Question

Which compound in each of the following pairs is most reactive to the conditions indicated?

A. A and C
B. B and C
C. A and D
D. B and D

Answer 1

Hint: Alcoholic KOH dissociates $R{{O}^{-}}$ into the water which acts as a strong base than $H{{O}^{-}}$ an ion, then it favours an elimination reaction. But sometimes the substitution reaction is favoured over elimination reaction due to unstable intermediate. In the presence of $NaN{{H}_{2}}$in $N{{H}_{3}}$, aromatic halide undergoes elimination –addition reaction.

Complete step by step solution:
In the case of a given aromatic halide, the elimination reaction is not favoured because an unstable carbanion is formed. A side chain substitution reaction (${{S}_{{{N}^{2}}}}$) is preferred over elimination which $H{{O}^{-}}$acts as a nucleophile. Primary carbon is more reactive than secondary carbon centres in ${{S}_{{{N}^{2}}}}$reaction. Compound (A) has a ${{1}^{{\mathrm O}}}$carbon centre, whereas in (B) has a ${{2}^{{\mathrm O}}}$ carbon centre. Hence (A) is more reactive than (B) under the given conditions.

Mechanism:

Under the given condition,$NaN{{H}_{2}}$in $N{{H}_{3}}$, compounds (C) undergo an elimination-addition reaction. The reaction starts with the removal of proton ortho to the leaving group $(C{{l}^{-}})$ by $NH_{2}^{-}$( strong base) followed by loss of $C{{l}^{-}}$. Thereby a very unstable benzyne intermediate is formed. In the next step, the corresponding intermediate is attacked by a nucleophile $NH_{2}^{-}$followed by protonation that produces the final product. There is no ortho hydrogen with respect to leaving the group $(C{{l}^{-}})$ in compound (D), so an elimination-addition reaction is not possible. Hence compound (c) is more reactive in elimination-addition reaction than (D) under given conditions.
Mechanism:

Thus, Option (A) is correct.

Note: Benzynes are very reactive and unstable intermediate. Like benzene, it is also an aromatic compound. It has a planar, cyclic and conjugated structure and contains $6\Pi $ electrons according to Huckel’s rule. It consists of one sigma and two pi bonds, where only one pi bond participates in conjugation and another pi bond is perpendicular to the first pi orbital.