Question

Deduce the identity of the following compound from the spectral data given.

C₉H₁₀O₂: ¹³C NMR, δ 18.06 (quartet), 45.40 (doublet), 127.32 (doublet), 127.55 (doublet), 128.61 (doublet), 139.70 (singlet), 180.98 (singlet); IR, broad 3500-2800, 1708 cm^-1

Answer 1

Molecular formula : C₉H₁₀O₂

Double bond equivalent (DBE) or IHD = 9+1-(10/2) = 10 - 5 = 5,

Which means 5 double bonds are present.

In 13C NMR: It should have 9 peaks, for the 9 carbon atoms present in the molecule.

Your data say

13C NMR, δ 18.06 (quartet), 45.40 (doublet), 127.32 (doublet), 127.55 (doublet), 128.61 (doublet), 139.70 (singlet), 180.98 (singlet)

180.98 - carbon atom of a carbonyl group.

Seven carbon. Two carbon data missing. In 13C NMR, there is no significance of quartet, doublet, singlet, etc.

IR Data:

3500-2800

3500–3200 (s,b) O–H stretch, H–bonded alcohols, phenols
3300–2500 (m) O–H stretch carboxylic acids
3330–3270 (n, s) –C≡C–H: C–H stretch alkynes (terminal)
3100–3000 (s) C–H stretch aromatics

2830–2695 (m) H–C=O: C–H stretch aldehydes

1708 cm^-1 - Indicates the carbonyl group is present.

Hence with these data, molecular structure can't be determine. Please give a 1H NMR spectra and a good 13C NMR spectra along with mass spectra.