Question

56. A CuH12O2 molecule shows IR absorptions at 3040, 2850, 2750, 1705, 1600, 1587, and 1230 cm! The ratio of carbon-bonded hydrogens in it's 'H-NMR spectrum is, 2:2:2:2:1:1:1:1, with corresponding signal multiplicities of d, d, d, d, ddt, dd, dd & t. The corresponding chemical shifts are 8 = 7.1, 6.9, 4.7, 3.7, 6.1, 5.4, 5.3 and 9.7. The proton decoupled "3C-NMR spectrum has 9 signals. Predict the structure & match the multiplicity and chemical shift of each nonequivalent set of hydrogens. (You must draw bond-line structure ONLY and match the multiplicities and chemical shifts on the structure for full credit).

Answer 1

Molecular formula C11H1202 Double bond equivalent (DBE) = C+1-H2=11 + 1 - 12/2 = 12 -6= 6 IR shows a peak at 1705 cm-1 which may be due to C=0 streching of an aldehyde which is also supported by a peak at 9.7 ppm in 1 H NMR 1 H NMR shows two sets doublet peak in the aromatic region at 7.1 & 6.9 ppm which indicates it is an para disubstituted aromatic aldehdy de Close analysis of 1 H NMR data suggests below structure C11H1202 1H NMR Estimation 5.3, 1H, dd 6.1, ,H1H, ddt 5.4, TH, dd H TO 2H, d 7.1, - 2H, d - H 2H. d 20 HC 3.7, 2H, d 3.1. H 9.7, 1H,