chapter 10 nmr in practice i.chemical equivalence a.using molecular symmetry 1)chemically equivalent...

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Chapter 10 NMR in Practice I. Chemical Equivalence A. Using Molecular Symmetry 1) Chemically equivalent protons have identical chemical shifts 2) It is not always easy to see if protons are equivalent 3) If protons are related by a mirror plane, they are equivalent 4) If protons are related by a rotational axis, they are equivalent 5) Rotational axis = line of molecular rotation producing identical structures

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Page 1: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

Chapter 10 NMR in Practice

I. Chemical EquivalenceA. Using Molecular Symmetry

1) Chemically equivalent protons have identical chemical shifts

2) It is not always easy to see if protons are equivalent

3) If protons are related by a mirror plane, they are equivalent

4) If protons are related by a rotational axis, they are equivalent

5) Rotational axis = line of molecular rotation producing identical structures

Page 2: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

B. Conformational Changes can cause equivalence

1) Chloroethane has 2 1H NMR peaks, not 3

a) Conformational rotation averages out gauche and anti positions

b) Must be fast on the NMR timescale (Lifetime < 1 s)

c) Cool down solution to –180 oC to see all three peaks

2) Cyclohexane has only one 1H NMR peak, not 2

a) Axial and equatorial protons have different chemical environments

b) Ring flip conformational change is fast on NMR timescale, averages out the axial and equatorial protons to one peak

c) Cool to –90 oC to see both peaks

Page 3: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

II. IntegrationA. The number of protons responsible for a peak determines the peak size

1) Integration = finding the area under a 1H NMR peak

2) The NMR computer will trace a line above each peak that has its length proportional to the area under the peak

3) We measure the length of each line (area under each peak) and compare them with each other

4) There can only be integer numbers of protons, so we normalize to whole #

Page 4: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

B. We can use chemical shift and integration to assign structure

1) Dichlorination of propane:

2) CH3CH2CHCl2 1H 2H 3H

3) ClCH2CH2CH2Cl 4H 2H

4) CH3CHClCH2Cl 1H 2H 3H

III. Spin-Spin SplittingA. Neighboring protons effect each other

1) Protons are tiny magnets, with and spin states in a magnetic field

2) Proton on the same carbon or on adjacent carbons influence the total magnetic field felt by their neighbor, just as e- magnetic field does

3) Simple case of one proton on each adjacent carbon

Page 5: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

4) This phenomenon is called Spin-Spin Splitting

5) Single peak is split into a Multiplet by spin-spin splitting

Singlet, Doublet, Triplet, Quartet, etc… to Multiplet

6) Coupling Constant = J (Hz) = how far apart the multiplet peaks are

7) Spin-spin splitting is only seen for:

a) Geminal Protons = chemically inequivalent protons on the same carbon (J up to 18 Hz)

b) Vicinal Protons = chemically inequivalent protons on adjacent carbons (J = 6-8 Hz)

c) Both protons are always split by same J. If Ha splits Hb, then Hb splits Ha by the same amount

d) Chemically equivalent protons do not split each other

B. Local Field Effects are Additive

1) What if a proton has multiple neighboring protons?

Ha

C C

Hb

CHa

Hb

Hc

C C

Hd

C

HcHd

Hd

Triplet, 1:2:1, 3H,1.5 ppm

Quartet, 1:3:3:1, 2H,1.3 ppm

Page 6: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

2) N + 1 Rule = N equivalent nuclei split a neighboring proton resonance into N + 1 peaks

a) CH—CH3 doublet(3H) and quartet(1H)

b) CH2—CH3 triplet(3H) and quartet(2H)

3) Pascal’s Triangle Predicts relative intensities of the peaks in a multiplet

Page 7: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

4) Example: Bromoethane

Page 8: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

5) Example: 2-Iodopropane

Page 9: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

6) Example: 1,1-dichloro-2,2-diethoxyethane

Page 10: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not
Page 11: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

D. Complications to Splitting Patterns

1) First Order Spectrum = >> J (easy to see all splitting)

2) Non-First Order Spectrum = = J (all smashed together)

3) You can go to higher magnetic field to spread it back out

90 MHz

500 MHz

Page 12: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

4) Many close peaks in same molecule = won’t see all of the splitting

Long Alkyl Chains are Notorious for this

Page 13: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

5) We must take into account all neighbors, when predicting splitting

a) Predict splitting based on one type of neighbor

b) Apply splitting by other type of neighbor to each of the split peaks from the first type of neighbor

Page 14: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

6) Sometimes the spectrum will appear First Order, but isn’t (J’s are same)

Page 15: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

7) Fast Proton Exchange “Decouples” some protons

a) RCH2OH protons exchange quickly with protic solvents

b) See an average RCH2O—H peak

c) Happens fast on the NMR time scale

d) See no coupling to the CH2 group, unless you cool solution down

E. NMR Solvents

1) You usually take NMR spectra of a sample dissolved in a solvent

2) The fast tumbling of molecules in solution is best for NMR

3) Solvent is at much greater concentration than sample, so you would see only the solvent protons in your spectrum

4) We use Deuterated solvents (2H) because they have the same chemical properties (solubilities) but have shifts outside the 1H range.

5) You can’t get rid of all 1H, so you usually still see a small solvent peaks

CH3OH

Page 16: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

IV. Carbon-13 NMRA. Abundance of 13C effects its use in NMR

1) 98.9% of carbon is 12C, 1.11% is 13C

2) Much weaker signal for carbon NMR, we must take many scans, ( time)

3) 13C-13C next to each other is statistically unlikely; no C-C splitting

4) Carbon peaks are split by the 1H’s attached to them

a) Useful to tell us how many H’s are attached (triplit = CH2)

b) Usually Decouple the protons with a broad constant proton pulse, which keeps protons flipping and gives no splitting

c) Decoupling produces all singlets in the 13C NMR spectrum

B. Chemical Shift in Carbon-13 NMR

1) Carbon resonances occur over a large range 0-250 ppm (TMS = 0 ppm)

2) This is very useful, because often proton NMR is smashed together

3) Every chemically inequivalent carbon gives a unique singlet

a) Alkyl groups: 5-50 ppm

b) Alkyl Halides: 25-50

c) Alcohols/Ethers: 50-90

d) Alkenes: 100-150

e) Carbonyl: 170-210

Page 17: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

Bromoethane Carbon-13 NMR

Page 18: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

C. DEPT (Distortionless Enhanced Polarization Transfer)

1) 13C NMR Experiment that tells you how many H’s are attached to each C

D. Using 13C NMR to assign structure

All C’s

CH only

+CH3 and –CH2 only

CH3CH2CHCl2 CH3CHClCH2Cl ClCH2CH2CH2Cl10.1

34.9

73.2 49.5

55.8

22.4 42.2

35.6

42.2

Page 19: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

COSY 2D NMR Spectrum

Page 20: Chapter 10 NMR in Practice I.Chemical Equivalence A.Using Molecular Symmetry 1)Chemically equivalent protons have identical chemical shifts 2)It is not

HETCOR 2D NMR Spectrum