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The correct structure of a compound that exhibits the following 1H-NMR spectral data is :

δ3.8(S, 3H), 3.9 (S, 3H) 6.82 (dd, J = 8 and 2 H2, 1H), 7.06 (t, J = 8 H2, 1H), 7.15 (dd, J = 8 and 2 H2, 1H)

  • A.Benzene ring with two methoxy groups and a chlorine in a 1,3,5 pattern (chlorine between the two ring positions flanking neither methoxy)
  • B.Benzene ring with two adjacent methoxy groups and a chlorine para to one of them
  • C.Benzene ring with two adjacent methoxy groups and a chlorine on the carbon next to the second methoxy
  • D.Benzene ring with two methoxy groups in a 1,3 relationship and a chlorine between para to one and ortho to the other

Answer: C

The compound is 1-chloro-2,3-dimethoxybenzene, C.

The two singlets at δ 3.8 and 3.9 (3H each) are two non-equivalent OCH3 groups. The aromatic region has three protons: a triplet with J = 8 Hz and two doublets of doublets with J = 8 and 2 Hz. A proton with two ortho neighbours gives the triplet, and the two flanking protons each have one ortho (8 Hz) and one meta (2 Hz) partner. Three adjacent ring protons mean a 1,2,3-trisubstituted benzene, which is C.

A (1,3,5) is symmetrical, with equivalent OCH3 groups and no ortho couplings. B and D are 1,2,4-trisubstituted, which would show an isolated proton as a small doublet (meta coupling only) and no triplet. The paper writes Hz as 'H2'.

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