Integrated Problem 17.52 Get help answering Molecular Drawing questions. Draw the structure of a compound with the molecular formula C9H10O2 that exhibits the following spectral data. (a) IR: 3005 cm-1, 1676 cm-1, 1603 cm-1 (b) 1H NMR: 2.6 ppm (singlet, I

The answer is attached and other details about the answer is also attached.

Explanation:

The molecular formula (C9H10O2) indicates five

degrees of unsaturation (see Section 14.16), which is

strongly suggestive of an aromatic ring, as well as one

additional double bond or ring. The signal just above

3000 cm-1 in the IR spectrum confirms the aromatic ring,

as does the signal just above 1600 cm-1. The 1

H NMR

spectrum exhibits two doublets between 6.9 and 7.9

ppm, each with an integration of 2. This is the

characteristic pattern of a disubstituted aromatic ring, in

which the two substituents are different from each other:

The singlet at 3.9 ppm (with an integration of 3)

represents a methyl group. The chemical shift is

downfield from the expected benchmark value of 0.9

ppm for a methyl group, indicating that it is likely next to

an oxygen atom:

The singlet at 2.6 ppm (with an integration of 3)

represents an isolated methyl group. The chemical shift

of this signal suggests that the methyl group is

neighboring a carbonyl group:

The carbonyl group accounts for one degree of

unsaturation, and together with the aromatic ring, this

would account for all five degrees of unsaturation. The

presence of a carbonyl group is also confirmed by the

signal at 196.6 ppm in the 13C NMR spectrum.

We have uncovered three pieces, which can only be

connected in one way, as shown:

This structure is consistent with the 13C NMR data: four

signals for the sp2 hybridized carbon atoms of the

aromatic ring, and two signals for the sp3 hybridized

carbon atoms (one of which is above 50 ppm because it

is next to an oxygen atom).

Also notice that the carbonyl group is conjugated to the

aromatic ring, which explains why the signal for the

C=O bond in the IR spectrum appears at 1676 cm-1,

rather than 1720 cm-1.