Question

What is the correct increasing order of extent of deprotonation for the following acids in their aqueous solutions?

(I) $(C{H}_3{)}_{2}CHCOOH$

(II) $C{H}_{3}C{H}_{2}COOH$

(III) $C{l}_{2}CHC{H}_{2}COOH$

(IV) $C{H}_{3}CC{l}_{2}COOH$

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

Answer 1

The correct option is C I < II < III < IV

Higher the stability of the conjugate base more the extent of deprotonation.
(i) $(C{H}_3{)}_{2}CHCO{O}^{-}$
(II) $C{H}_{3}C{H}_{2}CO{O}^{-}$
(III) $C{l}_{2}CHC{H}_{2}CO{O}^{-}$
(IV)$C{H}_{3}CC{l}_{2}CO{O}^{-}$
In compound (IV) two $Cl$ molecules withdraw the negative charge from the conjugate base and make it the most stable among the others.
Similarly, in compound (III) two $Cl$ molecules withdraw the negative charge from the conjugate base making it more stable than the other two.
Comparing (III) and (IV),
-I effect depends on the distance at which -I group present from the negative charge. Since in (IV), Cl is present at shorter distance than in (III) its conjugate base is more stable than (III).
In compound (I) two $C{H}_3$ molecules show +I effect increasing the negative charge on the conjugate base hence making it less stable than (II).