+ All Categories
Home > Documents > 1 Nuclear Magnetic Resonance (NMR) Spectroscopy basic theory 1. properties of nucleus spin of...

1 Nuclear Magnetic Resonance (NMR) Spectroscopy basic theory 1. properties of nucleus spin of...

Date post: 21-Dec-2015
Category:
View: 222 times
Download: 1 times
Share this document with a friend
52
1 Nuclear Magnetic Resonance Nuclear Magnetic Resonance (NMR) Spectroscopy (NMR) Spectroscopy basic theory 1. properties of nucleus spin of nucleus nuclear spin quantum number I = n/2 n : integ er atomic mass number number Z A I example 1 even even 0 12 C, 16 O, 28 Si, 56 Fe odd even n : eve n 2 H, 10 B, 14 N, 50 V odd odd n : od d 1 H, 13 C, 19 F, 55 Mn * NMR properties of some nuclei with I = 1/2 * NMR properties of some quadrup olar nuclei (I > 1/2) number of possible spin states = 2I + 1 magnetic quantum number
Transcript

1

Nuclear Magnetic Resonance (NMR) Nuclear Magnetic Resonance (NMR) SpectroscopySpectroscopy

basic theory1. properties of nucleus

spin of nucleus nuclear spin quantum numberI = n/2 n : integer

atomic mass number number Z A I example 1

even even 0 12C, 16O, 28Si, 56Fe odd even n : even 2H, 10B, 14N, 50V odd odd n : odd 1H, 13C, 19F, 55Mn * NMR properties of some nuclei with I = 1/2* NMR properties of some quadrupolar nuclei

(I > 1/2)number of possible spin states = 2I + 1magnetic quantum number

m = +I, +(I-1), ……., -Iwithout a magnetic field, the spin states are degenerate nucleus I No. of states m values 1

1H 1/2 2 +1/2, -1/211B 3/2 4 +3/2, +1/2, -1/2, -3/

212C 0 1 014N 1 3 +1, 0, -1 1

2

NMR properties of some nuclei with I = 1/2

3

NMR properties of some quadrupolar nuclei (I > 1/2)

4

2. nuclear Zeeman effecta nucleus with I≠0 in a magnetic field, 2I+1 spin states are not degenerate; they separate in energy with the largest positive m value corresponding to the lowest-energy state ex. I = 1/2

m = -1/2

Bo E E

m = +1/2

Bo

spin state energy hEi = -miBo —— : magnetogyric ratio 2

transition m = -1for a nucleus with I = 1/2, the energy difference

h BoE = ————

2precession – some sort of uniform periodic

motion, the magnetic moment wobble around the axis of applied field

Lamar frequency = Bo

linear Lamar frequency = /2 = Bo/2

5

Boltzmann distribution Pm=-1/2 hBo——— = e –E/kT E = ——— Pm=+1/2 2

If Bo = 2.35 T E = 6.63 x 10-26 JPm= -1/2 Pm=-1/2 =0.4999959

———— = 0.999984 Pm= +1/2 Pm=+1/2 =0.5000041

experimental considerationssample solution solid (magic-angle spin)

magnet radio-frequency transmitter

spectrometer receiver decoupler recording device

6

magnet : permanent magnet (1 – 2 T) electromagnet (1.8 – 2.3 T) superconducting magnet (up to 13

T)two important characteristics of magnet

• stability – sensitive to temperature • homogeneity

continuous wave experiment1. frequency-sweep

2. field-sweep

7

Fourier transform technique

relaxation processesspin-lattice relaxation T1

-t/T1 Peq – P = (Peq – Po) e

spin-spin relaxation T2

much faster than spin-lattice relaxationT2 < T1

8

information from NMR spectrum(1) chemical shift the nuclei are screen from the magnetic field Bo, the net field effective at a nucleus is

Beff = Bo (1 – ) : the shielding constant

each chemically distinct nucleus is associated with a characteristic frequency

ex. B10H14 4 distinct B nuclei

chemical shift relative to a standard for the isotope concerned

obs - ref = 106 × ——————————

spectrometer frequencyunit: ppm

a shift to higher frequency than standard ==> positive

decrease in shielding ≡ increase in chemical shift

9

relative NMR frequency (MHz) standard compound common nucleus (B0 = 4.7 T) reference range (ppm) 1

1H 200.0 (CH3)4Si -30 – 20 13C 50.2 (CH3)4Si -100 – 400 19F 188.2 CFCl3 -200 – 200 29Si 39.8 (CH3)4Si -350 – 40 31P 81.0 85% aq. H3PO4 -100 – 250 77Se 38.2 (CH3)2Se -300 – 200 119Sn 74.5 (CH3)4Sn -1000 – 8000 195Pt 43.0 [Pt(CN)6]2- -200 – 15000

(2) intensity integration of the areanot for 13C

(3) spin-spin coupling non-equivalent magnetically active nuclei couple each other

chemically equivalent magnetically equivalent

10

notation >> J A, X, M, Q small A, B, C

splitting pattern 2nI + 1coupling constant J

ex. 1H, 13C NMR spectra of H13CO2- 

    

  (i) first-order (ii) satellites(iii) second-order

(4) exchange

11

12

number of lines splitting determined by Pascal’s triangle number of equivalent name of coupling nuclei pattern ratio of integration

0 singlet 11 doublet 1 1

2 triplet 1 2 1

3 quartet 1 3 3 1

4 quintet 1 4 6 4 1

5 sextet 1 5 10 10 5 1

13

AX2 ??classification of the nuclei

• I = 1/2, 100%abundance 1H, 31P, 19F, 103Rh 

• I = 1/2, low abundance 13C, 15N, 29Si, 77Se, 109Ag, 119Sn, 125Te, 183W, 195Pt, 199Hg

• I > 1/2, 100% abundance 14N, 27Al, 51V, 59Co

• I > 1/2, low abundance 11B, 121Sb, 193Ir

(I) 1H

14

15

Sn(CH3)4

1H 119Sn

expanded 1H 13C

52 Hz

54 Hz

1J117Sn-13C = 317 Hz

1J119Sn-13C = 329 Hz

16

GeH4

Si2H6 (29Si I = ½, 4.7%)

17

CH3–CH2–S–PF2

K[BH4]

18

1H NMR spectrum of PF215NHSiH3

3JPH = 8 Hz, 3JHH = 4 Hz, 2JNH = 2 Hz, 4JFH = 2 Hz

1H{15N} NMR spectrum of PF215NHSiH3

19

[HV(CO)5]2-

doublet of doublet of triplet

J Pt-H = ? J Pc-H = ?

20

for CH3 (or C(CH3)3) group in tertiary phosphine complexes, doublet 1H spectra indicate mutually cis arrangements and triplet spectra mutually trans

PtCl2(PMe2Ph)2

PMe2Ph

21

22

23

24

L KHMCl3‧xH2O MCl3L3 MHCl2L3 (I)

EtOH, 1h

(M: Rh, Ir) (L: PR3, AsR3) MHCl2L3 (II)

(i) Ir, PEt2Ph

158 Hz

19 Hz

18 Hz

12 Hz

25

(ii) Rh, AsMePh2

9 Hz

4 Hz

26

(iii) Rh, PMePh2

206 Hz

2JHP = 14.5, 9 Hz1JRhH = 13.5 Hz

2JHP = 9 Hz1JRhH = 4 Hz

27

(II) 13C

M-CO

M=C

28

1JPt-C = 35 Hz2JPt-C = 0 Hz

1JPt-C = 1936 Hz2JPt-C = 180 Hz

29

DEPT

30

[Ti(13CO)6]2-

13C

47,49 Ti

20 lines 5 lines

31

(III) 31P

32

31P NMR spectrum of P(OMe)3

31P NMR spectrum of [Cu(PMe)3]+

31P NMR spectrum of PHF2(15NH2)2

33

31P NMR spectra of the mixed products from the reaction of trans-[PtCl4(PEt3)2] + trans-[PtBr4(PEt3)2]

[PxFy]- x = ? y = ?

34

(IV) 19F

19F NMR spectrum of the products from the reaction: UF6 + Me3SiCl (halogen exchange)

35

(V) 29Si29Si NMR spectrum of SiMe4

(VI) 195Pt

36

2-D NMR1. correlated spectroscopy (COSY)

provide information about couplings between nuclei of a single isotopes

the off-diagonal peak at a frequency (f1, f2)

implies that there is a coupling between the nuclei resonating at f1 and f2

ex. COSY 11B spectrum for B10H14

        

B(2), B(4) (-35 ppm)B(6), B(9) (11 ppm)B(1), B(3) (13 ppm)B(5), B(7), B(8), B(10)(1 ppm)

     

37

ex. COSY 11B spectrum for B9H11NH 

       

2. heteronuclear correlation spectroscopy (HETCOR or HCOR)

ex. HETCOR 11B/1H spectra for B10H14 

              

38

3. nuclear Overhauser effect spectroscopy (NOSEY)identify a NOE which arises from the proximity of the two nuclei in spaceheteronuclear NOSEY (HOSEY)

ex. 2D 1H/6Li HOSEY spectrum for tmeda adduct of 2-lithio-1-phenylpyrrole 

 

39

ex. homonuclear 2D 13C scalar coupling (COSY) and chemical exchange (NOSEY) spectra for [Os3H2(CO)10]

ex. CH2CH2Br

O=P OCH2CH2

OCH2CH2

expanded 1H NMR spectra

40

 

COSY (1H—1H) COSY (1H—13C)

41

exchange reactionsex. 1 31P{1H} NMR spectrum of the products derived from [Rh4(CO)9{P(OPh)3}3]

under 400 atm of CO at 300 K

Rh4 cluster broke down to 2 dinuclear

complexes

42

ex. 2     19F NMR at 180 K

chemical shift pattern intensity 68 ppm doublet of triplets 2

of doublets -61 ppm triplet of doublets 1

of narrow triplets 68 ppm triplets of quartets 1

230 K two higher-frequency resonances broaden and lose detail

300 K coalesced to a single broad linethe lowest-frequency peak remained unchanged

43

ex. 3 13C{1H} spectrum of Rh5(13CO)15-

under pressure of 13CO (5 bar)

44

ex. 4 13C{1H} spectrum of the CO region of (5-C5H5)2Rh2(13CO)3

45

ex. 5 31P{1H} NMR spectra of [Ru2Cl5(PEtPh2)4•Ag(PEtPh2)]

46

ex. 6 trans-[IrCl(CO)(PMe3)2] + SF4 ―→

Cl P

F Ir CO P

SF3

68 -61 -383 ppm

(2F) (1F) (1F)

47

ex. 7 variable-temperature 1H NMR spectra of [Ru3W(C5H5)(CO)11H3]

48

ex. 8 EXSY 2D 1H NMR of Pt complex

49

solid state NMR spectroscopydifficulties – immobility of the nuclei in

solids(i) dipolar coupling are not averaged to zero

==> very broad resonance(ii) chemical shift anisotropy in solids is not

averaged out==> line broadening

(iii) relaxation time T1 is very long==> good signal-to-noise ratio is difficult to get

solution:(i) magic angle sample spinning (MASS,

MAS) techniquean angle = 54.7o to the magnetic field,the effect of chemically anisotropy can be averaged out

(ii) cross-polarization (CP) techniqueovercome the problem of long relaxation time

50

ex. 1 13C NMR spectra of 2Ca(CH3CO2)2•H2O

ex. 2 119Sn chemical shift of Ph3SnOH in

solution –80 ppm

==> 4-coordinated, tetrahedral

in solid phase –298 ppm

==> 5-coordinated similar to Me3SnF

51

ex. 3 31P chemical shift of phospha-alkene complex

in solution –31 ppm, JPt-P = 498 Hz==> -bonded ligand

in solid phase 247 ppm, JPt-P = 4720 Hz==> -bonded ligand

52


Recommended