Each ether in List-I is heated with concentrated HI. Match List-I (et…

The correct answer is Option (2) → A-II, B-III, C-IV, D-I

Concentrated HI first protonates the ether oxygen. The iodide ion then breaks one C–O bond, giving an alkyl iodide and an alcohol (or phenol). Which bond breaks, and by which mechanism, depends on the groups on the oxygen.

(A) Methoxyethane, with 1 mol of HI — This matches (II) Of the two alkyl groups, the lower alkyl group forms the alkyl iodide. Both groups are primary, so I⁻ attacks the less substituted methyl carbon by S_N2:

    \( \displaystyle \ce{CH3-O-C2H5 + HI -> CH3I + C2H5OH} \)

(B) 2-Methoxy-2-methylpropane — This matches (III) The reaction follows the S_N1 mechanism through a stable carbocation. The leaving group (CH₃OH) departs to give the stable tertiary carbocation, so the tertiary group, not the methyl group, ends up as the iodide:

    \( \displaystyle \ce{(CH3)3C-O-CH3 + HI -> (CH3)3C-I + CH3OH} \)

(C) Methoxybenzene (anisole) — This matches (IV) The O–alkyl bond breaks because the other C–O bond has partial double bond character. The O–C₆H₅ bond is stronger (sp² carbon, partial double bond), so the weaker O–CH₃ bond breaks. The phenol formed does not react further, because its sp² carbon cannot undergo nucleophilic substitution:

    \( \displaystyle \ce{C6H5-O-CH3 + HI -> C6H5OH + CH3I} \)

(D) Ethoxyethane, with excess HI at high temperature — This matches (I) The hydroxy compound formed first reacts with another molecule of HI to give a second iodide. The ethanol formed in the first step is converted to ethyl iodide, so two molecules of ethyl iodide are obtained:

    \( \displaystyle \ce{C2H5-O-C2H5 + HI -> C2H5I + C2H5OH} \)

    \( \displaystyle \ce{C2H5OH + HI -> C2H5I + H2O} \)

NCERT (2026-27 edition):
7.6.3 Chemical Reactions – “…Thus, in the cleavage of mixed ethers with two different alkyl groups, the alcohol and alkyl iodide formed, depend on the nature of alkyl groups. When primary or secondary alkyl groups are present, it is the lower alkyl group that forms alkyl iodide (S_N2 reaction).”

“However, when one of the alkyl group is a tertiary group, the halide formed is a tertiary halide.” … “It is because in step 2 of the reaction, the departure of leaving group (HO–CH₃) creates a more stable carbocation [(CH₃)₃C⁺], and the reaction follows S_N1 mechanism.”

“In case of anisole, methylphenyl oxonium ion, … is formed by protonation of ether. The bond between O–CH₃ is weaker than the bond between O–C₆H₅ because the carbon of phenyl group is sp² hybridised and there is a partial double bond character. Therefore the attack by I^– ion breaks O–CH₃ bond to form CH₃I. Phenols do not react further to give halides because the sp² hybridised carbon of phenol cannot undergo nucleophilic substitution reaction needed for conversion to the halide.”

“When HI is in excess and the reaction is carried out at high temperature, ethanol reacts with another molecule of HI and is converted to ethyl iodide.”