2‑Methyl‑1‑butanol (CH_3CH(CH_3)CH_2CH_2OH) is heated with concentrated H_2SO_4 at 170°C. Predict the major alkene product formed, give its IUPAC name, and explain why it is favoured over other possible alkenes.

2‑Methyl‑1‑butanol ( $C H_{3} CH (C H_{3}) C H_{2} C H_{2} OH$ ) is heated with concentrated $H_{2} S O_{4}$ at $17 0^{\circ} C$ . Predict the major alkene product formed, give its IUPAC name, and explain why it is favoured over other possible alkenes.

hard 5 marks Class 12 · Chemistry · Alcohols Phenols and Ethers

Answer

2‑Methyl‑1‑butene, IUPAC name: 2‑methyl‑but‑1‑ene

The correct answer is 2‑Methyl‑1‑butene, IUPAC name: 2‑methyl‑but‑1‑ene — IUPAC name: 2‑methyl‑but‑1‑ene.

Identify the type of reaction: acid‑catalysed dehydration of a primary alcohol.
Protonate the –OH group to make it a good leaving group, forming water.
Loss of water generates a primary carbocation at C‑1: $C H_{3} CH (C H_{3}) C H_{2} C H_{2}^{+}$ .
A primary carbocation is unstable; a 1,2‑hydride shift from the adjacent carbon (C‑2) occurs, converting it to a more stable secondary carbocation $C H_{3} C^{+} (C H_{3}) C H_{2} C H_{3}$ .
From this secondary carbocation, elimination of a proton from the adjacent carbon (C‑1) gives the double bond between C‑1 and C‑2, producing $C H_{3} CH = CHC H_{2} C H_{3}$ which is 2‑methyl‑1‑butene.
Alternative elimination from C‑3 would give 3‑methyl‑1‑butene, a less substituted (trisubstituted) alkene, which is less stable according to Zaitsev’s rule.
Hence the major product is the more substituted alkene, 2‑methyl‑but‑1‑ene.
IUPAC name: 2‑methyl‑but‑1‑ene.

Common mistake: Assuming the double bond forms between the most substituted carbons without considering carbocation rearrangement.

Concept tested: Acid‑catalysed dehydration – Zaitsev’s rule