Question

Carbonyl compounds on reduction with selective reducing agents give alcohols. The structure of alcohol formed depends upon the nature of reducing agents. $LiAlH _{4}$, $NaBH _{4}$, sodium alcohol, $Mg ( Hg ) H _{2} O$ etc can be used.
When $CH _{3}- CH = CH - CHO$ is reduced with $NaBH _{4}$, the product formed is

• A. $CH _{3}- CH = CH - CH _{2} OH$
• B. $CH _{3}- CH _{2}- CH _{2}- CHO$
• C. $CH _{3}- CH _{2}- CH _{2}- CH _{3}$
• D. $CH _{3}- CH _{2}- CH _{2}- CH _{2} OH$

Answer 1

Answer: (A)

$NaBH _{4}$ does not reduce $C = C$ or $C \equiv C$ bond ; only $> C = O$ group is reduced to alcoholic group.
$CH _{3}- CH = CH - CHO \xrightarrow{NaBH_4}CH _{3}- CH = CH - CH _{2} OH$