Inept

Figure 7.14

radians). The upfield component of the proton doublet, which arises from protons attached to 13C nuclei in the ft state, rotates in the opposite direction in the X-y' plane (clockwise) toward the +X axis with angular frequency —J/2 Hz (—nJ radians). After a period of time equal to 1/(2J), the two components are exactly opposite to each other, with the downfield (H = a) component on the —x axis and the upfield (H = ft) component on the +x axis. This special condition is called antiphase magnetization and is represented as 2Ix Sz, where I stands for the 1H spins and S stands for the 13C spins. This can be read as "I spin magnetization along the — X axis, antiphase with respect to the z orientation of the individual S spins." The downfield component of the doublet (which we will always show as the a component) points along the indicated axis (—x from the — 2IX part of the product) in the vector model, and the upfield (ft) component points along the opposite (+x) axis (Fig. 7.14).

Acquisition of data at this point would yield a spectrum with a proton doublet, in which the upfield component of the doublet is of opposite phase (upside down) with respect to the downfield component. If we choose +x as our reference axis (receiver phase), the downfield component will be negative absorptive and the upfield component will be positive absorptive in the spectrum (Fig. 7.14). Instead of starting acquisition at this point, however, we deliver two simultaneous 90o pulses: one at the 1H frequency and the other at the 13C frequency (Fig. 7.13). This is where the magic happens. The 1H magnetization, which is antiphase with respect to the attached 13C nucleus, is converted into 13C magnetization, which is antiphase with respect to its attached 1H nucleus. The observable (x-y plane) magnetization has jumped from the 1H to the 13C attached to it as a result of the simultaneous 90o pulses on the 1H and 13C channels. This coherence transfer is possible only if we have a J coupling between the two nuclei, and if the magnetization is in this antiphase state at the time of the pulses. In terms of product operators, the proton Ix component is rotated by the 90o proton pulse on the y' axis onto the — z axis, whereas the 13C Sz component is rotated by the 90o 13C pulse on the y' axis onto the +x! axis:

Sz component ^ Sx by 90o 13 C pulse on/ axis

—2IxSz ^ — 2(—Iz)Sx = 2SxIz by simultaneous 90o 1H and 13C pulses on / axis

Figure 7.15

This is just following the rules for each individual operator in the product (Fig. 7.15). The resulting product operator, 2SXIz, can be described as 13C magnetization on the X axis, antiphase with respect to the z magnetization (a or state) of its attached 1H nucleus. Acquisition at this time would yield an FID at the 13C frequency, and Fourier transformation with phase reference would give a carbon doublet with a downfield component of normal phase and an upfield component of opposite phase (upside-down). This is what you will see in your INEPT spectrum: an antiphase 13 C doublet. Later, we will see that this spectrum is actually four times as intense as the normal 13C spectrum: this is the enhancement part of INEPT.

As with the APT experiment (Chapter 6), we have the problem that chemical shift differences during the 1/(27) delay will lead to a hopelessly confused pattern of phases in the final spectrum. To eliminate the chemical shift evolution, we use the same strategy we used in the APT experiment: convert the 1/(2J) delay into a spin echo. By placing a 180° 1H pulse in the center of the 1/(2J) delay, we refocus the phase shifts due to 1H chemical shift differences. To make sure that the J coupling effect from the attached 13C nucleus is not refocused as well, a 180° 13C pulse is delivered simultaneously with the 180° 1H pulse. This is a more sophisticated method than the decoupler switching method used in the APT sequence. In this case to use the APT approach, we would have to decouple the 13C nuclei, which requires nonstandard hardware. The result is the same: the 1H magnetization evolves into antiphase under the influence of the J coupling from the attached 13 C nucleus, but the 1H chemical shift differences are refocused by the spin echo. The complete INEPT sequence is shown in Figure 7.16.

There are two important consequences of this method. Both result from the fact that the magnetization that is being observed (13C antiphase doublet) arises from magnetization that is rotated from its equilibrium state along the z axis by the first 90o proton pulse. The first consequence arises because the 1H population difference at equilibrium (sometimes called "polarization") is four times the carbon population difference at equilibrium. This results from the larger energy separation between the a and states for protons:

AP (1H) = (yh Bo h/4nkT )N AP (13C) = (yc Bo h/4nkT)N

where N is the number of identical nuclei in the sample and T is the temperature in kelvin. This means that, compared to a normal13 C experiment in which the observed signal originates from AP(13C), the signal will be four times as large! This increase in sensitivity is extremely important for nuclei such as 13C that have a weak nuclear magnet (small y). The second consequence is that with repetitive scans it is the 1H T1 value, and not the 13C T1 value, which determines how long the relaxation delay needs to be, as we start the experiment with AP(1H) rather than AP(13C). The 13C z magnetization can be completely saturated and this experiment will still work! Because the 1H T1 values are often shorter than those of13 C, and may be dramatically shorter than those of more "exotic" heavy nuclei, the reduction in experiment time can be significant.

Figure 7.17 shows the INEPT spectrum of sucrose (top) compared to the nondecoupled 13C spectrum (bottom) acquired with the same number of scans. This is run with the phase cycle to remove all coherence that comes from the 13C z magnetization (Sz), so the result is pure antiphase coherence on 13C. Note that the CH resonances give antiphase doublets with each component four times as intense as the corresponding component in the 13 C spectrum. The lone quaternary carbon (C-f2) is missing in the INEPT spectrum, and the CH2 resonances give rise to triplets with the central line missing (1, 0, -1). These CH2 patterns are antiphase with respect to only one of the two protons of the CH2 group (the

one from which the coherence was transferred), so the intensities result from two antiphase doublets (1, —1 and 1, -1) for which the two inner peaks are overlapped and cancel. For example, if the coherence comes from H1 to C, we have a1 a2 (most downfield peak) antiphase with respect to p1 a2 (center peak): 1, —1; and a1 /32 (center peak) antiphase with respect to p1 p2 (most upfield peak): 1, —1. These four lines form the triplet and the two inner lines (p1 a2 and a1 p2) have the same frequency, so the intensities are 1, 0, —1 for the triplet. It is easy to see that as it is, this is not a very practical experiment: without decoupling the peaks are spread out into complex overlapping multiplets (doublets, triplets, quartets), and with decoupling there is no intensity because each multiplet adds up to intensity zero. In Section 7.13, we will see how a refocusing delay brings the antiphase terms back to in-phase so we can apply 1H decoupling and have singlet peaks for each 13C resonance in the spectrum.

7.8 SELECTIVE POPULATION TRANSFER (SPT) AS A WAY OF UNDERSTANDING INEPT COHERENCE TRANSFER

At this point you may understand the math of the product operator "switch" that occurs with the two simultaneous 90o pulses, but you might be feeling unsatisfied and in need of an explanation of what is really happening. Of course, the sooner you accept the product operator math as an explanation the easier the things will be for you, but there is a vector model way of explaining INEPT coherence transfer that involves keeping track of populations. Consider a "thought experiment" in which we apply a 180o (inversion) pulse to only one component of the 1H doublet. This can actually be done using shaped (selective) pulses, as we will see in Chapter 8. In the vector model, we start with both vectors (representing the two components of the 1H net magnetization in the 1H-13C pair) pointing along the +z axis. This is the equilibrium state. Now we apply a selective 180o pulse to the downfield (13C = a) component of the 1H doublet. The vector labeled "a" will rotate from the +z axis to the — z axis, giving us an antiphase state along the z axis (Fig. 7.18). We can describe this state as —2Iz Sz, as the Sz "multiplier" represents +1/2 for the pairs with 13C in the a state and —1/2 for the pairs with 13C in the /3 state. The "a" vector is on the —z axis (—2Iz [1/2] = —Iz) and the vector is on the +z axis (—2Iz [—1/2] = Iz). We will see in the next section that this state is a useful "intermediate" in coherence transfer.

Figure 7.19

Usually when we are talking about nonequilibrium populations, we use the population diagrams (energy states with filled and open circles) to describe the changes. Let's draw an energy diagram for the 1H-13C pair, just like we did for the 1Ha-1Hb pair in our discussion of the NOE difference experiment (Chapter 5, Section 5.12). As with the homonuclear case, there will be four energy states: aHaC, aH pC, pHaC, and pHpC. The difference is that the energy gap for the 13C = a to 13C = p transition is one-fourth of the energy gap for the1H = a to1H = p transition because yH = 4 yC. In other words, the much stronger1H nuclear magnet requires four times more energy to turn it against the Bo field. So we need to draw the energy diagram to reflect this difference in energy gaps. From the center of the diagram (E = 0) we go up in energy 4 units for 1H = p, down in energy 4 units for 1H = a, up 1 unit for 13 C = p and down 1 unit for 13C = a. This puts the aHaC state at E =—4 — 1 = -5, the aHpC state at E = — 4 + 1 = —3, the pHaC state at E = +4 — 1 = 3 and the pHpC state at E = +4 +1 = +5 units (Fig. 7.19). The energy unit here is AEC/2, where AEC is the energy gap for a 13C transition. We can identify the 1H transitions as aa ^ pa (1H transition with 13 C = a) and ap ^ pp (1H transition with 13C = p). The 13C transitions, which have an energy difference one-fourth of the 1H transitions, are aa ^ ap (13C transition with 1H = a) and pa ^ pp (13C transition with 1H = p). The two 1H transitions correspond to the two components of the doublet in the 1H spectrum (downfield component is aa ^ pa, upfield component is ap ^ pp) and the two 13 C transitions correspond to the two components of the doublet in the 13C spectrum (downfield component is aa ^ ap, upfield component is pa ^ pp).

Now we need to write in the population of each level at equilibrium (Fig. 7.20). According to the Boltzmann distribution, the population of each level will be N/4 times the exponential factor:

The deviation in population from an equal distribution between all four states (N/4) is thus proportional to the energy. For the aHaC state, we draw five filled circles (E = —5) to indicate a population of N/4 + 53 where 8 is (N/4)(AEC/8kT). For the aHpC state we draw three filled circles (E = —3), for the pHaC state we draw three open circles (E = +3), and for the pHpC state we draw five open circles (E = +5) representing a population deficit of 58 (Fig. 7.20). Now look at the population differences at equilibrium: for the 1H transitions,

Figure 7.20

we have AP = 88 [+55 - (-38) for the 13C = a transition and +35 - (-58) for the 13C = P transition] and for the 13 C transitions we have AP = 28 [+58 - (+38) for the 1H = a transition and -38 - (-58) for the 1H = P transition]. Note that population difference is always defined as the population in the lower energy state minus the population in the higher energy state. As expected, the 1H population differences are four times as large as the 13 C population differences (88 vs. 28).

We are finally ready to consider the effect of the selective 180° 1H pulse on the downfield component of the 1H doublet (aHaC ^ pHaC transition). We know that the 180° 1H pulse reverses ("inverts") the spin state of each proton in the sample: every proton in the a state is converted to the P state and every proton in the P state is converted to the a state. Looking at the energy diagram for the 1H-13C pair, we see that the 180° 1H pulse swaps the entire population across the 1H transition: each pair in the aa state is now Pa, and each pair in the Pa state is now aa. The population of the aa state, which was N/4 + 58, is now N/4 - 38, and the population of the aP state, which was N/4 - 38, is now N/4 + 58 (Fig. 7.21). The redistribution of populations is the result of every single pair in the Pa state moving down to the aa state, and every single pair in the aa state moving up to the Pa state. Consider the analogy of two towns on opposite sides of a river. East Podunk has a population of 51, and West Podunk has a population of 49. If two people move from East Podunk to West Podunk, we now have reversed the populations of the two towns. But this is not analogous to NMR inversion: the correct analogy is that all 51 residents of East Podunk move to West Podunk, and all 49 residents of West Podunk move to East Podunk, reversing the populations.

The spin state diagrammed in Figure 7.21 is the -2IzSz state, an antiphase state on the z axis. It can be represented in a vector diagram with the 1H net magnetization (13C = a) vector pointing down along the -z axis and the 1H net magnetization (13C = P) vector pointing up along the +z axis (Fig. 7.18). Note that spin I (1H) and spin S (13C) play equal roles in the product; we could also write it as -2Sz Iz because neither operator is observable and the rule that observable operators go first does not apply. Writing it with Sz first implies that we have net S spin (13C) magnetization on the z axis, and the Iz term is acting as a multiplier (+1/2 for 1H in the a state and -1/2 for 1H in the P state). Now look at this spin state from the point of view of population differences for the 13 C transitions (Fig. 7.21). The aa ^ aP transition (downfield or "1H = a" component of the 13 C doublet) has a population difference of -68 (-38 - (+38)), or -3 times the equilibrium difference of 28, and the Pa ^

ooooo AP= 108

ooooo AP= 108

Figure 7.21

PP transition (upfield or '4lH = P" component of the13 C doublet) has a population difference of 103 (+58 — (-53)), or 5 times the equilibrium difference of 28. In the vector model, we can represent the 13 C net magnetization (on a separate coordinate system from the 1H net magnetization) as a vector labeled "1H = a" pointing down along the — z axis with length 3 (three times the normal length for 13 C) and another vector labeled "1H = ft" pointing up along the +z axis with length 5 (five times the normal length for13C (Fig. 7.22). We can think of this as the sum of the equilibrium 13 C net magnetization (both "a" and "ft" vectors on +z with length 1) and the antiphase z magnetization transferred from 1H ("a" vector on — z and "P" vector on +z, both four times as long as the normal 13C equilibrium magnetization). The equilibrium 13C z magnetization is still there because we have not perturbed it: only a selective 1H pulse has been delivered to the sample. The transferred magnetization is four times as large as normal13 C magnetization because it came from the equilibrium population difference ("polarization") of the proton, which is four times as large as 13C. This is the origin of the enhancement by polarization transfer part (EPT) of the name INEPT. The proton population difference has been converted into a 13 C population difference of the same magnitude. The product operator view of the process thus far is as follows:

Our equilibrium starting point is both Iz (equilibrium proton net magnetization) and Sz (equilibrium 13C carbon magnetization). Adding these two together (Iz + Sz) simply means that both are present; it is like a list of the types of net magnetization in the sample. This is very different from multiplying them together (—2Iz Sz); in this case the first operator is net magnetization and the second is a multiplier that makes it antiphase (+1/2 for 13C = a, —1/2 for 13C = P). Note that there is no systematic way of calculating the effect of the selective 1H pulse on Iz; we have to reason it out using the vector model and then name the final state — 2Iz Sz based on our understanding of how to represent antiphase magnetization using the product operators. When we switch the order of operators from 2Iz Sz to 2Sz Iz we put in a factor of 4 to indicate that net13 C magnetization is now four times larger than equilibrium13 C magnetization (Sz). Because we are viewing 2SzIz as 13C net magnetization modified by the multiplier Iz, the point of comparison for the magnitude of this magnetization is now Sz and not Iz.

At this point in order to observe a 13 C spectrum, we need to rotate the 13 C net magne-

Figure 7.23

tization vectors into the x-y plane with a 90° 13 C pulse. If we set the phase of the pulse to y (B1 field on the y' axis in the rotating frame), the 1H = j vector, which is on the +z axis, rotates 90° ccw to the xX axis and the 1H = a vector, which is on the -z axis, rotates 90° ccw to the —xX axis (Fig. 7.23). If we begin acquiring the FID at this moment, with the x! axis as our phase reference, we will see a 13 C doublet with the downfield component (1H = a) upside-down with intensity three times the normal13 C spectrum, and the upfield component (1H = j) positive absorptive with intensity five times the normal 13C spectrum. In product operator terms, we have

The vector representation of the final state is the sum of antiphase 13 C magnetization on the -x axis (1H = a vector of length 4 on -x', 1H = j vector of length 4 on X) and in-phase13 C magnetization on the +x axis (1H = a vector of length 1 on x and 1H = j vector of length 1 on x'). The sum can also be represented as the sum of two spectra: an antiphase 13C doublet four times as intense as a normal 13 C spectrum (downfield component upside-down) plus a normal (in-phase) 13C spectrum (Fig. 7.24). The antiphase part was transferred from 1H equilibrium net magnetization (enhancement by polarization transfer), and the in-phase part is just the normal13 C spectrum derived from 13 C equilibrium net magnetization.

What does this "thought experiment" have to do with the INEPT experiment? The same spin state that results from the selective 1H 180° pulse on the downfield component of the 1H doublet (-2IzSz + Sz) can be obtained in a more practical way by a nonselective 90° 1H pulse followed by a delay of 1/(27) and another nonselective 1H 90° pulse:

This may seem more complicated, but it is much more practical because all of the pulses are nonselective ("hard") pulses. If we add the final 13C 90° pulse used in the SPT experiment to bring the magnetization to the x-y plane, we have the INEPT experiment:

The final 90° pulses on 1H and 13 C can be applied simultaneously or one immediately after the other, as shown above.

Exercise: The equilibrium 13C z magnetization makes the SPT analysis more complicated than it has to be. Go through the SPT thought experiment again showing the population diagrams and the final 13C spectrum if the 13C resonances are first saturated by continuous low-power RF irradiation at both 13C transition frequencies. Draw the population diagram after saturation of the 13C transitions. Then start the SPT experiment with the selective 180° 1H pulse on the 1H (13C = a) component of the 1H doublet.

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