Organic Chemistry

Section 14.18 13 CNMR Spectroscopy 565

PROBLEM 36

Answer the following questions for each of the compounds:

a. How many signals are in the NMR spectrum?

b. Which signal is at the lowest frequency?

1. 
   
   
   
   5. 
      
      
      
      9. 
         
      
      
      
      
   
   
   
   


2. 
   
   
   
   6. 
      
      
      
      10. 
          
      
      
      
      
   
   
   
   


3. 
   
   
   
   7. 
      
   
   
   
   


4. 
   
   
   
   8. CH 3 CCH 2 CH 2 CCH 3
   
   
   
   

OO

CH 2 “CHBr

CH 3 CH 2 OCH 3 Cl

CH 3 CHCH 3

Br

CH 3 CHCH

CH 3

O

(CH 3 ) 2 C“CH 2

CH 3 COCH 3

CH 3

CH 3

CH 3 CH 2 COCH 3

O

CH 3 CH 2 CH 2 Br

13 C

Figure 14.34
Proton-decoupled 13 C NMRspectrum of 2,2-dimethylbutane.

The signals in a NMR spectrum can be split by nearby hydrogens. However,
this splitting is not usually observed because the spectra are recorded using spin-de-
coupling, which obliterates the carbon–proton interactions. Thus, all the signals are
singlets in an ordinary NMR spectrum (Figure 14.32).
If the spectrometer is run in a proton-coupledmode, the signals show spin–spin split-
ting. The splitting is not caused by adjacent carbons, but by the hydrogens bonded to the
carbon that produces the signal. The multiplicity of the signal is determined by the
rule. The proton-coupled NMR spectrumof 2-butanol is shown in Figure 14.33.
(The triplet at 78 ppm is produced by the solvent, ) The signals for the methyl car-
bons are each split into a quartet because each methyl carbon is bonded to three hydrogens
The signal for the methylene carbon is split into a triplet
and the signal for the carbon bonded to the OH group is split into a doublet
The NMR spectrum of 2,2-dimethylbutane is shown in Figure 14.34. The three
methyl groups at one end of the molecule are equivalent, so they all appear at the same
chemical shift. The intensity of a signal is somewhat related to the number of carbons
giving rise to the signal; consequently, the signal for these three methyl groups is the
most intense signal in the spectrum. The tiny signal is for the quaternary carbon;
carbons that are not attached to hydrogens give very small signals. Becasue they are
attached to a quaternary carbon, their signal appears at a higher frequency than that of
the other methyl group (Table 14.4).

13 C

11 + 1 = 22.

13 + 1 = 42. 12 + 1 = 32 ,

CDCl 3.

13 C

N+ 1

13 C

13 C

Missing Art