NMR Nuclear Magnetic Resonance Spectroscopy p. 83 a hydrogen nucleus (a proton) has a charge, spread over the surface a spinning charge produces a magnetic moment  (a vector = direction + magnitude) along rot axis

which produces a magnetic field, just like a bar magnet p. 83

this moment interacts with an external field, just like a magnet

N

B0

S

BUT only certain orientations are allowed, which depends upon the nuclear spin quantum number I p. 83 The magnetic moment, is proportional to the spin quantum number I

 =  I (h/2) where  = Magnetogyric ratio = magnetic moment angular momentum

 is different for each nucleus (a constant)

I is allowed only certain values and these values can range from mI = –I to +I in integer steps ONLY

Our proton is simple as it has I = ½, so only two values of mI are permitted, +1/2 and -1/2 In ‘bar magnet’ terms, these values of mI correspond to the magnet aligned with and opposed to the external field

or B0 =

mI = +1/2 -1/2

However, these two states ONLY have different energy in a non-zero external field, B0

E

B0 E = h

B0 = 0

p. 83/84 p. 84

B0 E = h so, if we put ‘our proton’ in a strong magnetic field,

B0, there are TWO nuclear spin states: lower energy = aligned with field (mI = +1/2) higher energy = opposed to field (mI = -1/2)

A transition will occur at a frequency  this is in the radiofrequency range, ~108Hz (100 MHz) and this is the basis of the NMR (or MRI) experiment we make a nucleus ‘flip’ its spin However, we do NOT really have little bar magnets!!!

p. 84

But this frequency,  is the same as what we have called the ‘flip frequency’

NOTE:  is proportional to B0 – more about that later p. 85 The spin number, I depends on the isotope

x = number of protons + neutrons = mass x E Y Y = number of protons = charge = Atomic #

1 e.g. 1H has mass of 1 and charge of 1 2 1H = D has mass 2, charge 1 mass 2 = 1 proton + 1 neutron

IF x and y are both even, I = 0, NOT NMR ACTIVE

12 16 4 e.g. 6C 8O 2He

Common isotopes p. 85

If charge (y) is odd and mass (x) is even, I = integer

2 14 e.g. 1D and 7N both have I = 1

If mass is odd, I = Integer/2

1 13 19 31 1H, 6C 9F 15P all have I = 1/2 IMPORTANT

35 17 17Cl I = 3/2; 8O I = 5/2 too difficult for now Number of spin states p. 85

Nuclei have spin states of magnetic quantum number

MI = +I, (I-1), (I-2)...-I where total number of states is then 2I+1

so if I = ½ , # states = 2, and MI = +1/2 and -1/2

if I = 1, # = 3, MI = +1, 0, -1 case for Deuterium

11 if I = 3/2, # = 4, MI = +3/2, +1/2, -1/2, -3/2 case for B p. 86

I = ½ 1H

I = 3/2 11B E p. 86

B0 E = h

B0 = 0

This frequency is the Larmor frequency, where:

B0  = 2 so irradiation of nucleus with frequency causes a transition from lower to upper state which we call

‘flipping the spin’

This frequency depends on B0 (field) and (nucleus type) p. 86/87 B 0 Earth’s magnetic field is  = ca. 0.5 G at surface 2

For 1H,  = 267.51 (radian MHz per Tesla)

Then when B0 = 2.35 Tesla,  = 100MHz

[1 Tesla = 10,000 gauss = 10kGauss]

so for each Tesla of B0,  = 42.55 MHz

so for each 100 MHz you need 2.35 Tesla of B0

Our largest departmental NMR is referred to as a 500 MHz instrument because this is the resonance frequency of 1H at its magnetic field of 11.74 Tesla (cost ~ 500K$) p. 86 Magnetogyric ratios

nucleus  rel (relative freq to 1H) 1H 267.1 1 19F 251.8 0.94 31P 108.40 0.40 11B 85.84 0.32 13C 67.28 0.25 29Si 53.15 0.19 2H 41.06 0.15

1 19 31 13 29 2 E H F P C Si D Magnetic field 2.35 Tesla

100 94 40 25 19 15 0 , MHz p. 87 For different nuclei:

1 = k 1 B1 2 = k 2 B2 k = 1/2

1 1 B1 so = 2 2 B2

So if know five of these we can calculate other

but note that B1 = B2 for the same instrument NMR Instruments p. 87

Known by what PROTON frequency they operate at e.g. we have 300, 360 and 500 MHz instruments

13 these all can run carbon spectra, and since rel for C = 0.25, these same instruments operate at 75, 90 and 125MHz respectively when running carbon spectra

[magnetic field strength is essentially fixed for each instrument]

If all protons absorbed at exactly B0  = the same frequency the NMR 2 experiment would NOT be very useful! p. 87 Fortunately, this is not the case: electrons have an associated magnetic field too so small differences in the electronic environment can cause one 1H nucleus to resonate at a slightly different than another

H H B0  = H C C O alcohol 2 H H H

Essentially, the applied field is slightly modified by the local electronic environment:  is a very sensitive probe of the local field B and we can measure radiofrequencies very accurately

Bactual = B0 + Blocal p. 88

SHIELDING: local electronic effects on Bnet

Bo Binduced e-

Electrons in bonds circulate in a magnetic field creating an opposing magnetic field Binduced that cancels part of Bo

- More e means a bigger Binduced and a smaller Bnet so to reach resonance, you would need to apply a higher Bo (or, if Bo is fixed, a lower ν is required) p. 88/89 CHEMICAL SHIFT, 

Variations in  caused by local effects are on the order of 1 part in 105 to 107 of the ‘normal’ 1H resonant 

We COULD just quote the absolute resonant frequencies  for a given 1H type BUT:

a) The differences would be tiny and unwieldy to quote: e.g. 250.000000 vs. 250.000002 MHz

b) Absolute frequencies vary linearly with B0: e.g. 250.000002 vs. 500.000004 on a 250 vs. 500 MHz machine p. 88 CHEMICAL SHIFT, 

A better way: measure all resonances RELATIVE to a standard and divide by B0 (or more simply the standard resonance frequency for a given nucleus at a particular B0) to remove the field dependence

The Standard: TMS = tetramethylsilane = Si(CH3)4 = 0 Sample peak  - TMS peak  Hz  = = = ppm Spectrometer  in MHz MHz

shift in Hz

 sample TMS p. 89

CHCl3 = 7.2 ppm

The shift from TMS is thus 7.2 x spectrometer operating frequency in MHz e.g. at 100MHz = 7.2 x 100 = 720 Hz at 300MHz = 7.2 x 300 = 2160 Hz at 500MHz = 7.2 x 500 = 3600 Hz

Thus the absolute shift depends upon the instrument, but the chemical shift does not (7.2 ppm)

Higher frequency (larger field) instruments are more sensitive, and more expensive! SUMMARY Nuclei have spin I (depends on # protons/neutrons) Spinning charges = magnetic field

Nuclear spin I quantized: only mI states from +I,..,-I allowed, each with a different magnetic moment mI states same energy unless placed in an external magnetic field:

+ mI (spin aligned with ext. field) lower energy than - mI NMR experiment probes  required to ‘flip’ spin: cause nuclear spin state to change (eg. mI = 1/2 to -1/2)  depends on ext. field strength but also on tiny variations in electron density (because e- are spinning charges too) in the locale of the nucleus: we can tell something about the nuclear environment from  p. 90

H H Deshielded Br O H Downfield H H Low field TMS

5.7 3.5

Cl H H O H Cl H

7.3 3.7 p. 90

Cl H H O H Cl H

 7.3 3.7

= integration = area under peak is proportional to NUMBER of H’s = 1:3

it is a RATIO, i.e 1:3 or 2:6 or 3:9 etc p. 90

H H Br O H H H

TMS

= integration = area under peak is proportional to NUMBER of H’s = 2:3

it is a RATIO, i.e 2:3 or 4:6 or 6:9 etc p. 91

H CH3 C OCH CH3 CH3 H

  3.5 size ratio 3:1 p. 92

SPIN-SPIN COUPLING F---PCl2

Both P and F have spin I = ½ (we can ignore Cl here)

high  low field high field low  THE COUPLING CONSTANT J p. 93

n JXY where n = number of bonds between coupled the nuclei X = nucleus that is being observed Y = nucleus that is coupling to X (often the order of XY or YX is arbitrary)

4 1 2 2 CH P 1 C C 3 1 C C 3 C HH F 2 F

3 4J 2 JHH FH JPF p. 93

 = chemical shift = centre of all lines (ppm)

J = coupling constant = separation of lines (Hz) s, d, t, q = quartet, pentet, sextet, septet, octet nonet, decet p. 93 Cl Br ** HA C C HB HA=HB Cl Br

**cannot see coupling between identical nuclei More nuclei –AX2 e.g. PF2Cl

The P can see both fluorines up

(twice as likely) one up, one down

both down so there are THREE lines, the middle one has twice the intensity p. 94 TREE DIAGRAMS

AX2 A

A split by X

JAX split by X again 121

J/2 J/2 J/2 J/2

Can always construct pattern as a tree, by splitting one nucleus at a time – when nuclei are same, e.g. X above, lines fall on top of each other and simplify pattern p. 94

11

121

1 1 1 46 4 1 AX The number of lines at any branch inthetreediagramrepresentthe AX2 number of degenerate possibilities 3 3 for the alignment of the nuclear spins. This translates to the area under the line in the multiplet. The quintet shown AX3 above would have relative areas of 1 : 4 : 6 : 4 : 1 AX n AX Spectrum of A: (2nI + 1) lines 4 4+1 = 5 lines I = spin of X; n = number of X atoms 1:4:6:4:1 For I = ½ , = n + 1 lines intensities p. 94 Numerically, we can get intensities from Pascal’s triangle:

1 11 12 1 1 331 1 46 41 1 5 10 10 5 1 1+5 1 5+10 10+10 10+5 5+1 1 =6 =15 =20 =15 =6

so, intensities of a septet are 1:6:15:20:15:6:1 Example A p. 95

3 neighbors, so 3+1=4 lines 1:3:3:1 2 neighbors, so Br----CH2-----CH3 2+1 = 3 lines 1:2:1 Example B p. 95

CHCl2—CHCl---CHCl2 2 identical neighbors, so triplet 1 neighbor, so doublet Example C p. 96 If J’s are different, best to draw a tree, e.g. PHFCl

1 1 You need to know J’s, assume JPF > JPH for now

In 31P P Spectrum: 1J PF P split by F

1J 1 PH JPH split again by H

 so spectrum is a doublet of doublets Example D p. 96 P PH2F PH F H H 2

1 1 F JPF > JPH

31P nmr

P P

P split by F 1 JPF

1 1 JPH JPH split again by 2H

1 1 1 1 get a doublet of JPH JPH JPH JPH triplets 112 1 2 1 p. 96 For AXYZ draw tree by starting with largest J then next largest J then smallest J then lines will cross over each other the least!

Consider CHF2CH3 H is triplet of quartets

13 H is doublet CF3CH2CH3 2 of quartets F C CCH 3 3 of quartets H2 SIZE OF COUPLING CONSTANTS Table, manual p. 97

Generally 1J >> 2J >> 3J > 4J > 5J > 6J Ball park figures!! 200Hz 50Hz 10Hz 3Hz 1Hz * only P, F,.... but there are plenty of exceptions

>CHAHB ~15 =CHAHB~2 0 Hz 1 Hz H H 2 3 2 3 4 4 * For H-H = 0, except 1 3 1 3 in pi-bonds HH HH 7Hz 10-18Hz Example E p. 98

C5H8F4O

C5H8F4O - C5H12O = 0 DBE Integration = 1:1:6 ---8H total

Peak at  6 is 1H t of t (triplet of triplets) p. 98 Integration = 1:1:6 ---8H total C5H8F4O

so t of t must be caused Other H signals by F’s are singlets arranged 2 + 2 There is a large coupling and a small coupling Large Coupling ~ 1ppm 2 =60Hz = J = H-C-F2 small coupling ~0.1ppm 3 =6Hz = J = H-C-C-F2

so we have CHF2-CF2- group not split by anything else p. 98

C5H8F4O

Integration = 1:1:6 ---8H total CHF2—CF2-- we have 3C’s, O, 7H’s 6H’s are identical and are to find NOT split by neighbors

X, Y not H or F otherwise X—C(CH3)2 would see coupling 3J

Y One of these must be CHF2CF2 p. 98

C5H8F4O

CHF —CF -- 2 2 Integration = 1:1:6 ---8H total

C—C(CH3)2 now groups left to attach

O are H and CHF2CF2-- so CHF2CF2—C(CH3)2 we will learn later that OH usually do not couple OH You can now do ASSIGNMENT 4