Question

The order of $E_2$ elimination for alkyl halide is

• A. $1^{\circ }>2^{\circ }>3^{\circ }$
• B. $3^{\circ }>2^{\circ }>1^{\circ }$
• C. $2^{\circ }>3^{\circ }>1^{\circ }$
• D. $3^{\circ }>1^{\circ }>2^{\circ }$

Answer 1

Answer: (B)

The order of $E_2$ elimination for alkyl halide follows as $3^{\circ }R-L>2^{\circ }R-L>1^{\circ }R-L$.
Where $L=$ any leaving group.
This is because
(i) In $3^{\circ }R-L$, there is more number of $\beta {-}^{\prime }{H}^{\prime }.$ So, the probablity of attack of base on $\beta {-}^{\prime }{H}^{\prime }$ is more and hence $E_2$ reactivity increases.
(ii) Secondly, the transition state of $E_2$ reaction is more stable in case of $E_2$ reaction is more hyperconjugation with its incipient double bond character and $\beta -$ ' $H$ '.
S.O. of $T_s$ of $E_2$ reaction.

So, $E_2$ reactivity order $3^{\circ }R-L>2^{\circ }R->1^{\circ }R-L$