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  4. When CH2=CHCOOH is reduced with LiAlH4 the compound obtained is

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Q. When $CH2=CHCOOH $is reduced with $LiAlH_4 $ the compound obtained is

Solution:

$LiAlH_4$ can reduce only COOH group to $CH_2OH$ group but cannot reduce double bond.
$\ce{CH_2 = CH - COOH ->[LiAIH_4] CH_2 = CH - CH_2O }$