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Discussion

The major product in the following reaction is :

A secondary alcohol two carbons from a p-methoxybenzyl ether on a branched carbon chain, treated with DDQ in CH2Cl2

  • A.A ketone with an isopropyl group on one side and, on the other, a CH2 and a CH(CH3) bearing an O-(4-methoxybenzyl) ether
  • B.A secondary alcohol with an isopropyl group, whose other end carries a CH(CH3) ester of 4-methoxybenzoic acid
  • C.A diol-derived chain with an isopropyl group, an O-(4-methoxybenzyl) ether on the carbon next to it and a secondary alcohol at the other end
  • D.A 1,3-dioxane ring bearing a 4-methoxyphenyl group at C2, a methyl group at C4 and an isopropyl group at C6

Answer: D

The product is (D), the cyclic p-methoxyphenyl acetal.

DDQ oxidizes p-methoxybenzyl (PMB) ethers by removing a hydride from the benzylic CH2. The p-methoxy group stabilizes the resulting oxocarbenium ion, Ar-CH=O+-R. In wet solvent water would attack it and release the free alcohol. Here the solvent is dichloromethane and the molecule has its own free OH two carbons away, in a 1,3 relationship, so that hydroxyl captures the cation intramolecularly. The six-membered ring formed is a 1,3-dioxane bearing the p-methoxyphenyl group on the acetal carbon: a PMP acetal.

(A) would need DDQ to oxidize an ordinary secondary alcohol to a ketone, which it does not do, since the alcohol is neither allylic nor benzylic. (B), an ester, would need a second oxidation at the benzylic carbon. (C) moves the PMB group to the other oxygen, which the mechanism gives no route for.

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