Chapter 14: Nuclear Magnetic Resonance Spectroscopy I. Introduction of of NM NMR Sp Spectroscopy (1 (14.1) II. 1H NMR: Number of Signals (14.2) III. 1H NMR: Position of Signals (14.3) IV. IV. Chemic emical al Shif Shiftt of of Pro Proto ton ns on on sp sp2 and sp Hybridized Carbons (14.4) V. 1H NMR: Intensity of Signals (14.5) VI. 1H NMR: Spin-Spin Splitting (14.6) VII. VII. More More Comp Complex lex Exam Example pless of Split Splittin ting g (14.7) VIII.Spin-Spin Splitting in Alkenes (14.8) IX. Other Facts About 1H NMR Spectroscopy (14.9) X. Using 1 NMR to Identify an Unknown (14.10) XI. 13C NMR Spectroscopy (14.11) 1
I. Introduction to NMR Spectroscopy 1H, 13C, 31P, 19F, 15N
Some nuclei have nuclear spin.
A. Basis of NMR Spectroscopy
A spinning proton produces a magnetic field.
Bo Absence of external magnetic field: spins oriented randomly
Presence of external magnetic field B o : more spins align with Bo 2
I. Introduction to NMR Spectroscopy 1H, 13C, 31P, 19F, 15N
Some nuclei have nuclear spin.
A. Basis of NMR Spectroscopy
A spinning proton produces a magnetic field.
Bo Absence of external magnetic field: spins oriented randomly
Presence of external magnetic field B o : more spins align with Bo 2
RF Radiation
∆E = h ν
h ν
Bo
• Two Two nucl nuclea earr spin spin sta state tess sepa separa rate ted d by ∆E • Lower energy state – nuclear spin magnetic magnetic moment aligned with Bo • Higher energy state – nuclear spin magnetic magnetic moment aligned against Bo • Applicat Application ion of of RF radiation radiation pulse pulse causes causes spin to flip from lower to higher energy state. ν are hertz • Units of Bo are telsa telsa (T); (T); unit unitss of of ν (Hz = s-1).
• ν is directly proportional to Bo: ν = γBo 3
Fig. 14.1
4
How Fourier transform NMR works: z
z
Apply RF pulse
y
x
along +x
y
x
Spins precess about +z. Net magnetization ( blue arrow) lies along +z.
After RF pulse, net magnetization relaxes into +z direction.
Signal is detected by looking along +y with an RF receiver: Fourier transform
time domain
frequency domain
5
B. 1H NMR Spectroscopy 1H
NMR spectrum of CH3OC(CH3)3 (CH3)3C –
upfield downfield
y t i s n e t n I
δ
CH3O – TMS
scale
10
8
6
4
2
0
Chemical Shift (ppm)
• TMS is tetramethylsilane, (CH3)4Si, a chemical shift reference. • t -Butyl methyl ether has sharp absorptions at 1.2 ppm and 3.2 ppm. •
δ (ppm)
ν (Hz) downfield from TMS =
ν of NMR spectrometer (MHz) 6
II. Number of Signals in 1H NMR Spectra • Protons in different chemical environments give different NMR signals. • Equivalents protons have identical NMR signals. Examples:
Figure 14.2: Number of NMR Signals of Representative Organic Compounds
7
Substitution test for chemical shift nonequivalence: substitute each H with another atom (e.g. Cl). Example: CH3
methylcyclobutane
Cl CH2Cl CH3 (chloromethyl)cyclobutane
1-chloro-1-methylcyclobutane
CH3
CH3
Cl
Cl
cis and trans-1-chloro-2-methylcyclobutanes
Cl
CH3
Cl
CH3
cis and trans-1-chloro-3-methylcyclobutanes 8
A. Determining Equivalent Protons in Alkenes and Cycloalkanes H
C N
Ha
H
C N
Cl
1,1-dicyanoethene H’s are equivalent 1 NMR signal H
H
H
Hb
H
Hc
cyclopropane All H’s equivalent 1 NMR signal
C N
1-chloro-2-cyanoethene Ha and Hb nonequivalent 2 NMR signals
H
H
Hb
Hb
C N
Ha Hc
cyanocyclopropane 3 types of H ’s 3 NMR signals
How many NMR signals does methoxyethene give? 9
B. Enantiotopic and Diastereotopic Protons
Use substitution test to determine whether CH2 protons are equivalent. 1. Enantiotopic Protons F H3C
F
Substitute each
H H
fluoroethane
H with Cl
F
+ H3C
H Cl
S -1-chloro-1fluoroethane
H3C
Cl H
R-1-chloro-1fluoroethane
enantiomers
CH2 protons of fluoroethane are enantiotopic. Enantiotopic protons are equivalent and give the same NMR signal.
10
2. Diastereotopic Protons
H
H
F
H H
H
S -3-fluorocyclobutene
Substitute each CH2 H with Cl
H
Cl
F
H
+
H H
H
H
3S ,4 R-3-chloro4-fluorocyclocyclobutene
F
H
H
Cl
3 R,4 R-3-chloro4-fluorocyclocyclobutene
diastereomers
CH2 protons of S -3-fluorocyclobutene are diastereotopic. Diastereotopic protons are always nonequivalent and give different NMR signals.
11
III. Position of Signals in 1NMR Spectrum A. Shielding and Deshielding Effects
Orbital motion of the electron creates a magnetic field that opposes B0. Electron shields the nucleus from B0.
Figure 14.3: How Chemical Shift is Affected by Electron DensityAround a Nucleus
• Shielding shifts absorptions upfield . • Deshielding shifts absorptions downfield . 12
B. Characteristic Chemical Shift Values Type of Proton
δ(ppm)
sp3 C-H RCH3 R2CH2 R3CH
0.9 – 2 ~0.9 ~1.3 ~1.7
Z
Z = C,
C
O, N
C
H
-C≡C-H
1.5 – 2.5 ~2.5
H
Z = N, O, X
2.5 – 4
C Z H C
4.5 – 6
C
6.5 – 8
H
R
O
O
C
C
H
R
OH
RO-H or R2N-H
9 – 10 10 – 12 1–5
13
IV. Chemical Shift of Protons on sp2 and sp Hybridized Carbons Circulating π electrons affect the chemical shifts of H’s bound to multiply bonded carbons. A. Protons on Benzene Rings Circulating π electrons
H
H
B0 induced magnetic field
• Circulating π electrons create a ring current. • Induced magnetic field reinforces B0 in vicinity of aromatic protons. • Aromatic protons are deshielded and absorb downfield at 7.3 ppm.
14
B. Protons on Carbon-Carbon Double Bonds
C
C
H H
Induced magnetic field reinforces B0 in vicinity of vinyl protons.
B0
• Vinyl protons are deshielded and absorb downfield at 4.5 – 6 ppm. C. Protons on Carbon-Carbon Triple Bonds H C
C
B0
R
• Induced magnetic field opposes B0 in vicinity of alkynyl proton. • Alkynyl protons are shielded and absorb upfield at ~2.5 ppm.
15
V. Intensity of 1H NMR Signals The area under an NMR signal is proportional to the number of protons that give rise to that signal. 1H
NMR spectrum of C5H12O
60
20
10
8
6 4 Chemical Shift (ppm)
TMS
2
0
Determining structure from integrals: 1. Add integrals: 60 + 20 = 80 2. Divide by number of H’s in formula: 80/12 = 6.7 16
3. Divide each integral by 6.7: 20/6.7 = 3.0 (3 protons) 60/6.7 = 9.0 (9 protons) 4. Assign substructures: 3.2 ppm signal arises from 3 equivalent protons (1 CH3 group). This group must be attached to O, thereby giving –OCH3. 1.2 ppm signal arises from 9 equivalent protons, consistent with 3 CH3 groups. Of the formula C5H12O, only C is left. 5. Draw the structure: CH3 H3C
O
C CH3 CH3
t -butyl methyl ether 17
VI. 1H NMR: Spin-Spin Splitting • Spin-spin splitting occurs between nonequivalent protons on the same carbon or adjacent carbons. • Protons that split each other’s NMR signals are coupled. The splitting of the signal (in Hz) is called the coupling constant. • Coupling with n adjacent protons splits a signal into n+1 peaks.
Chemical Shift (ppm) 18
A. Why Spin-Spin Splitting Occurs Number of Adjacent Protons
Adjacent Proton Spins
Splitting
0
none
singlet
1
doublet
2
triplet
3
quartet
Conclusion: n adjacent protons split NMR signal into n+1 peaks.
19
B. More on Spin-Spin Splitting and the n + 1 Rule n
n+1
Peak Pattern
Name
0
1
1
singlet
1
2
2
3
3
4
4
5
5
6
6
7
1 1 1 1 1 1
2 3
4 5
1
doublet 1
3 6
triplet 1
4
10 10
6 15 20 15
The degree to which an NMR signal is split by spinspin coupling is measured by the coupling constant, J .
quartet 1
5
quintet 1
6
sextet 1
septet
J J
20
C. More Features of Spin-spin Splitting
1. Equivalent protons do not spin couple.
Br-CH2-CH2-Br
One singlet only
2. Splitting is observed for nonequivalent protons on the same carbon or adjacent carbons. Ha
CH3
Ha
Hb
C N
Cl
Hb C N
Ha splits Hb into a doublet, and vice versa.
3. Splitting is usually not observed if the protons are separated by more than three σ bonds. Ha and Hb are not coupled.
O H2C Ha
C
2 proton quartet 3 proton triplet CHCH3 Hb
3 proton singlet 21
VII. More Complex Spin-Spin Splitting Patterns More complex splitting occurs when the absorbing proton is coupled to nonequivalent protons on two (or more) adjacent carbons.
Figure 14.6: 1H NMR Spectrum of 2-Bromopropane
When adjacent nonequivalent protons form an equivalent set, the n+1 rule still holds. 22
Figure 14.7: 1H NMR Spectrum of 1-Bromopropane
When n protons on one adjacent carbon and m protons on another adjacent carbon are not equivalent, the signal is split into (n+1)(m+1) peaks.
23
Figure 14.8: Splitting Diagram for Hb Protons of 1-Bromopropane
• Hb signal is split into 12 peaks, a triplet of quartets. • If J ab >> J bc, all 12 peaks are observed. • If J ab ~ J bc, peaks overlap and fewer peaks are observed.
24
VIII. Spin-Spin Splitting in Alkenes Characteristic coupling constants for disubstituted alkenes: Ha C
Hb
Ha
C
C
Ha
C
C
Hb geminal H’s J geminal 0-3 Hz
C Hb
<
cis H’s J cis 5-10 Hz
<
trans H’s J trans 11-18 Hz
Figure 14.9: 1H NMR Spectra for Alkenyl Protons of ( E)- and ( Z)-3-Chloropropenoic Acids 25
Monosubstituted alkenes give more complex splitting patterns.
Figure 14.10: The 1H NMR Spectrum of Vinyl Acetate
26
Hd
Hc C Hb
C
O O C CH3
J bc = 1.2 Hz (geminal) Jcd = 6.5 Hz (cis) J bd = 14 Hz (trans)
Figure 14.11: Splitting Diagram for Alkenyl Protons of Vinyl Acetate
27
IX. More on 1NMR Spectroscopy A. OH Protons
Figure 14.12: 1H NMR Spectrum of Ethanol
• OH proton of an alcohol usually does not split NMR signal of adjacent protons. • Signal due to an OH proton not split by adjacent protons. • OH protons exchange rapidly among ROH molecules.
28
B. Cyclohexane Conformations H H H H
Rapid conformational interconversion averages axial and equatorial proton chemical shifts. C. Protons on Benzene Rings
Figure 14.13: 1H NMR Spectra of Aromatic Protons
29
X. Using 1H NMR to Identify an Unknown Problem 14.24: Propose a structure for a compound of formula C7H14O2 with an IR absorption at 1740 cm-1 and the NMR data in the table.
Absorption
δ
(ppm)
Integral
singlet
1.2
26
triplet
1.3
10
quartet
4.1
6
Step 1: Identify functional group(s) and number of different types of protons.
1740 cm-1 IR absorption: C=O NMR data: 3 sets of nonequivalent protons
30
Step 2: Determine number of protons that give rise to the three NMR absorptions.
Integral total = 26 + 10 + 6 = 42 units Divide by number of protons: 42/14 = 3.0 integral units per H Divide integral for each absorption by this number and round to nearest digit: 26/3.0 = 8.7 10/3.0 = 3.3 6/3.0 = 2.0
9 H’s at 1.2 ppm 3 H’s at 1.3 ppm 2 H’s at 4.1 ppm
Step 3: Use splitting patterns to determine which C’s are bonded to each other.
9 proton singlet at 1.2 ppm arises from 3 CH3 groups bonded to C. 3 proton triplet at 1.3 ppm must be CH3 bonded to CH2. 2 proton quartet at 4.1 ppm must be CH2 bonded to CH3.
31
Step 4: Draw out the pieces and put them together to get the structure.
-O-
CH3CH2CH3
H3C
C
C O
CH3
CH3 O H3C
C
C OCH2CH3
CH3
32
XI.
13C
NMR Spectroscopy
• Physical basis same as 1H NMR spectroscopy. •
13C
natural abundance is 1.1%.
• 1H - 13C spin splitting suppressed. • All 13C peaks are singlets. • Broader chemical shift range than 1H NMR. • Peak intensity not proportional to number of absorbing carbons. A. Number of Signals in 13C NMR Spectra
Number of signals equals number of different types of carbon in a molecule. O
5 signals
7 signals
3 signals 33
B. Position of Signals in 13C NMR Spectra Type of Carbon sp3 C-H
C≡C―
―
δ
(ppm) 5 – 45
65 – 100
H Z = N, O, X
C
30 - 80
Z C
100 – 140
C
120 – 150
C O
160 – 210
C
34
