Which of the following reactions takes place most readily by the SN2…

The correct answer is Option (1) → \( \ce{CH3Br + OH^- -> CH3OH + Br^-} \)

• An S_N2 reaction (substitution nucleophilic bimolecular) takes place in one step. The nucleophile attacks the carbon from the side opposite to the leaving group, and the new bond forms as the old one breaks. The rate depends on both the alkyl halide and the nucleophile:

      \( \displaystyle \text{Rate}=k[\ce{RX}][\ce{Nu^-}] \)

• Because the nucleophile must reach the carbon bearing the halogen, bulky groups on that carbon slow the reaction (steric hindrance). The fewer and smaller the groups, the faster the S_N2 reaction:

      \( \displaystyle \ce{CH3X} \gt 1^\circ \gt 2^\circ \gt 3^\circ \)

• In bromomethane the carbon carries only three small hydrogen atoms, so the hydroxide ion reaches it most easily:

      \( \displaystyle \ce{CH3Br + OH^- -> CH3OH + Br^-} \)

Why the other options are incorrect

Option (2) 2-bromopropane + OH^– → propan-2-ol + Br^– — Incorrect.

• 2-Bromopropane is a secondary halide: the carbon bearing bromine also carries two methyl groups.

• It can react by S_N2, but the two methyl groups crowd the path of the nucleophile, so it reacts far more slowly than bromomethane.

      \( \displaystyle \ce{CH3-CH(Br)-CH3 + OH^- -> CH3-CH(OH)-CH3 + Br^-} \)

Option (3) ethanol → ethene + H₂O (dehydration) — Incorrect. This is an elimination reaction, not a substitution. Ethanol loses a molecule of water when heated with concentrated sulphuric acid at 443 K, and a double bond forms.

    \( \displaystyle \ce{CH3CH2OH} \xrightarrow[\text{443 K}]{\text{conc. }\ce{H2SO4}} \ce{CH2=CH2 + H2O} \)

Option (4) 2-bromo-2-methylpropane + OH^– → 2-methylpropan-2-ol + Br^– — Incorrect.

• 2-Bromo-2-methylpropane is a tertiary halide. Three methyl groups block the carbon, so S_N2 attack is almost impossible.

• This reaction goes by the S_N1 route instead: the C–Br bond first breaks to give a tertiary carbocation, which then combines with \( \ce{OH^-} \).

      \( \displaystyle \ce{(CH3)3CBr + OH^- -> (CH3)3COH + Br^-} \)

NCERT (2026-27 edition):
(a) Substitution nucleophilic bimolecular (S_N2) – “… Since this reaction requires the approach of the nucleophile to the carbon bearing the leaving group, the presence of bulky substituents on or near the carbon atom have a dramatic inhibiting effect. Of the simple alkyl halides, methyl halides react most rapidly in S_N2 reactions because there are only three small hydrogen atoms.”