Heteronuclear correlation - HETCOR

Transcription

1 Heteronuclear correlation - HETCOR Last time we saw how the second dimension comes to be, and we analyzed how the COSY experiment (homonuclear correlation) works. In a similar fashion we can perform a 2D experiment in which we analyze heteronuclear connectivity, that is, which 1 H is connected to which 13 C. This is called HETCOR, for HETeronuclear CORrelation spectroscopy. The pulse sequence in this case involves both 13 C and 1 H, because we have to somehow label the intensities of the 13 C with what we do to the populations of 1 H. The basic sequence is as follows: C: H: t 1 { 1 H}

2 HETCOR (continued) We first analyze what happens to the 1 H proton (that is, we ll see how the 1 H populations are affected), and then see how the 13 C signal is affected. For different t 1 values we have: y z 90, t 1 = 0 x 90 x y y z 90, t 1 = J / 4 x 90 x y y z 90, t 1 = 3J / 4 x 90 x y

3 HETCOR ( ) As was the case for COSY, we see that depending on the t 1 time we use, we have a variation of the population inversion of the proton. We can clearly see that the amount of inversion depends on the J CH coupling. Although we did it on-resonance for simplicity, we can easily show that it will also depend on the 1 H frequency (δ). From what we know from SPI and INEPT, we can tell that the periodic variation on the 1 H population inversion will have the same periodic effect on the polarization transfer to the 13 C. In this case, the two-spin energy diagram is 1 H- 13 C: αβ 2 13 C 4 ββ 1 H 1,2 3,4 1 H αα 1 13 C 3 βα 1,3 2,4 I S Now, since the intensity of the 13 C signal that we detect on t 2 is modulated by the frequency of the proton coupled to it, the 13 C FID will have information on the 13 C and 1 H frequencies.

4 HETCOR ( ) Again, the intensity of the 13 C lines will depend on the 1 H population inversion, thus on ω 1H. If we use a stacked plot for different t 1 times, we get: t 1 (ω 1H ) The intensity of the two 13 C lines will vary with the ω 1H and J CH between +5 and -3 as it did in the INEPT sequence. ω 13C f 2 (t 2 ) Mathematically, the intensity of one of the 13 C lines from the multiplet will be an equation that depends on ω 13C on t 2 and ω 1H on t 1, as well as J CH on both time domains: A 13C (t 1, t 2 ) trig(ω 1H t 1 ) * trig(ω 13C t 2 ) * trig(j CH t 1 ) * trig(j CH t 2 )

5 HETCOR ( ) Again, Fourier transformation on both time domains gives us the 2D correlation spectrum, in this case as a contour plot: ω 13C J CH ω 1H f 1 f 2 The main difference in this case is that the 2D spectrum is not symmetrical, because one axis has 13 C frequencies and the other 1 H frequencies. Pretty cool. Now, we still have the J CH coupling splitting all the signals of the 2D spectrum in little squares. The J CH are in the Hz range, so we can start having overlap of cross-peaks from different CH spin systems. We ll see how we can get rid of them without decoupling (if we decouple we won t see 1 H to 13 C polarization transfer ).

6 HETCOR with no J CH coupling The idea behind it is pretty much the same stuff we did with the refocused INEPT experiment t 1 / 2 t 1 / 2 13 C: H: t { 1 H} ω 1H f 1 f 2 ω 13C

7 HETCOR (continued) The 1 and 2 delays are such that we maximize antiphase 13 C magnetization for 1 J CH couplings. That is, 1 and 2 are in the 2 to 5 ms range (the average 1 J CH is ~ 150 Hz, and the 1 and 2 delays were 1 / 2J). This is fine to see CH correlations between carbons and protons which are directly attached ( 1 J CH ). Lets see what this means for camphor, which we discussed briefly in class: H 3 C CH 3 H a b H 3 C O An expansion of the HETCOR spectrum for carbons a and b would look like: f 2 ( 13 C) H b f 1 ( 1 H) H a H c C a C b

8 Long range HETCOR ( ) A problem here is that both carbons a and b are pretty similar chemically and magnetically: From this data alone we would not be able to determine which one is which. It would be nice if we could somehow determine which of the two carbons is the one closer to the proton at C c, because we would unambiguously assign these carbons in camphor: H 3 C b c CH 3 H H b H a a H 3 C O H c C a C b How can we do this? There is, in principle, a very simple experiment that relies on long-range CH couplings. Apart from 1 J CH couplings, carbons and protons will show long-range couplings, which can be across two or three bonds (either 2 J CH or 3 J CH ). Their values are a lot smaller than the direct couplings, but are still considerably large, in the order of 5 to 20 Hz. Now, how can we twitch the HETCOR pulse sequence to show us nuclei correlated through long-range couplings?

9 Long range HETCOR ( ) t 1 / 2 t 1 / 2 13 C: H: t { 1 H} We set the 1 and 2 delays to 1 / 2 * 2 J CH Since 2 J CH is around 10 Hz, 1 and 2 will be in the order of 50 ms instead of 5 ms in normal HETCOR

10 Long range HETCOR ( ) If we take our HETCOR using longer values for 1 and 2, we get: H 3 C b c CH 3 H H b H a a H c H 3 C O C a C b We can now see our long-range 1 H- 13 C coupling, and we can now determine which CH 2 carbon is which in camphor.

11 Summary The HETCOR sequence reports on which carbon is attached to what proton and shows them both - Great for natural products and organic molecules. The way this is done is by inverting 1 H population and varying the transfer of 1 H polarization to 13C during the variable t1. We can obtain a decoupled version by simply lumping in an refocusing echo in the middle. By suitable choice of delay times, a long-range HETCOR experiment can show connectivities of carbons with protons 2, 3 and 4 bonds away. Modern versions of the two above experiments are called HMQC and HMBC, respectively.

12 Connections through space : NOE So far we have analyzed experiments that work because the system has scalar couplings. COSY, and HETCOR, tell us about the chemical structure of the system, but nothing about the conformation or stereochemistry. We had seen at the very beginning that if we saturate a proton in the sample, it will relax by either zero- or doublequantum processes, giving energy (enhancing) the signals of protons dipolarly coupled to it (protons close by ). This was the nuclear Overhauser effect (NOE): ββ ( ) W 1I W 1S ( ) αβ W 0IS W 2IS βα ( ) W 1S W 1I αα ( ) We have seen that relaxation by either W 2IS or W 0IS will occur depending on the size of the molecule, actually its rate of tumbling, or correlation time, τ c.

13 The two-spin system case We have established a relationship between the different rates or probabilities and the magnitude of the NOE. If we limit everything to 1 H- 1 H (γ I / γ S = 1), we have: η I = W 2IS -W 0IS 2W 1S + W 2IS + W 0IS = f I {S} f I {S} is the NOE enhancement of nucleus I when we saturate nucleus S. You may also find this equation written in another way in some textbooks. We define the numerator as the cross-relaxation rate constant, σ IS, and the denominator as the dipolar longitudinal relaxation, ρ IS. σ IS = IS W 2IS -W 2IS -W 0IS 0IS ρ IS = IS 2W 2W 1S + 1S W 2IS + 2IS W 0IS 0IS η I = I σ IS / IS / ρ IS =f IS =f I {S} I {S} It has been found that the sign of the NOE will depend on the tumbling rate (size) of the molecule. There are two distinct cases, one in which ω * τ c << 1, or extreme narrowing limit, and ω * τ c >> 1, or diffusion limit.

14 Relaxation rates W and internuclear distance r W 0IS γ 0IS γ I2 γ I2 γ S2 r -6 S2 r IS -6 τ IS τ c / c /[[ 1 + (ω (ω I - I -ω S ) S ) 2 2 τ 2 τ c 2 c ]] W 2IS γ 2IS γ I2 γ I2 γ S2 r -6 S2 r IS -6 τ IS τ c / c /[[ 1 + (ω (ω I + I ω S ) S ) 2 2 τ 2 τ c 2 c ]] W 1S γ 1S γ I2 γ I2 γ S2 r -6 S2 r IS -6 τ IS τ c / c /[[ 1 + ω S2 τ 2 S2 τ c 2 c ]] W 1I γ 1I γ I2 γ I2 γ S2 r -6 S2 r IS -6 τ IS τ c / c /[[ 1 + ω I2 τ 2 I2 τ c 2 c ]] The relationship is to the inverse sixth power of r IS, which means that the NOE decays very fast as we pull the two nuclei away from each other. For protons, this means that we can see things which are at most 5 to 6 Å apart in the molecule (under ideal conditions ).

15 Steady-state NOE In small, rigid molecules the following relationships are valid: (ω (ω I - I -ω S ) S )** ττ c << c << 1 ω S * S * ττ c << c << 1 ω S * S * ττ c << c << 1 (ω (ω I + I ω S ) S )** ττ c << c << 1 Basically, they move really fast. There is a big simplification of the probability equations (the W stuff), and we end with a simple dependency for the NOE enhancement: η I = 0.5. Bummer. We had the r IS -6 dependency there and now it s gone. The problem is that we have only two spins, and this basically means no geometry. This is normally the case for 1 H- 13 C. Fortunately, when we look at the proton enhancements in a molecule, we are always looking at more than one proton. This means that if we irradiate a proton, it will be dipolarly coupled to several protons at the same time. b a c r ba r ac

16 Steady-state NOE (continued) What happens is that we have competing relaxation mechanisms for the proton we are saturating (two or more protons in the surroundings). Now the rates of the relaxation with the different protons becomes important (the respective Ws). The equations get really complicated, but if we are still in the extreme narrowing limit, we can simplify things quite a bit. In the end, we can establish a simple relationship between the NOE enhancement and the internuclear distances: ff I {S} I {S} = η max * max * r -6 r IS -6 - IS -Σ X f X f X {S} X {S}** r -6 r IX -6 IX r -6 r IS -6 + IS Σ X r -6 X r IX -6 IX In order to estimate distances between protons in a molecule we can saturate one nuclei and analyze the relative enhancements of other protons. This is know as steady-state NOE. We take two spectra. The first spectrum is taken without any irradiation, and the second with. The two are subtracted, and the difference gives us the enhancement from which we can estimate distances...

17 NOE difference spectroscopy If our molecule has three protons, two of them at a fixed distance (a CH 2 ), we have: Ha Hb Hc Hb Ha C Hc _ = η ab η ac Since we have a reference pair of protons (Ha and Hb) for which we know the distance, we can establish a reference NOE effect for it, and then calculate the distance between the other protons (Ha and Hc in this example): η ab ab r -6 r ab -6 ab η ac ac r -6 r ac -6 ac rr ac =r ac =r ab * ab *(( η ab / ab / η ac ) ac ) -1/6-1/6 The sizes of the peaks are measured by integration. Since the effect is not symmetric, we usually do the I S and then the S I enhancement and take an average value.

18 Transient NOE One of the problems of steady-state NOE is that we are continuously giving power to the system (saturation). This works well for small molecules, because W2 processes (double-quantum) are dominant and we have few protons. However, as the size and τ c increase, other processes are more important (normal single-quantum spin-spin relaxation and zero-quantum transitions). Additionally, there are more protons in the surroundings of a larger molecule, and we have to start considering a process called spin diffusion: I S Basically, the energy transferred from I to S then diffuses to other nuclei in the molecule. We can see an enhancement of a certain proton even if it is really far away from the center we are irradiating, which would give us ambiguous results. Therefore, we need to control the amount of time we saturate the system. The longer we irradiate, the more spin diffusion

19 Transient NOE (continued) There are also some technical difficulties if we try to do this by selective saturation. Since what we need is to see how a system returns to equilibrium through cross-relaxation, we can selectively invert one transition and then see how the NOE grows with time: ( ) αβ ββ ( ) W 1I W 1S W 2IS βα () W 0IS W 1S W 1I αα ( ) A pulse sequence to do this could be the following: 180 s 90 t m selective inversion The last π / 2 pulse is usually called a read pulse, and its only job is to convert whatever magnetization is in <z> after t m into <xy> magnetization (detectable). All the equations are the same, but the NOE will also depend on the mixing time, t m.

20 Transient NOE ( ) If we do it for different t m values, we get NOE build-up curves, which in the case of two isolated protons and ideal conditions are exponentials that grow until they reach η max. η max If we also take into account the T 1 /T 2 relaxation, the NOE grows and then falls to zero (all the magnetization returns to the <z> axis ): t m η max t m

21 Transient NOE and NOESY If we had more spins and spin diffusion, the curve (for three spins, I, S, and X) would look like this: η max Diffusion to X If we want to measure distances more or less accurately, we have to find a compromise between the mixing time t m that we use and the spin-diffusion we may have. Doing one proton at a time takes forever, so we obviously use a 2D experiment in which we study all protons at the same time. The sequence is called NOESY (NOE SpectroscopY), and it s shown below: t m t 1 t m inversion

22 NOESY (continued) The first two pulses (with the t 1 variable time) are basically an inversion for all protons, in which we label everything with chemical shifts and couplings (no re-focusing). What ends up in the <-z> axis evolves during the t m mixing time, and dipolarly coupled spins will suffer NOE crossrelaxation according to all the equations we ve seen. The 2D spectrum after 2D FT has chemical shifts in both f 1 and f 2 (t 2 has all the chemical shifts also) and cross-peaks for systems that are dipolarly coupled: δ (f 1 ) Hb Ha C Hc Hb Ha Hc δ (f 2 ) The size of the cross-peak will depend on the internuclear distances. These are measured relative to a cross-peak for which we know a distance, using the cross-peak volumes.

23 NOESY ( ) One thing we have to consider is the sign of the crosspeaks. This will depend on the correlation time τ c, in the same way that the NOE enhancement does. We therefore have positive cross-peaks for ω * τ c << 1 and negative crosspeaks for ω * τ c << 1. As we said earlier, there is a range of ω * τ c values for which the enhancement can be zero. If we express the NOE enhancement as a function of ω * τ c we get a sigmoid function, and we solve it for η I = 0: η I = 5 + ω 2 τ c2-4ω 4 τ 4 c η I = 0 for ω * τ c = ω 2 τ c2 + 4ω 4 τ 4 c The options are to change τ c (kind of hard), ω (the magnet is pretty hard to change ), or trick the system to think that we are working at a very low field, where ω * τ c << 1 always.

24 Summary The NOE enhancement is, among other things, proportional to r -6, so it allows us to make determination of internuclear distances: stereochemistry and conformation. Depending on the size/rigidity (τ c ) of the molecule, steadystate or transient NOE experiments will give more accurate results: Small organic compounds: Steady-state NOE. Proteins, nucleic acids, saccharides: Transient NOE, build-up curves. We can do everything at once using NOESY (transient NOE).

25 Spin-locking - Working at lower fields So far all the NMR experiments that we have studied work at the magnetic field of the magnet, which is pretty big. We want this because it increases sensitivity and resolution. However, there are certain cases in which a lower magnetic field would come in real handy. For example, we saw that in certain cases having a fixed B o and a molecule with a particular τ c precludes the use of NOESY. Ideally we would like to have the resolution and sensitivity that are associated with B o, but study the behavior of the spin system (polarization transfer, coupling, cross-correlation and relaxation) at a different field. In a 2D experiment this means that the preparation, evolution, and acquisition periods are carried out at B o, but the mixing is done at a lower field. The technique normally used to achieve this is called spinlocking. The idea is that we take the spins (magnetization) away from the effects of B o by locking them with a different field, i. e., we make it precess at a new B magnetic field not aligned with <z>.

26 Spin-locking - Theory In order to spin-lock the magnetization we first have to take it away from the <z> direction (away from B o ). This normally means to put it along the <x> or <y> axis, i. e., a π / 2 pulse. Now comes the locking part. Once the magnetization is in the <xy> plane, we have to hold it there. As we said before, this involves having it precessing around a new magnetic field aligned with <x> or <y>. This is done either by applying a continuous wave field or a composite pulse (a train of pulses ) that has the same effect than CW irradiation: z z y x 90 B SL (x) y B SL x Once we achieve this condition (called the Hartmann-Hahn condition), the frequencies of all the transitions of our system will be proportional to B SL : before before SL: SL: ω o = o γγb o o after after SL: SL: ω SL = SL γγb SL SL

27 Spin-locking - Theory (continued) One thing we have to keep in mind is that B SL is a fluctuating magnetic field, applied at (or near) the resonant condition of the spins in our sample. Since it is static in the rotating frame, we only worry about its intensity. This is why these experiments are commonly called rotating-frame experiments. There are different ways to generate the B SL. One of them is simply to use a CW field that we turn on and leave on for the time we want to spin-lock the spin system. The main problem is the spectral width we can cover with CW excitation. We will spin-lock properly only spins whose ω o is close to the ω SL frequency. To cover things to the side we have to increase power a lot. ω SL The problem with this is heat. We need a pretty soupedup B SL to achieve spin locking of the whole spin system. The transmitter coil will get hot and passes heat to the sample.

28 Spin-locking - MLEV We can use short RF pulses and obtain the same results. These are usually called composite-pulses, because they are a collection of short (µs) pulses spaced over the whole mixing time period that will have the same net effect as CW irradiation. The most common one used for spin-locking is called MLEV, for Malcom LEVitt s decoupling cycle. A common variation is called DIPSI (Decoupling In the Presence of Scalar Interactions). These sequences are decoupling schemes (after all, a CW B SL can be consider as a decoupler), and we have to understand how composite pulses (CPs) work. A CP is basically a bunch of pulses lumped together that we can use repeatedly. Two typical ones are R = (π/2) x (π) y (π/2) x or R = (π/2) x (3π/2 π/2) y (π/2) x. We can see what they do to spins that are slightly non-aligned with <xy>: z z x R x y y

29 TOCSY To make a very long story short, we have thorough mixing of all states in the system, and coherence from a certain spin in a coupled system will be transferred to all other spins in it. In other words, this spin correlates to all others in the system: A B C D X The maximum transfer between two spins with a coupling of J Hz is optimal when t m = 1 / 2J. Longer t m s allow transfer to weakly coupled spins: We go deeper in the spin system. The 1D-TOCSY needs a selective π / 2 pulse to affect only the spin we want to spin-lock: 90 s t m During t m, coherence from the locked spin will spill over to all connected spins

30 TOCSY ( ) If we spin-lock different nuclei from spin systems in a molecule like this, we would get: * * A B C * Locked spin: BB AA CC

31 TOCSY ( ) Again, this is fine for a small molecule with not much stuff on it. Other problems with this sequence is the use of selective pulses, which in practice are never as selective as we need them to be. We use non-selective excitation (a hard π / 2 pulse) in a 2D technique. The pulse sequence looks like this: t 1 t m The two pulses before and after the mixing period are called trimming pulses, and are there to make the spin-lock work when we use CP hard pulses like MLEV or DISPI As for NOESY (and all homonuclear 2D experiments for that matter), during the variable time t 1 the system is modulated by chemical shifts and couplings, so we have them in f 1. During t m, we have mixing of all spins belonging to the same spin system in the molecule, so correlations between all spins that belong to the same spin system are created. In t 2 (f 2 ), we also detect δ and J, so we get a symmetrical 2D plot

32 TOCSY ( ) In the 2D plot we get all spins from a particular spin system in the same line. For the example used before: δ Taking a TOCSY is routine for peptides and polysaccharides, because the spins systems of different units are independent.