Question

What is the correct increasing order of extent of deprotonation of the following acids in its aqueous solution?
(I) $\left( CH _{3}\right)_{2} CHCOOH$
(II) $CH _{3} CH _{2} COOH$
(III) $Cl _{2} CHCH _{2} COOH$
(IV) $CH _{3} CCl _{2} COOH$

• A. $II < I < III < IV$
• B. $IV < III < I < II$
• C. $I < IV < III < II$
• D. $I < II < II < IV$

Answer 1

Answer: (D)

Electron withdrawing group $(Cl)$ increases acidic strength while electron donating group $\left(- CH _{3}\right)$ decreases acidic strength. Also magnitude of inductive effect decreases drastically on increasing distance, $IV$ is stronger acid than $III$.



Two methyl groups (in $ I$) gives greater $+I$ effect than one methyl group (in $II$), hence I is a weaker acid than $II$.



Two $-Cl $ from closer position in $IV$ exert greater-I effect than the same from larger distance in $III$, hence $IV$ is stronger acid than $III$.