When $C H_{2} = C H - C O O H$ is reduced with $L i A l H_{4}$ , the compound obtained will be :

C H_{2} = C H - C H_{2} O H

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C H_{3} - C H_{2} - C H_{2} O H

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C H_{3} - C H_{2} - C H O

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C H_{3} - C H_{2} - C O O H

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Solution

The correct option is B $C H_{2} = C H - C H_{2} O H$
$L i A l H_{4}$ can reduce the carboxylic acid group without affecting the double bond because alkene is electron-rich species.
$C H_{2} = C H - C O O H L i A l H_{4} - --- \to C H_{2} = C H - C H_{2} - O H$

Hence, option A is correct.

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Similar questions

C H_{2} = C H - C H_{2} O H \to C H_{2} = C H - C H O

C H_{3} - C O O H \to C H_{3} C O N H_{2}

(A) C H_{3} - C H_{2} - C | C O O H H - - C H - C H_{3}

(B) C H_{3} - C | C O O H H - C H_{2} - - C H - C H_{3}

(C) C | C O O H H_{2} - C H_{2} - C H_{2} - - C H - C H_{3}

C H_{3} - C H_{2} - C H | C O O H - C H_{2} - C l | C H - C H_{2} - C H_{3}

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