Question

1. Use the 'H NMR, 13C NMR, mass spectrometry and IR dat of a compound to determine its structure. IR Spectrum Mass Spectrometry Data relative abund 77 105 117 133 147 148 100 1H NMR Data 2H 24 ) H !! 13C NMR Data There are two unresolved lines between 130 and 131 ppa. TTTTTTTT 160 140 120 100 80 60 20 PPM 0 DEPT 135: positive: 18.4, 55.1, 113.9, 123.3, 126.9, 130.5 ppm DEPT 90: 113.9, 123.3, 126.9, 130.5 ppm DEPT 45: 18.4, 55.1, 113.9, 123.3, 126.9, 130.5 ppm Decoupled 18.4.551. 113.9. 1233. 126.9 12305.130.9. 158 ppm

Answer 1

IR spectrum

There are sp3 and sp2 C-H stretching presents around ~ 2950 cm-1 and 3050cm-1.

There is no IR active functional group are present apart from sp3 and sp2 C-H, and C-C stretching.

1H NMR

1H NMR shows two doublets from 6.7 - 8 ppm that are aromatic protons. Since we are getting only two peaks in aromatic that tells aromatic ring is 1,4 disubstituted.

Other two peaks from 6 -7 ppm belongs to alkene protons.

The peak at ~3.8 ppm belongs to aliphatic proton. The intergration is three. It can be a methyl group which may be attached to oxygen.

The peak 1.8 ppm belong to another methyl group which attached to alkene.

13C NMR

There are 6 peaks from 100 -160 ppm. They belong to aromatic and alkene proton. Another two peaks from 15-60 ppm observed which are belong to aliphatic carbon

Using the information, we can deduce the structure.

(E)-1-methoxy-4-(prop-1-en-1-yl)benzene Chemical Formula: C10H420 Exact Mass: 148.09 ChemNMR 1H Estimation 7.09 6.06 H 7.62 3.81 7.09 2.05 7.62 6.44 ŽE PPM ChemNMR 13C Estimation 114.2 130.2 159.8 55.8 124.4 114.2 128.7 186 120. 11142 18.8 130.5 130.2 160 140 120 100 80 60 40 20 PPM