Question

Which is the correct order for the rate of sulphonation of benzene?
(a) ${K_{{C_6}{H_6}}} > {K_{{C_6}{D_6}}} > {K_{{C_6}{T_6}}}$
(b) ${K_{{C_6}{H_6}}} < {K_{{C_6}{D_6}}} < {K_{{C_6}{T_6}}}$
(c) ${K_{{C_6}{H_6}}} = {K_{{C_6}{D_6}}} = {K_{{C_6}{T_6}}}$
(d) ${K_{{C_6}{H_6}}} > {K_{{C_6}{D_6}}} < {K_{{C_6}{T_6}}}$

Answer 1

Hint : The order of bond strength will be $C - H < C - D < C - T$ . Deuterium and tritium are the isotopes of hydrogen atoms. We know that tritium is heavier than deuterium and deuterium is heavier than hydrogen. The bond with the carbon atom will experience less vibratory motion if the isotope is heavy. So, $C - D$ has less energy than $C - H$ making the $C - D$ bond more stable and stronger than $C - H$ .

Complete Step By Step Answer:
The rate of the electrophilic aromatic substitution decreases as the hydrogen atom of benzene is successively substituted with deuterium and tritium. The concentrated sulphuric acid releases a molecule of $S{O_3}$ . The electrophile, sulphur trioxide, attacks the benzene ring to form a sigma complex. The abstraction of ${H^ + }$ will be the rate determining step. The order of bond strength will be $C - H < C - D < C - T$ . So, the rate of reaction decreases in the order ${C_6}{H_6} > {C_6}{D_6} > {C_6}{T_6}$ . The correct order of the rate of sulphonation of benzene is ${K_{{C_6}{H_6}}} > {K_{{C_6}{D_6}}} > {K_{{C_6}{T_6}}}$ .
Therefore, the correct answer is option A.

Note :
Sulphur trioxide is the electrophile here because it is a highly polar molecule with a fair amount of positive charge on sulphur. The p electrons of the aromatic $C = C$ act as nucleophile and attack the electrophilic sulphur, pushing the charge onto the electronegative oxygen atom. The loss of the proton from the $s{p^3}$ carbon atom bearing the sulfonyl group reforms the $C = C$ and the aromatic system. In sulphonation, the protonation of the conjugate base of the sulfonic acid by sulphuric acid produces sulfonic acid.