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The strongest acid amongst the following compounds is:
A. $C{H_3}COOH$
B. $HCOOH$
C. $C{H_3}C{H_2}CH\left( {Cl} \right)COOH$
D. $ClC{H_2}C{H_2}C{H_2}COOH$

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Last updated date: 16th Jul 2024
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Answer
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Hint: We have to know that a solid corrosive is one that is totally separated, or ionized in a watery arrangement. It is a compound animal category with a high ability to lose a proton. Diprotic and polyprotic acids may lose more than one proton, however the solid corrosive $pKa$ worth and response allude just to the deficiency of the primary proton. Most solid acids are destructive, yet a portion of the super acid is definitely not. Conversely, a portion of the powerless acids can be profoundly destructive.

Complete answer:
We have to know, $C{H_3}COOH$ is a powerless acid and separates part of the way in answer for structure ${H^ + }$ and $C{H_3}CO{O^ - }$ particles. Since this is a reversible interaction, $C{H_3}CO{O^ - }$ can acknowledge ${H^ + }$ to frame back $C{H_3}COOH$ . Along these lines the idea of $C{H_3}CO{O^ - }$ is fundamental and we call $C{H_3}CO{O^ - }$ the form base of $C{H_3}COOH$.
Therefore, option (A) is incorrect.
We have to know, a weak base is a proton acceptor that when placed in water will just somewhat separate. We should take a gander at the case of a feeble corrosive, formic acid, $HCOOH$ . This is a carboxylic acid that has the accompanying design. The acidic hydrogen is the one that is named in red.
Therefore, option (B) is incorrect.
We have to know that $C{H_3}C{H_2}CH\left( {Cl} \right)COOH$ is the strongest acid. Because of the $ - I$ effect of $Cl$ . The common name of $C{H_3}C{H_2}CH\left( {Cl} \right)COOH$ is $2$ -chlorobutanoic acid.
Therefore, option (C) is correct.
We have to know the $ClC{H_2}C{H_2}C{H_2}COOH$ a weak acid. The common name of $ClC{H_2}C{H_2}C{H_2}COOH$ is $4$ -chlorobutanoic acid.
Therefore, option (D) is incorrect.

Note:
We have to know, the $ + I$ bunches decrease the positive charge on the carbon by giving negative charge thickness through a positive inductive impact. While, the $ - I$ bunches destabilize the carbocations as they increment the positive charge by pulling out electron thickness.