Consider the following structures:I. CH_3-CH_ | vedclass

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(A) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion) formed after the loss of a proton $(H^+)$.E

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$II > III > I > IV$

Vedclass · Answer

(A) The acidity of carboxylic acids is determined by the stability of the conjugate base (carboxylate ion) formed after the loss of a proton $(H^+)$.Electron-withdrawing groups (EWGs) stabilize the carboxylate ion through the inductive effect ($-I$ effect),thereby increasing acidity.$1$. $II$ $(CH_3-CH(Cl)-COOH)$: The $Cl$ atom is on the $\alpha$-carbon,exerting a strong $-I$ effect,which significantly stabilizes the conjugate base.$2$. $III$ $(Cl-CH_2-CH_2-COOH)$: The $Cl$ atom is on the $\beta$-carbon. The $-I$ effect decreases with distance,so it is weaker than in $II$.$3$. $I$ $(CH_3-CH_2-COOH)$: This is propanoic acid,which has no $EWG$. The alkyl group $(CH_3-CH_2-)$ is electron-donating ($+I$ effect),which destabilizes the conjugate base compared to $II$ and $III$.$4$. $IV$ $(CH_3-CH_2-CH_2-COOH)$: This is butanoic acid. The longer alkyl chain $(CH_3-CH_2-CH_2-)$ has a slightly stronger $+I$ effect than the ethyl group in $I$,making it the least acidic.Therefore,the correct order of acidity is $II > III > I > IV$.

Consider the following structures:
$I.$ $CH_3-CH_2-COOH$
$II.$ $CH_3-CH(Cl)-COOH$
$III.$ $Cl-CH_2-CH_2-COOH$
$IV.$ $CH_3-CH_2-CH_2-COOH$
The correct order of their acidity is

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Consider the following structures:$I.$ $CH_3-CH_2-COOH$$II.$ $CH_3-CH(Cl)-COOH$$III.$ $Cl-CH_2-CH_2-COOH$$IV.$ $CH_3-CH_2-CH_2-COOH$The correct order of their acidity is