Question

Identify the product obtained when tert-Butyl methyl ether is heated with hot HI.

• A. \(\text{CH}_{3}\text{I}+(\text{CH}_{3})_{3}\text{COH}\)
• B. \(\text{CH}_{3}\text{OH}+(\text{CH}_{3})_{3}\text{C-I}\)
• C. \(\text{CH}_{3}\text{OH}+(\text{CH}_{3})_{3}\text{C-OH}\)
• D. \(\text{CH}_{3}\text{I}+(\text{CH}_{3})_{3}\text{C-I}\)

Hint:

With excess hot HI, an ether is completely cleaved and both alkyl groups end up as alkyl iodides.

Answer 1

The Correct Option is D

Step 1: Hot HI means complete cleavage
Heating with HI converts every alkyl group of the ether into an alkyl iodide and the oxygen into water.

Step 2: Write the overall reaction
$(\text{CH}_3)_3\text{C-O-CH}_3+2\text{HI}\rightarrow(\text{CH}_3)_3\text{C-I}+\text{CH}_3\text{I}+\text{H}_2\text{O}$

Step 3: Match
This agrees with the option giving two alkyl iodides.

Final Answer:
Both groups end up as iodides, namely CH3I and tert-butyl iodide. \[ \boxed{\text{(D) }\text{CH}_{3}\text{I}+(\text{CH}_{3})_{3}\text{C-I}} \]