Draw the structures of the individual mutarotating a and b anomers of maltose

Draw the mirror images of glucose and fructose. Are glucose and fructose chiral? Do you expect them to be optically active?

(a) How many asymmetric carbon atoms are there in an aldotetrose? Draw all the aldotetrose stereoisomers. (b) How many asymmetric carbons are there in a ketotetrose? Draw all the ketotetrose stereoisomers. (c) How many asymmetric carbons and stereoisomers are there for an aldohexose? For a ketohexose?

(a) There is only one ketotriose, called dihydroxyacetone. Draw its structure. (b) There is only one aldotriose, called glyceraldehyde. Draw the two enantiomers of glyceraldehyde.

Draw and name the enantiomers of the sugars shown in Figure 23-2. Give the relative configuration (d or l) and the sign of the rotation in each case.

Which configuration (R or S) does the bottom asymmetric carbon have for the d series of sugars? Which configuration for the l series?

(a) Draw d-allose, the C3 epimer of glucose. (b) Draw d-talose, the C2 epimer of d-galactose. (c) Draw d-idose, the C3 epimer of d-talose. Now compare your answers with Figure 23-3. (d) Draw the C4 epimer of d-xylose. Notice that this epimer is actually an l-series sugar, and we have seen its enantiomer. Give the correct name for this l-series sugar.

Draw the Haworth projection for the cyclic structure of d-mannose by laying down the Fischer projection.

Allose is the C3 epimer of glucose. Draw the cyclic hemiacetal form of d-allose, first in the chair conformation and then in the Haworth projection.

Talose is the C4 epimer of mannose. Draw the chair conformation of d-talopyranose.

(a) Figure 23-2 shows that the degradation of D-glucose gives D-arabinose, an aldopentose. Arabinose is most stable in its furanose form. Draw d-arabinofuranose. (b) Ribose, the C2 epimer of arabinose, is most stable in its furanose form. Draw D-ribofuranose.

The carbonyl group in d-galactose may be isomerized from C1 to C2 by brief treatment with dilute base (by the enediol rearrangement, Section 23-7). The product is the C4 epimer of fructose. Draw the furanose structure of the product.

Draw the following monosaccharides, using chair conformations for the pyranoses and Haworth projections for the furanoses. (a) \(\alpha\)-D-mannopyranose (C2 epimer of glucose) (b) \(\beta\)-D-galactopyranose (C4 epimer of glucose) (c) \(\beta\)-D-allopyranose (C3 epimer of glucose) (d) \(\alpha\)-D-arabinofuranose (e) \(\beta\)-D-ribofuranose (C2 epimer of arabinose)

Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of \(\alpha\)-D-galactopyranose is +\(150.7^{\circ}\), and that of the \(\beta\) anomer is +\(52.8^{\circ}\). When either of the pure anomers dissolves in water, the specific rotation gradually changes to +\(80.2^{\circ}\). Determine the percentages of the two anomers present at equilibrium.

When d-glucose is reduced with sodium borohydride, optically active glucitol results. When optically active d-galactose is reduced, however, the product is optically inactive. Explain this loss of optical activity.

Emil Fischer synthesized L-gulose, an unusual aldohexose that reduces to give d-glucitol. Suggest a structure for this L sugar, and show how L-gulose gives the same alditol as D-glucose. (Hint: D-Glucitol has \(-CH_2OH\) groups at both ends. Either of these primary alcohol groups might have come from reduction of an aldehyde.)

Draw and name the products of bromine water oxidation of (a) d-mannose (b) d-galactose (c) d-fructose

Draw and name the products of nitric acid oxidation of (a) D-mannose (b) D-galactose

Two sugars, A and B, are known to be glucose and galactose, but it is not certain which one is which. On treatment with nitric acid, A gives an optically inactive aldaric acid, while B gives an optically active aldaric acid. Which sugar is glucose, and which is galactose?

Except for the Tollens test, basic aqueous conditions are generally avoided with sugars because they lead to fast isomerizations. (a) Under basic conditions, the proton alpha to the aldehyde (or ketone) carbonyl group is reversibly removed, and the resulting enolate ion is no longer asymmetric. Reprotonation can occur on either face of the enolate, giving either the original structure or its epimer. Because a mixture of epimers results, this process is called epimerization. Propose a mechanism for the base-catalyzed equilibration of glucose to a mixture of glucose and its C2 epimer, mannose. *(b) Propose a mechanism for the isomerization of a ketose to an aldose, via the enediol intermediate, shown immediately above. Note that the enediol has two -OH protons, and removing one or the other gives two different enolate ions.

Which of the following are reducing sugars? Comment on the common name sucrose for table sugar. (a) methyl a-d-galactopyranoside (b) \(\beta\)-L-idopyranose (an aldohexose) (c) \(\alpha\)-D-allopyranose (d) ethyl \(\beta\)-D-ribofuranoside

Draw the structures of the compounds named in Problem 23-20 parts (a), (c), and (d). Allose is the C3 epimer of glucose, and ribose is the C2 epimer of arabinose.

The mechanism of glycoside formation is the same as the second part of the mechanism for acetal formation. Propose a mechanism for the formation of methyl b-d-glucopyranoside.

Show the products that result from hydrolysis of amygdalin in dilute acid. Can you suggest why amygdalin might be toxic to tumor (and possibly other) cells?

Treatment of either anomer of fructose with excess ethanol in the presence of a trace of HCl gives a mixture of the \(\alpha\) and \(\beta\) anomers of ethyl-D-fructofuranoside. Draw the starting materials, reagents, and products for this reaction. Circle the aglycone in each product.

(a) Show the product that results when fructose is treated with an excess of methyl iodide and silver oxide. (b) Show what happens when the product of part (a) is hydrolyzed using dilute acid. (c) What do the results of parts (a) and (b) imply about the hemiacetal structure of fructose?

Propose a mechanism for methylation of any one of the hydroxy groups of methyl \(\alpha\)-D-glucopyranoside, using NaOH and dimethyl sulfate.

Draw the expected product of the reaction of the following sugars with excess methyl iodide and silver oxide. (a) a-d-fructofuranose (b) b-d-galactopyranose

Predict the products formed when the following sugars react with excess acetic anhydride and pyridine. (a) a-d-glucopyranose (b) b-d-ribofuranose

Show that Ruff degradation of d-mannose gives the same aldopentose (d-arabinose) as does d-glucose.

d-Lyxose is formed by Ruff degradation of galactose. Give the structure of d-lyxose. Ruff degradation of d-lyxose gives d-threose. Give the structure of d-threose.

d-Altrose is an aldohexose. Ruff degradation of d-altrose gives the same aldopentose as does degradation of d-allose, the C3 epimer of glucose. Give the structure of d-altrose.

Ruff degradation of D-arabinose gives d-erythrose. The Kiliani–Fischer synthesis converts D-erythrose to a mixture of D-arabinose and D-ribose. Draw out these reactions, and give the structure of D-ribose.

The Wohl degradation, an alternative to the Ruff degradation, is nearly the reverse of the Kiliani–Fischer synthesis. The aldose carbonyl group is converted to the oxime, which is dehydrated by acetic anhydride to the nitrile (a cyanohydrin). Cyanohydrin formation is reversible, and a basic hydrolysis allows the cyanohydrin to lose HCN. Using the following sequence of reagents, give equations for the individual reactions in the Wohl degradation of D-arabinose to D-erythrose. Mechanisms are not required. (1) hydroxylamine hydrochloride (2) acetic anhydride (3) \(^{-}OH\), \(H_2O\)

Aldohexoses A and B both undergo Ruff degradation to give aldopentose C. On treatment with warm nitric acid, aldopentose C gives an optically active aldaric acid. B also reacts with warm nitric acid to give an optically active aldaric acid, but A reacts to give an optically inactive aldaric acid. Aldopentose C is degraded to aldotetrose D, which gives optically active tartaric acid when it is treated with nitric acid. Aldotetrose D is degraded to (+)@glyceraldehyde. Deduce the structures of sugars A, B, C, and D, and use Figure 23-3 to determine the correct names of these sugars.

In 1891, Emil Fischer determined the structures of glucose and the seven other d-aldohexoses using only simple chemical reactions and clever reasoning about stereochemistry and symmetry. He received the Nobel Prize for this work in 1902. Fischer had determined that D-glucose is an aldohexose, and he used Ruff degradations to degrade it to (+)-glyceraldehyde. Therefore, the eight D-aldohexose structures shown in Figure 23-3 are the possible structures for glucose. Pretend that no names are shown in Figure 23-3 except for glyceraldehyde, and use the following results to prove which of these structures represent glucose, mannose, arabinose, and erythrose. (a) Upon Ruff degradation, glucose and mannose give the same aldopentose: arabinose. Nitric acid oxidation of arabinose gives an optically active aldaric acid. What are the two possible structures of arabinose? (b) Upon Ruff degradation, arabinose gives the aldotetrose erythrose. Nitric acid oxidation of erythrose gives an optically inactive aldaric acid, meso-tartaric acid. What is the structure of erythrose? (c) Which of the two possible structures of arabinose is correct? What are the possible structures of glucose and mannose? (d) Fischer’s genius was needed to distinguish between glucose and mannose. He developed a series of reactions to convert the aldehyde group of an aldose to an alcohol while converting the terminal alcohol to an aldehyde. In effect, he swapped the functional groups on the ends. When he interchanged the functional groups on d-mannose, he was astonished to find that the product was still D-mannose. Show how this information completes the proof of the mannose structure, and show how it implies the correct glucose structure. (e) When Fischer interchanged the functional groups on D-glucose, the product was an unnatural L sugar. Show which unnatural sugar he must have formed, and show how it completes the proof of the glucose structure.

Draw the structures of the individual mutarotating \(\alpha \text { and } \beta\) anomers of maltose.

Give an equation to show the reduction of Tollens reagent by maltose.

Does lactose mutarotate? Is it a reducing sugar? Explain. Draw the two anomeric forms of lactose.

Is gentiobiose a reducing sugar? Does it mutarotate? Explain your reasoning.

Trehalose is a nonreducing disaccharide (C12H22O11) isolated from the poisonous mushroom Amanita muscaria. Treatment with an a@glucosidase converts trehalose to two molecules of glucose, but no reaction occurs when trehalose is treated with a b@glucosidase. When trehalose is methylated by dimethyl sulfate in mild base and then hydrolyzed, the only product is 2,3,4,6-tetra-O-methylglucose. Propose a complete structure and systematic name for trehalose.

Raffinose is a trisaccharide (\(C_{18}H_{32}O_{16}\)) isolated from cottonseed meal. Raffinose does not reduce Tollens reagent, and it does not mutarotate. Complete hydrolysis of raffinose gives D-glucose, D-fructose, and D-galactose. When raffinose is treated with invertase, the products are d-fructose and a reducing disaccharide called melibiose. Raffinose is unaffected by treatment with a \(\beta\)-galactosidase, but an \(\alpha\)-galactosidase hydrolyzes it to D-galactose and sucrose. When raffinose is treated with dimethyl sulfate and base followed by hydrolysis, the products are 2,3,4-tri-O-methylglucose, 1,3,4,6-tetra- O-methylfructose, and 2,3,4,6-tetra-O-methylgalactose. Determine the complete structures of raffinose and melibiose, and give a systematic name for melibiose.

Cellulose is converted to cellulose acetate by treatment with acetic anhydride and pyridine. Cellulose acetate is soluble in common organic solvents, and it is easily dissolved and spun into fibers. Show the structure of cellulose acetate.

Cytosine, uracil, and guanine have tautomeric forms with aromatic hydroxy groups. Draw these tautomeric forms.

(a) An aliphatic aminoglycoside is relatively stable to base, but it is quickly hydrolyzed by dilute acid. Propose a mechanism for the acid-catalyzed hydrolysis. (b) Ribonucleosides are not so easily hydrolyzed, requiring relatively strong acid. Using your mechanism for part (a), show why cytidine and adenosine (for example) are not so readily hydrolyzed. Explain why this stability is important for living organisms.

All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance forms of the amide groups are drawn, as is done for thymine at left. Redraw the hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in Figure 23-24, using the polar resonance forms of the amides. Show how these forms help to explain why the hydrogen bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises between an electron-deficient hydrogen atom and an electron-rich pair of nonbonding electrons.

Draw the Fischer projections and the chair conformations of the anomers and epimers of glucose from memory. Identify and name these sugars based on how they differ from the structure of glucose. (Problems 23-46 and 47)

Correctly name monosaccharides and disaccharides, and draw their structures from their names.

Predict which carbohydrates mutarotate, which reduce Tollens reagent, and which undergo epimerization and isomerization under basic conditions. (Those with free hemiacetals will, but glycosides with full acetals will not.)

Predict the reactions of carbohydrates with oxidizing and reducing agents. Predict the reactions that convert their hydroxy groups to ethers or esters, and their carbonyl groups to acetals. (Problems 23-50, 51, and 52)

Determine the structure of an unknown carbohydrate based on its reactions. Determine its ring size from methylation followed by hydrolysis. (Problems 23-51, 57, 58, 59, and 60)

Draw the common types of glycosidic linkages, and identify these linkages in disaccharides and polysaccharides.

Recognize the structures of DNA and RNA, and draw the structures of the common ribonucleotides and deoxyribonucleotides. (Problems 23-63, 64, and 66)

Glucose is the most abundant monosaccharide. From memory, draw glucose in (a) the Fischer projection of the open chain. (b) the most stable chair conformation of the most stable pyranose anomer. (c) the Haworth projection of the most stable pyranose anomer.

Without referring to the chapter, draw the chair conformations of (a) b-d-mannopyranose (the C2 epimer of glucose). (b) a-d-allopyranose (the C3 epimer of glucose). (c) b-d-galactopyranose (the C4 epimer of glucose). (d) N-acetylglucosamine, glucose with the C2 oxygen

Use Figure 23-3 (the d family of aldoses) to name the following aldoses. (a) the C2 epimer of d-arabinose (b) the C3 epimer of d-mannose (c) the C3 epimer of d-threose (d) the enantiomer of d-galactose (e) the C5 epimer of d-glucose

Classify the following monosaccharides. (Examples: d-aldohexose, l-ketotetrose.) (a) (+)@glucose (b) (-)@arabinose (c) l-fructose CHO OH H H CH2OH OH OH H H OH (+)-gulose (d) C H CH2OH OH ()-ribulose H2OH H OH C O (e) (f) HO CH2OH H (+)-threose CHO H OH CHO H OH CH2OH H H H NHCOCH3 N-acetylglucosamine OH HO (g)

(a) Give the products expected when (+)-glyceraldehyde reacts with HCN. (b) What is the relationship between the products? How might they be separated? (c) Are the products optically active? Explain.

The relative configurations of the stereoisomers of tartaric acid were established by the following syntheses: (1) D-(+)-glyceraldehyde \(\stackrel{\mathrm{HCN}}{\longrightarrow}\) diastereomers A and B (separated) (2) Hydrolysis of A and B using aqueous \(Ba(OH)_2\) gave C and D, respectively. (3) \(HNO_3\) oxidation of C and D gave (-)-tartaric acid and meso-tartaric acid, respectively. (a) You know the absolute configuration of D-(+)-glyceraldehyde. Use Fischer projections to show the absolute configurations of products A, B, C, and D. (b) Show the absolute configurations of the three stereoisomers of tartaric acid: (+)-tartaric acid, (-)-tartaric acid, and meso-tartaric acid.

Predict the products obtained when d-galactose reacts with each reagent. (a) \(Br_2\) and \(H_2O\) (b) NaOH, \(H_2O\) (c) \(CH_3OH\), \(H^{+}\) (d) \(Ag(NH_3)_2 ^{+}\ ^{-}OH\) (e) \(H_2\), Ni (f) excess \(Ac_2O\) and pyridine (g) excess \(CH_3\) I, \(Ag_2O\) (h) \(NaBH_4\) (i) \(Br_2\), \(H_2O\), then \(H_2O_2\) and \(Fe_2(SO_4)_3\) (j) (1) KCN/HCN; (2) \(H_2\), \(Pd/BaSO_4\); (3) \(H_3O^{+}\) (k) excess \(HIO_4\)

Draw the following sugar derivatives. (a) methyl b-d-glucopyranoside (b) 2,3,4,6-tetra-O-methyl-d-mannopyranose (c) 1,3,6-tri-O-methyl-d-fructofuranose (d) methyl 2,3,4,6@tetra@O@methyl@b-d-galactopyranoside

Draw the structures (using chair conformations of pyranoses) of the following disaccharides. (a) 4-O-(a-d-glucopyranosyl)-d-galactopyranose (b) a-d-fructofuranosyl-b-d-mannopyranoside (c) 6-O-(b-d-galactopyranosyl)-d-glucopyranose

Erwin Chargaff’s discovery that DNA contains equimolar amounts of guanine and cytosine and also equimolar amounts of adenine and thymine has come to be known as Chargaff’s rule: G = C and A = T (a) Does Chargaff’s rule imply that equal amounts of guanine and adenine are present in DNA? That is, does G = A? (b) Does Chargaff’s rule imply that the sum of the purine residues equals the sum of the pyrimidine residues? That is, does A + G = C + T? (c) Does Chargaff’s rule apply only to double-stranded DNA, or would it also apply to each individual strand if the double helical strand were separated into its two complementary strands?

Which of the sugars mentioned in 23-53 and 23-54 are reducing sugars? Which ones would undergo mutarotation?

After a series of Kiliani-Fischer syntheses on (+)-glyceraldehyde, an unknown sugar is isolated from the reaction mixture. The following experimental information is obtained: (1) Molecular formula \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\) (2) Undergoes mutarotation. (3) Reacts with bromine water to give an aldonic acid. (4) Reacts with \(\mathrm{HNO}_3\) to give an optically active aldaric acid. (5) Ruff degradation followed by \(\mathrm{HNO}_3\) oxidation gives an optically inactive aldaric acid. (6) Two Ruff degradations followed by \(\mathrm{HNO}_3\) oxidation give meso-tartaric acid. (7) When the original sugar is treated with \(\mathrm{CH}_3 \mathrm{I}\) and \(\mathrm{Ag}_2 \mathrm{O}\), a pentamethyl derivative is formed. Hydrolysis gives a tetramethyl derivative with a free hydroxy group on C5. (a) Draw a Fischer projection for the open-chain form of this unknown sugar. Use Figure 23-3 to name the sugar. (b) Draw the most stable conformation of the most stable cyclic hemiacetal form of this sugar, and give the structure a complete systematic name.

An unknown reducing disaccharide is found to be unaffected by invertase enzymes. Treatment with an \(\alpha\)-galactosidase cleaves the disaccharide to give one molecule of D-fructose and one molecule of D-galactose. When the disaccharide is treated with excess iodomethane and silver oxide and then hydrolyzed in dilute acid, the products are 2,3,4,6-tetra-O-methylgalactose and 1,3,4-tri-O-methylfructose. Propose a structure for this disaccharide, and give its complete systematic name.

(a) Which of the d-aldopentoses will give optically active aldaric acids on oxidation with HNO3 ? (b) Which of the d-aldotetroses will give optically active aldaric acids on oxidation with HNO3 ? (c) Sugar X is known to be a d-aldohexose. On oxidation with HNO3, X gives an optically inactive aldaric acid. When X is degraded to an aldopentose, oxidation of the aldopentose gives an optically active aldaric acid. Determine the structure of X. (d) Even though sugar X gives an optically inactive aldaric acid, the pentose formed by degradation gives an optically active aldaric acid. Does this finding contradict the principle that optically inactive reagents cannot form optically active products? (e) Show what product results if the aldopentose formed from degradation of X is further degraded to an aldotetrose. Does HNO3 oxidize this aldotetrose to an optically active aldaric acid?

When the gum of the shrub Sterculia setigera is subjected to acidic hydrolysis, one of the water-soluble components of the hydrolysate is found to be tagatose. The following information is known about tagatose: (1) Molecular formula C6H12O6 (2) Undergoes mutarotation. (3) Does not react with bromine water. (4) Reduces Tollens reagent to give d-galactonic acid and d-talonic acid. (5) Methylation of tagatose (using excess CH3 I and Ag2O) followed by acidic hydrolysis gives 1,3,4,5-tetra-O-methyltagatose. (a) Draw a Fischer projection structure for the open-chain form of tagatose. (b) Draw the most stable conformation of the most stable cyclic hemiacetal form of tagatose.

Some protecting groups can block two OH groups of a carbohydrate at the same time. One such group is shown here, protecting the 4-OH and 6-OH groups of \(\beta\)-D-glucose. (a) What type of functional group is involved in this blocking group? (b) What did glucose react with to form this protected compound? (c) When this blocking group is added to glucose, a new chiral center is formed. Where is it? Draw the stereoisomer that has the other configuration at this chiral center. What is the relationship between these two stereoisomers of the protected compound? (d) Which of the two stereoisomers in part (c) do you expect to be the major product? Why? (e) A similar protecting group, called an acetonide, can block reaction at the 2? and 3? oxygens of a ribonucleoside. This protected derivative is formed by the reaction of the nucleoside with acetone under acid catalysis. From this information, draw the protected product formed by the reaction.

An important protecting group developed specifically for polyhydroxy compounds like nucleosides is the tetraisopropyldisiloxanyl group, abbreviated TIPDS, that can protect two alcohol groups in a molecule. base TIPDS chloride a ribonucleoside H H H H OH OH HO O i-Pr i-Pr Si i-Pr Cl Si + Cl i-Pr O Et3N (a) The TIPDS group is somewhat hindered around the Si atoms by the isopropyl groups. Which OH is more likely to react first with TIPDS chloride? Show the product with the TIPDS group on one oxygen. (b) Once the TIPDS group is attached at the first oxygen, it reaches around to the next closest oxygen. Show the final product with two oxygens protected. (c) The unprotected hydroxy group can now undergo reactions without affecting the protected oxygens. Show the product after the protected nucleoside from (b) is treated with tosyl chloride and pyridine, followed by NaBr, ending with deprotection with Bu4NF.

Draw the structures of the following nucleotides. (a) guanosine triphosphate (GTP) (b) deoxycytidine monophosphate (dCMP) (c) cyclic guanosine monophosphate (cGMP)

Draw the structure of a four-residue segment of DNA with the following sequence. (3’ end) G-T-A-C (5’ end)

Retroviruses like HIV, the pathogen responsible for AIDS, incorporate an RNA template that is copied into DNA during infection. The reverse transcriptase enzyme that copies RNA into DNA is relatively nonselective and error-prone, leading to a high mutation rate. Its lack of selectivity is exploited by the anti-HIV drug AZT (3?-azido-2?,3?-dideoxythymidine), which becomes phosphorylated and is incorporated by reverse transcriptase into DNA, where it acts as a chain terminator. Mammalian DNA polymerases are more selective, having a low affinity for AZT, so its toxicity is relatively low. (a) Draw the structures of AZT and natural deoxythymidine. (b) Draw the structure of AZT 5?-triphosphate, the derivative that inhibits reverse transcriptase.

Exposure to nitrous acid (see Section 19-16), sometimes found in cells, can convert cytosine to uracil. (a) Propose a mechanism for this conversion. (b) Explain how this conversion would be mutagenic upon replication. (c) DNA generally includes thymine, rather than uracil (found in RNA). Based on this fact, explain why the nitrous-acidinduced mutation of cytosine to uracil is more easily repaired in DNA than it is in RNA.

H. G. Khorana won the Nobel Prize in Medicine in 1968 for developing the synthesis of DNA and RNA and for helping to unravel the genetic code. Part of the chemistry he developed was the use of selective protecting groups for the 5 OH group of nucleosides. HO Ph H3CO C Ph Ph C OR OR OR base OHOH O Ph Ph H3CO OCH3 C Ph trityl, triphenylmethyl ether MMT, monomethoxytrityl ether DMT, dimethoxytrityl ether 5 3 2 The trityl ether derivative of just the 5 OH group is obtained by reaction of the nucleoside with trityl chloride, MMT chloride, or DMT chloride and a base like Et3N. The trityl ether derivative can be removed in dilute aqueous acid. DMT derivatives hydrolyze fastest, followed by MMT derivatives, and trityl derivatives slowest. (a) Draw the product with the trityl derivative on the 5 oxygen. (b) Explain why the trityl derivative is selective for the 5 OH group. Why doesnt it react at 2 or 3? (c) Why is the DMT group easiest to remove under dilute acid conditions? Why does the solution instantly turn orange when acid is added to a DMT derivative?