The halide which will undergo SN^2 reaction easily

The halide which will undergo \( SN^2 \) reaction easily

Answer & explanation

Correct answer: option 2

The correct answer is Option (2) → Ethyl Chloride.

Note: The given answer is as per the official NTA Answer Key. However, the correct answer should be Option (1) Benzyl Chloride, as explained below.

An \( \mathrm{S_N2} \) reaction takes place in one step: the nucleophile attacks the carbon carrying the halogen from the side opposite to the halogen, and the C–X bond breaks as the new bond forms. Two things decide how fast this happens:

Steric hindrance: in the transition state five groups crowd round the carbon, so the rate falls from methyl to primary to secondary to tertiary halides.
Stabilisation of the transition state: when the carbon under attack is next to a C=C bond or a benzene ring (allylic and benzylic halides), the \( \pi \) electrons overlap with the orbitals of the transition state and lower its energy. Such halides react faster than ordinary primary halides.

Option (1) Benzyl Chloride. — Correct as per NCERT. \( \ce{C6H5CH2Cl} \) is a primary halide, so it is no more hindered than ethyl chloride, and in addition its \( \mathrm{S_N2} \) transition state is stabilised by the benzene ring. So it reacts more easily than ethyl chloride.
Option (2) Ethyl Chloride. — The answer given by NTA, but incorrect. A simple primary halide reacts readily by \( \mathrm{S_N2} \), but it has no neighbouring π system to stabilise the transition state:
\( \displaystyle \ce{OH^- + CH3CH2-Cl -> CH3CH2-OH + Cl^-} \)
Option (3) Tertiary Butyl Chloride. — Incorrect. Three methyl groups block the back of the carbon, so \( \ce{(CH3)3C-Cl} \) does not react by \( \mathrm{S_N2} \); it reacts by \( \mathrm{S_N1} \).
Option (4) Allyl Chloride — Incorrect. Its neighbouring C=C bond also stabilises the transition state, so it reacts faster than ethyl chloride, but less readily than benzyl chloride, whose benzene ring spreads the electron density over a larger π system.

The halide which will undergo \( SN^2 \) reaction easily

Answer & explanation

Correct answer: option 2

The correct answer is Option (2) → Ethyl Chloride.

Note: The given answer is as per the official NTA Answer Key. However, the correct answer should be Option (1) Benzyl Chloride, as explained below.

Steric hindrance: in the transition state five groups crowd round the carbon, so the rate falls from methyl to primary to secondary to tertiary halides.
Stabilisation of the transition state: when the carbon under attack is next to a C=C bond or a benzene ring (allylic and benzylic halides), the \( \pi \) electrons overlap with the orbitals of the transition state and lower its energy. Such halides react faster than ordinary primary halides.

Option (1) Benzyl Chloride. — Correct as per NCERT. \( \ce{C6H5CH2Cl} \) is a primary halide, so it is no more hindered than ethyl chloride, and in addition its \( \mathrm{S_N2} \) transition state is stabilised by the benzene ring. So it reacts more easily than ethyl chloride.
Option (2) Ethyl Chloride. — The answer given by NTA, but incorrect. A simple primary halide reacts readily by \( \mathrm{S_N2} \), but it has no neighbouring π system to stabilise the transition state:
\( \displaystyle \ce{OH^- + CH3CH2-Cl -> CH3CH2-OH + Cl^-} \)
Option (3) Tertiary Butyl Chloride. — Incorrect. Three methyl groups block the back of the carbon, so \( \ce{(CH3)3C-Cl} \) does not react by \( \mathrm{S_N2} \); it reacts by \( \mathrm{S_N1} \).
Option (4) Allyl Chloride — Incorrect. Its neighbouring C=C bond also stabilises the transition state, so it reacts faster than ethyl chloride, but less readily than benzyl chloride, whose benzene ring spreads the electron density over a larger π system.