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Which of the following orders is correct regarding the acidity of carboxylic group?

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${CH}_{3} {CH}_{2} CH (Cl) COOH < {CH}_{3} CH (Cl) {CH}_{2} COOH < {ClCH}_{2} {CH}_{2} {CH}_{2} COOH$

${CH}_{3} {CH}_{2} CH (Cl) COOH < {CH}_{3} CH (Cl) {CH}_{2} COOH > {ClCH}_{2} {CH}_{2} {CH}_{2} COOH$

${CH}_{3} {CH}_{2} CH (Cl) COOH > {CH}_{3} CH (Cl) {CH}_{2} COOH > {ClCH}_{2} {CH}_{2} {CH}_{2} COOH$

${CH}_{3} {CH}_{2} CH (Cl) COOH > {CH}_{3} CH (Cl) {CH}_{2} COOH < {ClCH}_{2} {CH}_{2} {CH}_{2} COOH$

answer is A.

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Detailed Solution

The farther away chlorine from $- COOH$ group, the weaker the acid.

The acidic nature of carboxylic acids is increased by the presence of -I-showing groups like Cl, and the acidity decreases with an increase in the distance between -COOH and -/-showing groups.

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