Question

Propose a structure for the compound that has the following spectra: NMR: δ 1.28 (3H, t, J = 7 Hz); δ 3.91 (2H, q, J = 7 Hz); δ 5.0 (1H, d, J = 4 Hz); δ 6.49 (1H, d, J = 4 Hz) ppm IR: 3100, 1644 (strong), 1104, 1166, 694 cm -1 (strong) no IR absorptions in the range 700-1100 cm -1 or above 3100 cm-1 Mass spectrum: m/z = 152, 150 (equal intensity; double molecular ion)

Answer 1

1) Mass spectral data:

i) Equal intensity molecular ion peaks at 150 and 152 indicate presence of 1 Bromine atom (Br exists as Br-81 and Br-79 inalmost equal natural abundance.

ii) With m/e =150 as M.W.=150, by rule of 13 lets derive formula. Divide M.W. 150 by 13 take quotient as C number and quotient+remainder as H number. Hence we get C11H18. Now to add Br (79) as it is present reduce formula by C6H7 (79) hence we get C5H11Br.

iii) As no IR absorption above 3100 cm^-1 and in the region 700-1100 cm^-1 , possibility of presence of O-H is ruled out. But still PMR indicate presence of O (16), so CH4 (16) is subtracted from molecular formula. Hence M.F. becomes C4H7OBr.

iv) The corresponding hydrocarbon is C4H8 (1 H compensated for 1Br). The saturated hydrocarbon is C4H10. This indicates presence of 1 unsaturation site.

2) IR data analysis:

i) 1644 strong band indicate presence of C=C .

ii) 694 absorption indicate aliphatic C-Br grouping.

3) PMR data.

i) δ 1.28 (3H, t, J = 7 Hz) indicate –CH3 grouping.

ii) δ 3.91 (2H, q, J = 7 Hz) indicate –CH2- group bonded to strong electron withdrawer like O . Hence O-CH2- group present. By coupling constant value its clear that it couple with above protons hence the fragment is –O-CH2-CH3.

iii) δ 5.0 (1H, d, J = 4 Hz) and δ 6.49 (1H, d, J = 4 Hz) are protons on olefinic C.These protons are mutually coupled and a coupling constant J=4 indicates cis arrangement.

iv) joining all fragments we get structure as, (Z)-1-bromo-2-ethoxyethene.

v) In digram we can see H_A as downfield over H_B. This is because of +R effect of O-CH2- grouping. As we can see resonance structure whrere H_B gets –ve charge. Still H_B is much downfield that’s becauses of –I of Br. Downfieldness of H_A is because of –I of –O-CH2-