Which alkyl iodide is formed when ethyl methyl ether is heated with an equimolar amount of concentrated hydrogen iodide (HI)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Methyl iodide
Ethers are cleaved by concentrated HI. The ether oxygen is first protonated; the iodide ion then attacks the least substituted carbon of the oxonium ion by an SN2 path and pushes out an alcohol molecule.
Protonation:
\( \displaystyle \ce{CH3-O-C2H5 + HI -> CH3-O^+(H)-C2H5 + I^-} \)
Attack by I−: the methyl carbon is less hindered than the ethyl carbon, so I− attacks the methyl group:
\( \displaystyle \ce{I^- + CH3-O^+(H)-C2H5 -> CH3I + C2H5OH} \)
With an equimolar amount of HI, the products are methyl iodide and ethanol: the lower alkyl group forms the alkyl iodide.
Why the other options are incorrect
Option (1) Ethyl iodide — Incorrect. The ethyl group leaves as ethanol; I− attacks the less substituted methyl carbon.
Option (3) Both methyl iodide and ethyl iodide — Incorrect. Ethanol is turned into ethyl iodide only when HI is in excess at high temperature; one equivalent of HI is used up in the first cleavage.
NCERT (2026-27 edition):
7.6.3 Chemical Reactions – “Iodide is a good nucleophile. It attacks the least substituted carbon of the oxonium ion formed in step 1 and displaces an alcohol molecule by SN2 mechanism. … When primary or secondary alkyl groups are present, it is the lower alkyl group that forms alkyl iodide (SN2 reaction).”