Question

1. label all spectra in IR

2. figure out structure. label a,b,c,d, etc in structure and corresponding letter on all peaks in HMNR and CNMR

3. show all calculations

IR/HNMR/CNMR MW=122: 124 AMU 1:1 200 180 160 340 120 100 0 0 0

Answer 1

Molecular formula mass is 122

The ratio of M:M+2 peak is 1:1

So the halogen atom is Bromine.

From 1H -NMR 6 Hydrogens doublet indicate the presence of isopropyl group

(CH3)2-CH-

1 Hydrogen atom multiplet at 3.9 ppm indicates the presence of following group.

(CH3)2-CH-Br

Hence the structure of the compound is

(CH3)2-CH-Br

IR spectral data:

Peak at ~2900 cm-1 indicates the presence of alkane C-H ( sp3 C-H)

Peak near ~800 cm-1 indicates the presence of C-Br bond

1H -NMR:

(CH3)2-CH-Br

a b

Protons a: 6 H doublet at 1.2 ppm

Protons b: 1 H multiplet at 3.9 ppm

13 C -NMR:

Carbon a: peak at 28

Carbon b : peat at 45