When ${\rm{C}}{{\rm{H}}_2} = CH-COOH$ is reduced by ${\rm{LiAl}}{{\rm{H}}_4}$, the compound obtained is:

${\rm{C}}{{\rm{H}}_2} = CH{\rm{C}}{{\rm{H}}_2}OH$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{COOH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{\rm{OH}}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CHO}}$

NEET UG CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

1 Easy

1 of 1

Application +4 | -1

When

\mathrm{C}{\mathrm{H}_2} = CH-COOH

is reduced by

\mathrm{LiAl}{\mathrm{H}_4}

, the compound obtained is:

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_2} = CH\mathrm{C}{\mathrm{H}_2}OH

Option B (

CH_2=CHCH_2OH

) is correct because

LiAlH_4

reduces the carboxylic acid group to a primary alcohol while leaving the alkene unaffected.