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The major product P formed in the following transformation is

1-(4-methoxyphenyl)propane-1,3-diol treated with MnO2 in acetone at 25 C, then PBr3 in CH2Cl2 at 0 C, then NaOEt in ethanol, gives P

  • A.4-Methoxyphenyl vinyl ketone, a 1-(4-methoxyphenyl)prop-2-en-1-one
  • B.4-Methoxycinnamaldehyde, E-3-(4-methoxyphenyl)prop-2-enal
  • C.2-Bromo-3-(4-methoxyphenyl)-3-oxopropanal, a bromo keto aldehyde
  • D.3-(4-Methoxyphenyl)prop-2-ynal, an arylpropynal

Answer: A

The product P is the aryl vinyl ketone, 1-(4-methoxyphenyl)prop-2-en-1-one, which is option (A).

MnO2 oxidises benzylic alcohols selectively, so the secondary benzylic OH becomes a ketone and the primary alcohol is left alone, giving ArCOCH2CH2OH. PBr3 converts the primary alcohol to the bromide, ArCOCH2CH2Br. NaOEt then removes the acidic alpha hydrogen and expels bromide (E1cB elimination), forming the C=C of the vinyl ketone.

Option (B) would need oxidation of the primary alcohol and loss of the benzylic oxygen. Option (C) needs an alpha bromination that these reagents do not do, and (D) needs an alkyne. None can come from the three steps.

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