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: Understanding the Concept
When an ether is heated with excess hot HI, the C-O bonds are cleaved and the alcohol formed in the first step is converted into the alkyl iodide.

Step 2: First cleavage
\((\text{CH}_3)_3\text{C-O-CH}_3+\text{HI}\rightarrow(\text{CH}_3)_3\text{C-I}+\text{CH}_3\text{OH}\)

Step 3: Further reaction
The methanol formed reacts with more hot HI: \(\text{CH}_3\text{OH}+\text{HI}\rightarrow\text{CH}_3\text{I}+\text{H}_2\text{O}\)

Step 4: Final products
Both alkyl groups are converted into iodides: \(\text{CH}_3\text{I}\) and \((\text{CH}_3)_3\text{C-I}\).

Final Answer:
The products are methyl iodide and tert-butyl iodide. \[ \boxed{\text{(D) }\text{CH}_{3}\text{I}+(\text{CH}_{3})_{3}\text{C-I}} \]