Among the following acids,the strongest acid | vedclass

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(C) The strength of an acid is determined by the stability of its conjugate base. Electron-withdrawing groups $(EWG)$ stabilize the carboxylate anion

Vedclass · Accepted Answer

$F_3CCOOH$

Vedclass · Answer

(C) The strength of an acid is determined by the stability of its conjugate base. Electron-withdrawing groups $(EWG)$ stabilize the carboxylate anion $(RCOO^-)$ through the inductive effect ($-I$ effect),thereby increasing the acidity of the parent carboxylic acid.Comparing the substituents:$1$. $NCCH_2-$: Contains a cyano group $(-CN)$,which has a $-I$ effect.$2$. $O_2NCH_2-$: Contains a nitro group $(-NO_2)$,which has a strong $-I$ effect.$3$. $F_3C-$: Contains three fluorine atoms. Fluorine is the most electronegative element,and the $-I$ effect of three fluorine atoms is significantly stronger than that of a single $-NO_2$ or $-CN$ group.$4$. $Cl_3C-$: Contains three chlorine atoms. Chlorine is less electronegative than fluorine,so the $-I$ effect of $Cl_3C-$ is weaker than that of $F_3C-$.Since fluorine is the most electronegative,$F_3CCOOH$ exhibits the strongest $-I$ effect,making its conjugate base the most stable and the acid the strongest among the given options.

Among the following acids,the strongest acid is

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