Question

Determine the structure of each compound for each 1H NMR spectrum. Draw the determined structure and annotate (correlate the equivalent protons with the corresponding NMR signal(s) in the spectrum).

Compound C, M = 102.13 g/mol PPM Compound D, M = 136.15 g/mol 12 10 8 6 PPM 4 2 0

Answer 1

Compound C

1H NMR

The peak 5.0 ppm belongs to CH which is attached to oxygen. The peak 1.2 ppm shows doublet which is attached to CH. Both peaks responsible for isopropyl group.

The peak at ~2.0 ppm belongs to methyl group. Because the intergration is three. It may be attached to carbonyl oxygen.

Using the information we can deduce the structure.

Compound D

1H NMR

1H NMR shows two doublets from 6-8 ppm that belongs to aromatic proton and the pattern confirms that the ring is substituted at 1,4 position.

The peak 12.5 ppm is responsible for COOH group.

The peak at ~2.0 ppm belongs to methyl group. Because the intergration is three. It may be attached to aromatic ring.  

Using the information we can deduce the structure.

isopropyl acetate Molecular Weight: 102.13 ChemNMR TH Estimation 1.23 4.99 zon To Lose 2.01 ó 1.23 PPM COOH CH3 4-methylbenzoic acid Molecular Weight: 136.15 ChemNMR *H Estimation O OH 12.71 8.04 8.04 7.30 V 7.30 CH3 2.41 PPM