Question

\( (CH_3)_3C - OC_2H_5 \) on reaction with HI gives

• A. \( (CH_3)_3C - I \) and \( C_2H_5 - I \)
• B. \( (CH_3)_3C - OH \) and \( C_2H_5 - I \)
• C. \( (CH_3)_3C - I \) and \( C_2H_5 - OH \)
• D. \( (CH_3)_3C - OH \) and \( C_2H_5 - OH \)

Hint:

For primary/secondary alkyl ethers, the reaction follows \( S_N2 \) and iodide goes to the smaller alkyl group. For tertiary alkyl ethers, it follows \( S_N1 \) and iodide goes to the tertiary group.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Reaction of ethers with HI depends on the nature of the alkyl groups. If one of the alkyl groups is tertiary (\( 3^\circ \)), the reaction follows an \( S_N1 \) mechanism.
Step 2: Detailed Explanation:
The ether is tert-butyl ethyl ether. Upon protonation by HI, the molecule is \( (CH_3)_3C-O^+(H)-C_2H_5 \).
Because the tert-butyl group can form a highly stable \( 3^\circ \) carbocation \( (CH_3)_3C^+ \), the C-O bond breaks via \( S_N1 \).
The iodide ion (\( I^- \)) then attacks this stable carbocation to form \( (CH_3)_3C-I \) (tert-butyl iodide).
The other product formed is ethanol (\( C_2H_5OH \)).
Step 3: Final Answer:
The products are \( (CH_3)_3C-I \) and \( C_2H_5OH \).