Consider the following statements about the cleavage of ethers by hydrogen halides.
(A) In the cleavage of ethers, the order of reactivity of hydrogen halides is HI > HBr > HCl.
(B) Anisole on heating with concentrated HI gives phenol and methyl iodide.
(C) Methyl tert-butyl ether on heating with HI gives methyl iodide and tert-butyl alcohol.
(D) Ethyl methyl ether on heating with HI gives methyl iodide and ethanol.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
HI first protonates the ether oxygen. The iodide ion then breaks one C–O bond; which bond breaks depends on the groups on the oxygen.
(A) In the cleavage of ethers, the order of reactivity of hydrogen halides is HI > HBr > HCl. — True. HI and HBr are acidic enough to protonate the ether, and I− is a good nucleophile.
(B) Anisole on heating with concentrated HI gives phenol and methyl iodide. — True. The O–CH3 bond is weaker than the O–C6H5 bond, which has partial double bond character, so the methyl group leaves as CH3I.
(C) Methyl tert-butyl ether on heating with HI gives methyl iodide and tert-butyl alcohol. — False. When one group is tertiary, the stable (CH3)3C+ ion forms (SN1), so the products are tert-butyl iodide and methanol:
\( \displaystyle \ce{(CH3)3C-O-CH3 + HI -> CH3OH + (CH3)3C-I} \)
(D) Ethyl methyl ether on heating with HI gives methyl iodide and ethanol. — True. With primary groups, I− attacks the least substituted carbon by SN2, so the lower alkyl group forms the alkyl iodide:
\( \displaystyle \ce{CH3-O-CH2CH3 + HI -> CH3I + CH3CH2OH} \)
NCERT (2026-27 edition):
7.6.3 Chemical Reactions – “Alkyl aryl ethers are cleaved at the alkyl-oxygen bond due to the more stable aryl-oxygen bond. The reaction yields phenol and alkyl halide. Ethers with two different alkyl groups are also cleaved in the same manner. The order of reactivity of hydrogen halides is as follows: HI > HBr > HCl.”
“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).”
“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 SN1 mechanism.”
“In case of anisole, … The bond between O–CH3 is weaker than the bond between O–C6H5 … Therefore the attack by I– ion breaks O–CH3 bond to form CH3I.”