When $\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH}$ is reduced with $\mathrm{LiAlH}_4$, the compound obtained…

When $\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH}$ is reduced with $\mathrm{LiAlH}_4$, the compound obtained will be
  1. $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH}$
  2. $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{OH}$
  3. $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CHO}$
  4. $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{COOH}$

Solution

$\mathrm{LiAl}^{\mathrm{H}}{ }_4$ can reduce $\mathrm{COOH}$ group and not the double bond $\mathrm{CH}_2=\mathrm{CH}-\mathrm{COOH} \underset{\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{OH}}{\stackrel{\mathrm{LAH}}{\longrightarrow}}$

Asked in: JEE Main 2003

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