When methyl iodide reacts with sodium hydroxide in water, what is the mechanism and product?

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Multiple Choice

When methyl iodide reacts with sodium hydroxide in water, what is the mechanism and product?

The key idea is nucleophilic substitution by SN2 on a primary alkyl halide. Methyl iodide is a very small, primary substrate, so it undergoes SN2 easily: the hydroxide ion attacks the carbon attached to iodine from the opposite side, the C–I bond breaks as the C–O bond forms in one concerted step. There’s no opportunity for a carbocation to form, so SN1 is not feasible here, especially since methyl cations are extremely unstable. E2 is not possible either because the molecule has no β-hydrogens to eliminate. The result is methanol (CH3OH) and iodide, with Na+ pairing to form NaI in solution.

The key idea is nucleophilic substitution by SN2 on a primary alkyl halide. Methyl iodide is a very small, primary substrate, so it undergoes SN2 easily: the hydroxide ion attacks the carbon attached to iodine from the opposite side, the C–I bond breaks as the C–O bond forms in one concerted step. There’s no opportunity for a carbocation to form, so SN1 is not feasible here, especially since methyl cations are extremely unstable. E2 is not possible either because the molecule has no β-hydrogens to eliminate. The result is methanol (CH3OH) and iodide, with Na+ pairing to form NaI in solution.

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