MCQ
When $CH_3-CH_2-COOH$ is reduced with $LiAlH_4$, the compound obtained will be
- A$CH_3CH_2COOH$
- B$CH_3CH_2CHO$
- ✓$CH_3CH_2CH_2OH$
- D$H_2C=CH-CH_2-OH$
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$ -\mathrm{OCH}_3,-\mathrm{NO}_2,-\mathrm{CN},-\mathrm{CH}_3-\mathrm{NHCOCH}_3, $
$ -\mathrm{COR},-\mathrm{OH},-\mathrm{COOH},-\mathrm{Cl}$