Question

This compound, (C10H12O3), shows strong, sharp absorption between 1700 and 1720 cm-1, and strong, broad absorption over the region 2500-3000 cm-1. Given this information and the following NMR data, please draw the structure of the compound in the box below.

1H-NMR: 2.49 ppm, t(2H); 2.80 ppm, t(2H); 3.72 ppm, s(3H); 6.78 ppm, d(2H); 7.11 ppm, d(2H); 12.4, s(1H) 13C-NMR: 173.89; 157.87; 132.62; 128.99; 113.55; 54.84; 35.75; 29.20

Answer 1

Step-by-step explanation:

The peak at 1700-17200 cm⁻¹ in the infrared spectrum shows the existence of a carbonyl group (C=O). The appearance of an OH peak is shown by the broad peak at 2500-3000 cm⁻¹. The peak at 12.4 ppm in 1H NMR identifies the existence of carboxylic acid. The peaks at 6.78 ppm and 7.11 ppm suggest that the prevalence existence of a benzene ring in the molecule. Because these peaks are doublets, the benzene is being para-substituted.

The existence of the CH2 group is shown by the peak at 2.49 ppm, which integrates 2 hydrogens. Provided that the breaks are doublets, It would be bound to another CH2 molecule as well as carbon that is devoid of hydrogen (benzene carbon).

The existence of the CH2 group, which is linked to the above-listed CH2 group, is also demonstrated by the next peak at 2.8 ppm. It must be bonded with carbonyl carbon so it breaks as a doublet. As a result, it has a greater chemical transition. The CH3 category is shown by the peak at 3.72 ppm. Its chemical change is strong, indicating that CH3 is linked to an oxygen atom that is linked to benzene.

NMR details at 13C. Carbonyl carbon of acid has a high (peak) at 173.89 ppm. Benzene carbon peaks range from 113.55 to 157.87 ppm. The peak at 64 is made up of CH3, while the other two are made up of CH2.

The structure of the diagram is shown below with its degree of unsaturation.