Discussion
An organic compound with molecular formula C8H12O2 exhibit the following spectral data IR : 1720 cm−1; 1H-NMR : δ 6.95 (d, J = 8.5 Hz, 1H) 5.9 (d, J = 8.5 Hz, 1H), 4.53 (q, J = 6Hz, 1H) 1.41 (d, J = 6Hz, 3H), 1.20 (S, 3H), 1.15 (S, 3H).
The correct structure of the compound is :
- A.
- B.
- C.
- D.
Answer: C
The compound is (C), 5,5,6-trimethyl-5,6-dihydro-2H-pyran-2-one.
C8H12O2 has three degrees of unsaturation: a ring, a C=C and a C=O. The IR band at 1720 cm−1 fits an α,β-unsaturated lactone carbonyl.
- δ 6.95 and 5.9, two doublets with the same J: the β and α vinyl protons of the enone, coupled only to each other. So the carbon beyond the β-CH has no hydrogen.
- δ 4.53, quartet, 1H, coupled to the 3H doublet at 1.41: an O-CH(CH3) group whose other neighbour has no hydrogen.
- δ 1.20 and 1.15, two 3H singlets: a gem-dimethyl carbon, its methyls made different by the nearby stereocentre.
Only (C) puts a quaternary CMe2 between the vinyl CH and the OCH(CH3). (A) has a methyl on the double bond, leaving only one vinyl H. In (B) the vinyl H would also couple to the ring CH, and there is no OCH quartet because the oxygen-bearing carbon carries two methyls. (D) is a pyranone whose O-CH=CH protons and ring CH next to the ketone do not fit a quartet at 4.53.
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