stable.

But for alc. $\ce{KOH}$ shouldn't we consider it as an $\mathrm{E2}$ mechanism?

By this I mean, shouldn't the reaction rate depend on the steric hindrances caused by the beta carbon group to the attacking base?

Is this wrong? What would be the correct explanation?

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  • $\begingroup$ Are you aware of the $\beta-$elimination mechanism? The base abstracts the H-atom here, followed by the leaving of the Br atom. It's easiest to remove the H-atom in A, which is why it's the fastest. Refer here. $\endgroup$ – Aniruddha Deb Jul 27 at 6:34
  • $\begingroup$ The RDS in an E$2$ mechanism is the transition state: The more stable the transition state, the faster the rate. So the more substituted the more number of $\alpha$-hydrogens, which makes it more stable. $\endgroup$ – Safdar Jul 27 at 7:04
  • $\begingroup$ bases have nothing to do with size, they can abstract a proton regardless of size as they do not have to attack. In $\ce{E2}$ a stronger base is used and stability is decided by transition state ($\ce{alpha}$-hydrogen) $\endgroup$ – shreya Jul 27 at 9:01
  • $\begingroup$ @Aniruddha Deb: Your terminology that "the base abstracts the H-atom" and "followed by the leaving of the Br atom". The description makes the process sound as though it is stepwise rather than concerted. $\endgroup$ – user55119 Jul 27 at 22:21

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