Question

Find the reactants which on heating with alcoholic $KOH$ produces the compound
$CH _{3}- CH _{2}- CH _{2}- CH = CH _{2}$
(i) $CH _{3}- \underset{\overset{|}{Br}}{C}H - CH _{2}- CH _{2}- CH _{3}$
(ii) $CH _{3}- CH _{2}- \underset{\overset{|}{Br}}{C}H - CH _{2}- CH _{3}$
(iii) $CH _{3}- CH _{2}- CH_{2} - \underset{\overset{|}{Br}}{C}H - CH _{2}Br$
(iv) $CH _{3}- CH _{2}- CH _{2}- CH _{2}- \underset{\overset{|}{Br}}{C}H _{2}$

• A. (ii), (iv)
• B. (i), (iv)
• C. (ii), (iii), (iv)
• D. (i), (ii)

Answer 1

Answer: (B)

On heating haloalkane with alcoholic $KOH$, gives $CH _{3} - CH _{2}- CH _{2}- CH = CH _{2}$ are:

(i) $CH _{3}- \underset{\overset{|}{Br}}{C}H - CH _{2}- CH _{2}- CH _{3}$

$\ce{->[{KOH{\text{(alc)}}}][{\text{Elimination}}]}CH _{2}= CH - CH _{2}- CH _{2}- CH _{3}$

[by using Hofmann’s rule].

Hence, the formation of given product is possible.

(ii) $CH _{3}- CH _{2}- \underset{\overset{|}{Br}}{C}H - CH _{2}- CH _{3}$

$\ce{->[{KOH(\text{alc})}]}CH _{3}- CH = CH - CH _{2}- CH _{3}$

Hence, the formation of given product is not possible.

(iii) $CH _{3}- CH _{2}- CH_{2} - \underset{\overset{|}{Br}}{C}H - CH _{2}Br$

$\ce{->[{KOH{\text{(alc)}}}][{\text{Elimination}}]}CH _{3}= CH_{2} - CH _{2}- CH - CHBr$

Hence, the formation of given product is not possible.

(iv) $CH _{3}- CH _{2}- CH _{2}- CH _{2}- \underset{\overset{|}{Br}}{C}H _{2}$

$\ce{->[{KOH{\text{(alc)}}}][{\text{Elimination}}]}CH _{3}= CH_{2} - CH _{2}- CH - CH _{2}$

Hence, the formation of given product is possible. Thus, $(i)$ and $(iv)$ are the reactants that can give the given product.