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Cmr
1. 13C NMR Spectroscopy
Gives structural and functional details of a compound
Very useful for determining the structure of unknown
compounds Shows the different magnetic environments of
carbon atoms in a compound
2. ABUNDANCE
13C is difficult to record because of
1. The most abundant isotope of carbon that is
12C (99.1%) is not detected by nmr because it has
an even no of protons and neutrons 12C is nmr
inactive.
2. Magnetic resonance of 13C is much weaker.
Moreover, gyromagnectic ratio of 13C being only
one fourth that of proton, so the resonance
frequency of 13C is one fourth that of proton nmr.
3. Advantages of 13C- NMR over 1H- NMR
1.13C- provides information about the backboneof
molecules rather than the periphery.
2.The chemical shifts range for 13C- NMR for most
organic compounds is 200 ppm compared to 10 –15
ppm for H, hence there is less overlap of peaks for
13C- NMR.
3. Homonuclear spin-spin coupling between carbon
atoms is not observed because the natural abundance
of 13C is too low for two 13C to be next to one
another.
Heteronuclear spin coupling between 13C and
12C does not occur because the spin quantum
number of 12C is zero.
4. There are a number of excellent methods for
decoupling the interaction between 13 C and 1H.
5. 13C NMR spectroscopy
13C NMR spectroscopy measures the difference in energy between the aligned
spin state and unaligned spin state
The difference in energy is measured in ppm (parts per million).
The lower the energy gap the easier it is to flip from one state to another
6. Chemical Environments
13C NMR spectroscopy shows peaks for each of the different chemical environments of the carbon atom
in a molecule.
The environment of a carbon atom can be determined by looking at the sequence of bonds the carbon
atom has to other atoms. If two carbon atoms have the same bond sequence they will have the same
environment.
1 1
2
7. Chemical Environments
The number of chemical environment a molecule has is the
number of peaks in the 13C NMR spectrum.
For example: IR has told you that you have an alcohol
functional group, but you don’t know if the alcohol is
propan-1-ol or propan-2-ol. The 13C NMR will allow you to
tell the difference.
Number of carbon chemical environments in propan-
1-ol: Number of carbon chemical environments in
propan-2-ol:
pg 18 and
19
8.
9. Do now:
How many carbon environments will the following
compounds
have?
3-methylbutanal
2,4-dichloropentane pentan-2-one
NH2
H3C CH CH2 CH3 C C C C C
H
H CH3 H
H
H
H
H
H
H
10. Predicting 13C Spectra
CH3 C H 3
C C
C C
CH3
C
plane of symmetry
4 lines
O
CH3
CH3
O
CH3
c
CH3
CH3
C
C
O
CH3 5 lines
CH3
CH3
CH3
C
C
H
5 lines
11. upfield
20
40
60
220 200 180 160 140 120 100 80 0
CH3
CH2
CH
C X (halogen)
C N
C O
C C
C N
C C
C O
13C Chemical shift
(
)
TMS
Aromatic C
downfield 13C chemical shift
(δ)
13C NMR spectroscopy
The position of the peak in a 13C NMR spectrum is determinedby the shielding
or deshielding effects of other atoms around it.
The less electrons (ie the more electronegative atoms around the
carbon) the higher magnetic field the nucleus is exposed to, hence
it is harder to flip and the ppm value is
higher.
12.
13. 13C NMR spectroscopy
Shielding and deshielding
Shielding and deshielding effects explain the chemical shifts that we observe
in the NMR. The closer the 13C nucleus is to an electronegative atom the more
deshielded it will be.
Electronegative atoms (like O, Cl, Br etc) or multiple bonds (C=C, C≡C) partially
remove electrons from the 13C nucleus. This leaves it exposed to more of a
magnetic field and it takes more energy to flip, resulting in a higher chemical shift.
It is not possible to predict the exact position of 13C peaks as they are
influenced by the environment around them but we can predict the approximate
location.
14. 13C NMR spectroscopy
Using this information could 13C NMR distinguish
between the following functional groups?
Acid chlorides and
aldehydes Ketones and
amides
15. 13C NMR spectroscopy
This information is only a guide. It is extremely difficult to
predict exactly the chemical shift of carbon atoms in a
compound.
Carbon atoms in the same environment will always have the
same
chemical shift.
16.
17.
18. SPIN –SPIN SPLITTING OF 13C
SIGNALS
Splitting take place acc. to 2nI+1 rule
Where n= no. of nuclei
I=spin quantum number
CH3 = 3+1=4 quartet
CH2 = 2+1=3 triplet
CH = 1+1=2 doublet
C = 0+1=1 singlet
CDCl3 gives three peaks because its I=1 so acc. to 2nI+1
2 1 1+1=3 so it gives 1:1:1 peaks
Solvents used are CDCl3, DMSO, d6acetone, d6 benzene
19. 13C NMR spectroscopy
Comparing hex-1-ene to hex-3-ene.
How many peaks would you expect to see in the 13C NMR
spectra of the above compounds? Why?
hex-1-
ene
hex-3-
ene
20. 13C NMR spectroscopy
How many different carbon environments would ethyl
ethanoate
have?
Where would you expect the peaks to be in the spectrum?
pg 28 -
34
21. Match the following six compounds to the following
six
13C NMR spectra.
propanoic acid propan-1-ol
propanone propan-2-ol
propanal methyl ethanoate
71. c) DEPT data
DEPT = distortionless enhancement by polarization
Distinguishes:
CH3 - methyl groups
-CH2- methylene groups
I
-CH- methine groups
I
-C-
I
4o carbons ( not detected by DEPT)
94. Examples (chemical shifts in ppm from TMS)
OH
O
sp3 hybridized carbon is more shielded than sp2
61
202
95. Examples (chemical shifts in ppm from TMS)
OH
23
an electronegative atom deshields the
carbon to which it is attached
61
96. Table 13.3 (p 513)
Type of carbon Chemical shift (),
ppm
Type of carbon Chemical shift (),
ppm
RCH3 0-35
CR2
R2C
65-90
CR
RC
R2CH2 15-40
R3CH 25-50
R4C 30-40
100-150
110-175
97. Table 13.3 (p 513)
Type of carbon Chemical shift (),
ppm
RCH2Br 20-40
RCH2Cl 25-50
35-50
RCH2NH2
50-65
RCH2OH
RCH2OR 50-65
98. Table 13.3 (p 513)
Type of carbon Chemical shift (),
ppm
Type of carbon Chemical shift (),
ppm
RCH2Br 20-40
RCH2Cl 25-50
35-50
RCH2NH2
50-65
RCH2OH
RCH2OR 50-65
RCOR
O
160-185
RCR
O
190-220
99. CH3
OH
Figure 13.21 (page 514)
Chemical shift (, ppm)
0
20
40
60
80
100
120
140
160
180
200
7 carbons give
7 signals, but
intensities are not
equal
100. Chemical shift (, ppm)
0
20
40
60
80
100
120
140
160
180
200
Figure 13.23 (a) (page 516)
O
C
C
CH
CH CH
CH2
CH2
CH2
CH3
CCH2CH2CH2CH3
O