The decreasing order of acidic strength for t | vedclass

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Question

(B) The acidic strength of carboxylic acids is determined by the stability of the carboxylate anion formed after the loss of a proton.Electron-Withdra

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$D > C > B > A$

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(B) The acidic strength of carboxylic acids is determined by the stability of the carboxylate anion formed after the loss of a proton.Electron-Withdrawing Groups $(EWG)$ like $Cl$ stabilize the carboxylate anion through the $-I$ effect,thereby increasing acidic strength.Electron-Donating Groups $(EDG)$ like $CH_3$ destabilize the anion,decreasing acidic strength.Analysis of the given acids:$(A)$ $CH_3COOH$: Contains a $CH_3$ group which shows a $+I$ effect,making it the least acidic.$(B)$ $CH_3CHClCH_2COOH$: Contains one $Cl$ atom at the $\beta$-position. The $-I$ effect is weaker due to the distance from the $-COOH$ group.$(C)$ $ClCH_2COOH$: Contains one $Cl$ atom at the $\alpha$-position. The $-I$ effect is stronger than in $(B)$.$(D)$ $Cl_2CHCOOH$: Contains two $Cl$ atoms at the $\alpha$-position. The cumulative $-I$ effect is the strongest.Therefore,the decreasing order of acidic strength is: $D > C > B > A$.

The decreasing order of acidic strength for the following acids is:
$(A)$ $CH_3COOH$
$(B)$ $CH_3CHClCH_2COOH$
$(C)$ $ClCH_2COOH$
$(D)$ $Cl_2CHCOOH$

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The decreasing order of acidic strength for the following acids is:$(A)$ $CH_3COOH$$(B)$ $CH_3CHClCH_2COOH$$(C)$ $ClCH_2COOH$$(D)$ $Cl_2CHCOOH$