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The compound that shows the following spectral data is,

IR : 1720 cm−1
1H-NMR (ppm) : 1.30 (t, J = 7Hz, 3H), 1.93(s, 3H), 4.19(q, J = 7Hz, 2H), 5.58 (d, J = 3Hz, 1H), 6.15(d, J = 3Hz, 1H)

  • A.CH\(_{3}\)C(=O)-O-C(=CH\(_{2}\))CH\(_{2}\)CH\(_{3}\)
  • B.CH\(_{2}\)=C(CH\(_{3}\))C(=O)-O-CH\(_{2}\)CH\(_{3}\)
  • C.CH\(_{2}\)=C(CH\(_{3}\))-O-C(=O)CH\(_{2}\)CH\(_{3}\)
  • D.CH\(_{3}\)C(=O)CH=CH-O-CH\(_{2}\)CH\(_{3}\)

Answer: B

The compound is ethyl methacrylate, CH2=C(CH3)COOCH2CH3.

Read the data piece by piece:

  • 1.30 (t, 3H) and 4.19 (q, 2H), both J = 7 Hz: an ethyl group on oxygen, O-CH2CH3, as in an ethyl ester.
  • 1.93 (s, 3H): a methyl on a carbon with no hydrogens, here the alkene carbon.
  • 5.58 and 6.15 (each d, 1H, J = 3 Hz): two non-equivalent =CH2 protons with a small geminal coupling.
  • IR 1720 cm−1: a conjugated ester C=O.

Option A is an enol acetate: its ethyl CH2 sits on carbon, near 2.2 ppm, not on oxygen at 4.19, and it has an acetyl singlet near 2.1 rather than a vinyl methyl. Option C, isopropenyl propanoate, likewise has its CH2 quartet near 2.4 ppm instead of an OCH2 at 4.19, and enol esters absorb near 1760 cm−1. Option D, an enol ether ketone, has two vinyl protons coupled trans with J of about 12 Hz, not two 3 Hz doublets.

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