Question

Identify 'Z' in the following reaction series,

$C{H}_{3}C{H}_{2}C{H}_{2}Br\stackrel{aq.NaOH}{\to }\left(x\right)\stackrel{A{l}_2{O}_3}{\to }\left(y\right)\stackrel{HOCl}{\to }\left(Z\right)$

• A. $C{H}_3\underset{Cl}{\underset{|}{C}}H-\underset{Cl}{\underset{|}{C}}{H}_2$ and $C{H}_3\underset{OH}{\underset{|}{C}H}-\underset{OH}{\underset{|}{C}{H}_2}$
• B. $C{H}_3-\underset{OH}{\underset{|}{C}H}-\underset{Cl}{\underset{|}{C}{H}_2}$
• C. $C{H}_3-\underset{Cl}{\underset{|}{C}H}-\underset{OH}{\underset{|}{C}{H}_2}$
• D. $C{H}_3-\underset{Cl}{\underset{|}{C}H}-\underset{Cl}{\underset{|}{C}{H}_2}$

Answer 1

The correct option is B $C{H}_3-\underset{OH}{\underset{|}{C}H}-\underset{Cl}{\underset{|}{C}{H}_2}$

The compound Z is represented by the option B.
$C{H}_{3}C{H}_{2}C{H}_{2}Br\stackrel{aq.NaOH}{\to }\underset{\left(x\right)}{C{H}_{3}C{H}_{2}C{H}_{2}OH}\stackrel{A{l}_2{O}_3}{\to }\underset{\left(y\right)}{C{H}_3-CH=C{H}_2}\stackrel{HOCl}{\to }\underset{\left(Z\right)}{C{H}_{3}CH\left(OH\right)-C{H}_{2}Cl}$
First step is the hydrolysis of $n$- propyl bromide in presence of aq $NaOH$ to form $n$- propyl alcohol. The second step is the dehydration of $n$- propyl alcohol in presence of alumina to obtain propene. Last step is the addition of hypochlorous acid to obtain propylene chlorohydrin. $-OH$ is attached to more substituted $C$ atom and $-Cl$ is attached to less substituted $C$ atom.