Question

Consider the following reaction ,
$CC{l}_{3}CH=C{H}_2\underset{H_2{O}_2/HOCl}{\overset{NaB{O}_4}{\to }}Product\left(P\right)$
$P$ is:

• A. $CC{l}_3\underset{\stackrel{|}{OH}}{CH}-C{H}_{2}Cl$
• B. $CC{l}_3\underset{\stackrel{|}{OH}}{CH}\underset{\stackrel{|}{OH}}{C{H}_2}$
• C. $CC{l}_3\underset{\stackrel{|}{Cl}}{CH}C{H}_{2}OH$
• D. $CC{l}_3\underset{\stackrel{|}{Cl}}{CH}\underset{\stackrel{|}{Cl}}{C{H}_2}$

Answer 1

The correct option is C $CC{l}_3\underset{\stackrel{|}{Cl}}{CH}C{H}_{2}OH$

Since, $CC{l}_3$ is an electron withdrawing group product $P$ is $CC{l}_3\underset{\stackrel{|}{Cl}}{CH}-C{H}_{2}OH$