Question

Learning Group Problem Chp 9, problem 1.

1. Given the following information, elucidate the structures of compounds A and B. Both compounds are soluble in dilute aqueous HCl, and both have the same molecular formula. The mass spectra of A and B have M+ 149. Other spectroscopic data for A and B are given below. Justify the structures you propose by assigning specific aspects of the data to the structures. Make sketches of the NMR spectra.

(a) The IR spectrum for compound A show two bands in the 3300-3500 cm-1 region. the broadband protons decoupled 13C NMR spectrum displayed for the following signals:
A: 13C NMR: 140 (C), 127 (C), 125 (CH), 118 (CH), 24 (CH2), 13 (CH3)

(b) The IR spectrum for compound A show no bands in the 3300-3500 cm-1 region. the broadband protons decoupled 13C NMR spectrum displayed for the following signals:
B: 13C NMR: 147 (C), 129 (CH), 115 (CH), 111 (CH), 44 (CH2), 13 (CH3)

Answer 1

Based on given information that both compounds A and B are soluble in aqueous HCl indicates that there must be a nirogen atom in the molecule whose ammonium chloride salt will be soluble in acidic medium. Presence of molecular ion peak at 149 indicates that it is an aromatic compound.

Compound A: In IR data presence of signals in 3300-3500 cm-1 shows N-H stretching and total 6 signals in 13C NMR is showing tha compound has symmetry in it. Pressnce of peak at 24 and 13 ppm belongs to CH2CH3 and this group is attached to aromatic ring (CH2 at 24 ppm). Futher signals at 140 (C) and 127 (C) shows that ring is multisubstituted.

Compound B: Molecule does not have free NH group as no peak at 3300-3500 cm-1. As this compound is an isomer of compound A, then few information exolained in A would be also helpful for B. Eg. it is also symmetrical and has -CH2CH3 unit but attached to Nitroge (CH2 at higher, 44 ppm). This ring is mono-stituted as there are two CH signals (129, 115) and only one C (147).

Based on these informations structure of compounds can be given as: