proton nmr spectroscopy
DESCRIPTION
Proton NMR Spectroscopy. The NMR Phenomenon. Most nuclei possess an intrinsic angular momentum , P . Any spinning charged particle generates a magnetic field. P = [I(I+1)] 1/2 h/2 p where I = spin quantum # I = 0, 1/2, 1, 3/2, 2, …. Which nuclei have a “spin”?. - PowerPoint PPT PresentationTRANSCRIPT
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Proton NMR Spectroscopy
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The NMR Phenomenon
• Most nuclei possess an intrinsic angular momentum, P.
• Any spinning charged particle generates a magnetic field.
P = [I(I+1)]1/2 h/2where spin quantum #
I = 0, 1/2, 1, 3/2, 2, …
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Which nuclei have a “spin”?• If mass # and atomic # are both even, I = 0 and the
nucleus has no spin. e.g. Carbon-12, Oxygen-16
• For each nucleus with a spin, the # of allowed spin states can be quantized:
• For a nucleus with I, there are 2I + 1 allowed spin states.
1H, 13C, 19F, 31P all have I = 1/2E = h/2)Bo
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Spin states split in the presence of B0
no field applied field
E
+1/2 parallel
-1/2 antiparallel
Bo
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When a nucleus aligned with a magnetic field, B0, absorbs radiation frequency (Rf), it can change spin orientation to a higher energy spin state. By relaxing back to the parallel (+1/2) spin state, the nucleus is said to be in resonance. Hence,
NMR
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Presence of Magnetic Field
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NMR instruments typically have a constant Rf and a variable B0.
A proton should absorb Rf of 60 MHz in a field of 14,093 Gauss (1.4093 T).
Each unique probe nucleus (1H perhaps) will come into resonance at a slightly different - and a very small percentage of - the Rf.
All protons come into resonance between 0 and 12/1,000,000 (0 – 12 ppm) of the B0.
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• Nuclei aligned with the magnetic field are lower in energy than those aligned against the field
• The nuclei aligned with the magnetic field can be flipped to align against it if the right amount of energy is added (DE)
• The amount of energy required depends on the strength of the external magnetic field
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Energy Difference (E) Between Two Different
Spin States of a Nucleus With I=1/2
+1/2
-1/2
E 400 MHz300 MHz200 MHz100 MHz
23,500 47,000 70,500 104,000
parallel
antiparallel
inc. magnetic field strength, Gauss
B0
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What Does an NMR Spectrum Tell You?
• # of chemically unique H’s in the molecule # of signals
• The types of H’s that are present e.g. aromatic, vinyl, aldehyde …
chemical shift• The number of each chemically unique H
integration• The H’s proximity to eachother
spin-spin splitting
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Chemical EquivalenceHow many signals in 1H NMR spectrum?
O OO
O
CH3
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Number of Equivalent Protons
O
O
OO
6 4
CH3
1 3 5 2 4
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Homotopic H’s– Homotopic Hydrogens
• Hydrogens are chemically equivalent or homotopic if replacing each one in turn by the same group would lead to an identical compound
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Enantiotopic H’s
• If replacement of each of two hydrogens by some group leads to enantiomers, those hydrogens are enantiotopic
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Diastereotopic H’s• If replacement of each of two hydrogens
by some group leads to diastereomers, the hydrogens are diastereotopic
– Diastereotopic hydrogens have different chemical shifts and will give different signals
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Vinyl Protons
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Typical 1H NMR Scale is 0-10 ppm
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The Scale
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Tetramethylsilane (TMS)
CH3SiCH3
CH3
CH3
TMS
Arbitrarily assigned a chemical shift
of 0.00
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Chemical Shift Ranges, ppm
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Diamagnetic AnisotropyShielding and Deshielding
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Deshielding in Alkenes
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Shielding in Alkynes
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Integration
CH3
CH3
CH3
Br
H H
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Methyl t-butyl ether (MTBE)
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Toluene at Higher Field
Splitting patterns in aromatic groups can be confusing
A monosubstituted aromatic ring can appear as an apparent singlet or a complex pattern of
peaks
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Integral Trace
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Spin-Spin Splitting
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The Doublet in 1H NMR
C C
HH
B0
a b
Ha splits into a 1:1 doublet peak
Hb is parallel or anti-parallel to B0
Ha is coupled to Hb
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Hb in 1,1,2-Tribromoethane
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The Triplet in 1H NMR
C C
H
H
H
B0
a b
Ha splits into a 1:2:1 triplet peak
Hb can both be parallel, anti-parallel
Ha is coupled to Hb and Hb
b
or one parallel and one anti-parallel
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Ha in 1,1,2-Tribromoethane
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1,1,2-Tribromoethane
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The Quartet in 1HMR
B0
C
H
C
H
H
H
proton splits into n+1
n = # adjacent H'squartet 1:3:3:1
shieldeddeshielded
Chemical Shift
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1,1-Dichloroethane
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Ethyl benzene
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CH3CH2OCH3
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Equivalent Protons do not Couple
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Pascal’s Triangle
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Methyl Isopropyl Ketone
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1-Nitropropane
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Differentiate using 1H NMR
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Coupling Constants (J values)
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Para Nitrotoluene
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Bromoethane
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para-Methoxypropiophenone
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Styrene
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Ha splitting in Styrene“Tree” Diagram
C C
H
HHa
b
c
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In the system below, Hb is split by two different sets of hydrogens : Ha and Hc
– Theortically Hb could be split into a triplet of quartets (12 peaks) but this complexity is rarely seen in aliphatic systems
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0.750.50
60 MHz
100 MHz
300 MHz
300 240 180 120 60 0 Hz
300 240 180 120 60 0 Hz
300 240 180 120 60 0 Hz
Why go to a higher field strength?
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0.75 (t, 2H, J=10)0.50 (t, 2H, J=10)
60 MHz
100 MHz
300 MHz
300 240 180 120 60 0 Hz
300 240 180 120 60 0 Hz
300 240 180 120 60 0 Hz
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C-13 NMR Spectroscopy
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C-13 chemical shifts
One signal for each chemically unique carbon
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Methyl Propanoate C-13 proton decoupled
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Ethyl Acrylate; C-13 prediction
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Ethyl Acrylate; C-13 NMR
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Coupling in C-13 NMR
methine group
B0
the doublet in C-13 NMR
C splits into a 1:1 doublet peak
H is parallel or anti-parallel to B0
C is coupled to H
C
H
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methylene group
C
H
HB0 a
the triplet in C-13 NMR
C splits into a 1:2:1 triplet peak
Ha & Hb can both be parallel, anti-parallel
C is coupled to Ha and Hb
b
or one parallel and one anti-parallel
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B0
C
H
H
H
the quartet in C-13 NMR
carbon splits into n+1
n = # attached H'squartet 1:3:3:1
shieldeddeshielded
methyl group
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Butanone - Coupled and Decoupled
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1,2,2-Trichloropropane1H and 13C NMR Spectra
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Coupled C-13 NMR Spectrum
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HC CCH2CH2CH2CH3
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CH2CO2H
SPA
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CH3CH2OCCH2CCH3
O O
coupled spectrum
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