Carbon-13 NMR
Only 1.1% of all carbon atoms are carbon-13 isotope, the only isotope of carbon
that displays the ability to create the magnetic field when its nucleus spins.
Taking a carbon NMR of 1-pentanol using the same number of scans as a proton
NMR:
Which are real and which peaks are “noise”?
Now, taken with multiple extra scans (can sometimes take HOURS):
Out of 100 carbon atoms only around 1 is a carbon-13 isotope. So – how is it
possible that we can study carbon-13 NMR? Consider a solution of 1-butanol
molecules:
1
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
CH3-CH2-CH2-CH2-OH
When averaged over an entire solution, probability and statistics will show that
every carbon position will have a carbon-13 in some molecule. This generally
requires more scans (more time) and a larger sample (more molecules!) than a
proton NMR, in order to get enough data acquired to identify true peaks away from
“noise” on the spectrum.
Calibration Peak at 0.0000 ppm – all four carbons of tetramethylsilane are
equivalent and given a value of 0.0000 ppm.
• Can also calibrate on the triple set of peaks caused by CHCl3, the center
peak of which appears at 77.0000 ppm.
Symmetry:
Every signal on the carbon NMR represents a different carbon atom in the
molecule.
Symmetrical carbons produce twice the data points during acquisition of a
spectrum so those peaks generally appear larger as a result.
How many different types of carbons?
CH3CH2CH2CH3
2
25
20
15
10
PPM
5
0
5
0
(CH3)3CH
25
20
15
PPM
10
(CH3)2CHCH2Br
45
40
35
30
25
PPM
20
15
10
5
0
Aromatic Compounds – Symmetry:
Monosubstituted:
Br
3
140
120
100
80
Disubstituted:
Para:
PPM
60
40
Br
Br
Br
Cl
20
0
1-hydroxy-4-methoxybenzene:
160
140
120
100
80
PPM
60
Meta:
Br
Br
40
Br
20
0
Cl
1-bromo-3-methylbenzene:
4
140
120
100
80
PPM
60
40
20
Ortho:
0
Br
Cl
Br
Br
1-bromo-2-methylbenzene
140
120
100
80
PPM
60
40
20
0
Same sort of analysis with trisubstituted aromatic rings:
1,2,3-substituted:
Br
Cl
Br
Br
Br
Cl
Which one is it?
5
140
120
100
80
PPM
60
1,2,4-substituted:
40
20
0
Cl
CH3
OH
160
140
120
100
80
PPM
60
40
20
0
1,3,5-substituted:
Br
Br
Br
Br
Br
Cl
Br
Cl
CH3
Which one is it?
6
140
120
100
80
PPM
60
40
20
0
and this time?
160
140
120
100
PPM
80
60
40
20
0
Chemical Shift – X-axis – 0-220 ppm
0-50 ppm
alkyl carbons
also benzylic C’s
45-90 ppm
C-O (N, X)
110-160 ppm
aromatic carbons (also alkene)
115-130 ppm
nitrile carbons
170-220 ppm
carbonyl carbons
Simple Problem: Consider each of the following and analyze for what you EXPECT
to find on the Carbon NMR?
7
Now: Match the Carbon NMRS
a.
140
120
100
80
PPM
60
40
20
0
b.
8
Integration:
Peak areas are not typically measured on carbon-13 NMR spectra so integration is
not performed. Instead, one can view the general peaks heights to determine
where on the spectrum signals containing symmetrical carbons are found.
2-Butanone has four carbons, all of which are different. Peak heights make no
difference.
The are two factors that affect the height of the peak:
1. The number of carbons producing the signal: the more carbons (symmetry!), the
stronger the signal, the taller the peak.
2. The presence of hydrogen atoms attached to carbon increase the strength of
the signal on the spectrum. Those carbons without a hydrogen attached tend to
have smaller, weaker signals.
9
For your Qual lab, you must:
1. Identify the number of carbons on your 13C spectrum and determine if you have
any symmetry
2. Identify what each type of carbon is, based on the chemical shift value.
Now: Puzzle Solving Using 13C NMR
DEPT-13C NMR Spectroscopy:
-“Distortionless Enhancement by Polarization Transfer”
-Distinguishes between CH3, CH2, CH and quarternary C’s
Ex: 6-methyl-5-penten-2-ol: Symmetry?
OH
A DEPT experiment is done in three stages:
Stage 1: run ordinary spectrum to find chemical shifts of all peaks
Peaks: 18, 23, 24, 25, 40, 64, 124, 132 ppm
Stage 2: Run a DEPT-90 spectrum – only CH peaks appear on the spectrum -
10
Peaks: 64, 124 ppm
Stage 3: Run a DEPT-135 spectrum – CH3 and CH peaks appear as positive peaks
(UP from baseline) and CH2 peaks appear as negative peaks (DOWN from baseline).
Positive: 18, 23, 25, 64, 124 ppm
Negative: 24, 40 ppm
Those not appearing on either the second or third spectrum are quaternary
carbons. Peak? 132 ppm.
Examples:
C6H13Br
50
40
30
PPM
20
10
0
DEPT-90: no peaks
DEPT-135: positive peak at 29 ppm; negative peaks at 28, 49 ppm
C6H12O
11
220
200
180
160
140
120
100
PPM
80
60
40
20
0
DEPT-90: 24 ppm
DEPT-135: positive peak at 23, 24, 31 ppm; negative peaks at 55 ppm
MORE Puzzle Solving:
Using Splitting Patterns:
In normal 13C NMR mode, splitting cannot occur between carbons because of the
low natural abundance of carbon-13 (not likely to find on adjacent positions).
In spin-coupled mode, the magnetic fields of the protons are allowed to interact
with the carbon atom magnetic fields and splitting can occur between the carbon13 atoms and the protons directly attached to those carbon-13 atoms.
12
N+1 Rule for Carbon-13 NMR:
The signal for a carbon (or equivalent carbons) with N hydrogen atoms attached
will be split into N+1 peaks in spin-coupled mode.
N
N+1
CH3
3
4 (quartet)
CH2
2
3 (triplet)
CH
1
2 (doublet)
C
0
1 (singlet)
See Problems online.
13