By a proper choice of reagents, both symmetrical and unsymmetrical ethers can be prepared by Williamson synthesis which involves the reaction between an alkyl halide and an alkoxide ion. The reverse process involving the cleavage of ethers to give back the original alkyl halide and the alcohol can be carried out by heating the ether with HI at 373K.
Consider the following statements about the mechanism of the cleavage of ethers by hydrogen halides, referred to in the passage.
(A) The reaction of an ether with concentrated HI starts with protonation of the ether molecule.
(B) The cleavage of ethers takes place with HBr or HI because these reagents are sufficiently acidic.
(C) When ethyl methyl ether is heated with excess HI at high temperature, the ethanol formed reacts with another molecule of HI and is converted to ethyl iodide.
(D) In the cleavage of 2-methoxy-2-methylpropane by HI, the departure of methanol creates a tertiary carbocation, and the reaction follows SN2 mechanism.
Which of the statements given above are correct? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
Ethers are among the least reactive functional groups; their C–O bond is broken under drastic conditions, with an excess of a hydrogen halide such as HI or HBr.
The acid first protonates the oxygen, and the halide ion then breaks a C–O bond. Whether this second step is SN2 or SN1 depends on the alkyl groups.
(A) The reaction of an ether with concentrated HI starts with protonation of the ether molecule. — True. The oxygen takes up H+ and an oxonium ion is formed, e.g. \( \ce{CH3-O-CH2CH3 + HI <=> CH3-O^+H-CH2CH3 + I^-} \).
(B) The cleavage of ethers takes place with HBr or HI because these reagents are sufficiently acidic. — True. The first step needs a strong acid that can protonate the ether.
(C) When ethyl methyl ether is heated with excess HI at high temperature, the ethanol formed reacts with another molecule of HI and is converted to ethyl iodide. — True. The first cleavage gives CH3I and ethanol; with excess HI the ethanol is converted to ethyl iodide as well.
(D) In the cleavage of 2-methoxy-2-methylpropane by HI, the departure of methanol creates a tertiary carbocation, and the reaction follows SN2 mechanism. — False. A reaction that goes through a carbocation is SN1. The stable (CH3)3C+ ion forms first and then takes up I−, so the products are methanol and tert-butyl iodide:
\( \displaystyle \ce{(CH3)3C-O-CH3 + HI -> CH3OH + (CH3)3C-I} \)
NCERT (2026-27 edition):
7.6.3 Chemical Reactions – “Ethers are the least reactive of the functional groups. The cleavage of C-O bond in ethers takes place under drastic conditions with excess of hydrogen halides.”
“The reaction of an ether with concentrated HI starts with protonation of ether molecule.” … “The reaction takes place with HBr or HI because these reagents are sufficiently acidic.”
“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.”
“However, when one of the alkyl group is a tertiary group, the halide formed is a tertiary halide.”
\( \displaystyle \ce{CH3-C(CH3)2-O-CH3 + HI -> CH3OH + CH3-C(CH3)2-I} \)
“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.”