Question

CH\(_3\)-O-CH\(_3\) when treated with excess HI gives:

• A. CH\(_3\)-OH + CH\(_3\)-I
• B. 2CH\(_3\)-OH
• C. 2CH\(_3\)-I
• D. CH\(_3\)-I + CH\(_4\)

Hint:

Ethers undergo cleavage when treated with excess HI. The bond between the oxygen and the alkyl group is broken, leading to the formation of alkyl iodide and alcohol.

Answer 1

The Correct Option is C

The reaction of ether (CH\(_3\)-O-CH\(_3\)) with excess hydroiodic acid (HI) results in the cleavage of the ether bond. This cleavage produces methyl iodide (CH\(_3\)-I) and methyl alcohol (CH\(_3\)-OH). However, with excess HI, both ether bonds are cleaved, producing 2 moles of CH\(_3\)-I and CH\(_3\)-OH. Thus, the correct answer is (C), where two moles of CH\(_3\)-I are formed.