$α -$ chloropropanoic acid on treatment with alcoholic KOH followed by acidification gives:

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

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

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

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

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Solution

The correct option is B $C H_{2} = C H - C O O H$
$α -$ chloropropanoic acid on treatment with alcoholic $K O H$ followed by acidification gives $C H_{2} = C H - C O O H$ . A molecule of $H C l$ is eliminated. It is an example of dehydrohalogenation reaction.

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

(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 O O H + B r_{2} P B r_{3} - -- \to C H_{3} - C | B r H - C O O H

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

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

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