701 CHAPTER 25 CARBOHYDRATES SOLUTIONS TO TEXT PROBLEMS 25.1 (b) Redraw the Fischer projection so as to show the orientation of the groups in three dimensions. Reorient the three-dimensional representation, putting the aldehyde group at the top and the primary alcohol at the bottom. What results is not equivalent to a proper Fischer projection, because the horizontal bonds are directed “back” when they should be “forward.” The opposite is true for the vertical bonds. To make the drawing correspond to a proper Fischer projection, we need to rotate it 180° around a vertical axis. Now, having the molecule arranged properly, we see that it is L-glyceraldehyde. is equivalent to CHO CH 2 OH HO H CHO CH 2 OH HO HC CHO CH 2 OH HOHC rotate 180H11034 H OH HOCH 2 CHOC CHO CH 2 OH HOHC turn 90H11034 is equivalent to H OH HOCH 2 CHO OH HOCH 2 C H CHO Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 702 CARBOHYDRATES (c) Again proceed by converting the Fischer projection into a three-dimensional representation. Look at the drawing from a perspective that permits you to see the carbon chain oriented ver- tically with the aldehyde at the top and the CH 2 OH at the bottom. Both groups should point away from you. When examined from this perspective, the hydrogen is to the left and the hydroxyl to the right with both pointing toward you. The molecule is D-glyceraldehyde. 25.2 Begin by drawing a perspective view of the molecular model shown in the problem. To view the compound as a Fischer projection, redraw it in an eclipsed conformation. The eclipsed conformation shown, when oriented so that the aldehyde carbon is at the top, vertical bonds back, and horizontal bonds pointing outward from their stereogenic centers, is readily trans- formed into the Fischer projection of L-erythrose. 25.3 L-Arabinose is the mirror image of D-arabinose, the structure of which is given in text Fig- ure 25.2. The configuration at each stereogenic center of D-arabinose must be reversed to trans- form it into L-arabinose. HO H HOH CHO HOH CH 2 OH d-(H11002)-Arabinose H HO HO H OH CHO H CH 2 OH l-(H11001)-Arabinose is equivalent to or CHO CH 2 OH CHO H CHO H L-Erythrose HO H HO H CHO CH 2 OH OHOH HOCH 2 H H CH O OH H HOCH 2 H HO CH O Staggered conformation Same molecule in eclipsed conformation OHOH HOCH 2 H H CH O is equivalent to CHO CH 2 OH HOHC CHO OH HOCH 2 HC is equivalent to CHO OH HOCH 2 H CHO OH HOCH 2 HC Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 25.4 The configuration at C-5 is opposite to that of D-( H11001 )-glyceraldehyde. This particular carbohydrate therefore belongs to the L series. Comparing it with the Fischer projection formulas of the eight D-aldohexoses reveals it to be in the mirror image of D-( H11001 )-talose; it is L-( H11002 )-talose 25.5 (b) The Fischer projection formula of D-arabinose may be found in text Figure 25.2. The Fischer projection and the eclipsed conformation corresponding to it are Cyclic hemiacetal formation between the carbonyl group and the C-4 hydroxyl yields the H9251- and H9252-furanose forms of D-arabinose. (c) The mirror image of D-arabinose [from part (b)] is L-arabinose. The C-4 atom of the eclipsed conformation of L-arabinose must be rotated 120° in a clock- wise sense so as to bring its hydroxyl group into the proper orientation for furanose ring formation. Original eclipsed conformation of l-arabinose HO H 4 OH H H OH 3 2 1 5 CH 2 OH Conformation suitable for furanose ring formation H HOCH 2 OH H H OH OH rotate about C-3 C-4 bond C O H C O H CHO CH 2 OH HO OHH OHH H d-Arabinose CHO CH 2 OH H HHO HO H OH l-Arabinose Eclipsed conformation of l-arabinose HO H 4 HO H H OH 3 2 1 5 CH 2 OH C O H H9252-d-Arabinofuranose H HOCH 2 OH H HHO HO H O H9251-d-Arabinofuranose H HOCH 2 H OH HHO HO H O d-Arabinose HOH HOH CHO CH 2 OH HO H rotate about C-3 C-4 bond Eclipsed conformation of d-arabinose H HO 4 H HO HO H 3 2 1 5 CH 2 OH Conformation suitable for furanose ring formation HOCH 2 H 4 H HO HO H C O H 3 2 1 5 OH C O H CARBOHYDRATES 703 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Cyclization gives the H9251- and H9252-furanose forms of L-arabinose. In the L series the anomeric hydroxyl is up in the H9251 isomer and down in the H9252 isomer. (d) The Fischer projection formula for D-threose is given in the text Figure 25.2. Reorientation of that projection into a form that illustrates its potential for cyclization is shown. Cyclization yields the two stereoisomeric furanose forms. 25.6 (b) The Fischer projection and Haworth formula for D-mannose are The Haworth formula is more realistically drawn as the following chair conformation: HO HO H H OH H H H OH HOCH 2 O H9252-d-Mannopyranose CHO CH 2 OH HO OHH OHH H HO H d-Mannose H HO HO OH H H HO H CH 2 OH H9252-d-Mannopyranose (Haworth formula) H HO OH H HO H HOCH 2 H HO H O C O H H HO OH H 4 3 2 O H 1 H11001 H9252-d-Threofuranose OH H HHO HO H O H9251-d-Threofuranose H OH HHO HO H O C O H H HO OH H 4 3 2 O H 1 d-Threose CH 2 OH HO OHH H C OH 1 2 3 4 is equivalent to C O H H9251-l-Arabinofuranose H HOCH 2 OH H OH H HOH O H9252-l-Arabinofuranose H HOCH 2 H OH OH H HOH O 704 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Mannose differs from glucose in configuration at C-2. All hydroxyl groups are equatorial in H9252-D-glucopyranose; the hydroxyl at C-2 is axial in H9252-D-mannopyranose. (c) The conformational depiction of H9252-L-mannopyranose begins in the same way as that of H9252-D-mannopyranose. L-Mannose is the mirror image of D-mannose. To rewrite the eclipsed conformation of L-mannose in a way that permits hemiacetal forma- tion between the carbonyl group and the C-5 hydroxyl, C-5 is rotated 120° in the clockwise sense. Translating the Haworth formula into a proper conformational depiction requires that a choice be made between the two chair conformations shown. (d) The Fischer projection formula for L-ribose is the mirror image of that for D-ribose. CHO CH 2 OH H H HOH OH OH d-Ribose CHO CH 2 OH HO HO HO H H H l-Ribose Eclipsed conformation of l-ribose is oriented properly for ring closure. HO H HO H HO H CH 2 OH Haworth formula of H9252-l-ribopyranose HO H HO H HO H H O OH C O H HO H H CH 2 OH HO H OH H H O OH Haworth formula of H9252-l-mannopyranose Less stable chair conformation; CH 2 OH is axial OH H H CH 2 OH H H HO H OH OH O More stable chair conformation; CH 2 OH is equatorial OH OH OH H H H H HO O HOCH 2 H HO H HO H HO HH OH CH 2 OH C H O O 1 2 3 4 55 6 HO H H CH 2 OH HO HH OH OH C H 1 2 3 4 H9252-l-Mannopyranose (remember, the anomeric hydroxyl is down in the l series) HO H H CH 2 OH HO HH OH H OH O rotate about C - 4 C - 5 bond CHO CH 2 OH H HHO HHO OH HOH l-Mannose CHO CH 2 OH HO OHH OHH H HO H d-Mannose Eclipsed conformation of l-mannose HO H H H HO HO H OH CH 2 OH C O H CARBOHYDRATES 705 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Of the two chair conformations of H9252-L-ribose, the one with the greater number of equatorial substituents is more stable. 25.7 The equation describing the equilibrium is Let A H11005 percent H9251 isomer; 100 H11002 A H11005 percent H9252 isomer. Then A(H1100129.3°) H11001 (100 H11002 A)(H1100217.0°) H11005 100(H1100114.2°) 46.3A H11005 3120 Percent H9251 isomer H11005 67% Percent H9252 isomer H11005 (100 H11002 A) H11005 33% 25.8 Review carbohydrate terminology by referring to text Table 25.1. A ketotetrose is a four-carbon ke- tose. Writing a Fischer projection for a four-carbon ketose reveals that only one stereogenic center is present, and thus there are only two ketotetroses. They are enantiomers of each other and are known as D- and L-erythrulose. 25.9 (b) Because L-fucose is 6-deoxy-L-galactose, first write the Fischer projection formula of D-galactose, and then transform it to its mirror image, L-galactose. Transform the C-6 CH 2 OH group to CH 3 to produce 6-deoxy-L-galactose. H HO HO H OH CHO H OH H CH 2 OH d-Galactose (from Figure 25.2) HO H HOH CHO HO H HOH CH 2 OH l-Galactose HO H HOH CHO HO H HOH CH 3 6-Deoxy-l-galactose (l-fucose) H CO OH CH 2 OH CH 2 OH d-Erythrulose HO CO H CH 2 OH CH 2 OH l-Erythrulose OH HOCH 2 HO HO OH O H9251-D-Mannopyranose [H9251] 20 H11001 29.3H11034 D OH HOCH 2 OHHO HO O H9252-D-Mannopyranose [H9251] 20 H11002 17.0H11034 D Open-chain form of D-mannose HOCH 2 HO HO CH O OH OH HO H HO H HO H H O OH Less stable chair conformation of H9252-l-ribopyranose OH HO OH OH O More stable chair conformation of H9252-l-ribopyranose OH OHHO O HO 706 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 25.10 Reaction of a carbohydrate with an alcohol in the presence of an acid catalyst gives mixed acetals at the anomeric position. 25.11 Acid-catalyzed addition of methanol to the glycal proceeds by regioselective protonation of the dou- ble bond in the direction that leads to the more stable carbocation. Here again, the more stable car- bocation is the one stabilized by the ring oxygen. Capture on either face of the carbocation by methanol yields the H9251 and H9252 methyl glycosides. 25.12 The hemiacetal opens to give an intermediate containing a free aldehyde function. Cyclization of this intermediate can produce either the H9251 or the H9252 configuration at this center. The axial and equa- torial orientations of the anomeric hydroxyl can best be seen by drawing maltose with the pyranose rings in chair conformations. HO CH 2 OH OH HO O H9252-Configuration of hemiacetal (equatorial) OH HO HO O HOCH 2 O Key intermediate formed by cleavage of hemiacetal OCH CH 2 OH HO HO OH OH HO HO O HOCH 2 O H9251-Configuration of hemiacetal (axial) CH 2 OH HO HO HO O OH HO HO O HOCH 2 O HOCH 2 HO H O HO HO HOCH 2 H HO O H H11001 HO HOCH 2 H O HO H11001 H11001 CHO CH 2 OH H H H H HO HO OH OH d-Galactose Methanol Methyl H9251-d-galactopyranoside Methyl H9252-d-galactopyranoside CH 3 OHH11001 H11001 HCl OH CH 2 OH H OH HO OCH 3 O OH H CH 2 OH OCH 3 OH HO O CARBOHYDRATES 707 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Only the configuration of the hemiacetal function is affected in this process. The H9251 configuration of the glycosidic linkage remains unchanged. 25.13 Write the chemical equation so that you can clearly relate the product to the starting material. Ribitol is a meso form; it is achiral and thus not optically active. A plane of symmetry passing through C-3 bisects the molecule. 25.14 (b) Arabinose is a reducing sugar; it will give a positive test with Benedict’s reagent, because its open-chain form has a free aldehyde group capable of being oxidized by copper(II) ion. (c) Benedict’s reagent reacts with H9251-hydroxy ketones by way of an isomerization process involv- ing an enediol intermediate. 1,3-Dihydroxyacetone gives a positive test with Benedict’s reagent. (d) D-Fructose is an H9251-hydroxy ketone and will give a positive test with Benedict’s reagent. (e) Lactose is a disaccharide and will give a positive test with Benedict’s reagent by way of an open-chain isomer of one of the rings. Lactose is a reducing sugar. HOCH 2 HO O O OHHO HOCH 2 O OH OH HO HOCH 2 HO O O OHHO HOCH 2 OH OH CHO HO positive test; Cu 2 O formed Lactose (structure presented in Section 25.14) Open-chain form Benedict’s reagent positive test; Cu 2 O formed HO H HOH CH 2 OH CHOH HOH d-Fructose HO H HOH CH 2 OH O CH 2 OH C HOH Benedict’s reagent C OH CO CH 2 OH CH 2 OH 1,3-Dihydroxyacetone Enediol Glyceraldehyde C CH 2 OH H OH C OH CHOH CH 2 OH Benedict’s reagent positive test; Cu 2 O formed C OH CHO CH 2 OH H H H OH OH OH d-Ribose NaBH 4 H 2 O Plane of symmetry Ribitol CH 2 OH CH 2 OH H H H OH OH OH 708 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website ( f ) Amylose is a polysaccharide. Its glycoside linkages are inert to Benedict’s reagent, but the ter- minal glucose residues at the ends of the chain and its branches are hemiacetals in equilibrium with open-chain structures. A positive test is expected. 25.15 Because the groups at both ends of the carbohydrate chain are oxidized to carboxylic acid functions, two combinations of one CH 2 OH with one CHO group are possible. L-Gulose yields the same aldaric acid on oxidation as does D-glucose. 25.16 In analogy with the D-fructose - D-glucose interconversion, dihydroxyacetone phosphate and D-glyceraldehyde 3-phosphate can equilibrate by way of an enediol intermediate. 25.17 (b) The points of cleavage of D-ribose on treatment with periodic acid are as indicated. Four moles of periodic acid per mole of D-ribose are required. Four moles of formic acid and one mole of formaldehyde are produced. (c) Write the structure of methyl H9252-D-glucopyranoside so as to identify the adjacent alcohol functions. HOCH 2 OCH 3 HO HO HO O HOCH 2 OCH 3 HCO 2 H HC HC O O O 2HIO 4 H11001 Methyl H9252-D-glucopyranoside HCO 2 H HCO 2 H HCO 2 H HCO 2 H HCH 4HIO 4 O H H H OH OH OH CH 2 OH d-Ribose C OH CO CH 2 OH CH 2 OP(OH) 2 Dihydroxyacetone phosphate Enediol d-Glyceraldehyde 3-phosphate C H O C O C CHOH OH CH 2 OP(OH) 2 O CH 2 OP(OH) 2 O triose phosphate isomerase triose phosphate isomerase OHH equivalent to CHO H HO H H OH H OH OH CH 2 OH HNO 3 heat HNO 3 heat H HO H H OH CHO H OH OH CH 2 OH d-Glucose HO H HHO CHO HO H HOH CH 2 OH l-Gulose H HO H H OH H OH OH CO 2 H CO 2 H d-Glucaric acid CARBOHYDRATES 709 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Two moles of periodic acid per mole of glycoside are required. One mole of formic acid is produced. (d) There are two independent vicinal diol functions in this glycoside. Two moles of periodic acid are required per mole of substrate. 25.18 (a) The structure shown in Figure 25.2 is D-(H11001)-xylose; therefore (H11002)-xylose must be its mirror image and has the L-configuration at C-4. (b) Alditols are the reduction products of carbohydrates; D-xylitol is derived from D-xylose by conversion of the terminal GCHO to GCH 2 OH. (c) Redraw the Fischer projection of D-xylose in its eclipsed conformation. HO H HOH CHO HOH CH 2 OH d-Xylose Eclipsed conformation of d-xylose Haworth formula of H9252-d-xylopyranose redrawn as H H HO OH H H OH H HO OH H H H OH OH CH 2 O CHO O d-Xylitol HOH HO H CH 2 OH CH 2 OH HOH d-(H11001)-Xylose HOH HO H CHO CH 2 OH HOH l-(H11002)-Xylose HO H HOH CHO CH 2 OH HO H 2HIO 4 H CH CH HC O O OCH 3 HCH H O O H11001 O H CH 2 OH OCH 3 H OH HHO H HOH O 710 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The pyranose form arises by closure to a six-membered cyclic hemiacetal, with the C-5 hydroxyl group undergoing nucleophilic addition to the carbonyl. In the H9252-pyranose form of D-xylose the anomeric hydroxyl group is up. The preferred conformation of H9252-D-xylopyranose is a chair with all the hydroxyl groups equatorial. (d) L-Xylose is the mirror image of D-xylose. To construct the furanose form of L-xylose, the hydroxyl at C-4 needs to be brought into the proper orientation to form a five-membered ring. The H9251 anomeric hydroxyl group is up in the L series. (e) Methyl H9251-L-xylofuranoside is the methyl glycoside corresponding to the structure just drawn. ( f ) Aldonic acids are derived from aldoses by oxidation of the terminal aldehyde to a carboxylic acid. d-Xylose HOH HO H CH 2 OH HOH d-Xylonic acid HOH HO H CH 2 OH HOH C OH C OHO OCH 3 H H HOCH 2 HHO HO H O HO H HO HO H H CH 2 OH CHO 1 2 3 4 5 H HOCH 2 HO HO H H O H C O H rotate about C-3 C-4 bond OH H H HOCH 2 HHO HO H O HO H HOH CHO HOH CH 2 OH H HO HO H OH H CH 2 OH CHO d-Xylose l-Xylose Eclipsed conformation of l-xylose HO H HO H H HO CH 2 OH CHO is better represented as H HO OH H OH H H OH O Haworth formula of H9252-d-xylopyranose Chair conformation of H9252-d-xylopyranose HO HO HO OH O CARBOHYDRATES 711 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (g) Aldonic acids tend to exist as lactones. A H9254-lactone has a six-membered ring. (h)AH9253-lactone has a five-membered ring. (i) Aldaric acids have carboxylic acid groups at both ends of the chain. 25.19 (a) Reduction of aldoses with sodium borohydride yields polyhydroxylic alcohols called alditols. Optically inactive alditols are those that have a plane of symmetry, that is, those that are meso forms. The D-aldohexoses that yield optically inactive alditols are D-allose and D-galactose. H H H OH OH CHO H OH OH CH 2 OH CHO H HO HO H H OH H OH CH 2 OH CH 2 OH CH 2 OH H HO HO H H OH H OH NaBH 4 NaBH 4 d-Allose Allitol (meso compound) Galactitol (meso compound) d-Galactose H H H OH OH H OH OH CH 2 OH CH 2 OH HO H HOH CHO HOH CH 2 OH HO H H OH H OH CO 2 H CO 2 H d-Xylose d-Xylaric acid HO H HO H H OH CH 2 OH C OH O H9252H9251 H9253 H HOCH 2 OH H H OH O H H9253-Lactone of d-xylonic acidd-Xylonic acid rotate about C-3 C-4 bond H HOCH 2 O OH H HOH O C OH O H9254-Lactone of D-xylonic acid D-Xylonic acid C OHO CH 2 OH HO OH OH H H H redrawn as Eclipsed conformation of D-xylonic acid H HO OH H H OH CH 2 O C OH O H9253 H9252 H9251 H9254 H intramolecular ester formation H HO OH H H OH O O 712 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) All the aldonic acids and their lactones obtained on oxidation of the aldohexoses are optically active. The presence of a carboxyl group at one end of the carbon chain and a CH 2 OH at the other precludes the existence of meso forms. (c) Nitric acid oxidation of aldoses converts them to aldaric acids. The same D-aldoses found to yield optically inactive alditols in part (a) yield optically inactive aldaric acids. (d) Aldoses that differ in configuration only at C-2 enolize to the same enediol. The stereogenic center at C-2 in the D-aldose becomes sp 2 -hybridized in the enediol. The other pairs of D-aldohexoses that form the same enediols are D-Glucose and D-mannose D-Gulose and D-idose D-Galactose and D-talose 25.20 (a) To unravel a pyranose form, locate the anomeric carbon and mentally convert the hemiacetal linkage to a carbonyl compound and a hydroxyl function. Convert the open-chain form to a Fischer projection. H OH CHO OHH H OH HO CH 2 OH HOCH 2 OH CH OHH H OH O H OH equivalent to HO H HO H CHO H OH HO H CH 2 OH rotate about C-4 C-5 bond H H HOCH 2 OH OH OH OH O HOCH 2 OH CH OHH H H OH O H OH HOH HOH CHO CH 2 OH HOH HOH d-Allose HOH HOH CHOH C CH 2 OH OH HOH Enediol HOH HOH CHO CH 2 OH HO H HOH d-Altrose H H H OH OH CHO H OH OH CH 2 OH CHO H HO HO H H OH H OH CH 2 OH CO 2 H CO 2 H H HO HO H H OH H OH HNO 3 HNO 3 d-Allose Allaric acid (meso compound) Galactaric acid (meso compound) d-Galactose H H H OH OH H OH OH CO 2 H CO 2 H CARBOHYDRATES 713 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Proceed in the same manner as in part (a) and unravel the furanose sugar by disconnecting the hemiacetal function. The Fischer projection is (c) By disconnecting and unraveling as before, the Fischer projection is revealed. (d) Remember in disconnecting cyclic hemiacetals that the anomeric carbon is the one that bears two oxygen substituents. HO HOCH 2 CH 2 OH HO HO OH O C CH 2 OH CH 2 OH H OH OH H H OHH OH O H H CH 2 OH HO HO H OH H HOCH 2 O rotate about C-5 C-6 bond H CH 2 OH HO HO OH CH 2 OH H H O H11013 OH OH HO H 3 C CHO H HO HO OH H CHO CH 2 OH H OH H H 3 C HO OH HO HO HO H 3 C OH O equivalent to HOH HOH CHO CH 2 OH HOH HH H HO H HO H OH CH 2 CH 2 OH C O CH H HO H OH OH CH OOH H H H CH 2 OH HH H CH 2 OH HH H HO OH O rotate about C-3 C-4 bond 714 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 25.21 Begin the problem by converting the Fischer projection of D-ribose to a perspective view. Remem- ber that the horizontal lines of a Fischer projection represent bonds coming toward you, and the ver- tical lines are going away from you. Rank the groups attached to each stereogenic center. Identify each stereogenic center as either R or S according to the methods described in Chapter 7. Remember that the proper orientation of the low- est ranked group (usually H) is away from you. Molecular models will be helpful here. Each of the stereogenic centers in D-ribose has the R configuration. The IUPAC name of D-ribose is (2R,3R,4R)- 2,3,4,5-tetrahydroxypentanal. 25.22 (a) The L sugars have the hydroxyl group to the left at the highest numbered stereogenic center in their Fischer projection. The L sugars are the ones in Problem 25.20a and c. (b) Deoxy sugars are those that lack an oxygen substituent on one of the carbons in the main chain. The carbohydrate in Problem 25.20b is a deoxy sugar. H H H OH OH CHO H H OH CH 2 OH No hydroxyl group at C-5 H CH 2 OH OH H H HH H HO OH O OH H 3 COH CHO CH 2 OH H HO H Highest numbered stereogenic center is l. H 3 C OH HO HO HO O OH HOCH 2 HO O OH OH H HOH CHO CH 2 OH H HO HO HO H Highest numbered stereogenic center is l. H H H OH OH CH OH CH 2 OH O C-2 is R C-3 is R C-4 is R H H H OH OH CH OH CH 2 OH O H H H OH OH CH OH CH 2 OH O is equivalent to CARBOHYDRATES 715 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) Branched-chain sugars have a carbon substituent attached to the main chain; the carbohydrate in Problem 25.20c fits this description. (d) Only the sugar in Problem 25.20d is a ketose. (e) A furanose ring is a five-membered cyclic hemiacetal. Only the compound in Problem 25.20b is a furanose form. ( f ) In D sugars, the H9251 configuration corresponds to the condition in which the hydroxyl group at the anomeric carbon is down. The H9251-D sugar is that in Problem 25.20d. In the H9251-L series the anomeric hydroxyl is up. Neither of the L sugars—namely, those of Problem 25.20a and c—is H9251; both are H9252. H9251-Pyranose form of a d-ketose Anomeric hydroxyl is down. HOCH 2 CH 2 OH HO HO HO OH O H CH 2 OH OH H H HH H HO OH O Ketone carbonyl in Fischer projection HOH HOH CH 2 OH O CH 2 OH C HOH HOH HOCH 2 CH 2 OH HO HO HO OH O OH H 3 COH CHO CH 2 OH H HO H Methyl group attached to main chainH 3 C OH HO HO HO O 716 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 25.23 There are seven possible pentuloses, that is, five-carbon ketoses. The ketone carbonyl can be located at either C-2 or C-3. When the carbonyl group is at C-2, there are two stereogenic centers, giving rise to four stereoisomers (two pairs of enantiomers). When the carbonyl group is located at C-3, there are only three stereoisomers, because one of them is a meso form and is superposable on its mirror image. 25.24 (a) Carbon-2 is the only stereogenic center in D-apiose. Carbon-3 is not a stereogenic center; it bears two identical CH 2 OH substituents. (b) The alditol obtained on reduction of D-apiose retains the stereogenic center. It is chiral and optically active. (c, d) Cyclic hemiacetal formation in D-apiose involves addition of a CH 2 OH hydroxyl group to the aldehyde carbonyl. H H CH 2 OH HO H OH OH C O H 4 32 1 H11001 4 3 2 1 OH H CH 2 OH H HO OH O 4 3 2 1 H OH CH 2 OH H HO OH O d-Apiose (optically active) HOH HOCH 2 OH CHO CH 2 OH d-Apiitol (optically active) HOH HOCH 2 OH CH 2 OH CH 2 OH NaBH 4 d-Apiose HOH HOCH 2 OH CHO CH 2 OH 1 2 3 4 HO H CH 2 OH CH 2 OH C O HOH HOH CH 2 OH CH 2 OH C O HO H HOH CH 2 OH CH 2 OH C O HOH HOH CH 2 OH C CH 2 OH O HOH d-Erythropentulose (d-ribulose) l-Erythropentulose (l-ribulose) HO H CH 2 OH C CH 2 OH O HO H d-Threopentulose (d-xylulose) HO H CH 2 OH C CH 2 OH O HOH l-Threopentulose (l-xylulose) HOH CH 2 OH C CH 2 OH O HO H CARBOHYDRATES 717 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Three stereogenic centers occur in the furanose form, namely, the anomeric carbon C-1 and the original stereogenic center C-2, as well as a new stereogenic center at C-3. In addition to the two furanose forms just shown, two more are possible. Instead of the reaction of the CH 2 OH group that was shown to form the cyclic hemiacetal, the other CH 2 OH group may add to the aldehyde carbonyl. 25.25 The most reasonable conclusion is that all four are methyl glycosides. Two are the methyl glycosides of the H9251- and H9252-pyranose forms of mannose and two are the methyl glycosides of the H9251- and H9252- furanose forms. In the case of the methyl glycosides of mannose, comparable amounts of pyranosides and furano- sides are formed. The major products are the H9251 isomers. 25.26 (a) Disaccharides, by definition, involve an acetal linkage at the anomeric position; thus all the disaccharides must involve C-1. The bond to C-1 can be H9251 or H9252. The available oxygen atoms in the second D-glucopyranosyl unit are located at C-1, C-2, C-3, C-4, and C-6. Thus, there are 11 possible disaccharides, including maltose and cellobiose, composed of D-glucopyranosyl units. HO HOCH 2 OH HO OCH 3 O Methyl H9251-d-mannopyranoside HO HOCH 2 OCH 3 O HO OH Methyl H9252-d-mannopyranoside Methyl H9251-d-mannofuranoside H CH 2 OH H OCH 3 OH HHO HO HH O Methyl H9252-d-mannofuranoside H CH 2 OH OCH 3 H OH HHO HO HH O two furanose forms shown on page 717 rotate C-3 120H11034 about the C-2 C-3 bond H11001 4 3 2 1 OH H OH H HOCH 2 OH O 4 3 2 1 H OH OH H HOCH 2 OH O OH HOCH 2 H OH OH C O H 3 2 1 4 H 2 C CH 2 OH HO H OH OH C O H 32 1 4 H 2 C 718 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website H9251,H9251(1,1) H9251,H9252(1,1) H9252,H9252(1,1) H9251(1,2) H9252(1,2) H9251(1,3) H9252(1,3) H9251(1,4) (maltose) H9252(1,4) (cellobiose) H9251(1,6) H9252(1,6) (b) To be a reducing sugar, one of the anomeric positions must be a free hemiacetal. All except H9251,H9251(1,1), H9251,H9252(1,1), and H9252,H9252(1,1) are reducing sugars. 25.27 Because gentiobiose undergoes mutarotation, it must have a free hemiacetal group. Formation of two molecules of D-glucose indicates that it is a disaccharide and because that hydrolysis is catalyzed by emulsin, the glycosidic linkage is H9252. The methylation data, summarized in the following equation, require that the glucose units be present in pyranose forms and be joined by a H9252(1,6)-glycoside bond. 25.28 Like other glycosides, cyanogenic glycosides are cleaved to a carbohydrate and an alcohol on hydrolysis. (a) In the case of linamarin the alcohol is recognizable as the cyanohydrin of acetone. Once formed, this cyanohydrin dissociates to hydrogen cyanide and acetone. d-Glucose HO HOCH 2 OH HO O HOH11001 H 2 O H11001 Linamarin HO HOCH 2 OC(CH 3 ) 2 CN HO O HO H H11001 or enzyme Acetone cyanohydrin HOCCH 3 CH 3 CN H11001 Hydrogen cyanide HCN Acetone CH 3 CCH 3 O RO RO RO ROCH 2 CH 2 O O RO RO OR H11001 O OR R H11005 H: gentiobiose R H11005 CH 3 : gentiobiose octamethyl ether H 3 O H11001 CH 3 O CH 3 O OCH 3 CH 3 OCH 2 O OH 2,3,4,6-Tetra-O-methyl-d-glucose CH 3 O CH 3 O OCH 3 HOCH 2 O OH 2,3,4-Tri-O-methyl-d-glucose Site subject to mutarotation when R H11005 H CARBOHYDRATES 719 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) Laetrile undergoes an analogous hydrolytic cleavage to yield the cyanohydrin of benzalde- hyde. 25.29 Comparing D-glucose, D-mannose, and D-galactose, it can be said that the configuration of C-2 has a substantial effect on the relative energies of the H9251- and H9252-pyranose forms, but that the configura- tion of C-4 has virtually no effect. With this observation in mind, write the structures of the pyranose forms of the carbohydrates given in each part. (a) The H9252-pyranose form of D-gulose is the same as that of D-galactose except for the configura- tion at C-3. The axial hydroxyl group at C-3 destabilizes the H9251-pyranose form more than the H9252 form be- cause of its repulsive interaction with the axially disposed anomeric hydroxyl group. There should be an even higher H9252H20862H9251 ratio in D-gulopyranose than in D-galactopyranose. This is so; the observed H9252H20862H9251 ratio is 88 : 12. (b) The H9252-pyranose form of D-talose is the same as that of D-mannose except for the configuration at C-4. Because the configuration at C-4 has little effect on the H9251- to H9252-pyranose ratio (compare D-glucose and D-galactose), we would expect that talose would behave very much like man- nose and that the H9251-pyranose form would be preferred at equilibrium. This is indeed the case; the H9251-pyranose form predominates at equilibrium, the observed H9251H20862H9252 ratio being 78 : 22. H9252-d-Talopyranose HOCH 2 HO OH OH OH 4 H9251-d-Talopyranose HOCH 2 HO OH OH OH 4 H9251-d-Mannopyranose (68% at equilibrium) HOCH 2 HO HO OH OH 4 OOO H9252-d-Galactopyranose (64% at equilibrium) 3 CH 2 OH HO HO HO OH H9252-d-Gulopyranose HO 3 CH 2 OH HO HO OH H9251-d-Gulopyranose (1,3 diaxial repulsion between hydroxyl groups) HO 3 CH 2 OH HO HO OH OO O H 2 OH11001H11001 HO HO CO 2 H O HO OCHC 6 H 5 CN Laetrile HO HO CO 2 H O HO OH D-Glucuronic acid Benzaldehyde cyanohydrin C 6 H 5 CHCN OH H11001 Hydrogen cyanide HCN Benzaldehyde O C 6 H 5 CH 720 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (c) The pyranose form of D-xylose is just like that of D-glucose except that it lacks a CH 2 OH group. We would expect the equilibrium between pyranose forms in D-xylose to be much like that in D-glucose and predict that the H9252-pyranose form would predominate. It is observed that the H9252H20862H9251 ratio in D-xylose is 64 : 36, exactly the same as in D-glucose. (d) The pyranose form of D-lyxose is like that of D-mannose except that it lacks a CH 2 OH group. As in D-mannopyranose, the H9251 form should predominate over the H9252. The observed H9251H20862H9252 distribution ratio in D-lyxopyranose is 73 : 27. 25.30 (a) The rate-determining step in glycoside hydrolysis is carbocation formation at the anomeric position. The carbocation formed from methyl H9251-D-fructofuranoside (compound A) is tertiary and therefore more stable than the one from methyl H9251-D-glucofuranoside (compound B), which is secondary. The more stable a carbocation is, the more rapidly it will be formed. Faster: Slower: H H11001 H11001 B H CH 2 OH H OCH 3 OH HHO H HOH O HCH 3 OH H11001 H11001 Secondary carbocation H CH 2 OH OH HHO H HOH O H H11001 H11001 A H HOCH 2 CH 2 OH OCH 3 HHO HO H O CH 2 OH CH 3 OH H11001 H11001 Tertiary carbocation H HOCH 2 HHO HO H O H9252-d-Lyxopyranose HO HO OH O OH H9251-d-Lyxopyranose HO HO OH O OH H9251-d-Mannopyranose (68% at equilibrium) HOCH 2 HO HO OH O OH H9252-d-Glucopyranose (64% at equilibrium) CH 2 OH HO OH HO OH H9252-d-Xylopyranose HO HO OH OH H9251-d-Xylopyranose HO HO OH OH O O O CARBOHYDRATES 721 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website (b) The carbocation formed from methyl H9252-D-glucopyranoside (compound D) is less stable than the one from its 2-deoxy analog (compound C) and is formed more slowly. It is destabilized by the electron-withdrawing inductive effect of the hydroxyl group at C-2. Faster: Slower: 25.31 D-Altrosan is a glycoside. The anomeric carbon—the one with two oxygen substituents—has an alkoxy group attached to it. Hydrolysis of D-altrosan follows the general mechanism for acetal hydrolysis. 25.32 Galactose has hydroxyl groups at carbons 2, 3, 4, 5, 6. Ten trimethyl ethers are therefore possible. 2,3,4 2,4,5 3,4,5 4,5,6 2,3,5 2,4,6 3,4,6 2,3,6 2,5,6 3,5,6 H11001 H H11001 HO HOCH 2 O HO OH H OH HO OH O O H OH HO OH O O H H11001 H OH HO HOCH 2 OH O H11001 HO HOCH 2 OH O OH H OH d-Altrose CHO CH 2 OH HO OHH OHH OHH H conformational change H 2 O H11001 H OH HO HO HOCH 2 O OCH 3 H H11001 H H11002CH 3 OH slow H H11001 , fast OH HO HO HOCH 2 O H11001 Less stable HO HO OH HOCH 2 O OCH 3 H D H HO HO HOCH 2 O OCH 3 H HO HO HOCH 2 O OCH 3 H H11001 H H11002CH 3 OH slow H H11001 , fast HO HO HOCH 2 O H H11001 More stableC 722 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website To find out which one of these is identical with the degradation product of compound A, carry com- pound A through the required transformations. 25.33 The fact that phlorizin is hydrolyzed to D-glucose and compound A by emulsin indicates that it is a H9252-glucoside in which D-glucose is attached to one of the phenolic hydroxyls of compound B. The methylation experiment reveals to which hydroxyl glucose is attached. Excess methyl iodide re- acts with all the available phenolic hydroxyl groups, but the glycosidic oxygen is not affected. Thus when the methylated phlorizin undergoes acid-catalyzed hydrolysis of its glycosidic bond, the oxy- gen in that bond is exposed as a phenolic hydroxyl group. This compound must arise by hydrolysis of CCH 2 CH 2 CH 3 O CH 3 O O H9252-d-glucose O OCH 3 Compound B CCH 2 CH 2 CH 3 O CH 3 O OH O OCH 3 H11001 d-Glucose (C 6 H 12 O 6 ) HOCH 2 OH HO HO HO O Compound A (C 15 H 14 O 5 ) CCH 2 CH 2 HO OH OH O OHH11001 H 2 OC 21 H 24 O 10 emulsin H 3 O H11001 (acetal hydrolysis) CH 3 I Ag 2 O O O OCH 3 OCH 3 CH 3 O Tri-O-methyl ether of compound A O O OH OH HO Compound A HO HO CHO H OCH 3 OCH 3 CH 3 O H OCH 3 CH 3 O H CHO HO H H OCH 3 CH 2 OH H11013 2,3,5-Tri-O-methyl- D-galactose CARBOHYDRATES 723 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website The structure of phlorizin is therefore 25.34 Consider all the individual pieces of information in the order in which they are presented. 1. Chain extension of the aldopentose (H11002)-arabinose by way of the derived cyanohydrin gave a mixture of (H11001)-glucose and (H11001)-mannose. Chain extension of aldoses takes place at the aldehyde end of the chain. The aldehyde function of an aldopentose becomes C-2 of an aldohexose, which normally results in two carbohydrates diastereomeric at C-2. Thus, (H11001)-glucose and (H11001)-mannose have the same configuration at C-3, C-4, and C-5; they have opposite configurations at C-2. The configu- ration at C-2, C-3, and C-4 of (H11002)-arabinose is the same as that at C-3, C-4, and C-5 of (H11001)- glucose and (H11001)-mannose. 2. Oxidation of (H11002)-arabinose with warm nitric acid gave an optically active aldaric acid. Because the hydroxyl group at C-4 of (H11002)-arabinose is at the right in a Fischer projection formula (evidence of step 1), the hydroxyl at C-2 must be to the left in order for the aldaric acid to be optically active. If the C-2 hydroxyl group had been to the right, an optically inactive meso aldaric acid would have been produced. Therefore we now know the configurations of C-3 and C-5 of (H11001)-glucose and (H11001)-mannose and that these two aldohexoses have opposite configurations at C-2, but the same (yet to be determined) configuration at C-4. CHO CH 2 OH H H HHO OH CHOH OH 1 2 3 4 5 6 CHO CH 2 OH HO H HHO H CHOH OH 1 2 3 4 5 6 [One of these is (H11001)-glucose, the other is (H11001)-mannose.] CHO CH 2 OH H OH CHOH H OH 1 2 3 4 5 CO 2 H CO 2 H H CHOH OHH OH Achiral meso form; cannot be optically active HNO 3 CHO CH 2 OH H H CHOH HO OH 1 2 3 4 5 Partial stereostructure of (H11002)-arabinose CO 2 H CO 2 H HO H CHOH H OH 1 2 3 4 5 Aldaric acid from (H11002)-arabinose; optically active irrespective of configuration at C-3 HNO 3 CH 2 CH 2 OH CH 2 OH O HO HO O OH OH OH C O 724 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website 3. Both (H11001)-glucose and (H11001)-mannose are oxidized to optically active aldaric acids with nitric acid. Because both (H11001)-glucose and (H11001)-mannose yield optically active aldaric acids and both have the same configuration at C-4, the hydroxyl group must lie at the right in the Fischer pro- jection at this carbon. The structures of the corresponding aldaric acids are Both are optically active. Had the C-4 hydroxyl group been to the left, one of the aldaric acids would have been a meso form. 4. There is another sugar, (H11001)-gulose, that gives the same aldaric acid on oxidation as does (H11001)-glucose. This is the last piece in the puzzle, the one that permits one of the Fischer projections shown in the first part of step 3 to be assigned to (H11001)-glucose and the other to (H11001)-mannose. Consider first the structure CHO CH 2 OH HO H HO equivalent to HO H H OH H CH 2 OH CHO H H H HHO OH OH OH CO 2 H CO 2 H HO HO H HHO H H OH CO 2 H CO 2 H H HO H HHO OH H OH (This aldaric acid is optically inactive.) CO 2 H CO 2 H H H H HHO OH OH OH CO 2 H CO 2 H HO H H HHO H OH OH CHO CH 2 OH H H H HHO OH OH OH 1 2 3 4 5 6 CHO CH 2 OH HO H H HHO H OH OH 1 2 3 4 5 6 [One of these is (H11001)-glucose, the other is (H11001)-mannose.] CARBOHYDRATES 725 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website Oxidation gives the aldaric acid This is the same aldaric acid as that provided by one of the structures given as either (H11001)-glucose or (H11001)-mannose. That Fischer projection therefore corresponds to (H11001)-glucose. The structure of (H11001)-mannose is therefore A sugar that yields the same aldaric acid is This is, in fact, not a different sugar but simply (H11001)-mannose rotated through an angle of 180°. CH 2 OH CHO HO H H HHO H OH OH CHO CH 2 OH HO H H HHO H OH OH CHO CH 2 OH H H H HHO OH OH OH This must be (H11001)-glucose. CO 2 H CO 2 H H H H HHO OH OH OH 726 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website SELF-TEST PART A A-1. Draw the structures indicated for each of the following: (a) The enantiomer of D-erythrose (b) A diastereomer of D-erythrose (c) The H9251-furanose form of D-erythrose (use a Haworth formula) (d) The anomer of the structure in part (c) (e) Assign the configuration of each stereogenic center of D-erythrose as either R or S. A-2. The structure of D-mannose is Using Fischer projections, draw the product of the reaction of D-mannose with (a) NaBH 4 in H 2 O (b) Benedict’s reagent (c) Excess periodic acid A-3. Referring to the structure of D-arabinose shown, draw the following: (a) The H9251-pyranose form of D-arabinose (b) The H9252-furanose form of D-arabinose (c) The H9252-pyranose form of L-arabinose A-4. Using text Figure 25.2, identify the following carbohydrate: HO H H H OH HO H CH 2 OH OH O H CHO CH 2 OH HO H H H OH OH d-Arabinose CHO CH 2 OH HO H H HHO H OH OH d-Mannose CHO CH 2 OH H H OH OH CARBOHYDRATES 727 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website A-5. Write structural formulas for the H9251- and H9252-methyl pyranosides formed from the reaction of D-mannose (see Problem A-2 for its structure) with methanol in the presence of hydrogen chloride. How are the two products related—are they enantiomers? Diastereomers? PART B B-1. Choose the response that provides the best match between the terms given and the structures shown. Diastereomers Enantiomers (a) 1, 3, and 4 1 and 3 (b) 1 and 2 1 and 3 (c) 1, 2, and 3 1 and 3 (d) 1 and 4 1 and 2 B-2. A D carbohydrate is (a) Always dextrorotatory (b) Always levorotatory (c) Always the anomer of the corresponding L carbohydrate (d) None of the above B-3. Two of the three compounds shown yield the same product on reaction with warm HNO 3 . The exception is (a)(b)(c)(d) None of these (all yield the same product) B-4. The optical rotation of the H9251 form of a pyranose is H11001150.7°; that of the H9252 form is H1100152.8°. In solution an equilibrium mixture of the anomers has an optical rotation of H1100180.2°. The per- centage of the H9251 form at equilibrium is (a) 28% (b) 32% (c) 68% (d) 72% CHO CH 2 OH H HO HO OH H H CHO CH 3 H HO HO OH H H CHO CH 2 OH H H HO OH OH H CHO CH 2 OH H H H OH OH OH 1 CHO CH 2 OH H H HO OH OH H 2 CHO CH 2 OH HO HO HO H H H 34 CH 2 OH CH 2 OH H H OH OH CO 728 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-5. Which of the following represents the anomer of the compound shown? (a)(c) (b)(d) None of these B-6. Which of the following aldoses yields an optically inactive substance on reaction with sodium borohydride? (a) 3 only (c) 2 and 3 (b) 1 and 4 (d) All (1, 2, 3, and 4) B-7. Which set of terms correctly identifies the carbohydrate shown? 1. Pentose 5. Aldose 2. Pentulose 6. Ketose 3. Hexulose 7. Pyranose 4. Hexose 8. Furanose (a) 2, 6, 8 (c) 1, 5, 8 (b) 2, 6, 7 (d) A set of terms other than these CH 2 OH OH HH HO OH O CHO CH 2 OH HO H H H OH OH 1 CHO CH 2 OH H HO H OH OH H 2 CHO CH 2 OH HO HO HO H H H 3 CHO CH 2 OH HO H H H OH OH 4 HOCH 2 H OH H OH HO HH O H HOCH 2 OH H HH HO OH O H HOCH 2 H OH OH H HOH O H HOCH 2 H OH HH HO OH O CARBOHYDRATES 729 Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website B-8. The structure of D-arabinose is shown in Problem A-3. Which of the following is L-arabinose? (a)(b)(c)(d)(e) B-9. Which one of the statements concerning the equilibrium shown is true? (a) The two structures are enantiomers of each other. They have equal but opposite opti- cal rotations and racemize slowly at room temperature. (b) The two structures are enantiomers of each other. They racemize too rapidly at room temperature for their optical rotations to be measured. (c) The two structures are diastereomers of each other. Their interconversion is called mutarotation. (d) The two structures are diastereomers of each other. Their interconversion does not require breaking and making bonds, only a change in conformation. (e) The two structures are diastereomers of each other. One is a furanose form, the other a pyranose form. B-10. The configurations of the stereogenic centers in D-threose (shown) are (a)2R,3R (b)2R,3S (c)2S,3R (d )2S,3S CHO CH 2 OH H HO OH H OH H HOCH 2 H HO HO H OH O OH H HOCH 2 H HO HO H H OH O H CHO CH 2 OH H H HO OH OH H CHO CH 2 OH H HO HO OH H H CHO CH 2 OH H H H H HO HO OH OH CHO CH 2 OH HO HO H H H OH CHO CH 2 OH H H H H HO HO OH OH 730 CARBOHYDRATES Back Forward Main Menu TOC Study Guide TOC Student OLC MHHE Website