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Figure 7.31
Figure 7.31
Sx is inphase with respect to both protons, 2SxIi is antiphase with respect to Hi and in
phase with respect to H2, and 4SxIî I? is antiphase with respect to both attached protons. In
Dx each case, the multiplication by I1 means that the i3C coherence is multiplied by 1/2 if Hi =
a and by —1/2 if H1 = ft. We can draw vector diagrams for any of these states, but we have to be more specific about the vector labels. We are already familiar with the inphase and antiphase states for a CH group (Fig. 7.31, top). For a CH2 group (triplet in the nondecoupled 13C spectrum), we have four states for the two protons (Fig. 7.31, center). Both can be in the a state (aa, giving rise to the downfield peak of intensity 1), one can be a and the other ft (aft or fta, giving rise to the central peak of intensity 2), or both can be ft (ftft, giving rise to the upfield peak of intensity 1). For the operator product 2Sy l1, we have four net magnetization vectors, each representing onefourth of the 13C nuclei, differentiated by the spin state of the two attached protons: aa, aft, fta, or ftft. Multiplication by l1 corresponds to multiplication by +1/2 if H1 is in the a state and by —1/2 if H1 is in the ft state, so the a1 a2 and a1 ft2 vectors lie on the +yf axis (a1) and the ft1 a2 and ft1 ft2 vectors lie on the — yf axis (ft 1, Fig. 7.31, center). In other words, it does not matter what state H2 is in, but the H1 state (a or ft) determines whether the vector is on +y' or — y'. Even though they are on the same axis, we cannot combine the aa and aft vectors into a single vector because they have different rotating frame frequencies: aa rotates at +J and aft rotates at 0 Hz for an onresonance triplet. The product —2SxI2 has the aa and fta vectors on —x and the aft and ftft vectors on +X: in this case only the H2 spin state affects the direction of the vector. The product 4Sy l112 represents 13 C coherence that is antiphase with respect to both attached protons. The aa and ftft vectors lie on the +yf axis (multiplication by +1/2 and +1/2 for aa and by —1/2 and —1/2 for ftft) and the aft and fta vectors lie on the — yf axis (multiplication by +1/2 and —1/2). These last two can be combined into one vector of twice the length as H1H2 = aft and H1H2 = fta have the same rotatingframe frequency (i.e., the same behavior during delays). Note that the normalization factor of 4 is needed to take care of the two factors of 1/2 coming from the Iz terms.
For a CH3 group (Fig. 7.31, bottom), we have eight different vectors: aaa (frequency = 3J/2); aaj, aja, and jaa (frequency = J/2); ajj, ¡Haft, and jja (frequency = J/2); and jjj (frequency = 3J/2). The product 2Sycan be drawn with all the 1H states starting with a on the +/ axis and all the 1H states starting with j on the —y' axis. The aaj and aja vectors can be combined, as can the jaj and jja vectors. Notice that the normalization factor of 2 is introduced for each antiphase relationship. Thus, for a CH3 group, we can have 8Syl1I2l3, which could be represented by a vector on the +y' axis for the H1H2H3 = aaa state (length 1), another vector on the +y' axis for the ajj, jja, and jaj states (length 3), a vector on the — y' axis for the aaj, aja, and jaa states (length 3), and a fourth vector on the — y axis for the jjj state (length 1). In all cases where the vector is on +/, we have an even number of protons in the j state (so the multiplication by —1/2 is cancelled), and in all cases where the vector lies on the — y' axis, we have an odd number of protons in the j state (so the multiplication by —1/2 takes effect). Again, we can group together vectors that are on the same axis and have the same rotation frequency. This product operator represents 13C coherence antiphase with respect to all three of the attached protons. A normalization factor of 8 is required because of three factors of ±1/2.
During a delay, all antiphase relationships evolve toward the inphase relationship and all inphase relationships evolve toward the antiphase relationship. For example, for a CH2 group during a delay A we have
2SxI.i cos(nJA) + Sy sin (nJA) H1 Jcoupling evolution
+ Sy sin (nJA)cos(nJA) — 2Sxl2 sin2 (nJA) H2 Jcoupling evolution
Here we are assuming that both J couplings (13CH1 and 13CH2) are the same. For simplicity, we are treating the A delay as two separate Jcoupling evolution periods—the first for Jcoupling evolution with respect to H1 only and the second for Jcoupling evolution with respect to H2 only. In the second step, the first two terms (in the upper line) come from the 2Sxl1cos(nJA) term and the last two terms (in the lower line) come from the Sy sin(nJA) term. So we see that, starting with 13C coherence antiphase with respect to H1, we get the unchanged product 2Sxl1 times the cosine term plus the inphase state with respect to H1, rotated 90° ccw in the x'y' plane (Sy) times the sine term. This last term represents refocusing. The effect of H2 Jcoupling evolution is to bring both of these terms toward the antiphase state with respect to H2 (defocusing) with a cosine term for the starting state and a sine term for the "destination" state.
Now we have the theoretical tools to look at the refocusing for CH2 and CH3 groups in the refocused INEPT experiment. For a CH2 group, we can start the experiment with either I1 or I2, the result is the same. So we will start with I1 and multiply the final result by 2, as the 13C coherence is coming equally from two different attached protons. Up until the refocusing delay, the product operator analysis is the same, resulting in the product 2SxI1 that represents coherence transfer from H1 to 13C. The refocusing process was already described above for this product, and yields
2SxI1 cos2(nJA) + 4SyI1I2 cos(nJA)sin(nJA) + Sy sin(nJA)cos(nJA) — 2SxI2 sin2(nJA)
With decoupling, all antiphase terms will give no signal, so we are interested only in the inphase term Sy. Its intensity is modulated by the factor sin(nJA)cos(nJA), which when we consider an equal amount of 13 C coherence coming from H2 can be increased to 2sin(nJA)cos(nJA). If we think of the product nJA as an angle 0, we would get a relative intensity of 1 for 0 = 45° (A = 1/(4J)), zero for 90o (A = 1/(2J)), and 1 for 135° (A = 3/(4J)). Considering that 2sin0cos0 = sin20, we can see that the maximum intensity occurs at 0 = 45° or 135° (20 = 90° or 270°).
We can also look at the refocusing of the CH2 group using vectors (Fig. 7.32). The vector representation of 2SxI. has the aa and aft vectors on the +X axis and the fta and ftft vectors on the — X axis. With 1H decoupling, this operator gives zero intensity, as all of the vectors have the same frequency and will always cancel each other out. Even without decoupling, the central peak (aft/fta) is gone because the two vectors are exactly opposed (a small x in Fig. 7.32 indicates cancellation). During a total refocusing delay of 1/(4J) (0 = 45°), the aa vector rotates 90° ccw and the ftft vector rotates 90° cw, bringing the two together on the +y' axis. The aft and fta vectors are frozen because their rotating frame frequencies are both zero relative to the chemical shift position. They will always cancel each other and we can never get that half of the intensity back! So the intensity of 1 that we observe (2sin0cos0 = 1) is onehalf of the intensity of an inphase triplet getting its coherence from two protons. After another delay of 1/(4J), the aa vector continues to rotate ccw and the ftft vector rotates cw until they are again opposite each other, with aa on the — X axis and ftft on the +X axis. At this point (total refocusing delay = 1/(2J), 0 = 90°), we have no intensity with 1H decoupling, and the spin state can be represented as —2Sxl2. Note that all states in which H2 = a are on the — X axis and all states in which H2 = ft are on the +X axis, as the spin state is on — X and antiphase with respect to H2. In the product operator expression, we see that nJA = nJ/(2J) = n/2, so each sine term is 1 and each cosine term is 0, leaving only the one sin2 term —2SxI2. This is the null point where only CH peaks are observed in the refocused INEPT spectrum. After a third delay of 1/(4J), for a total delay of 3/(4J), the aa vector rotates another 90° ccw and the ftft vector rotates another 90° cw, so both land on the — y' axis. Again we have onehalf of the full intensity we would have with all four vectors aligned, and the observable intensity is opposite in sign (on the — y' axis). With 0 = 135°, the factor 2sin0cos0 equals —1, and we see upsidedown peaks in the 1Hdecoupled spectrum for all CH2 groups.
Finally, let's look at the CH3 group. We will start with L[ and assume that starting with I2 or 13 would give the same result, so the final result will be increased by a factor of 3 to represent coherence transfer from all three attached protons. We now know that only the inphase 13C coherence will be observable with !H decoupling, so we can ignore any terms that will lead to antiphase terms at the end of the refocusing period. To get an inphase operator, we need to "unwind" the antiphase relationship with H1 and avoid "winding up" the antiphase state with respect to H2 or H3. We can therefore ignore any term that does not do what we want, while considering the three coupling relationships in three separate delays of length A:
2SxI.! ^ Sy sin(nJA) full Jcoupling evolution with respect to H1
^ Sy sin(nJA) cos(nJA) no Jcoupling evolution with respect to H2
^ Sy sin(nJA) cos(nJA) cos(nJA) no Jcoupling evolution with respect to H3
Putting in the factor of 3, we have a final observable spin state of Sy [3sin©cos2©]. This factor has a value of 1.061 for © = 45°, 0 for © = 90o, and 1.061 for © = 135°, compared to 3 if it were possible to fully refocus all coherence coming from the three protons. Just for fun, the vector diagram is shown in Figure 7.33. For both © = 45° and © = 135°, the "short" vectors aaa, Pfifi, afifi, and fiaa are at four opposite corners, canceling out, and the two "long" vectors ftaft/ftfta and aafi/afia are halfway between the +X and +y' axes and halfway between the — X and +y' axes. Each of the "long" vectors represents onefourth of the intensity of a fully inphase CH3 peak coming from one proton, so the vector sum has magnitude V2/4 = 0.3536. Multiplying by 3 for the three protons, we get 1.061 for the peak intensity. For © = 90°, we have the two long vectors on the +y' axis and the four short vectors on the — yf axis, leading to complete cancellation and a peak intensity of zero. The maximum value of the function 3sin©cos2© occurs at © = 35.26° (or 144.74°), corresponding to a A delay of about 1/(57) and an inphase intensity of 1.155.
So you see that refocusing of 13C is always complicated by the fact that the optimal refocusing time is different for a CH, CH2, or CH3 group. Of course, we can exploit these differences to get spectral editing (making peak phase—up, down, or missing—tells us about
the number of attached protons), which is even more informative than the APT experiment. If we do three refocused INEPT experiments using the three delay times A = 1/(47), 1/(2J), and 3/(47), we get the following inphase intensities:

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