Organic Chemistry

δ (ppm)

9876543210

2.52.6 2.7 2.8 2.9 3 3.5 4 4.5 5 5.5 6 7 8 9 10 11 12 13 14 15 16

4000380036003400320030002800260024002200 2000 1800 1600 1400 1200 1000 800 600

Wavelength (μm)

Wavenumber (cm−^1 )

% Transmittance

frequency

Figure 14.22
IR and 1 HNMR spectra for this problem-solving strategy.

Section 14.12 Coupling Constants 553

2. The position of a signal indicates the kind of proton(s) responsible for the signal
   (methyl, methylene, methine, allylic, vinylic, aromatic, etc.) and the kinds of
   neighboring substituents.


3. The integration of the signal tells the relative number of protons responsible for
   the signal.


4. The multiplicity of the signal tells the number of protons (N) bonded to
   adjacent carbons.


5. The coupling constants identify coupled protons.

PROBLEM-SOLVING STRATEGY

Identify the compound with molecular formula that gives the IR and

spectra in Figure 14.22.

C 9 H 10 O 1 H NMR

1 N+ 12

The best way to approach this kind of problem is to identify whatever structural features

you can from the molecular formula and IR spectrum and then use the information from

the spectrum to expand on that knowledge. From the molecular formula and IR

spectrum, we learn that the compound is a ketone: It has a carbonyl group at

only one oxygen, and no absorption bands at and that would indicate

an aldehyde. That the carbonyl group absorption is at a lower frequency than normal sug-

gests that it has partial single-bond character as a result of electron delocalization—

indicating that it is attached to an carbon. The compound contains a benzene ring

( and ), and it has hydrogens bonded to carbons

163000 cm-^12 .In the NMR spectrum, the triplet at '1.2 ppmand the quartet at '3.0 ppm

7 3000 cm-^1 , ' 1600 cm-^1 , 1440 cm-^1 sp^3

sp^2

' 2820 ' 2720 cm-^1

' 1680 cm-^1 ,

1 H NMR