Q1(c). Methyl iodide with sodium azide: DMF against methanol

Answer in brief

The reaction is SN2. Methanol hydrogen-bonds to N₃⁻ and lowers its energy; DMF solvates only Na⁺ and leaves a bare, high-energy azide, so ΔG‡ is smaller in DMF.

Key points

• SN2: rate = k[CH₃I][N₃⁻]; the nucleophile is in the slow step.
• Methanol (protic): O–H···N hydrogen-bond cage round the small anion; cage must partly break to reach the transition state: large ΔG‡.
• DMF (polar aprotic): no O–H or N–H; C=O oxygen solvates Na⁺.
• Azide left unsolvated (naked anion): higher ground-state energy, more reactive.
• Result: 4 × 10⁴ times faster in DMF; DMSO, acetone, CH₃CN behave alike.