Question

Arrange the following in decreasing rate in $E2$ reaction

• A. $1>2>3$
• B. $3>1>2$
• C. $3>2>1$
• D. $2<3<1$

Answer 1

The correct option is B $3>1>2$

E2 elimination reaction is an anti-periplanar reaction. Abstraction of proton is a rate determining step, hence, the availability of proton at anti-planar position will increase the rate of E2 elimination reaction.
In compound (2), there is no $H$ atom at anti-periplanar position of $Br$. Hence, it is the least reactive.
In compound (3), there is 2 $H$ atoms at anti-periplanar position of $Br$ and in compound (1) there is 1 $H$ atom at anti-periplanar position of $Br$,
Hence, the order of decrease rate of E2 reactions is
$3>1>2$