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(C) Acidic strength depends on the $-I$ effect of the electron-withdrawing groups. The $-I$ effect increases with the electronegativity of the halogen

Vedclass · Accepted Answer

$C > B > A > D$

Vedclass · Answer

(C) Acidic strength depends on the $-I$ effect of the electron-withdrawing groups. The $-I$ effect increases with the electronegativity of the halogen $(F > Cl > Br)$ and decreases with increasing distance from the $-COOH$ group. $1.$ In $C$ $(F-CH_2-COOH)$,the $F$ atom is at the $\alpha$-position,making it the strongest acid. $2.$ In $B$ $(Cl-CH(Cl)-CH_2-COOH)$,there are two $Cl$ atoms at the $\beta$-position,which exert a stronger collective $-I$ effect than a single $F$ or $Br$ at the same position. $3.$ In $A$ $(F-CH_2-CH_2-COOH)$ and $D$ $(Br-CH_2-CH_2-COOH)$,the halogens are at the $\beta$-position. Since $F$ is more electronegative than $Br$,$A$ is more acidic than $D$. Therefore,the decreasing order of acidic strength is $C > B > A > D$.

Arrange the following in decreasing order of acidic strength:
$A. F-CH_2-CH_2-COOH$
$B. Cl-CH(Cl)-CH_2-COOH$
$C. F-CH_2-COOH$
$D. Br-CH_2-CH_2-COOH$

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Arrange the following in decreasing order of acidic strength: $A. F-CH_2-CH_2-COOH$ $B. Cl-CH(Cl)-CH_2-COOH$ $C. F-CH_2-COOH$ $D. Br-CH_2-CH_2-COOH$