Question

Identify (Z) in the following reaction series,

$C_{2}H_{5}I\underset{KOH}{\overset{\quad Alcoholic\quad}{\rightarrow}}(X)\overset{\quad Br_{2}\quad}{\rightarrow}(Y)\overset{\quad\quad}{\rightarrow}(Z)$

• A. $CH_{3} - CH_{2} - CN$
• B. $\underset{\underset{CN}{|}}{C}H_{2} - \underset{\underset{CN}{|}}{C}H_{2}$
• C. $\underset{\underset{Br}{|}}{C}H_{2} - \underset{\underset{CN}{|}}{C}H_{2}$
• D. $\underset{\underset{Br}{|}}{C}H = \underset{\underset{CN}{|}}{C}H$

Answer 1

Answer: (D)

$\underset{\underset{Br}{|}}{C}H = \underset{\underset{CN}{|}}{C}H$

Answer 2

It follows E2 mechanism.

Hughes and In gold proposed that bimolecular elimination reaction take place when the two groups to be eliminated are trans i.e. E2 reaction are stereoselectively trans