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The compound which gives a fragment at \(m/z\) = 124[M+H]+ is

  • A.1,3-Dimethyl purine-2,6-dione with the imidazole ring partly saturated at the ring fusion carbons, N9 bearing H
  • B.Theophylline, 1,3-dimethylxanthine, with a fused aromatic imidazole ring and N-H
  • C.1,3-Dimethyluric acid, a purine-2,6-dione with a carbonyl at C8 and two N-H groups in the five-membered ring
  • D.1,3-Dimethyl purine-2,6-dione with a fully saturated five-membered ring containing two N-H groups

Answer: B

The compound is (B), theophylline (1,3-dimethylxanthine).

Theophylline is C7H8N4O2, mass 180, so its protonated molecule [M+H]+ appears at m/z 181. The characteristic fragmentation of the 1,3-dimethylxanthine ring is loss of methyl isocyanate (CH3NCO, 57) from the pyrimidinedione ring, which gives the ion at

\[ 181 - 57 = 124 \]

The same loss from the other structures gives different masses: (A), with two extra hydrogens at the ring fusion (C7H10N4O2, MH+ 183), gives 126; the uric acid (C), C7H8N4O3 (MH+ 197), gives 140; and the fully saturated (D), C7H12N4O2 (MH+ 185), gives 128.

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