Firstorder Splitting Patterns

All of this assumes that the proton in question is only coupled to other protons that are far away in chemical shift, so that its coupling pattern is simple ("first order" or "weak coupling"). If it is coupled to nearby peaks, distortions of the peak intensities and more complex patterns can result, and this effect is strongest at lower field strengths ("second order" or "strong coupling"). To state this more precisely, the J coupling in hertz between two spins must be much less than the chemical shift difference in hertz to see a simple first-order pattern. We could write this as

Av/J > 5 for first-order (weak) coupling where we arbitrarily divide it where J is one fifth of the chemical shift difference. Note that the chemical shift difference (Av) has to be expressed in hertz in order to directly compare it to the J coupling. This means that the criterion depends on field strength: a pattern that is second order at low field can be resolved into a simple first-order pattern at high field— yet another reason to spend the big bucks. For example, if two protons are coupled with a 7.0-Hz coupling constant and have a chemical shift difference of 0.1 ppm, they would be in a "second-order" splitting pattern at 200 MHz (Av = 20 Hz, Av/J = 2.86) and a "first-order" pattern at 600 MHz (Av = 60 Hz, Av/J = 8.57). Of course, the transition from first order to second order is a gradual process, so the cutoff of a factor of 5 is arbitrary, but you can see that higher field means fewer problems with distorted and more complex splitting patterns.

First-order patterns are easy to analyze because each splitting by a proton divides the pattern into two equal patterns separated by the coupling constant, J. To predict the splitting pattern, you can draw a diagram starting with the chemical shift position of the resonance. Arrange the coupling constants in descending order and write next to each one its multiplicity (doublet, d; triplet, t; quartet, q). For example, if a proton Ha has a chemical shift of 3.56 ppm and has two "neighbors," Hb and Hc, with coupling constants Jab = 10.0 Hz and Jac = 4.0 Hz, we have

We first divide the resonance position (3.56 ppm) into two equal peaks (1:1 ratio) by moving left 5.0 Hz (J/2) and right 5.0 Hz (J/2) (Fig. 2.5). Then each of these peaks is divided again into two equal peaks (1:1 ratio) by the 4.0-Hz coupling: 2.0 Hz to the left and 2.0 Hz to the right. This results in a pattern we call a doublet of doublets or (more concisely) a double doublet (abbreviated "dd"). In the literature we would report the peak like this: 53.56 (dd, 10.0,4.0). To "deconstruct" (analyze) the pattern, we first note that all four peaks are of the same height, and because 4 is a power of 2 (22 = 4), we assume that there is no overlap of

Double-doublet: J 10.0,4.0

3.56 ppm

Double-doublet: J 10.0,4.0

3.56 ppm

peaks. This implies that there are two coupling constants. The smaller one (Jac = 4.0 Hz) can be measured as the separation of peaks 1 and 2 (left-hand side pair, numbering from left to right) or peaks 3 and 4 (right-hand side pair). The larger coupling (Jab = 10.0 Hz) can be measured as the separation between peaks 1 and 3 or between peaks 2 and 4. The full "footprint" of the pattern (separation between peaks 1 and 4) is the sum of all coupling constants: 10 + 4 = 14 Hz. The more the time you spend diagraming coupling patterns, the easier it will be to recognize and deconstruct these patterns in real NMR spectra. In a more theoretical sense, we can see that the effect on Ha of Hb being in the a state is a downfield shift of its resonant frequency by 5.0 Hz (Jab/2). The effect on Ha of Hb being in the p state is an upfield shift by 5.0 Hz. The effect on Ha of Hc being in the a state is a downfield shift of 2.0 Hz (Jac/2), and the effect of Hc being in the p state is a upfield shift of 2.0 Hz. Thus, each component of the multiplet pattern can be viewed as a particular spin state of Hb and Hc and its effect on the resonant frequency of Ha. Peak 1 (leftmost) is labeled aa (Hb = a, Hc = a), peak 2 is labeled ap (Hb = a, Hc = p), and so on, and we can calculate the position of each peak relative to the center by adding J/2 for a and subtracting J/2 for p. For example, peak 2 (+3.0 Hz) represents the Ha resonance where Hb is in the a state (+10.0/2 = 5.0 Hz) and Hc is in the p state (-4.0/2 = -2.0 Hz). Adding the two effects, we get 5.0-2.0 = 3.0 Hz.

Note: In this book we will use the convention that the a state leads to a downfield shift (higher resonant frequency) for all coupled spins and the p state leads to an upfield shift. In fact, this may be reversed depending on the sign of the coupling constant J and the sign of the magnetogyric ratios, y. J couplings can be either negative or positive, as can magnetogyric ratios. For simplicity, we will ignore this detail.

A triplet can be viewed as a special case of a double doublet where Jab = Jac. In this case the two inner peaks (peaks 2 and 3) have the same resonant frequency and combine

Figure 2.6

to form a single peak with twice the intensity of the outer peaks. This can occur by true equivalence (i.e., by molecular symmetry making the Ha-Hb relationship identical to the Ha-Hc relationship) or by coincidence: the two coupling constants may just happen to be nearly or exactly the same. In the case of the triplet (1:2:1 ratio), the central peaks represent both the ap and the pa states of the Hb/Hc system, and we have less than 2n peaks (where n is the number of coupled spins affecting Ha) because of overlap.

More complex coupling patterns are dealt with in the same way. Generally, it is the overlap or near overlap of the 2n original components that makes the pattern complex and more challenging to take apart. Figure 2.6 shows a double triplet with coupling constants of 10.0 Hz (doublet coupling) and 5.0 Hz (triplet coupling). This means that Ha (2.17 ppm) is coupled to Hb (Jab = 10.0 Hz) and to two equivalent protons, Hc and Hd, each with a coupling to Ha of 5.0 Hz. The two triplet patterns meet in the center so that an intensity 1 peak of the left-hand triplet combines with an intensity 1 peak of the right-hand triplet to give a peak of intensity 2. Thus, the overall pattern is five equally spaced peaks with intensity ratio 1:2:2:2:1. This might be mistaken for a quintet with a single coupling of 5.0 Hz, but that would give an intensity pattern of 1:4:6:4:1, very different from the pattern we observe. To analyze this pattern, first note that there are five peaks in the multiplet, so we must have at least three couplings (22 = 4; 23 = 8). If there are three couplings, then we have eight peaks that are reduced to five peaks by overlap. The intensities are clearly not all the same: the outer peaks are smaller and the inner peaks all look the same. So 1:2:2:2:1 would be a good estimate of relative intensities. These numbers add up to eight, confirming that there are three couplings. Measuring frequency differences from the

Figure 2.7

outermost peaks, we see that peak 2 is 5.0 Hz from peak 1, and peak 3 is 10.0 Hz from peak 1. This suggests that we have couplings of 5 and 10 Hz. The equal spacing and the 1:2 ratio of peaks 1 and 2 suggest a triplet, and two triplets that meet in the center explain the pattern. Diagraming the coupling pattern gives us the complete story: a double triplet with couplings of 10.0 and 5.0 Hz. We report this as 8 2.17 (di, J = 10.0, 5.0). The simpler splitting comes first (d), and the coupling constants are listed in the same order as the coupling patterns, so we know that the 10.0-Hz coupling goes with the "d" and the 5.0-Hz coupling goes with the "t".

Figure 2.7 shows a triple-triplet pattern where one triplet coupling is exactly twice the other triplet coupling. As always, we start the diagram with the larger coupling and then split each of the peaks again with the smaller coupling. The three narrow triplet patterns grow out of the three peaks of the wider triplet, and we write the intensities according to the intensities of the "parent" peaks they grow out of: 1:2:1 for the outer triplets and 2:4:2 for the inner triplet derived from the intensity 2 peak of the wide triplet. The narrow triplets overlap in two places, and we add the intensities of the two peaks that combine in each case. The final intensity pattern is 1:2:3:4:3:2:1. This is distinct from a septet (splitting by six equivalent protons), which has intensity pattern 1:5:10:15:10:5:1. To analyze this pattern, we have to rule out the possibility that there are only three couplings (23 = 8), even though there are less than eight peaks. With three couplings, there would only be one overlap and the intensity ratio would have to be 1:1:1:2:1:1:1. With four couplings (24 = 16) we would have nine overlaps. In fact, the intensity ratio 1:2:3:4:3:2:1 adds up to 16, accounting for all

Figure 2.8

of the overlap. The "triangular" shape of the multiplet envelope suggests the 1:2:3:4:3:2:1 intensity ratio. Measuring in from the outer peaks, we see couplings of 3.0 Hz (peak 1 to peak 2) and 6.0 Hz (peak 1 to peak 3). One can visualize a triplet on each edge of the pattern (1:2:1) and a more intense triplet in the center (2:4:2), and the even spacing, leading to overlap, explains the anomolous intensity 3 peaks.

Figure 2.8 shows a double-double triplet (ddt) pattern. We see 10 peaks, not all evenly spaced. With four couplings (24 = 16) we would have six overlaps, and because we see four small peaks and six larger peaks, the intensity ratio might be 1:2:2:1:2:2:1:2:2:1, adding up to 16. Starting from the leftmost peak, we measure separations of 1.5, 3.0, and 4.0 Hz. The 1:2 ratio at the left edge suggests a triplet, so we know that the third peak (intensity 2) has another peak of intensity 1 from another triplet, suggested by the peak of intensity 1 (the fourth peak from the right-hand side) located 3.0 Hz to its right. It is useful when a component such as the 1:2:1 triplet is identified to mark the pattern with three lines on a piece of paper and then move the paper around on the multiplet to see if there are other identical patterns in the multiplet. In this case, we can locate four 1.5-Hz triplets, and their centers describe the double-doublet (J = 4.0, 3.0) that we see in the upper part of the diagram. Another way to look at this pattern is the superposition of two doublet-triplets (see Fig. 2.6, 1:2:2:2:1 ratio) with a separation of 4.0 Hz. Notice also what happens when the separation between two "lines" (components of the multiplet) comes close together: the smaller peak "climbs" up the larger one and becomes a "shoulder" on the side of the peak. We do not have "baseline separation" between the fourth and fifth peaks (1 and 0.5 Hz on the scale)

because their linewidths (about 0.5 Hz) are similar to their separations (0.5 Hz). Baseline separation means that the intensity level comes all the way to zero (to the noise baseline of the spectrum) between the peaks. When there is partial overlap, the separation of the two peaks in hertz will be a bit less than their actual frequency difference, and this can lead to errors in measuring coupling constants. If there are multiple places to measure a frequency difference, always avoid measuring it to a partially overlapped peak and choose two peaks that are baseline separated.

Some real-world examples will help to reinforce these concepts. Figure 2.9 shows a one-proton multiplet from the 600 MHz *H spectrum of a testosterone (steroid) metabolite. The chemical shift is measured at the precise center of the symmetrical pattern: 4.153 ppm, in the region of singly oxygenated CH groups. The spectrum on the right is processed with resolution enhancement: a sine-bell function starting at zero and ending at 180° of the sine function. The spectrum on the left is processed without resolution enhancement. The intensity ratio appears to be 1:1:1:2:1:1:1, which adds up to eight and suggests three couplings (23 = 8) with one overlap (seven peaks). If you mark the two leftmost peak positions on a piece of paper, you can see that the two rightmost peaks have the same spacing. Measuring from the third peak, we can make a mark on the right shoulder of the fourth (center) peak with the same spacing. These four lines (1-4 on the right-side spectrum) make up a double doublet. The coupling constants can be extracted from the line frequencies in hertz: J\ = 2501.2 — 2498.4 = 2.8 Hz; J2 = 2501.2 — 2495.4 = 5.8 Hz. This entire four-line pattern can then be transferred to the other side of the multiplet from your piece of paper, lining up the rightmost line with the rightmost peak (Fig. 2.9, right, 1'-4'). When this is done, the two lines in the middle do not perfectly overlap, resulting in a broader peak in the center with an intensity more like 1.5 rather than 2. It is still possible, however, to extract the third coupling constant by measuring the separation of the "second" lines (2 and 2') in the two four-peak (dd) patterns, or the separation of the two "third" lines (3 and 3'). The first (1 to 1') and fourth (4 to 4') peak separations require measurement to an overlapped pair of peaks (1' and 4) that are not perfectly aligned, so we avoid these measurements. Using the line frequencies, we have from the second peaks (lines 2 and 2'): J3 = 2498.4 — 2490.9 = 7.5 Hz; or from the third peaks (lines 3 and 3'): J3 = 2495.4 — 2487.9 = 7.5 Hz. The complete diagram is shown on the left-side spectrum: 8 4.153 (ddd, J = 7.5, 5.8, 2.8). Note that the method of measuring

Figure 2.10

the separation from the outermost peak fails in this case: The first two couplings can be obtained in this way but the third, which would come from the separation of peaks 1 and 1'/4, would be incorrect because peak 1'/4 is not a perfectly aligned overlap of two peaks. It is always better to completely analyze and understand the coupling diagram and measure from the resolved, single line peaks as much as possible.

A more complex pattern with some overlap is shown in Figure 2.10, another steroid metabolite. First-order splitting patterns are always symmetric about the center, so this cannot be a single resonance. Integration shows two protons, and the centers of the two symmetric patterns can be measured as 2.523 ppm (left-hand side pattern with 14 lines) and 2.474 ppm (right-hand side pattern with four lines). These two peaks are barely resolved at 600 MHz; at lower field the multiplets would expand about the same two chemical shift positions, and it would be very difficult to analyze the two overlapping patterns. The right-hand side multiplet is easy to analyze: it is a double-doublet with couplings of 13.7 and 5.7 Hz. The left-hand side pattern is more challenging. It has 14 lines (or "multiplet components") that suggests four couplings (24 = 16) and two overlaps: 1:1:1:1:1:2:1:1:2:1:1:1:1:1. Peaks 1-4 can be diagrammed as a double-doublet because the 1-2 spacing is the same as the 3-4 spacing. Marking down this pattern, we can transfer it to peaks 11-14 exactly. The center pattern can be viewed as two of these double-doublet patterns offset by the 1-2 separation. This gives overlap at peaks 6 and 9, which are twice the intensity of the others. We can trace peaks 1-4,5,6,8, and 9 onto a piece of paper and verify that the pattern lines up with peaks 6, 7,9,10, and 11-14. This confirms that we have a double-double-double-doublet (dddd), and we only have to measure the four coupling constants. The smallest coupling can be most easily measured as the difference between 1 and 2,3 and 4,11 and 12, or 13 and 14. In each case, however, the two peaks are far from baseline resolved, so the difference we measure, even in the resolution-enhanced spectrum, is an underestimate of the true coupling constant. The two peaks "ride up" on each other, and the measured separation is reduced by the overlap. A more accurate method is to make a computer simulation of two ideal shaped (Lorentzian) peaks added together, and vary the peak width and J coupling to get the best fit to the data in the lower trace (because of the resolution enhancement, the upper trace peaks do not have ideal peak shape). This analysis, using a non linear least-squares fit, gives a linewidth of

1.46 Hz and a J coupling of 1.47 Hz. This is slightly larger than the difference measured on the lower spectrum (1.37 Hz) but very close to the distance measured on the resolution-enhanced spectrum (1.51 Hz). Computer-generated peak lists can also be inaccurate due to the distortion of peak shapes by "grainy" digitization: It is better to use cursors positioned in the center of a peak by "eyeball" than to rely on an algorithm that simply looks for the highest intensity data point in the peak.

To measure the larger couplings, we measure from one peak of the double doublet to the corresponding peak of another double doublet. For example, we can measure the second-to-largest coupling between peaks 1 and 5, 2 and 6, 3 and 8, or 4 and 9. But it is best to avoid measuring to the "shoulder" peaks (5, 7, 8, and 10) because they are distorted toward the nearby taller (more intense) peak they are riding on. So we can measure this coupling best between peaks 2 and 6 or between 4 and 9. Peaks 6 and 9 are not distorted because they are pulled equally toward the two shoulders on either side. This separation gives us a coupling of 12.1 Hz, and measuring between the first and third group of four we get 13.5 Hz (distance between 1 and 6 or between 3 and 9). The peak can be reported as 8 2.523 (dddd, J = 13.5, 12.1, 5.0, 1.5). It is important to note that none of the couplings are between the two nearly overlapped resonances. If any coupling were between two peaks with such a small chemical shift difference (Av = (2.523 - 2.474) x 600 = 29.4 Hz), we would see some distortion of the peak intensities and possibly some additional, weak lines due to strong coupling (second-order pattern). In fact, the peaks can be assigned to H-6p (left-hand side peak) and H-1a (right-hand side peak), which are far apart in the steroid structure and have no mutual couplings. In this case we can often fully analyze two resonances even if they are overlapped, as long as we can recognize the coupling patterns and assign each individual line to one or the other resonance.

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