Q.Write the structure of the product in the following reaction: $CH_3CH_2OCH_2CH_3 + HI \xrightarrow{373\ K} ?$

HX cleaves ethers at the C-O bond; with dimethyl ether and 1 equivalent of cold HBr, one alkyl-oxygen bond breaks to give an alkyl halide and an alcohol.

Ethers undergo cleavage when treated with excess strong hydrohalic acids (HI, HBr) — the ether oxygen is first protonated, then a nucleophilic halide ion attacks one of the alkyl carbons (SN2 for methyl/1° groups), breaking the C-O bond:

$CH_3-O-CH_3 + HBr \rightarrow CH_3Br + CH_3OH$

…

HI cleaves the ether C–O bond by nucleophilic substitution; with excess HI and heat, both alkyl fragments end up as ethyl iodide.

Mechanism

Step 1 — the ether oxygen is first protonated by HI, making it a better leaving group; the iodide ion then attacks one of the (equivalent, primary) ethyl carbons in an $S_N2$ fashion, displacing ethanol:

$CH_3CH_2-O-CH_2CH_3 + HI \rightarrow CH_3CH_2I + CH_3CH_2OH$

Step 2 — under the excess HI and the elevated temperature (373 K) of these conditions, the ethanol liberated in step 1 itself reacts further with HI (again by $S_N2$ on the primary carbon) to give a second mole of ethyl iodide:

$CH_3CH_2OH + HI \rightarrow CH_3CH_2I + H_2O$

Overall, both alkyl groups of the (symmetrical) ether end up as ethyl iodide: …