Question

Identify X, Y, Z in the following series:

$C_2{H}_{5}I\stackrel{alcKOH}{\to }X\stackrel{B{r}_2}{\to }Y\stackrel{ExcessofKCN}{\to }Z$ :

• A. $BrC{H}_{2}C{H}_{2}CN$
• B. $C_2{H}_4$
• C. $NCC{H}_{2}C{H}_{2}CN$
• D. $BrC{H}_{2}C{H}_{2}Br$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $C_2{H}_4$
C $NCC{H}_{2}C{H}_{2}CN$
D $BrC{H}_{2}C{H}_{2}Br$

Alc. KOH is dehydrohalogenating agent so at first alkene forms which reacts further with bromine to form a dibromo product.

Then in presence of KCN, bromine is substituted by the cyanide group.

Reaction is :

$C_2{H}_{5}I\stackrel{alc.KOH}{\to }\underset{\left(x\right)}{C_2{H}_4}\stackrel{B{r}_2}{\to }\underset{\left(y\right)}{Br-C{H}_2-C{H}_2-Br}\stackrel{KCN}{\to }\underset{\left(z\right)}{CN-C{H}_2-C{H}_2-CN}$