When 3‑methyl‑1‑butanol ($\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_2\mathrm{OH}$) is heated with concentrated ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ at $170^{\circ }\text{C}$, predict the major alkene product. Explain the mechanism and apply Zaitsev’s rule.

Step-by-step solution

1. Step 1 – Protonation: The hydroxyl group is protonated by ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, converting $–OH$ to $–O{\mathrm{H}}_2^{+}$, making it a good leaving group.
2. Step 2 – Loss of water: Water departs, forming a primary carbocation at C‑1 ($\mathrm{C}{\mathrm{H}}_2^{+}\mathrm{C}{\mathrm{H}}_2\mathrm{CH}(\mathrm{C}{\mathrm{H}}_3)\mathrm{C}{\mathrm{H}}_3$).
3. Step 3 – Carbocation rearrangement: A hydride shift from the adjacent C‑2 to C‑1 converts the primary carbocation into a more stable secondary carbocation at C‑2.
4. Step 4 – Further stabilization: A subsequent 1,2‑methyl shift from C‑3 to C‑2 generates a tertiary carbocation at C‑3 ($(\mathrm{C}{\mathrm{H}}_3{)}_2{\mathrm{C}}^{+}\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$), which is the most stable intermediate.
5. Step 5 – Elimination: A base (conjugate base of ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$) abstracts a β‑hydrogen from C‑4, and the electrons form the double bond between C‑3 and C‑4.
6. Step 6 – Product formation: The resulting alkene is $\mathrm{C}{\mathrm{H}}_3\mathrm{C}(\mathrm{C}{\mathrm{H}}_3)=\mathrm{CHC}{\mathrm{H}}_3$, which is named 2‑methyl‑2‑butene.
7. Application of Zaitsev’s rule: Among possible alkenes, the most substituted (tetra‑substituted) alkene, 2‑methyl‑2‑butene, is favoured, confirming it as the major product.

Concept tested: Acid‑catalyzed dehydration & Zaitsev’s rule

Published 6 September 2026 · How this solution was prepared and checked