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The major product W formed in the following reaction is

Anisole treated with Na in liquid NH3 and with Et2O and EtOD gives product W

  • A.Cyclohexa-1,4-diene with OMe and D on one sp3 carbon and H and D on the opposite sp3 carbon
  • B.Methoxycyclohexa-1,3-diene type ring with OMe on an alkene carbon, a CHD group beside it and a second CHD group across the ring
  • C.Methoxycyclohexadiene with OMe on an alkene carbon and two adjacent CHD groups
  • D.Methoxycyclohexa-1,4-diene with OMe on an alkene carbon, with CHD groups at the two sp3 ring positions

Answer: B

The major product W is the 1-methoxycyclohexa-1,4-diene carrying a hydrogen and a deuterium on each of its two CH2 carbons, which is option (B).

Na in liquid NH3 with an alcohol is the Birch reduction. The methoxy group is electron donating, so the radical anion is protonated at the ortho and meta positions, leaving the methoxy carbon on a double bond in the product. EtOD is the proton source, so the two new hydrogens are deuterium. Two carbons become sp3: the one ortho to OMe and the one meta to it (para to the first), and each ends up as CHD, because one H was already present and one D is added.

Option (A) puts D on the methoxy carbon and reduces the para positions, which is the pattern for an electron-withdrawing substituent. Options (C) and (D) put the new sp3 carbons on the wrong positions, so they are not the 1,4-diene.

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