# Products of the reaction between ethoxyethene and HI

### 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) $$\ce{CH3CH(I)OCH2CH3}$$

(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).

• d is formed reversibly, then CH3CH(+)OCH2CH3 is cleaved by iodide. Apr 29 at 17:56
• 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. Apr 30 at 5:34
• @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? Apr 30 at 5:48
• $$\ce{H2C=CH-O-CH2CH3 ->[H+] H3C-CH=O^+\!-CH2CH3 ->[I-] H3C-CHO + I-CH2CH3}$$ The intermediate is more stable by resonance as shown. Apr 30 at 5:50