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Write Brief Answer · Q2

Q.Draw the major product formed when 1-ethoxyprop-1-ene (CH3-CH=CH-O-C2H5) is heated with one equivalent of HI.

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Step 1. 1-Ethoxyprop-1-ene, CH3-CH=CH-O-C2H5, is a vinyl (enol) ether: the ether oxygen is attached directly to one of the alkene's sp2 carbons.

Step 2. Protonation by HI occurs at the BETA-carbon (the one bearing the methyl group, away from oxygen), because this places the resulting positive charge on the ALPHA-carbon (the one attached to oxygen), where it is strongly stabilised as an OXOCARBENIUM ion by donation of a lone pair from the adjacent ether oxygen.

Step 3. Iodide ion then attacks this stabilised cationic alpha-carbon, giving the alpha-iodo ether CH3-CH2-CHI-O-C2H5 (1-ethoxy-1-iodopropane) as the major product.

Step 4. This is the vinyl-ether analogue of Markovnikov addition: the positive charge (and hence the incoming nucleophile) ends up on the carbon that CAN be resonance-stabilised by the adjacent oxygen, exactly as a simple alkene's Markovnikov addition favours the more substituted/stabilised carbocation.

✓Final answer

CH3-CH2-CHI-O-C2H5 (1-ethoxy-1-iodopropane), formed by Markovnikov-style addition of HI across the enol ether double bond

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