The reaction of CH₃Cl with NaOH is an example of:

Solution

Chemistry ma’am · 4 steps

Methyl chloride has no crowding and no beta hydrogen, so hydroxide just walks in from the back.

1. No elimination possible. E1 and E2 both remove a hydrogen from the carbon next to the one carrying Cl (a beta hydrogen). CH₃Cl has only one carbon, so it has no beta hydrogen and cannot form an alkene.
2. No SN1 either. SN1 needs the C–Cl bond to break first and leave a carbocation. A methyl cation CH₃⁺ has no alkyl groups to stabilise it, so it is far too unstable to form.
3. Backside attack. With only three small hydrogens around the carbon, OH⁻ attacks from the side opposite to Cl. The C–O bond forms as the C–Cl bond breaks, in one step, so the rate depends on both reactants.

   HO⁻ + CH₃Cl → [HO… CH₃ … Cl]⁻ → CH₃OH + Cl⁻

4. So. The reaction is bimolecular and goes through a single transition state with rate = k[CH₃Cl][OH⁻]. It is an SN2 reaction, option (b).

The idea to keep. Less crowded carbon means faster SN2: methyl > 1° > 2° > 3°. More stable carbocation means faster SN1, the reverse order. No beta hydrogen means no elimination at all.

Answer from the NTA final answer key. Solution by PrepWiser.