The correct order of pK_{a} of the following | vedclass

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(D) The acidity of carboxylic acids is directly proportional to the stability of their conjugate bases,which is enhanced by electron-withdrawing group

Vedclass · Accepted Answer

$(II) < (I) < (III) < (IV)$

Vedclass · Answer

(D) The acidity of carboxylic acids is directly proportional to the stability of their conjugate bases,which is enhanced by electron-withdrawing groups ($-I$ effect).$pK_a$ is inversely proportional to acidity.Comparing the compounds:$II. CF_3COOH$ (Strongest acid due to three highly electronegative $F$ atoms)$I. CCl_3COOH$ (Strong acid,but $Cl$ is less electronegative than $F$)$III. NO_2CH_2COOH$ (Stronger acid than $IV$ because $-NO_2$ is a stronger electron-withdrawing group than $-CN$)$IV. NCCH_2COOH$Acidity order: $II > I > III > IV$Since $pK_a = -\log(K_a)$,the order of $pK_a$ is the reverse of the acidity order:$pK_a$ order: $IV > III > I > II$This corresponds to $(II) < (I) < (III) < (IV)$.

The correct order of $pK_{a}$ of the following compounds is:
$I. CCl_3COOH$
$II. CF_3COOH$
$III. NO_2CH_2COOH$
$IV. NCCH_2COOH$

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The correct order of $pK_{a}$ of the following compounds is:$I. CCl_3COOH$$II. CF_3COOH$$III. NO_2CH_2COOH$$IV. NCCH_2COOH$