Question

$CH \equiv \equiv CH \xrightarrow[H_2O]{Hg^{2+}/H^{+}} A \xrightarrow{LiAlH_4} B \xrightarrow {P/Br_2} C$, Final product $C$ is

• A. $CH _{3} CH _{2} OH$
• B. $CH _{3} CH _{2}- Br$
• C. $CH _{3}- CH _{2}- I$
• D. $\underset{\overset{|}{Br}}{C}H_2 - CH_2 - OH$

Answer 1

Answer: (B)

The given road map problem is :
$CH \equiv \equiv CH \xrightarrow[H_2O]{Hg^{2+}/H^{+}} A \xrightarrow{LiAlH_4} B \xrightarrow {P/Br_2} C$,
The reaction takes place as follows :

Thus, $C$ is $CH_3CH_2 - Br$.