Furanose Proton J Values

Open-chain aldehyde form a-Glucose

Figure 1.12

The third category is rare in carbohydrates because the bulky OH groups prefer the equatorial position, pushing the H into the axial position.

In this ring form, the anomeric carbon (C1) of an aldehyde sugar (aldose) has one bond to the oxygen of the ring and another to an OH group external to the ring. Also external to the ring is the CH2OH group of the last carbon in the chain. The anomeric OH group can either be cis or trans to the external CH2OH group, depending on which side of the aldehyde or ketone group the OH group is added to. If it is cis, we call this isomer the p-anomer, and if it is trans we call it the a-anomer. When a crystalline monosaccharide is dissolved in water, these two ring forms rapidly form an equilibrium mixture of a and ¡5 anomers with very little of the open-ring aldehyde existing in solution (Fig. 1.12).

It is possible to link a monosaccharide to an alcohol at the anomeric carbon, so that instead of an OH group the anomeric carbon is connected to an OR group (e.g., OCH3) that is external to the ring. This is called a "glycoside," and the anomeric carbon is now a full acetal or ketal. The ring can no longer freely open into the open-chain aldehyde or ketone, so there is no equilibration of a and ¡5 forms. Thus a p-glycoside (OR group cis to the CH2OH group) will remain locked in the ¡5 form when dissolved in water. If the alcohol used to form the glycoside is the alcohol of another monosaccharide, we have formed a disaccharide with the two monosaccharides connected by a glycosidic linkage (Figure 1.13). Usually the alcohol comes from one of the alcohol carbons of the second sugar, but it is also possible to form a glycosidic linkage to the anomeric carbon of the second sugar. In this case we have a linkage C-O-C from one anomeric carbon to another, and both monosaccharides are "locked" with no possibility of opening to the aldehyde or ketone form.

1.2.2 NMR of Carbohydrates: Chemical Shifts

NMR chemical shifts give us information about the proximity of electronegative atoms (e.g., oxygen) and unsaturated groups (double bonds and aromatic rings). In this discussion we will ignore the protons attached directly to oxygen (OH) because they provide little

Figure 1.13

chemical information in NMR and are exchanged for deuterium by the solvent if we use deuterated water (D2O). In the case of carbohydrates, nearly all of the protons attached to carbon are in a similar environment: one oxygen attached to the carbon (CHOH or CH2OH). These protons all have similar chemical shifts, in the range of 3.3-4.1 ppm, so there is often a great deal of overlap of these signals in the 1H NMR of carbohydrates, even at the highest magnetic fields achievable. For this reason carbohydrate NMR (and NMR of nucleic acids RNA and DNA, which have a sugar-phosphate backbone) has been limited to relatively small molecules because the complexity of overlapping signals is limiting. The anomeric proton, however, is in a unique position because the carbon it is attached to has two bonds to oxygen. This additional inductive pull of electron density away from the hydrogen atom leads to a further downfield shift of the NMR signal, so that anomeric protons resonate in a distinct region at 5-6 ppm. A similar effect is seen for anomeric carbons, which have 13 C chemical shifts in the range of 90-110 ppm, whereas their neighbors with only one bond to oxygen resonate in the normal alcohol region of 60-80 ppm. Because each monosaccharide unit in a complex carbohydrate has only one anomeric carbon, we can count up the number of monosaccharide building blocks by simply counting the number of NMR signals in this anomeric region. Thus the analysis of carbohydrate NMR spectra is greatly simplified if we focus on the anomeric region of the 1H or 13C spectrum. The "alcohol" (nonanomeric) carbons of a sugar (H-C-O or H2C-O) are sensitive to steric crowding, so that the CH2OH carbons appear at higher field (60-70 ppm) than the more crowded CHOH carbons (70-80 ppm). This steric effect is also seen at the alcohol side of a glycosidic linkage (-O-CH-O-CH-C): this carbon is shifted downfield by as much as 10 ppm from the rest of the "alcohol" carbons (HO-CH-C) that are not involved in glycosidic linkages.

1.2.3 XH NMR: Coupling Constants

In the proton NMR spectrum, each signal is "split" into a multiple peak pattern by the influence of its "neighbors," the protons attached to the next carbon in the chain. These protons are three bonds away from the proton being considered and are sometimes called "vicinal" protons. For example, the anomeric proton in a cyclic aldose has only one neighbor: the proton on the next carbon in the chain (carbon 2). Note that because of rapid exchange processes or deuterium replacement in D2O, we seldom see splitting by the OH protons. Because it has only one neighbor, the anomeric proton will always appear as a doublet in the NMR spectrum. Also, because of its unique chemical shift position (5-6 ppm) and relatively rare occurrence (only one anomeric position per monosaccharide unit), the anomeric proton signal is usually not overlapped so we can see its splitting pattern clearly. The distance

(J, in frequency units of Hz) between the two component peaks of the doublet is a measure of the intensity of the splitting (or J coupling) interaction. For vicinal ("next-door neighbor") protons the value of J depends on the dihedral angle of the C-C bond between them. This angle is fixed in six-membered ring (pyranose) sugars because the ring adopts a stable chair conformation. For many common sugars (glucose, galactose, etc.) all or nearly all of the bulky groups on the ring (OH or CH2OH) can be oriented in the less crowded equatorial position in one of the two chair forms. Thus the sugar ring is effectively "locked" in this one chair form and we can talk about each proton on the ring as being in an axial or equatorial orientation. This is important for NMR because two neighboring (vicinal) protons that are both in axial positions ("trans-diaxial" relationship) have a dihedral angle at the maximum value of 180°, and this leads to the maximum value of the coupling constant J (about 10 Hz separation of the two peaks of the doublet). This does not make intuitive sense because in this arrangement the two protons are as far apart as possible; however, it is the parallel alignment of the two C-H bonds that leads to the strong coupling because the J-coupling (splitting) interaction is transmitted through bonds and not through space. Two vicinal protons in a locked chair with an axial-equatorial or an equatorial-equatorial relationship will have a much smaller coupling constant (much narrower pair of peaks in the doublet) in the range of 4 Hz. Thus we can use NMR coupling constants to determine the stereochemistry of sugars.

Here is how we can use this in the analysis of carbohydrate 1H NMR spectra: most naturally occurring sugars have an equatorial OH at the 2 position (numbering starts with the anomeric position as number 1), so the proton at carbon 2 is axial in a six-membered ring sugar. In addition, the CH2OH group is also equatorial in most pyranose sugars. So if the anomeric proton is axial, we should see it in the *H NMR spectrum as a doublet with a large coupling (10 Hz), because the Hi-H2 relationship is axial-axial. If the anomeric proton is axial, then the anomeric OH or OR substituent is equatorial and the sugar is in the j configuration (anomeric OH or OR cis to the CH2OH group at C5). If we see an anomeric proton with a small (4 Hz) coupling, then the anomeric proton is equatorial, the OH or OR group is axial, and we have an a sugar (anomeric OH or OR trans to the CH2OH group at C5). This reasoning works only if we are dealing with an aldopyranose (six-membered ring sugar based on an open-chain aldehyde) with an equatorial OH at C2; fortunately, nature seems to favor this situation.

1.2.4 Reducing Sugars

If the anomeric carbon of a sugar in the ring form bears an OH substituent instead of OC (glycosidic linkage), it will have the possibility of opening to the open-chain aldehyde or ketone form and reclosing in either the a or the j configuration. This is called a "reducing sugar" because the open-chain aldehyde form is accessible and can be oxidized to the carboxylic acid. The two isomers (a and j ) are in equilibrium and we usually see about a 2:1 ratio of j to a forms. The equilibration is slow on the NMR timescale (milliseconds) and so we see two distinct NMR peaks for the two isomers. The anomeric proton for the major j form will be a doublet with a large coupling constant (10 Hz) and for the minor form a doublet at a different chemical shift with a small coupling constant (4 Hz). The ratio of integrals for these two peaks will be about 2:1 (0.67:0.33 for normalized integrals). This pattern is a dead giveaway that you have a free (reducing) aldopyranose sugar. This monosaccharide could still be linked to other sugars by formation of a glycosidic linkage with one of the nonanomeric OH groups.

Sucrose

H rm Anomeric (a)

H rm Anomeric (a)

Figure 1.14

1.2.5 Keto Sugars

A ketose or keto sugar is a sugar based on a ketone rather than aldehyde functional group for its anomeric carbon. In this case the anomeric carbon is not Ci and there is no proton attached to the anomeric carbon (i.e., it is a quaternary carbon). The most common naturally occurring ketose is fructose, a 6 carbon sugar with the anomeric (ketone) carbon at position 2 in the chain. It forms a five-membered ring hemiketal (furanose) with the Ci and C6 CH2OH groups external to the ring. For a keto sugar you will not see an anomeric proton signal in the 1H NMR because the anomeric carbon has no hydrogen bonded to it. The only evidence will be the quaternary carbon in the 13 C spectrum that appears at the typical chemical shift (90-110 ppm) for an anomeric carbon (two bonds to oxygen). Furanose (five-membered ring) sugars pose another problem for NMR analysis: five-membered rings are generally flexible and do not adopt a stable chair-type conformation. For this reason we cannot speak of "axial" and "equatorial" protons or substituents in a furanose, so that stereochemical analysis by 1H NMR is very difficult.

1.2.6 Sucrose

A classic example of a keto sugar occurs in sucrose, a disaccharide formed from glucose in a six-membered ring linked to fructose in a five-membered ring, with the glycosidic linkage between the anomeric carbon of glucose (a configuration) and the anomeric carbon of fructose (ft configuration) (Fig. 1.14). In the 1H spectrum of sucrose (Fig. 1.15) we see the "alcohol" CH protons in the chemical shift range 3.4-4.2 ppm and the glucose anomeric proton at about 5.4 ppm. Fructose has no anomeric proton signal because the anomeric carbon is quaternary (keto sugar). The gl (glucose position 1) proton signal occurs as a doublet (coupled only to g2) with a small coupling constant (3.8 Hz) indicating that it is in the equatorial position (equatorial-axial coupling). This confirms that the glucose configuration is a because the glycosidic oxygen is pointing "down," opposite to the g6 CH2OH group. There is a double doublet at 3.5 ppm that can be broken down into two couplings: a doublet coupling of 10.0 Hz is further split by another doublet coupling of 3.8 Hz. The 3.8 Hz coupling matches the H-g1 doublet (also 3.8 Hz), so we can assign this peak to H-g2. Because the other coupling (to H-g2's other neighbor H-g3) is large, we know that H-g3 is axial and we confirm that H-g2 is also axial, further confirming that H-g1 is equatorial. There are three triplets with large coupling constants (3.4, 3.7, and

Figure 1.15

4.0 ppm), and it is likely that they represent axial protons in a cyclohexane chair structure with an axial proton on each side. Because all of the OH groups and the CH2OH group are in equatorial positions in the glucose portion, the nonanomeric H's are in axial positions, and we expect triplets with large couplings (~10 Hz) for H-g3 and H-g4 because both are in axial positions with one neighbor on each side in an axial position. These two large (axial-axial) couplings, if identical, would lead to a triplet pattern. Because we see three such triplets in the 1H spectrum, each one with normalized integral area 1, one of them must belong to the fructose part. Only H-f4 can be a triplet because it is the only fructose position with a single neighbor on each side. The doublet at 4.2 ppm (J = 8.8) can be assigned to H-f3 because it is next to the quaternary (anomeric) carbon C-f2 and therefore has only one coupling partner: H-f4. Note that this is the only doublet besides H-g1, which can be assigned because of its chemical shift in the anomeric region. Of the three resolved triplets, careful examination of the coupling constants reveals that one has a slightly smaller J value (8.5 Hz) that closely matches the H-f3 doublet splitting. Thus we can assign this triplet at 4.0 ppm to H-f4. A sharp singlet at 3.6 ppm (integral area 2) corresponds to the only CH2 group (H-f1) that is isolated from coupling by the quaternary carbon (C-f2). Because this is a chiral molecule, the two protons of CH2-f1 could have different chemical shifts, leading to a pair of doublets, but in this case they coincidentally have the same chemical shift and give a singlet. Two protons of the same carbon atom (CH2) are called "geminal" (twins), and if they have the same chemical shift in a chiral molecule they are called "degenerate." The overlapped group of signals between 3.75 and 3.9 ppm integrates to six protons and must contain the glucose CH2OH (H-g6), the other fructose CH2OH (H-f6), and the more complex H-g5 and H-f5 signals (each with one coupling partner at position 4 and two at position 6). Thus the only ambiguity remains the two resolved (not overlapped) triplet signals at 3.4 and 3.7 ppm that correspond to H-g3 and H-g4. To solve this puzzle, we will

Figure 1.16

need more information from more advanced NMR experiments such as two-dimensional NMR.

The 13C spectrum of sucrose is shown in Figure 1.16. Because it is proton decoupled, we see only one peak for each unique carbon in the molecule: 12 peaks for the C12H22O12 molecule of sucrose. We see two peaks in the anomeric (90-110 ppm) region, and we can assign the more substituted C-f2 (two bonds to carbon) to the more downfield of the two at 103.7 ppm. The less substituted C-g1 (one bond to carbon) appears at 92.2 ppm, about 10 ppm upfield of C-f2. This is a rule of thumb: about 10 ppm downfield shift each time an H is replaced with a C in the four bonds to a carbon atom. We see a tight group of three peaks at 60-63 ppm; these are the three CH2OH groups C-g6, C-f1 and C-f6. The remaining peaks are more spread out over the range 69-82 ppm; these are the nonanomeric "alcohol" or H-C-O carbons that constitute the majority of sugar positions. Again we see the roughly 10 ppm downfield shift due to substitution of an H with a C on the carbon atom of interest: CH2OH to C-CH-OH. How can we be sure that the CH2 and CH carbons are so neatly divided into chemical shift regions? More advanced one-dimensional 13C experiments called APT and DEPT allow us to determine the precise number of hydrogens attached to each carbon in the spectrum. To specifically assign the carbons within these three categories will require two-dimensional experiments.

1.2.6.1 Two-Dimensional Experiments A full NMR analysis of a carbohydrate, in which each 1H and 13C peak in the spectrum is assigned to a particular position in the molecule, requires the use of two-dimensional (2D) NMR. In a 2D spectrum, there are two chemical shift scales (horizontal and vertical) and a "spot" appears in the graph at the intersection of two chemical shifts when two nuclei (1Hor13C) in the molecule are close to each other in the structure. For example, one type of 2D spectrum called an HSQC spectrum presents the 1H chemical shift scale on the horizontal (x) axis and the 13 C chemical shift scale on the vertical (y) axis. If proton Ha is directly bonded to carbon Ca, there will be a spot at the intersection of the 1H chemical shift of Ha (horizontal axis) and the 13 C chemical shift of Ca (vertical axis). Because the peaks are spread out into two dimensions, the chances of overlap of peaks are much less and we can count up the number of anomeric and

Figure 1.17

nonanomeric peaks very quickly. The HSQC spectrum of sucrose is shown in Figure 1.17. There are 11 "spots" representing the 11 carbons that have at least one hydrogen attached. Quaternary carbons do not show up in the spectrum because the H has to be directly bonded to the C to generate a "spot." Note that the crosspeaks ("spots") fall roughly on a diagonal line extending from the lower left to the upper right. This is because there is a rough correlation between 1H chemical shifts and 13C chemical shifts: the same things that lead to downfield or upfield shifts of protons also affect the carbon they are attached to in the same way. We can also see that the small "triangle" of CH2OH peaks at the top is shifted "up" from the other nonaromatic peaks, due to the reduced steric crowding of the less-substituted CH2 (methylene) carbon compared to CH (methine) carbons. The 1H chemical shifts fall in the range of 3.5-4.2 ppm regardless of the degree of substitution.

A variation of this experiment, called HMBC (MB stands for multiple bond), shows spots only when the carbon and the proton are separated by two or three bonds in the structure. For example, for a monosaccharide we would see a spot at the chemical shift of the anomeric proton (H-1, horizontal axis) and the chemical shift of the C-3 carbon (vertical axis). Working together with data from the HSQC and HMBC 2D spectra, we can "walk" through the bonding structure of a carbohydrate, even "jumping" across the glycosidic linkages and establishing the points of connection of each monosaccharide unit.

Figure 1.18 shows a portion of the 1H spectrum of the trisaccharide D-raffinose in D2O. From just this portion we can conclude that, most probably, one of the sugars is a keto sugar and the other two are aldoses locked in the a configuration. The presence of two anomeric protons, each with a small doublet coupling (3.6 Hz) indicates that two of the sugars have the anomeric proton in the equatorial orientation. This assumes that we have the common pyranose arrangement with H-2 axial and the CH2OH group equatorial. The exact 1:1 ratio of integrals and the absence of major and minor (ft and a) anomeric peaks prove that these anomeric centers are locked in a glycosidic linkage. The absence of a third proton in the anomeric region means that the third sugar is most likely a keto sugar, with a quaternary anomeric carbon.

Figure 1.18

1.2.7 Terpenoids

A vast variety of plant and animal natural products are based on a repeating 5-carbon unit called isoprene: C-C(-C)-C-C. The end of the chain nearest the branch can be called the "head" and the other end is the "tail." Two isoprene units connected together make up a "monoterpene" or 10 carbon natural product (e.g., menthol, Fig. 1.19). Six isoprene units make a "triterpene" with 30 carbons. Cholesterol loses three of these in the biosynthetic process to give a 27 carbon "steroid" with four rings (Fig. 1.20). The trans ring junctures and the planar olefin "lock" the cyclohexane chairs into a single rigid conformation with well-defined axial and equatorial positions, just as we saw for the glucose ring in sucrose. Another triterpene skeleton that retains all 30 carbons is shown in Figure 1.21; the D and E rings are also locked in cyclohexane chair conformations.

1.2.8 Menthol

Menthol (Fig. 1.19) is a monoterpene natural product obtained from peppermint oil. Typical of terpenoids, menthol is only slightly soluble in water and is soluble in most organic solvents. The trans arrangement of the methyl and isopropyl substituents on the cyclohexane

Monoterpene
Figure 1.20

ring lock the ring in a single chair conformation with all of the substituents in the equatorial position.

The 250 MHz spectrum of menthol is shown in Figure 1.22. We see that even at 250 MHz a number of single proton signals are resolved (i.e., not overlapped with any other signals): "h," "l," "m," and "n." Integral values (normalized to one for the smallest resolved peaks) add up to 19.88 or 20 protons, consistent with the molecular formula C10H20O. The tall, sharp peaks at the right-hand side ("a," "b," and "c") represent the methyl groups, which usually give the most intense peaks because there are three equivalent protons. The most downfield signal ("ft") corresponds to the proton closest to the single functional group, the H-C-OH proton. The OH proton chemical shift depends on concentration because of hydrogen bonding with the OH oxygen of other menthol molecules in solution—looking at different samples it can be identified as the singlet peak at 1.55 ppm. It is a singlet because /-coupling interactions are averaged to zero by exchange: a particular OH proton on one menthol molecule jumps to another menthol molecule rapidly so it is constantly exposed to different H-C-OH protons at position 1, some in the a state and some in the p state, so it sees only a blur and appears as a singlet instead of a doublet. The H-1 proton at 3.37 ppm appears at a chemical shift typical for "alcohol" protons: protons attached to an sp3 hybridized (i.e., tetrahedral) carbon with a single bond to oxygen (3-4 ppm). Its coupling

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  • eugenia
    How do pyranose and furanose proton NMR differ?
    6 months ago

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