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?