NMR course at the FMP: NMR of organic compounds and small biomolecules - III-

Transcription

1 NMR course at the FMP: NMR of organic compounds and small biomolecules - III

2 The program 2/82 The product operator formalism (the PROF ) Basic principles Building blocks Two-dimensional NMR: The COSY

3 3/82 The produkt operator formalism (The PROF )

4 The PROF 4/82 The PROF is a quantummechanical description of NMR-experiments based on the density matrix formalism In the same way as quantum mechanics it can be introduced in an axiomatic way and does then consist of a set of rules O.W. Sørensen et al. Prog. NMR. Spectrosc. 16, (1983)

5 The PROF 5/82 When using those rules the mathemetics mainly consists of simple trigonometry and addition/subtraction. It is far more important to keep track of the calculations in order not make trivial mistakes. cos 2 α + sin 2 α = 1 trigonometric formula (1) sin 2 α = 2 sinα cosα cos 2 α = cos2α sin2α exp ±iα = cos α ± isinα

6 The PROF 6/82 sin(α+β) = sinα cosβ + cosα sinβ sin(α-β) = sinα cosβ -cosα sinβ cos(α+β) = cosα cosβ -sinα sinβ trigonometric formula (2) cos(α-β) = cosα cosβ + sinα sinβ sinα cosβ = ½ [sin(α+β) + sin(α-β)] cosα sinβ = ½ [sin(α+β) - sin(α-β)] cosα cosβ = ½ [cos(α+β) + cos(α-β)] sinα sinβ = ½ [cos(α-β) - cos(α+β)]

7 The PROF 7/82 In most calculations product operators are expressed in catesian coordinates At the beginning of an experiment we have H z Type of nucleus (here hydrogen, in more general equations I or S) H z direction (here cartesian coordinates)

8 The PROF 8/82 x,y-magnetization is then expressed as H x or H y or as C x and C y During a calculation operators of chemical shift or coupling act on the operators the represent magnetization. According to certain rules other operators are thus created. A βb Ccosβ + D sinβ

9 The PROF 9/82

10 The PROF 10/82 The name of the operators is derived from the type of magnetization they represent H z = longitudinal magnetization H x, H y = in-phase magnetization H 1x H 2z = anti-phase magnetization H 1x H 2y = multiple quantum magnetization } transverse magnetization Beside those operators based on cartesian coordinates there are others which are not of interest for now

11 The PROF 11/82 Chemical shift and J-coupling (as long as it is weak coupling) can be calculated independently and in an arbitrary order. That makes the calculation of building blocks possible: the effect of these building blocks is then known and does not have to be re-calculated in more complex experiments

12 The PROF 12/82 z y We use a righthanded coordinate system x

13 The PROF 13/82 RF(radio-frequency)-pulses: the x-puls Remember: Using the vector model 90 and 180 pulses hitting z-magnetization were described like this

14 The PROF 14/82 The rule for the x-pulse looks like that I z βi x I z cosβ -I y sinβ β = 90 : cos β = 0, sin β = 1, the result is -I y β = 180 : cos β = -1, sin β = 0, the result ist -I z

15 The PROF 15/82 z y Rules for pulses from other directions x are then easily derived I z I z I z βi x βi y βi -x I z cosβ -I y sinβ I z cosβ + I x sinβ I z cosβ + I y sinβ I z βi -y I z cosβ -I x sinβ

16 The PROF 16/82 Pulses do also act on transverse magnetization but not if both point in the z y x same direction I x I y βi y βi x I x cosβ -I z sinβ I y cosβ + I z sinβ βi x I x I y βi y I x I y } no effect!!

17 The PROF 17/82 chemical shift: remember: chemical shift Ω 0 acts for a time τ I x I z Ω 0 τ I x cos Ω 0 τ + I y sin Ω 0 τ =I x cos 2πδτ +I y sin 2πδτ angular frequency Ω 0 normal frequency δ (in hertz)

18 The PROF 18/82 z y Chemical shift acts only on transverse x magnetization, I z remains untouched, That is why it is sometimes called z-pulses I x I y I z I z Ωτ I z Ωτ I z Ωτ I x cosωτ + I y sinωτ I y cosωτ -I x sinωτ I z

19 The PROF 19/82 z y Scalar coupling leads to operators in which x several operators are multiplied with each other: product operators I 1x I 1z I 2z πj 12 τ I 1x cosπj 12 τ + 2I 1y I 2z sinπj 12 τ in-phase magnetization: This part stays untouched anti-phase magnetization: coupling causes an modulation of transverse magnetization due to the coupling partner

20 The PROF 20/82 z y Scalar coupling acts on various kinds of x transverse magnetization... I 1x I 1y 2I 1x I 2z 2I 1y I 2z I 1z I 2z πj 12 τ I 1z I 2z πj 12 τ I 1z I 2z πj 12 τ I 1z I 2z πj 12 τ I 1x cosπj 12 τ + 2I 1y I 2z sinπj 12 τ I 1y cosπj 12 τ - 2I 1x I 2z sinπj 12 τ 2I 1x I 2z cosπj 12 τ + I 1y sinπj 12 τ 2I 1y I 2z cosπj 12 τ - I 1x sinπj 12 τ

21 The PROF 21/82 z y... but not on longitudinal magnetization or on multiple quantum magnetization x I 1z 2I 1x I 2y 2I 1x I 2y I 1z I 2z πj 12 τ I 1z I 2z πj 12 τ I 1z I 3z πj 13 τ I 1z 2I 1x I 2y 2I1x I2y cosπj12 τ + 2I1y I2y I3z sinπj12 τ

22 The PROF 22/82 The evolution due to scalar coupling can be visiualized with cartesian coordinates z y x

23 The PROF 23/82 Aim of the calculation is to derive what happens during a pulse sequence. Then we must ask what types of product operators lead to detectable signals during the acquisition 1 H n X Δ Δ Δ Decoupling What is present at that point that is detectable?

24 The PROF 24/82 Not detectable are longitudinal magentization (I 1z ) or multiple quantum magnetization (z.b. I 1x I 2y ) since the usual selection rules are still valid Only transverse in-phase (I 1x ) magnetization is detectable, anti-phase magnetization (e.g. I 1x I 2z ) can evolve into something detectable during the acquisition time

25 The PROF 25/82 Let us now calculate what results from in-phase magnetization during the acquistion time under the influence of chemical shift and scalar coupling? I 1x I z Ω 1 t aq I 1x cosω 1 t aq + I 1y sin Ω 1 t aq I 1z I 2z πj 12 t aq I 1x cosω 1 t aq cosπj 12 t aq + 2I 1y I 2z cosω 1 t aq sinπj 12 t aq + I 1y sinω 1 t aq cosπj 12 t aq - 2I 1x I 2z sinω 1 t aq sinπj 12 t aq = I 1x cosω 1 t aq cosπj 12 t aq + I 1y sinω 1 t aq cosπj 12 t aq

26 The PROF 26/82 We use Ω 1 = 2πδ 1 and we get (using our trigonometiric formulas) I 1x ½[cos2π(δ 1 + J 12 /2)t aq + cos2π(δ 1 -J 12 /2)t aq ] + I 1y ½[sin2π(δ 1 + J 12 /2)t aq + sin2π(δ 1 -J 12 /2)t aq ] Imaginärteil! = I 1 ½[expi 2π(δ 1 + J 12 /2)t aq + expi 2π(δ 1 -J 12 /2)t aq ] in-phase magnetization results in two lines with the same sign, seperated by J Hz and centered around δ 1 Hz

27 The PROF 27/82 J 12 I 1x δ 1 Hence the name in-phase magnetization, since the result is an in-phase doublett

28 The PROF 28/82 This is what we get in a conventional 1D-NMRexperiment 90 -y ppm

29 The PROF 29/82 How about anti-phase magnetization? 2I 1x I 2z I z Ω 1 t aq 2I 1x I 2z cosω 1 t aq + 2I 1y I 2z sin Ω 1 t aq I 1z I 2z πj 12 t aq 2I 1x I 2z cosω 1 t aq cosπj 12 t aq + I 1y cosω 1 t aq sinπj 12 t aq + 2I 1y I 2z sinω 1 t aq cosπj 12 t aq -I 1x sinω 1 t aq sinπj 12 t aq = - I 1x sinω 1 t aq sinπj 12 t aq + I 1y cosω 1 t aq sinπj 12 t aq

30 The PROF 30/82 Again we use Ω 1 = 2πδ 1 and get I 1x ½[cos2π(δ 1 + J 12 /2)t aq -cos2π (δ 1 -J 12 /2)t aq ] + I 1y ½[sin2π(δ 1 + J 12 /2)t aq -sin2π (δ 1 -J 12 /2)t aq ] Imaginärteil! = I 1 ½[expi 2π(δ 1 + J 12 /2)t aq -exp2i π(δ 1 -J 12 /2)t aq ] anti-phase magnetization thus results in two lines with oposite sign, seperated by J Hz and centered around δ 1 Hz

31 The PROF 31/82 J 12 2I 1x I 2z δ 1 Hence the name anti-phase magnetization, since we obtain an anti-phase doublet To create that type of magnetization we need a bit more then just one pulse!!

32 The PROF 32/82 a first summary Produkt operators are used for a convinient calculation of complex pulse sequences Usually the calculation is performed up to the end of the pulse sequence right before the acquistion. The detectable operators that are present at that point give rise to well know signals during detection. The calculation thus describes the result of the pulse sequence

33 The PROF 33/82 Since the calculation can get confusing quite quickly follow those rules rule 1: calculate carfully, avoid sign or writing mistakes rule 2: always try to calculate seperable interactions seperately, i.e. chemical shift seperatly from scalar coupling rule 3: do not re-caclulate building blocks

34 34/82 Building blocks

35 Building blocks 35/82 Now we know what kind of signals result from the detectable types of magnetization, now we can start to calculate building blocks δ H J HH H H α,β J HC C α,β C

36 Building blocks z y 36/82 x 90 y Δ H 1z 90 H x chemical shift: δ H H x δ H Δ H x cos Ω H Δ + H y sin Ω H Δ δ H t aq H x cos Ω H Δ cos Ω H t aq + H y cos Ω H Δ sin Ω H t aq H y sin Ω H Δ cos Ω H t aq -H x sin Ω H Δ sin Ω H t aq

37 Building blocks z y 37/82 x H x cos Ω H Δ cos Ω H t aq + H y cos Ω H Δ sin Ω H t aq H y sin Ω H Δ cos Ω H t aq -H x sin Ω H Δ sin Ω H t aq = H x cos Ω H (Δ + t aq ) + H y sin Ω H (Δ + t aq ) = H expi Ω H (Δ + t aq ) = H expi (Ω H Δ) expi (Ω H t aq ) Obviously all signals acquire a phase that is dependent on the chemical shift, no phase correction is thus possible in a simple 1D spectrum

38 Building blocks z y 38/82 90 y x Δ H 1z 90 H x This is solved by keeping Δ short. exp(ω H Δ) = 1 + Ω H Δ for small values of Ω H Δ We see that in this case a linear phase correction (also called first order phase correction) is sufficient.

39 Building blocks z y 39/82 x 90 x 180 x Δ 2 Δ 1 H 1z 90 H x This sequence has already been inspected using the vector model, the result should be the same: No chemical shift, homonuclear but no heteronuclear scalar coupling

40 Building blocks z y 40/82 x First we do the calculation Δ 1 90 x Δ x H 1z 90 H x for chemical shift δ H 90 H x H z -H y δ H Δ 1 -H y cos δ H Δ 1 + H x sin δ H Δ H x H y cos δ H Δ 1 + H x sin δ H Δ 1 δ H Δ 2 Hy cos δ H Δ 1 cos δ H Δ 2 -H x cos δ H Δ 1 sin δ H Δ 2 H x sin δ H Δ 1 cos δ H Δ 2 + H y sin δ H Δ 1 sin δ H Δ 2

41 Building blocks H y cos δ H Δ 1 cos δ H Δ 2 -H x cos δ H Δ 1 sin δ H Δ 2 H x sin δ H Δ 1 cos δ H Δ 2 + H y sin δ H Δ 1 sin δ H Δ 2 =H y cos δ H (Δ 1 - Δ 2 ) + H x sin δ H (Δ 1 - Δ 2 ) if Δ 1 = Δ 2 = Δ =H y z y 41/82 x i.e. chemical shift has vanished in the end, it has been refocussed! (as we have seen in the vector model)

42 90 x 180 x Δ 1 90 H H Δ x 1z 2 Building blocks Now we calculate homonuclear coupling J HH z y 42/82 x 90 H x H 1z -H 1y πj HH Δ 1 -H 1y cos πj HH Δ 1 + 2H 1x H 2z sin πj HH Δ H x H 1y cos πj HH Δ 1-2H 1x H 2z sin πj HH Δ 1 πj HH Δ 2 H 1y cos πj HH Δ 1 cos πj HH Δ 2-2H 1x H 2z cos πj HH Δ 1 sin πj HH Δ 2-2H 1x H 2z sin πj HH Δ 1 cos πj HH Δ 2 -H 1y sin πj HH Δ 1 sin πj HH Δ 2

43 Building blocks 43/82 H 1y cos πj HH Δ 1 cos πj HH Δ 2-2H 1x H 2z cos πj HH Δ 1 sin πj HH Δ 2-2H 1x H 2z sin πj HH Δ 1 cos πj HH Δ 2 -H 1y sin πj HH Δ 1 sin πj HH Δ 2 = H 1y cos πj HH (Δ 1 + Δ 2 ) - 2H 1x H 2z sin πj HH (Δ 1 + Δ 2 ) if Δ 1 = Δ 2 = Δ H 1y cos πj HH 2Δ -2H 1x H 2z sin πj HH 2Δ i.e. homonuclear coupling is still present at the end of the spin echo it has not been refocussed!

44 90 x 180 x Δ 1 90 H H Δ x 1z 2 Building blocks Finally heteronuclear coupling J HX z y 44/82 x 90 H x H z -H y πj HX Δ 1 -H y cos πj HX Δ 1 + 2H x X z sin πj HX Δ H x Hy cos πj HX Δ 1 + 2H x X z sin πj HX Δ 1 πj HX Δ 2 H y cos πj HX Δ 1 cos πj HX Δ 2-2H x X z cos πj HX Δ 1 sin πj HX Δ 2 + 2H x H z sin πj HX Δ 1 cos πj HX Δ 2 +H y sin πj HX Δ 1 sin πj HX Δ 2

45 Building blocks 45/82 H y cos πj HX Δ 1 cos πj HX Δ 2-2H x X z cos πj HX Δ 1 sin πj HX Δ 2 + 2H x H z sin πj HX Δ 1 cos πj HX Δ 2 +H y sin πj HX Δ 1 sin πj HX Δ 2 = H y cos πj HX (Δ 1 - Δ 2 ) - 2H x X z sin πj HX (Δ 1 - Δ 2 ) if Δ 1 = Δ 2 = Δ H y i.e. heteronuclear coupling has vanished, it has been refocussed!

46 Δ 1 Δ 2 chemical shift: δ H Building blocks 46/82 1 H n X scalar (J-) coupling: J HH,J HX δ H and J HH are already known δ H refocussed if Δ 1 = Δ 2 J HH not refocussed if Δ 1 = Δ 2

47 J HX? Building blocks z y 47/82 x 90 H x H z -H y πj HX Δ 1 -H y cos πj HX Δ 1 + 2H x X z sin πj HX Δ H x H y cos πj HX Δ 1-2H x X z sin πj HX Δ 1 πj HX Δ X x H y cos π J HX Δ 1 cos π J HX Δ 2-2H x X z cos π J HX Δ 1 sin π J HX Δ 2-2H x H z sin π J HX Δ 1 cos π J HX Δ 2 -H y sin π J HX Δ 1 sin π J HX Δ 2 =H y cos π J HX (Δ 1 + Δ 2 ) - 2H x X z sin π J HX (Δ 1 + Δ 2 ) if Δ 1 = Δ 2 = Δ : H 1y cos πj HX 2Δ -2H 1x H 2z sin πj HX 2Δ i.e. heteronuclear coupling is not refocussed any more!

48 Building blocks 48/82 H y cos πj HX 2Δ -2H x X z sin πj HX 2Δ If Δ = 1/4J HX is choosen 2H x X z is obtained, i.e. an anti-phase signal, that is quite often utilized in complex pulse sequences If Δ = 1/2J HX is choosen H 1y is obtained, i.e. an in-phase signal J HX J HX

49 Building blocks 49/82 but H y cos πj HX 2Δ -2H x X z sin πj HX 2Δ If Δ is set to a value short relative to 1/2J, then the coupling does hardly have an effect J = 5 Hz, 2Δ = 5 msec << 1/2J = 100 msec cos πj HX 2Δ = 0.99 sin πj HX 2Δ = 0.08 i.e. the coupling has not evolved noticably

50 Building blocks 50/82 1 H Δ 1 Δ 2 chemical shift: δ H n X scalar (J-) coupling: J HH,J HX δ H and J HH are already known δ H not refocussed if Δ 1 = Δ 2 J HH not refocussed if Δ 1 = Δ 2

51 Building blocks z y 51/82 J HX? x 90 H x H z -H y πj HX Δ 1 -H y cos πj HX Δ 1 + 2H x X z sin πj HX Δ X x -H y cos πj HX Δ 1-2H x X z sin πj HX Δ 1 πj HX Δ 2 -H y cos πj HX Δ 1 cos πj HX Δ 2 + 2H x X z cos πj HX Δ 1 sin πj HX Δ 2-2H x H z sin πj HX Δ 1 cos πj HX Δ 2 -H y sin πj HX Δ 1 sin πj HX Δ 2 = - H y cos πj HX (Δ 1 - Δ 2 ) - 2H x X z sin πj HX (Δ 1 - Δ 2 ) if Δ 1 = Δ 2 = Δ : - H y i.e. heteronucleare coupling is again refocussed!

52 Building blocks 52/82 Those building blocks are now known and we do not need to recalculate them later on 1 H n X Δ 1 Δ 2 1 H Δ 1 Δ 2 1 H Δ 1 Δ 2 n X n X

53 53/82 Two-dimensional NMR-spectroscopy: The COSY

54 2D NMR: COSY 54/82 2D-NMR sequences contain two new elements: evolution time and mixing time preparation evolution mixing detection (t 1 ) (t 2 ) evolution: here a second time domain is created by indirect detection mixing: transfer of magentization from spin to spin via interactions between spins

55 2D NMR: COSY 55/82 The NMR signale that is recorded during the acquisition is a damped cosine (we look only at the real part) s(t)= exp(-t/t 2 ) exp(iω 0 t) s(t)= exp((iω 0-1/T 2 )t)

56 2D NMR: COSY 56/82 To analyze the signal we need to digitze it s(t)= exp((iω 0-1/T 2 )t) analog in digital out s(kδt)= exp((iω 0-1/T 2 )kδt)

57 2D NMR: COSY 57/82 1,5 intensity 1 0,5 0-0,5 75 Hz The FID is converted into a series of data points ,005 0,01 0,015 0,02 0,025 0,03 0,035 0,04 time (sec) s(11δt)= exp((iω 0-1/T 2 )11Δt) s(2δt)= exp((iω 0-1/T 2 )2Δt) s(0δt) = s(0) = exp((iω 0-1/T 2 )0) = 1

58 2D NMR: COSY 58/82 1, Hz 0,8 intensity 0,6 0,4 0, frequency A digital Fourier transformation (DFT) converts that into another series of data points the spectrum

59 2D NMR: COSY 59/82 The simplest two-dimensional spectrum is the COSY 90 x t 1 90 Y t 2 preparation evolution mixing detection

60 2D NMR: COSY 60/82 90 x 90 Y t t 2 1 We consider two protons with Ω H1 = 2πδ H1 and Ω H1 = 2πδ H2 90 H x H 1z -H 1y 2πδ H t 1 -H 1y cos 2πδ H1 t 1 + H 1x sin 2πδ H1 t 1 not detectable 90 H y -H 1y cos 2πδ H1 t 1 -H 1z sin 2πδ H1 t 1 2πδ H t 2 -H 1y cos 2πδ H1 t 1 cos 2πδ H1 t 2 + H 1x cos 2πδ H1 t 1 sin 2πδ H1 t 2 -H 1 cos 2πδ H1 t 1 exp 2πδ H1 t 2 Quadraturdetektion

61 2D NMR: COSY 61/82 90 y 90 Y t 1 t 2 To achieve quadrature detection in the indirect dimension we do a second experiment 90 H y H 1z H 1x 2πδ H t 1 H 1x cos 2πδ H1 t 1 + H 1y sin 2πδ H1 t 1 90 H y -H 1z cos 2πδ H1 t 1 + H 1y sin 2πδ H1 t 1 2πδ H t 2 not detectable H 1y sin 2πδ H1 t 1 cos 2πδ H1 t 2 + H 1x sin 2πδ H1 t 1 sin 2πδ H1 t 2 H 1 sin 2πδ H1 t 1 exp 2πδ H1 t 2

62 2D NMR: COSY 62/82 Taken together we have a hypercomplex signal H 1 cos 2πδ H1 t 1 exp 2πδ H1 t 2 und H 1 sin 2πδ H1 t 1 exp 2πδ H1 t 2 = H 1 exp 2πδ H1 t 1 exp 2πδ H1 t 2 = H 1 exp 2πδ H1 (k Δt 1 ) xexp2πδ H1 (k Δt 2 ) 90 x kδt 1 90 y kδt2 90 x kδt 1 90 y kδt2 This is done by the incrementation of Δt 1 This is done by the ADC 90 x kδt 1 90 y kδt 2 90 x kδt 1 90 y kδt 2 We obtain a two-dimensional area of data points...

63 2D NMR: COSY 63/82... a two-dimensional FID

64 2D NMR: COSY 64/82 After two FTs we get our well known 2D spectrum

65 2D NMR: COSY 65/82 But up to know we have the same information on both axes H 1 exp 2πδ H1 t 1 exp 2πδ H1 t 2 = H 1 exp 2πδ H1 (k Δt 1 ) exp 2πδ H1 (k Δt 2 ) (the same for H 2 ) We have created a second dimension by incrementing during the evolution time but our spectrum has only a diagonal δ H (ppm) H 1 H 2 δh (ppm) δ 2 δ 1 δ 2 δ 1

66 2D NMR: COSY 66/82 Now we remember that mixing is achieved via coupling and we calculate for J HH as well 90 x 90 Y t 1 t 2 90 H x H 1z -H 1y 2πδ H t 1 -H 1y cos 2πδ H1 t 1 + H 1x sin 2πδ H1 t 1 Watch this operator!! πj HH t 1 -H 1y cos 2πδ H1 t 1 cos πj HH t 1 + 2H 1x H 2z cos 2πδ H1 t 1 sin πj HH t 1 + H 1x sin 2πδ H1 t 1 cos πj HH t 1 + 2H 1y H 2z sin 2πδ H1 t 1 sin πj HH t 1

67 2D NMR: COSY 67/82 90 x 90 Y t t 2 1 Then the second 90 pulse Here is where the transfer takes place!! 90 H y -H 1y cos 2πδ H1 t 1 cos πj HH t 1-2H 1z H 2x cos 2πδ H1 t 1 sin πj HH t 1 -H 1z sin 2πδ H1 t 1 cos πj HH t 1 + 2H 1y H 2x sin 2πδ H1 t 1 sin πj HH t 1 not detectable two types of dectable magnetization remain -H 1y cos 2πδ H1 t 1 cos πj HH t 1-2H 1z H 2x cos 2πδ H1 t 1 sin πj HH t 1

68 2D NMR: COSY 68/82 90 x 90 Y t t 2 1 Then we start the acqusition -H 1y cos 2πδ H1 t 1 cos πj HH t 1-2H 1z H 2x cos 2πδ H1 t 1 sin πj HH t 1 δ H t 2 -H 1y cos 2π δ H1 t 1 cos πj HH t 1 cos 2π δ H1 t 2 + H 1x cos 2π δ H1 t 1 cos πj HH t 1 sin 2π δ H1 t 2 H1! -2H 1z H 2x cos 2π δ H1 t 1 sin πj HH t 1 cos 2π δ H2 t 2-2H 1z H 2y cos 2π δ H1 t 1 sin πj HH t 1 sin 2π δ H2 t 2 H2!

69 2D NMR: COSY 69/82 90 x 90 Y t t 2 1 we focus on detectable magnetization πj HH t 2 -H 1y cos 2πδ H1 t 1 cos πj HH t 1 cos 2πδ H1 t 2 cos πj HH t 2 + H 1x cos 2πδ H1 t 1 cos πj HH t 1 sin 2πδ H1 t 2 cos πj HH t 2 -H 2y cos 2πδ H1 t 1 sin πj HH t 1 cos 2πδ H2 t 2 sin πj HH t 2 + H 2x cos 2πδ H1 t 1 sin πj HH t 1 sin 2πδ H2 t 2 sin πj HH t 2 Im! Im! The caclulation for the second FID (1. pulse 90 y ) is omitted!

70 2D NMR: COSY 70/82 90 x 90 Y t t 2 1 As a result of our calculation we get: (only the real part is given) -H 1y cos 2πδ H1 t 1 cos πj HH t 1 cos 2πδ H1 t 2 cos πj HH t 2 -H 2y cos 2πδ H1 t 1 sin πj HH t 1 cos 2πδ H2 t 2 sin πj HH t 2 In t 1 immer die Verschiebung von H1 In t 2 die Verschiebung von H1 oder H2

71 2D NMR: COSY 71/82 Now something new has shown up: There are signals that have been labeled with different chemical shifts in both dimensions: the crosspeaks A transfer of magnetization has Transfer H 1 H 2 δh (ppm) δ 1 taken place δ H (ppm) δ 1 δ 2 δ 2

72 2D NMR: COSY 72/82 90 x t 1 90 Y t 2 COrrelation SpectroscopY = COSY The mixing time is a simple 90 pulse in this case. It causes a transfer of magnetization. Cross peaks in a 2D spectrum thus indicate the presence of a scalar coupling between the two nuclei whose frequencies intersect at the position of the cross peak.

73 2D NMR: COSY 73/82 We will now take a look at the fine structure of the peaks

74 2D NMR: COSY 74/82 -H 1y cos 2πδ H1 t 1 cos πj HH t 1 cos 2πδ H1 t 2 cos πj HH t 2 -H 2y cos 2πδ H1 t 1 sin πj HH t 1 cos 2πδ H2 t 2 sin πj HH t 2 Let s use our trigonometric formulas = -H 1y ½ [cos2π(δ H1 + J HH /2)t 1 + cos2π(δ H1 -J HH /2)t 1 ] x ½ [cos2π(δ H1 + J HH /2)t 2 + cos2π(δ H1 -J HH /2)t 2 ] -H 2y ½ [sin2π(δ H1 + J HH /2)t 1 -sin2π(2πδ H1 -J HH /2)t 1 ] x ½ [sin2π(δ H2 + J HH /2)t 2 -sin2π(δ H2 -J HH /2)t 2 ]

75 2D NMR: COSY 75/82 = -H 1y ½ [cos2π(δ H1 + J HH /2)t 1 + cos2π(δ H1 -J HH /2)t 1 ] x ½ [cos2π(δ H1 + J HH /2)t 2 + cos2π(δ H1 -J HH /2)t 2 ] -H 2y ½ [sin2π(δ H1 + J HH /2)t 1 -sin2π(2πδ H1 -J HH /2)t 1 ] x ½ [sin2π(δ H2 + J HH /2)t 2 -sin2π(δ H2 -J HH /2)t 2 ] We get a total of 8 peaks, 4 with a cosine and with δ H1 in both dimensions, 4 with a sine and with different chemical shifts in both dimensions

76 2D NMR: COSY 76/82 The first four are all positive (inphase), they have δ H1 in both dimensions (diagonal signal) and all are resulting from cosine functions J HH J HH δ H1 δ H1 + H 1 cos2π(δ H1 + J HH /2)t 1 cos2π(δ H1 + J HH /2)t 2 + H 1 cos2π(δ H1 + J HH /2)t 1 cos2π(δ H1 -J HH /2)t 2 + H 1 cos2π(δ H1 -J HH /2)t 1 cos2π(δ H1 + J HH /2)t 2 + H 1 cos2π(δ H1 -J HH /2)t 1 cos2π(δ H1 -J HH /2)t 2

77 2D NMR: COSY 77/82 The second four have alternating signs (anti-phase), different chemical shifts (δ H1 and δ H2 ) in the two dimensions (cross peaks) and all result from sine functions J HH J HH δ H2 δ H1 + H 2 sin2π(δ H1 + J HH /2)t 1 sin2π(δ H2 + J HH /2)t 2 -H 2 sin2π(δ H1 + J HH /2)t 1 sin2π(δ H2 -J HH /2)t 2 -H 2 sin2π(δ H1 -J HH /2)t 1 sin2π(δ H2 + J HH /2)t 2 + H 2 sin2π(δ H1 -J HH /2)t 1 sin2π(δ H2 -J HH /2)t 2

78 2D NMR: COSY 78/82 We thus have a problem with phase in-phase anti-phase Δφ = 90 cos vs. sin absorbtiv dispersiv

79 2D NMR: COSY 79/82 either or diagonal peak absorptiv, cross peak dispersiv diagonal peak dispersiv, cross peak absorptiv cross peak diagonal peak

80 2D NMR: COSY 80/82 That means: either the diagonal is braod an the cross peaks can vanish because of overlap or the cross peaks are braod and the signals with different signs cancel

81 2D NMR: COSY 81/82 In the regular COSY the solution is a magnitude calculation We have seen already that the DQF-COSY is another solution for that problem

82 82/82 That s it