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

C_{6} H_{6} = C_{6} D_{6} = C_{6} T_{6}

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C_{6} H_{6} > C_{6} D_{6} > C_{6} T_{6}

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C_{6} H_{6} = C_{6} D_{6} > C_{6} T_{6}

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C_{6} H_{6} < C_{6} D_{6} < C_{6} T_{6}

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Solution

The correct option is A $C_{6} H_{6} = C_{6} D_{6} = C_{6} T_{6}$
Nitration of benzene by mixed acid the rate of reaction in this case will be $C_{6} H_{6} = C_{6} D_{6} = C_{6} T_{6}$
in nitration of benzene the source of nitronium ion is rhe protonation of nitric acid by sulphonic acid which causes the loss of water molecule & formation of nitronium ion. nitration of benzene is an electrophilic substitution reaction and the rate depends upon the electron density in the aromatic ring . Since the electrons is same in all the isotopes of hydrogen so rate of their nitration is same.

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C_{6} H_{6} ≃ C_{6} D_{6} ≃ C_{6} T_{6}

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(i) C_{6} H_{6} + I C l A n h y d . A l C l_{3} - ------- \to iodobenzene

(i i) C_{6} H_{6} + I_{2} A n h y d . A l C l_{3} - ------- \to iodobenzene

C_{6} H_{6}

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