Question

$C_5{H}_{11}Br+NaCN\to A\stackrel{H_{3}O}{\to }B+NaOH\stackrel{CaO}{\to }C$
${}^{\prime }{C}^{\prime }$ has the formula $C_5{H}_{12}$ which can give four structural isomeric monochloro derivative . What is the structure of $C_5{H}_{11}Br$?

• A. $C{H}_3-\underset{C{H}_3}{\underset{|}{C}H}-C{H}_2-C{H}_2-Br$
• B. $C{H}_3-C{H}_2-\stackrel{Br}{\stackrel{|}{C}H}-C{H}_2-C{H}_3$
• C. $C{H}_3-C{H}_2-C{H}_2-\underset{Br}{\underset{|}{C}H}-C{H}_3$
• D. $C{H}_2-\stackrel{C{H}_3}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_2-Br$

Answer 1

The correct option is A $C{H}_3-\underset{C{H}_3}{\underset{|}{C}H}-C{H}_2-C{H}_2-Br$

in last step decarboxylation occur.

As $C$ can give$4$ isomers on monochlorination so c can be option A.

Also $C_5{H}_{11}Br$ is same.