tert-Butyl ethyl ether is treated with one equivalent of anhydrous HI at 0 degC. Taking the conventional SN1 outcome with substitution by iodide, the products are

MolSynapse / Question bank / Organic Reaction Mechanisms III: Nucleophilic Substitution / Reactions of ethers and epoxides

JEE Main levelSingle correct optionDifficulty 2 of 3o3189

tert-Butyl ethyl ether is treated with one equivalent of anhydrous HI at $0^{\circ} C$ . Taking the conventional SN1 outcome with substitution by iodide, the products are

A $(CH X_{3}) X_{3} C - I$ and $CH X_{3} CH X_{2} - OH$
B $(CH X_{3}) X_{3} C - I$ and $CH X_{3} CH X_{2} - I$
C $(CH X_{3}) X_{3} C - OH$ and $CH X_{3} CH X_{2} - I$
D $CH X_{2} = CH X_{2}$ and $(CH X_{3}) X_{3} C - OH$

Answer: A

The protonated ether can lose ethanol to form the tertiary cation, which is easier than SN2 attack at the ethyl carbon next to a bulky tert-butyl-oxygen centre. Iodide captures the cation to give tert-butyl iodide. Ethanol is the leaving group and stays as the alcohol since only one equivalent of HI is used.

Practice Organic Reaction Mechanisms III: Nucleophilic Substitution More on Reactions of ethers and epoxides

Previous questionNext question

tert-Butyl ethyl ether is treated with one equivalent of anhydrous HI at 0 ∘C. Taking the conventional SN1 outcome with substitution by iodide, the products are

Answer: A

More questions on Reactions of ethers and epoxides

tert-Butyl ethyl ether is treated with one equivalent of anhydrous HI at $0^{\circ} C$ . Taking the conventional SN1 outcome with substitution by iodide, the products are