An aromatic chiral compound G( C 8 H 8 Br 2) on treatment with aqueous NaOH gives H( C 8 H 9 BrO ). On heating G with potassium tert-butoxide, I( C 8 H 7 Br ) is formed. With one equivalent of methyl magnesium bromide in ether p -bromoisopropylbenzene is formed. Which observation/s is/are correct about these reactions?

Question

An aromatic chiral compound G( C 8 H 8 Br 2) on treatment with aqueous NaOH gives H( C 8 H 9 BrO ). On heating G with potassium tert-butoxide, I( C 8 H 7 Br ) is formed. With one equivalent of methyl magnesium bromide in ether p -bromoisopropylbenzene is formed. Which observation/s is/are correct about these reactions? (A) (I) is an optically inactive aromatic alkene and (H) is an optically active alcohol. (B)  (C) Benzylic halide gives S N 1 reaction at a faster rate than alkyl halide. (D) All are correct.

Tardigrade Admin · Accepted Answer

Correct answer is (d) All are correct.

$(I)$ is an optically inactive aromatic alkene and $(H)$ is an optically active alcohol.

Benzylic halide gives $S_{N} 1$ reaction at a faster rate than alkyl halide.

All are correct.

Correct answer is (d) All are correct.

(I) is an optically inactive aromatic alkene and (H) is an optically active alcohol.

Benzylic halide gives SN​1 reaction at a faster rate than alkyl halide.

All are correct.

Correct answer is (d) All are correct.

$(I)$ is an optically inactive aromatic alkene and $(H)$ is an optically active alcohol.

Benzylic halide gives $S_{N} 1$ reaction at a faster rate than alkyl halide.