Question

Propose a structure that is consistent with the spectral data below. (The spectra and data provided were obtained from a pure organic molecule. For 'H NMR spectra, the integral is given in number of hydrogens (#H) or as a relative ratio.) Mass Spectrum (not shown): [M] = 162 m/z (100%) IR Spectrum (not shown): 3022, 2980, 1685, 1600, 1510 cm (all listed are strong (s) unless otherwise indicated) 'H NMR Spectrum (300 MHz, DMSO, 25 °C) dd (J= 2 Hz) (J = 2 Hz) 1 t (J = 8 Hz) PPM 4 2H 1H 1H 13C NMR Spectrum proton-decoupled (100 MHz, DMSO, 25 °C) 200 180 160 140 120 100 80 60 40 20

Answer 1

Analyzing 1HNMR

2 x CH3 groups are present without any neighbouring protons (at 2.5 ppm). If there were neighbouring protons, then the signal will not be a singlet. Instead, the signal would have been split. Since all 6 protons give a single peak, it means that both the CH3 groups are equivalent.

There are 4 aromatic protons (range from 7 to 9 ppm). Out of these 2 protons are equivalent which are seen as a dd signal at 8.2 ppm. Other 2 protons are not equivalent.

Analyzing 13CNMR

There are 6 unique carbon peaks. Out of these 6 peaks, 4 peaks are twice the size of other 2 peaks. That means that these 4 peaks are from 2 equivalent carbons each. So there are total of 10 carbons (2 x 4 + 2).

There is a peak for carbonyl carbon at aroun 200ppm. This peak is due to 2 carbons. So the molecule has 2 carbonyl groups that are equivalent.

There is an aromatic carbon peak at around 136 ppm made up of 2 carbons

There is a peak at 133 ppm made up of 2 carbons

There is a peak at around 128 ppm from 1 carbon

There is a peak at around 127ppm from1 carbon

There is a peak at around 27 ppm made up of 2 carbons.

Now we have the information that there are 10 carbons, 10 hydrogens, and 2 oxygens (from carbonyl groups) in the molecule. The mass should be 10 x 12 + 1x 10 + 2 x 16 = 162. This is the mass obtained from mass spectrum.

Analyzing the IR data, there are aliphatic C -H stretching frequency (3022), =C-H stretching frequency (2980), C=O stretching frequency (1685), C=C frequencies (1600 and 1500).

Based on this information, the molecule will have the following structure

за 24 — О 3, 2, бо Hppm the 7-8 (1) Carbon 30ppm 1 137 2а,2, 137 3,3, 133 4 128 5а,Sь 197 6,6 37 8.2 7. s (ан) (ін) as (69)