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Problem

The products formed in the reaction

$$\ce{CH2=CHOCH2CH3 + HI ->}$$

are

(a) $\ce{CH2=CHI}$ and $\ce{CH3CH2I}$
(b) $\ce{CH2=CHI}$ and $\ce{CH3CH2OH}$
(c) $\ce{CH3CHO}$ and $\ce{CH3CH2I}$
(d) $\ce{CH3CH(I)OCH2CH3}$

Answer

(c) $\ce{CH3CHO}$ and $\ce{CH3CH2I}$

Question

Does ether cleavage occur or electrophilic addition of $\ce{HI}$ to the double bond take place or both? I'm equally inclined towards both (c) and (d).

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    $\begingroup$ d is formed reversibly, then CH3CH(+)OCH2CH3 is cleaved by iodide. $\endgroup$
    – user55119
    Apr 29, 2021 at 17:56
  • $\begingroup$ The substrate is an enol ether, which is very sensitive to acid. And HI is a very strong acid, which can cleave usually stable normal ether linkages. $\endgroup$ Apr 30, 2021 at 5:34
  • $\begingroup$ @MathewMahindaratne Why is the cleavage reaction favored over the electrophilic addition reaction? The cleavage reaction will proceed via breaking of a sigma bond whereas E+ addition proceeds via breaking a pi bond, so shouldn't E+ addition give a thermodynamically more stable product? $\endgroup$
    – Siddhant
    Apr 30, 2021 at 5:48
  • $\begingroup$ $$\ce{H2C=CH-O-CH2CH3 ->[H+] H3C-CH=O^+\!-CH2CH3 ->[I-] H3C-CHO + I-CH2CH3}$$ The intermediate is more stable by resonance as shown. $\endgroup$ Apr 30, 2021 at 5:50

1 Answer 1

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Here is a mechanism for the cleavage. After proton transfer, the carbocation has a resonance contributor. Nucleophilic attack can occur on the ethyl group because there is a good leaving group.

enter image description here

The electrophilic addition can happen, with iodide attacking the carbocation. However, it is reversible. Iodide is a good leaving group, and the carbocation is resonance stabilized. The cleavage reaction is much less reversible.

enter image description here

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  • $\begingroup$ Thanks Ben, your answer made things absolutely clear $\endgroup$
    – Siddhant
    May 7, 2021 at 13:17

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