Q58 / 180 · KCET 2021 · Chemistry
The order of reactivity of the compounds $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right) \mathrm{Br}, \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}\left(\mathrm{CH}_{3}\right) \mathrm{Br}$ and $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{C}\left(\mathrm{CH}_{3}\right)\left(\mathrm{C}_{6} \mathrm{H}_{5}\right) \mathrm{Br}$ in $\mathrm{S}_{\mathrm{N}} 2$ reaction is
- Correct answerA
- B
- C
- D
Solution
The order of reactivity of given compounds in $\mathrm{S}_{\mathrm{N}}$ 2 reaction is
For $\mathrm{S}_{\mathrm{N}} 2$ reaction, the order of reaction is inversely proportional to steric hinderance cause by bulky groups. In
there are two bulky groups are present i.e. benzene group and one methyl group which provide steric hinderance for the attack of nucleophile. Hence, it is least reactive while in $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}\left(1^{\circ}\right.$ halide ), the attack of nucleophile is easier due to less steric hinderance.
For $\mathrm{S}_{\mathrm{N}} 2$ reaction, the order of reaction is inversely proportional to steric hinderance cause by bulky groups. In
there are two bulky groups are present i.e. benzene group and one methyl group which provide steric hinderance for the attack of nucleophile. Hence, it is least reactive while in $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}\left(1^{\circ}\right.$ halide ), the attack of nucleophile is easier due to less steric hinderance.
Hence, the correct answer is option: 'A'.