Question

Problem 19 IR Spectrum 2000 1600 1200 800 100 Mass Spectrum No significant UV absorption above 220 nm 40 M+ 147149 40 80 120 160 200 240 280 m/e 3C NMR Spectrum (100 MHE, CDC, solution) proton decoupled 200 160 120 40 0 8(ppm) H NMR Spectrum 22 ppm 8 5 2 δ(ppm)

Answer 1

Given the chemical formula is C4H6NBr (C = 4, H = 6, X = 1, N = 1)

Degree of unsaturation, DU = C/2 - H/2 - X/2 + N/2 + 1 = 4 - 6/2 - 1/2 + 1/2 + 1 = 2

Analysis of IR spectroscopy:

The frequency at 2249 cm^-1 indicates the presence of a C-C or C-N triple bond.

Since DU is 2, the is no other double/triple bond/ring except the C-C or C-N triple bond.

Analysis from C13:

The presence of chemical shift around120 ppm conforms the presence of CN(nitrile) group.

Analysis form 1H NMR spectrum:

The prence of two triplet and one multiplet(pentet) conforms that the carbon chain is linear with the -Br attached at the terminal.

Hence the structure of the compound is Br-CH2-CH2-CH2-CN