Lesson eLearning

3-Pentanol (CH₃CH₂CH₂CH(OH)CH₃) is treated with conc. H_2SO_4 at 170°C. Identify the major alkene product formed and explain why it is favored.

medium 3 marks Class 12 · Chemistry · Alcohols Phenols and Ethers
Answer
2‑Pentene (CH_3CH_2CH=CHCH_3)

The correct answer is 2‑Pentene (CH_3CH_2CH=CHCH_3) — 4. By Zaitsev’s rule the more substituted alkene, 2‑pentene, is favoured.

Step-by-step solution

  1. 1. Protonation of the hydroxyl oxygen gives an oxonium ion, making the alcohol a good leaving group.
  2. 2. Loss of water generates a secondary carbocation at the 4‑carbon centre.
  3. 3. Deprotonation from an adjacent carbon forms a double bond; possible alkenes are 2‑pentene (more substituted) and 3‑pentene (less substituted).
  4. 4. By Zaitsev’s rule the more substituted alkene, 2‑pentene, is favoured.
Common mistake: Choosing 3‑pentene as the major product

Concept tested: Dehydration of alcohols (Zaitsev rule)

Published · How this solution was prepared and checked

← All Alcohols Phenols and Ethers questions