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An Introduction to NMR Spectroscopy

1H NMR

13C NMR

http://www.chem.ucla.edu/%7Ewebspectra/

The types of information accessible via high resolution NMR include:

1. Functional group analysis (chemical shifts) 2. Bonding connectivity and orientation (J coupling) 3. Through space connectivity (Overhauser effect) 4. Molecular conformations, DNA, peptide and enzyme sequence and structure 5. Chemical dynamics (lineshapes, relaxation phenomena)

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2Tuesday, March 20, 2012

1 of 40 CHEM 0310 Dr. P. Wipf University of Pittsburgh

Chem 1310/2370

Atom Structure

Electron spins:!electron waves (or particles) are characterized by 4 quantum numbers: n, l, m, sThe electron spin quantum numbers s can assume only two values: +1/2 or –1/2.

Bo

s = +1/2 s = -1/2higher energy

lowerenergy

externalmagnetic field

particle spinning on its axis spin angular momentum

charged particlestiny magnetic moment

Nuclear Spin:!The proton is a spinning charged particle and has also a magnetic moment.

Because nuclear charge is the opposite of electron charge, a nucleus whose magnetic moment is parallel to the magnetic field has the lower energy.

m = +1/2higher energy

lowerenergy

m = -1/2

B

1H: - nuclear spin quantum number m = 1/2

- such a nucleus is described as having a nuclear spin I of 1/2

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Chem 1310/2370

The difference in energy is given by:!ΔΕ = hγB0/2π

γ = magnetogyric ratio (a constant, typical for a nucleus, which essentially reflects the strength of the nuclear magnet)

B0 = strength of the applied magnetic field

h = Planckʼs constant (3.99 x 10-13 kJ s mol-1)

Note that as the field strength increases, the difference in energy between any two spin states increases proportionally.

0

0degenerate

!

"!

m = -1/2

m = +1/2

nuclear spin quantum #

Ei = -mh#B0/2$= -mµNB0

B0

[values of ±1/2 were picked for m, so that the difference in energy between two neighboring states will always be an integer multiple of B0(γh/2π)].

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Chem 1310/2370

The number of nuclei in the low energy state (N) and the number in the high energy state (N) will differ by an amount determined by the Boltzmann distribution:

Nβ/Nα = e(-ΔE/kT)! ! ! ! ! k = 1.381 x 10-23JK-1

When a radio frequency (RF) signal is applied, this distribution is charged if the radio frequency matches.

! ΔE = hν = hγ B0/2 ν = resonance frequency = γ B0/2

ν is therefore dependent upon both the applied field strength and the nature of the nucleus.

1H nuclei (protons) exhibit two possible magnetic spin orientations. What about other nuclei with additional protons and neutrons?

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Chem 1310/2370

Rules: - A nucleus with an even number of protons (Z) and neutrons(N) will have a nuclear spin I = 0 (for ex. 12C, 16O, 18O, 32S).

- cannot be detected by NMR

- both Z and N odd leads to integer values of I (2H (I=1), 10B(I=3), 14N (I=1))

- detectable by NMR

- Z even, N odd, or Z odd, N even will have I values of n/2: 1H(I=1/2), 11B (I =3/2), 13C (I =1/2), 15N (I =1/2), 17O (I=5/2), 19F(I=1/2), 31P (I=1/2)

- detectable by NMR

The total number of possible spin states (values of m) is determined directly by the number of I.

Ε -1

+1

Number of possible spin states:

0

(-I, -I +1, ...I -1, I)

m

m

ΔΕ = hγ B0/2π

2I +1

B0

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Chem 1310/2370

1H: in a 2.35 T field (earth magnetic field = 0.00006 T) = 0.999984

~ 1 in 106! ! ( ν = 100 MHz)

The difference in population of the two states is exceedingly small, in the order of few parts per million (even smaller in 13C, because γ is smaller).

Relatively low sensitivity of NMR compared to IR or UVLarge Bo needed to increase the population difference (Bo is usually

given in MHz of 1H resonance frequency).

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Chem 1310/2370

SUMMARY

- Nuclear spin is a property characteristic of each isotope and is a function of Z and N.

- Each isotope with I ≠ 0 has a characteristic magnetogyric ratio (γ) that determines the frequency of its precession in a magnetic field of strength B0

ν =2π

γB0

It is this frequency that must be matched by the incident electromagnetic radiation for absorption to occur.

- When a collection of nuclei with I ≠ 0 is immersed in a strong magnetic field, the nuclei distribute themselves among 2I + 1 spin states, the relative population of which is determined by the Boltzmann distribution, usually being near unity

= e(-ΔE/kT)

Nα

Nβ

- If two (or more) spin state populations become equal, the system is said to be saturated.

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Chem 1310/2370

Obtaining an NMR Spectrum

MagnetSource of RF radiationDetector + amplifierPlotter, sample

The magnet:

permanent electromagnet superconducting

cheap, stable,fixed field1.4T

more expensive,stronger, variablefield

expensive, stronger, variablefield18T (24T)

Strength of magnetic field shifts: lock necessary (= substance with strong, defined NMRsignal) Older: referenceinternal, external CDCl3TMS: 0.0 ppm singlett

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Chem 1310/2370

Once a stable field is established, the question remains as to whetherthat field is completely homogeneous throughout the region betweenthe pole faces of the magnet.

Sample

lines of magnetic flux not uniform

sample has to be placed near the center of the pole gap

N S

!For 2.35 T, to achieve a precision of +/- 1 Hz (10 ppb at 100 MHz) the field must be homogeneous to the extent of +/- 2.35 x 10-8 T! Such a phenomenal uniformity, even at the center of the field , can be achieved only by means of two additional techniques:

Spinning of the sample ("averages" out small inhomogeneities)

Variation of the contour of the field by passing extremely small currents through shim coils wound around the magnet itself: Shimming (manually, automatically)

Paradox: Large sample in order to have as many nuclei as possible, small sample to increase uniformity of the field.

→ narrow bore tubes

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Chem 1310/2370

The Pulsed Fourier Transform Technique

Further advances in S/N ratio improvement had to await the development of faster computer microprocessors: ~1970ʼs.

- RF radiation is supplied by a brief but powerful pulse of RF current through the transmitter coil. The spectral width of the pulse is chosen to cover absorption of all nuclei of interest.

Optimum tp are obtained by trial and error and are usually in the order of 10 µs for α = 90° for best S/N ratio.

RF

2SW

intensity

νo

The duration of the pulse (tp) determinesthe frequency range covered (Heisenberg's uncertainty principle: ΔΕ Δt h)>_

Frequency

SW ~ tp-1 ; tp >_(4SW)-1

The next step in the PFT process is to monitor the induced ACreceiver signal. Digital data collection gives us the modulated freeinduction decay (function of Mxy). FID because the current intensitydecreases with time. This decay is the result of T2 (spin-spin)relaxation.

M

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Chem 1310/2370

Time

Voltage

t0t0

td

the microprocessor samples thevoltage in the receiver coil at a regularinterval, called dwell time, td.

td > (2SW)-1

The frequency of the cosine wave is unaffected by the exponentialdecay:

(in the date acquisition process νo is subtracted electronically from the observed signal prior to digitization). → time to frequency transformation of the data.

ν =1t0

= νprecession - νo

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Chem 1310/2370

In a set of nuclei with different ν precession and T1/T2, the digital FID curve becomes very complex:

f( x) = ao + an cos2πnso x + bn sin 2πnsox{ }n=1

∞

∑

CH3

Δν1

frequency

Δν2 Δν3

where:a0 = constant; an = amplitude; x = period; so = fundamental frequency; xo = 1/so; andn = order of harmonic

The parent function is constructedby summing together a series of sine waves.

line width: uncertainty principle: ΔνΔt > 1

ν1/2 > 1T2*

Nuclei that are slow to relax give sharp signals,nuclei that relax rapidly give broad signals (solids).

paramagnetic residues line broadening

At this point it becomes necessary for the computer torecognize the patterns mathematically and extract thesignal frequencies and relative intensities for each set of nuclei. This analysis is performed by a Fourier transfor-mation of the FID date.

0

Summary:A typical example of the generation of a PFT spectrum:

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Chem 1310/2370

!tp:! pulse time (µsec)!tacq: the length of the time a given FID signal is actually monitored ! !(resolution, the ability to distinguish two nearby signals, is inversely !! prop. to tacq. R = (tacq)-1 3 sec → 0.3 Hz

tw: delay time, to allow for equilibrium distribution

tw = 3T1 - tacq.

+ dead time ( phasing necessary)

(for 1H no waiting time)

phase correction

adding up, FT, spectrum

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Chem 1310/2370

Taking an NMR – Practical Consideration

- Use 5 mm tube filled with ~ 0.5 mL of solution containing 1-5 mg of sample (1H NMR).

- common deuterated solvents:CCl4, CDCl3, C6D6, DMSO-d6, D2O, CD3CN, CD2Cl2, d6-acetone, CD3OD (because of HCl formation, do not leave sample in CDCl3!)

- peak listings in ppm and/or Hz.

- paramagnetic metal ion → broad peaks

Chemical shift

νH depends on Bo ⇒ therefore relative frequencies are reported:

1.39 ppm = δ = downfield from TMS.

2

1 0ppm (δ)

downfield

deshielded

upfield

shielded

ν =νact. −νTMS

νo=83.4Hz60x106

= 1.39x10−6 =1.39 ppm

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Chem 1310/2370

Integration

! Area under absorption peak ~ # of nuclei resonating at that ν

But:!nuclei must relax to equilibrium between pulses, not generally true of 13C NMR!

Correlating Proton chemical Shifts with Molecular Structure

Shielding and Deshielding

!That actual magnetic field (Bo) experienced by a given nucleus is diminished slightly by an opposing local magnetic field (BN) which results from circulation of nearby electron, induced by Bo:

The circulation of e- follows the right hand rule and magnetic fieldlines follow the left hand rule.

H

BN

Thus:Bo

deshieldinge

shielding

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Chem 1310/2370

Consequence:!The greater the magnitude of BN, the less the magnitude of B required to effect excitation. In other words, the frequency of excitation correlates with the e- density surrounding the nucleus in question.

Thus structural features which deplete e--density around a nucleus will cause the resonance to occur at relatively low field (downfield), and the nucleus is said to be deshielded. → increase of e—density → upshield shift: nucleus is said to be shielded.

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Chem 1310/2370

Specific Effects

Inductive Effects:

e- withdrawing groups move C, H resonances downfield; e—donatinggroups move C, H resonances upfield.

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Chem 1310/2370

Hʼs attached directly to Si + Metals → very upfield.

CHOCH3

CO2H

Ex.1.51.50.81.35.1

-O-R-COOH-Phenyl

(Exp.: 4.8)

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Chem 1310/2370

Anisotropic EffectsNon-spherical electron-distribution around bond.

Alkenes:

CH

sp2-hybridized and more electronegative

C C

H

C C

BN

CH

Bo

Lies in de-shielding region of local opposingmagnetic field induced by BO/π−e- interaction.Thus, both inductive withdrawal of e- density and anisotropic effects move H downfield

δc = 123.3 ppmδ = 5.25 ppm for ethylene

downfield

shielded upfield

sp2-hybridized and more electronegative

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Chem 1310/2370

Resonance effects can be superimposed on this to explain relativechemical shifts.

Hb

CO2CH3Hc

Haa b Obs.

5.80 6.05 5.82 6.206.38

c: 6.43

a: 5.25 + 0.55 = 5.8b: 5.25 + 0.8 = 6.05c: 5.25 + 1.18 = 6.43

O

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Chem 1310/2370

HB0

H

H

Benzene:

Allylic &benzylicprotonsmoveddownfield

deshielded regionδH = 7.27 ppmδC = 128.5 ppm

>

>

>

> >>

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Chem 1310/2370

First-Order Spin-Spin-Coupling

From previous discussions one could have gotten the impression thata typical 1H NMR spectrum exhibits just one signal for each set ofequivalent 1H-nuclei and that the same thing is true for 13C spectra,as well as for spectra of any other isotope. However, there are manymore lines in a spectrum, and while these extra lines do make aspectrum more complex, they also offer valuable structuralinformation that complements the chemical shift data.

1H - 1H

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Chem 1310/2370

CH3CH2OH:!!The spin states of two hydrogens (methylene group):

→ For methyl hydrogens the net experienced field will depend on the magnetization of the neighboring methylene group!

→ The methyl signal will be split into three lines with intensity ratio 1 : 2 : 1 (= spin-spin-coupling, homonuclear coupling because the coupling is between nuclei of the same isotope). → Triplet.

B0

Total MagnetizationM=-1 0 0 1

Three spin states with population ratio of 1:2:1

Accordingly, for the methylene signal, the possible spin states of themethyl group determine its multiplicity (number of lines in thesignal).

The spin states of three hydrogens:

M=3/2

M=1/2 M=-1/2

M=-3/24 spin states with populationratio of 1 : 3 : 3 : 1

Quartet

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Chem 1310/2370

Accordingly, a doublet is observed for hydrogens that are coupled toa methine (CH) proton.

The multiplicity of a given resonance = n+1 (n=# of neighboringequivalent nuclei). The relative intensities of the multiplet followPascalʼs triangle.

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Chem 1310/2370

J = spacing between lines.

The slight difference in energy between the resonances is thecoupling constant J [Hz]. Jʼs are independent of instrumentalparameters!

J J

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Chem 1310/2370

The magnitude of J depends on the through-bond distance and theangle between Ha + Hb.

Ha C Hb two-bond coupling 2J "geminal"

C CHbHa three-bond coupling 3J "vicinal"

C (C)n CHH long range coupling n+3J

HHΦ

Consider a 3-spin system with/without equivalent nuclei:

Ha will be a doublet with 3Jab

Hb will be a “doublet of doublets” with 3Jab and 3Jbc

Hc will be a doublet with 3Jbc

CHa

CHb

CHc

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Chem 1310/2370

Jab

Ha Hb Hc

Jbc

Jbc

Jab

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Chem 1310/2370

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Chem 1310/2370

Heteronuclear Spin-Spin-Coupling

Any magnetic (I ≠ 0)(nuclear spin) nuclei can lead to spin-coupling interactions.

2NI + 1 = multiplicity

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Chem 1310/2370

Symmetry/Chirality

Before we are able to understand and predict the appearance ofhomonuclear NMR spectra, we must be able to recognize whennuclei (and atoms) in a given structure will be distinguishable andwhen they will not. The test of distinguishability is based onsymmetry relations among the nuclei.

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Chem 1310/2370

Test of distinguishability:

OHCH3

CH3

heterotopic nuclei different constitutional environment

CH HAHB

Nuclei that are equivalent by virtue of a rotationalaxis are said to be homotopic.

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Chem 1310/2370

CCl

Br DH Nuclei that are equivalent by virtue of a reflection

in the mirror plane are said to be enantiotopic.

If we have difficulty deciding whether two nuclei are related by amirror plane, we can use the isotope substitution test.

CCl

Br DH

CCl

Br HD

enantiomers

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Chem 1310/2370

C CCl

H3CH

HH

Cl: C C

Cl

H3CH

HD

ClC C

Cl

H3CH

DH

Cl

diastereomersdiastereotopic H

AAʼBBʼ

A/Aʼ are chemically equivalent, but not magnetically equivalent

OHHA'

HB'NO2

HB

HA

homotopic

homotopic

but: different J's

JAB ≠ JABʼ 2nd order effect

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Chem 1310/2370

HHSi OR

H H

A set of nuclei that is magnetically equivalent must have identical coupling constants to all other nuclei! (Coupling between magnetically equivalent nuclei occurs, but does not show up in the spectrum).

Homotopic/enantiotopic H:!chemically equivalent (same chemical shift); magnetically equivalent or not (identical or non-identical Jʼs)

Heterotopic/diastereotopic H:!chemically and magnetically non-equivalent.

Chemical equivalency (same chemical shift) is necessary but not sufficient for magnetic equivalency.

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Chem 1310/2370

Accidental equivalence:

It occasionally happens that two nuclei that are not symmetry-equivalent in any way accidentally precess at exactly the samefrequency, and, hence, give rise to a single NMR signal.

C CH CH3δ = 1.80

13C NMR

• 12C (98%) has I = 0 → no NMR

• 13C (1.1.%) has I = 1/2 → NMR

•

• Observe typically 0 – 230 ppm (rel. to TMS)

(Ε = hν = γhBo/2π ⇒ gyromagnetic ratio such that νobs ≅ 1/4 that of 1H (300 MHz → 75 !MHz)

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Chem 1310/2370

Coupling in 13C – NMR

p = 0.33 sp2C [165]

p = 0.50 spC [250]

2. 1JC1H = 500 • p; p = 0.25 sp3C [125]

2JC1H = +60 - -10 Hz

3JC1H = +/-1 Hz

1. Low natural abundance means 13C-13C couplings are rare.

Typically decouple the protons by saturating them with a secondbroadband RF puls (double resonance technique, secondtransmitter coil; “white noise” → if the irradiating field is strongenough, not only will the 1H nuclei approach saturation, but virtuallyall the 1H magnetization will be tipped into the x1y plane. Since the 1Hnuclei are no longer aligned with (or against) the applied field (whichis along the z axis) they can no longer augment or diminish themagnetic field experienced by the carbons. As a result, the couplinginteraction disappears, and each 13C multiplet collapses to a singlet!(D coupling not effected!)

→ causes all 13C resonances to be singlets → affords Nuclear Overhauser Effect → makes integration of 13C spectra unreliable

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Chem 1310/2370

13C-NMR Chemical Shifts

CH4 CH3-CH3CH2(CH3)2CH(CH3)3C(CH3)4

CH3-CH3CH3-CH2-CH3CH3-CH(CH3)2CH3-C(CH3)3

CH3CH2CH3CH3CH2CH2-CH3CH3CH2CH(CH3)2CH3CH2C(CH3)3

α

-2.15.916.125.227.9

β

γ

5.915.624.331.5

15.613.211.58.7

α − effect: +9 ppm for each added C

β - effect: +9 ppm for each added C

γ - effect: -2.5 ppm for each added C

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Chem 1310/2370

29.5 23.1

32.4 14.2

Alkenes:

139.2

114.2

C C C C C C C C*

-1.5 +7.2

+10.6 -7.9

-1.8 +1.5

thus:

δC3 = 123.2 +10.6 + 7.2 -7.9 - 1.8 = 131.4 (exp.: 130.3)

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Chem 1310/2370

Alkynes:

75 to 95 ppmC

C

84.5

68.1

H

Aromatics: not effected by ring current

substituted Cʼs are typically of lower intensity

128.5 ppm

176.8 176.8

++

209.0 102.1

- -

85.3

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Chem 1310/2370

Carbonyls:

O O

H

O

OH

O

OEt

O

ClRO OR

O

NH

CH3

O

I

205.1 199.6 177.3 169.5

168.6~ 100 - 110

Heavy atom effect.174.9 158.9

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Tuesday, March 20, 2012


Chem 1310/2370

nmr spectroscopy - university of...

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