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(A) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ stabil

Vedclass · Accepted Answer

$F > E > D > B > C > A$

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(A) The acidity of phenols depends on the stability of the phenoxide ion formed after the loss of a proton. Electron-withdrawing groups $(EWG)$ stabilize the phenoxide ion,increasing acidity,while electron-donating groups $(EDG)$ destabilize it,decreasing acidity.$1$. $A$ (Cyclohexanol): The conjugate base is an alkoxide ion,which is less stable than the phenoxide ion. Thus,$A$ is the least acidic.$2$. $B$ (Phenol): The reference compound.$3$. $C$ (p-Methoxyphenol): $-OCH_3$ is an $EDG$ by resonance (+$M$ effect),which destabilizes the phenoxide ion. Thus,$C $4$. $D$ (p-Hydroxyacetophenone): $-COCH_3$ is an $EWG$ (-$M$ effect),which stabilizes the phenoxide ion. Thus,$D > B$.$5$. $E$ (m-Nitrophenol): $-NO_2$ is a strong $EWG$ (-$I$ and -$M$ effects). At the meta position,only the -$I$ effect operates. Thus,$E > B$.$6$. $F$ (p-Nitrophenol): $-NO_2$ at the para position exerts both -$I$ and -$M$ effects,providing maximum stabilization to the phenoxide ion. Thus,$F > E$.Comparing the $EWG$ strengths: $p-NO_2$ $(F)$ > $m-NO_2$ $(E)$ > $p-COCH_3$ $(D)$.Comparing the $EDG$ strengths: $p-OCH_3$ $(C)$ destabilizes more than $B$. $A$ is the least acidic.The correct order is $F > E > D > B > C > A$.

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