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The major products M and N formed in the following reactions are

2-Hydroxypyridine treated with CH3I and NaOH gives M, and with CH2N2 gives N

  • A.M is 5-methylpyridin-2-ol (5-methyl-2-hydroxypyridine); N is 2-methoxypyridine
  • B.M is 2-methoxypyridine; N is N-methylpyridin-2-one
  • C.M is N-methylpyridin-2-one; N is 2-methoxypyridine
  • D.M is N-methylpyridin-2-one; N is 5-methylpyridin-2-ol

Answer: C

The products are (C): M is N-methyl-2-pyridone and N is 2-methoxypyridine.

2-Hydroxypyridine exists mainly as its tautomer 2-pyridone, and its anion is ambident, able to react at nitrogen or oxygen. With CH3I and NaOH, the anion is alkylated at the softer, more nucleophilic nitrogen, giving the N-methyl pyridone M.

Diazomethane methylates acidic O-H groups: it takes the proton from the OH form to give a methyldiazonium ion, which the resulting oxygen anion attacks directly, so the product N is the O-methyl ether, 2-methoxypyridine.

(B) swaps the two products. (A) and (D) put a methyl group on a ring carbon, which neither reagent does.

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