Question

Can someone label and explain the peaks for both NMR and IR for an unknown solid. Theres no moleuclar or carbon splitting given.

%Transmillance 4000 3500 3000 3053 B4 2500 Wavenumbers (cm-1) 2000 IDITOS 1663.82 1612 23 1500 1442.10 1265 25 1155.45 1000 749.07 699.46 5001

Answer 1

From the combined data of IR and NMR spectra it seems to be an aromatic compound, it is not possible to exactly know the compound structure from this two graphs themselves.

In IR spectra the peak at 3053.84 cm-1 seems to be medium size to small size this is C-H stretching peak of aromatic alkene which matches with the prediction that it is an aromatic compound.

The peak at 1663.82 cm-1 seems to be C=O stretching of carbonyl group in conjugation with aromatic benzene this corresponds to aromatic compound being present.

The peak at 1612 cm-1 indicate it is again C=C stretching of conjugated alkene, this also corresponds to aromatic compound being present.

The strong peak at 749 cm-1 indicate that it is C-H bending peak corresponding to 1,2-disubstituted alkene.

All these peaks from IR confirm you have a conjugated substituted alkene (an aromatic compound)

Coming to NMR data:

The peak at 10.37 ppm is strongly indicative of proton peak of an aldehyde -CHO

the peaks ranging from 7.95 to 7.5 ppm indicate they are aromatic C-H protons with over all integration of 2.

the peak at 7.25 ppm is CDCl3 reference peak.

the peaks ranging from 7.02 to 6.92 ppm indicate they are aromatic C-H protons with overall integration corresponding to 2 protons.

The peak at 3.52 ppm is singlet with integration of 1.4, this is hard to identify the functional group it belongs to without further info.

So overall the compound appears to be an aromatic aldehyde.