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 $\beta-' H ' .$ So, the probablity of attack of base on $\beta-' H '$ 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 $3^{\circ} R - L >2^{\circ} R ->1^{\circ} R - L$