Reduction of carboxylic acids gives

Alcohol with hydrogen in presence of palladium

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Alcohol with

L i A l H_{4}

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Aldehyde with

L i A l H_{4}

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All of the above

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Solution

The correct option is B Alcohol with $L i A l H_{4}$
The $L i A l H_{4}$ is strong reducing agent, it reduces the acid into alcohol.
$R C O O H L i A l H_{4} - --- \to R C H_{2} O H$

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L i A l H_{4}

H^{-}

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