Question

In the given reactions, the rate of reaction of $\left(I\right)$ and $\left(II\right)$ are same. Both reactions proceed by which mechanism?

• A. $E_1$
• B. $E_2$
• C. $E_{1}cb$
• D. Anti-elimination

Answer 1

The correct option is A $E_1$

Following the $\beta$-elimination reaction mechanism we can determine the reaction pathway.

In E1 mechanism, cabocation forms heterolytic cleavage of C-Br bond. This is a slow and rate determining step. Followed by fast step of proton abstraction from adjacent carbon resulting in alkene formation. Thus only halide affects the rate of reaction

In E2 mechanism, both $Br$ and $D/H$ leaves simultaneously, therefore both halide and D/H affects the rate.

In E1cb mechanism, first step of proton abstraction resulting in carbanion formation and second step of halide/ leaving group cleavage. Since it is two step mechanism, therefore isotope effect will be present.

Anti-elimination and E2 are same

Therefore, option A is correct.