Answer:
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.
