3. When 2‑methyl‑2‑butanol ((CH_3)_3CCH_2OH) is heated with concentrated H_2SO_4 at 170°C, predict the major alkene product and explain the mechanism using Zaitsev’s rule.

3. When 2‑methyl‑2‑butanol ( $(C H_{3})_{3} CC H_{2} OH$ ) is heated with concentrated $H_{2} S O_{4}$ at $17 0^{\circ} C$ , predict the major alkene product and explain the mechanism using Zaitsev’s rule.

medium 3 marks Class 12 · Chemistry · Alcohols Phenols and Ethers

Answer

The major product is 2‑methyl‑2‑butene (

(C H_{3})_{2} C = CHC H_{3}

The correct answer is The major product is 2‑methyl‑2‑butene ( $(C H_{3})_{2} C = CHC H_{3}$ ). — State the final product.

Write the structure of 2‑methyl‑2‑butanol: $(C H_{3})_{3} CC H_{2} OH$ . The –OH is attached to a tertiary carbon.
In acid‑catalysed dehydration, the –OH is first protonated to $- O H_{2}^{+}$ , making it a good leaving group.
Loss of water gives a tertiary carbocation at the carbon bearing the –OH group: $(C H_{3})_{3} C^{+} C H_{3}$ . This carbocation is already stable.
Identify possible β‑hydrogen atoms that can be removed to form a double bond: the adjacent carbon (the –CH $_{2}$ group) has two hydrogens; removal of one gives a double bond between the tertiary carbon and the adjacent carbon.
Two possible alkenes can form: 2‑methyl‑2‑butene (more substituted, trisubstituted) and 2‑methyl‑1‑butene (disubstituted).
Apply Zaitsev’s rule: the more substituted alkene is favoured, i.e., 2‑methyl‑2‑butene.
Show the de‑protonation step: a base (sulphate ion) removes a β‑hydrogen, forming the C=C bond and regenerating the acid catalyst.
State the final product.

Common mistake: Selecting the less substituted alkene (2‑methyl‑1‑butene) as the major product.

Concept tested: Acid‑catalysed dehydration of alcohols