Question

Arrange the following organic compounds in descending order of their reactivity towards S_N1 reaction.

C₆H₅CH₂Br, C₆H₅CH(C₆H₅)Br, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br

विकल्प

• C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH₂Br > C₆H₅CH(CH₃)Br

• C₆H₅C(CH₃)(C₆H₅)Br > C₃H₅CH(CH₃)Br > C₆H₅CH₂Br > C₆H₅CH(C₆H₅)Br

• C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₂)Br > C₆H₅CH₂Br > C₆H₅CH(CH₃)Br

• C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH(CH₃)Br > C₆H₅CH₂Br

Answer 1

C₆H₅C(CH₃)(C₆H₅)Br > C₆H₅CH(C₆H₅)Br > C₆H₅CH(CH₃)Br > C₆H₅CH₂Br

Explanation:

The nucleophilic substitution in S_N1 occurs in two steps. In step, I, the polarised C-Br bond undergoes slow cleavage to produce a carbocation and a bromide ion. This is the slow rate-determining step. Thus, the stability of the carbocation formed determines the rate of reaction. The carbocation thus formed is then attacked by nucleophile in step II to complete the substitution reaction.

In C₆H₅C(CH₃)(C₆H₅)Br carbocation will be tertiary and will be resonance stabilized. So the rate of S_N1 reaction is greatest in this case. In the case of C₆H₅CH(C₆H₅)Br, the carbocation formed will be secondary and hence less stable than the previous one, so the reaction will be slower in this case. In C₆H₅CH(CH₃)Br, the carbocation formed will be less stable than that formed in the case of C₆H₅CH(C₆H₅)Br because in the latter the carbocation is stabilized by two phenyl groups due to resonance. In C₆H₅CH₂Br, the carbocation formed will be primary and hence least stable causing the rate of reaction to be least in this case.