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(A) In electrophilic aromatic substitution,the rate-determining step is the formation of the sigma complex (arenium ion) by the attack of the electrop

Vedclass · Accepted Answer

$k_{C_6H_6} = k_{C_6D_6} = k_{C_6T_6}$

Vedclass · Answer

(A) In electrophilic aromatic substitution,the rate-determining step is the formation of the sigma complex (arenium ion) by the attack of the electrophile on the aromatic ring. Since the $C-H$ (or $C-D$ or $C-T$) bond is not broken in the rate-determining step,there is no primary kinetic isotope effect. Therefore,the rates of nitration for benzene $(C_6H_6)$,hexadeuterobenzene $(C_6D_6)$,and hexatritiobenzene $(C_6T_6)$ are equal. Thus,$k_{C_6H_6} = k_{C_6D_6} = k_{C_6T_6}$.

For the electrophilic substitution reaction involving nitration,which of the following sequences regarding the rate of reaction is true?

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