In the nitration of benzene by mixed acid,the | vedclass

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Question

(A) The nitration of benzene proceeds via an electrophilic aromatic substitution mechanism. The rate-determining step is the formation of the $\sigma$

Vedclass · Accepted Answer

$C_6H_6 = C_6D_6 = C_6T_6$

Vedclass · Answer

(A) The nitration of benzene proceeds via an electrophilic aromatic substitution mechanism. The rate-determining step is the formation of the $\sigma$-complex (Wheland intermediate) by the attack of the electrophile $(NO_2^+)$ on the benzene ring. The subsequent step involves the loss of a proton ($H^+$,$D^+$,or $T^+$) from the $\sigma$-complex to restore aromaticity. Since the $C$-$H$ bond breaking occurs after the rate-determining step,there is no primary kinetic isotope effect. Therefore,the rate of reaction for $C_6H_6$,$C_6D_6$,and $C_6T_6$ is the same. Thus,$C_6H_6 = C_6D_6 = C_6T_6$.

In the nitration of benzene by mixed acid,the rate of reaction will be:

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