1559T_Ch24_423-43810/20/054:54APa9e423 EQA 24 Carbohydrates:Polyfunctional Compounds in Nature ou'll begin to a of "real world"molecule es some spciastcrcochcnicaltcms.O can.then this chapter shoudotthe puze posd by some of the problem.If you to be good at deductive reasoning Outline of the Chapter 24-1,242,24-3 Names and Structures of Carbohydrates Be prepared for a lot of new terminology. 24-4 through 24-8 Polyfunctional Chemistry of Sugars The basics:mostly (but not entirely)review material 249,2410epbrseebniadDegrateionaf5ugors 24-11,24-12 Complex Sugars in Nature Keys to the Chapter The naming system pre organized into a semioffic amework that is universally used.S o.sugars are all ketones or alde ydes con Fischer projections.ith the carbon chain vertical and the carbony!gro nearest the top.If the stereocenter nearest thebottom of the Fischer projection has theHon th right,it has on,and th 24-l and242 On Is on the lent,you n. of one he Non ofheomers)of any of heeid the mirror im 423
24 Carbohydrates: Polyfunctional Compounds in Nature In this chapter you’ll begin to apply the material you’ve just seen to a major class of “real world” molecules, carbohydrates (sugars). For a change, nomenclature will play a more central role: The names of sugars and sugar derivatives follow their own independent system, which includes some special stereochemical terms. On the other hand, most of the reactions are old ones and are needed for only a limited number of purposes such as structure determination, interconversion of derivatives, and synthesis of one sugar from another. You need to be good at deductive reasoning so that you can solve the puzzles posed by some of the problems. If you can, then this chapter should not be too hard for you. Outline of the Chapter 24-1, 24-2, 24-3 Names and Structures of Carbohydrates Be prepared for a lot of new terminology. 24-4 through 24-8 Polyfunctional Chemistry of Sugars The basics: mostly (but not entirely) review material. 24-9, 24-10 Step-by-Step Buildup and Degradation of Sugars Application in synthesis and structure determination. 24-11, 24-12 Complex Sugars in Nature Keys to the Chapter 24-1. Names and Structures of Carbohydrates The naming system presented in this section comes from an assortment of historically derived common names organized into a semiofficial framework that is universally used. So, sugars are all ketones or aldehydes containing alcohol groups, and their names all end in -ose. Almost all of them have stereocenters; and they are usually drawn in Fischer projections, with the carbon chain vertical and the carbonyl group nearest the top. If the stereocenter nearest the bottom of the Fischer projection has the OH on the right, it has an R configuration, and the sugar is said to belong to the D family. If this OH is on the left, you have an S configuration and an L-sugar. Look at Figures 24-1 and 24-2. Each horizontal row contains structures that are all diastereomers of one another. None of the mirror-image structures (enantiomers) of any of these is illustrated; the mirror image of any D-sugar is just an L-sugar with the same name. 423 1559T_ch24_423-438 10/20/05 4:54 AM Page 423
1559Tch24423-43810/20/054:54 M Page424 EQA 424 chapter 24 CARBOHYDRATES:POLYFUNCTIONAL COMPOUNDS IN NATURE 24-2and24-3. Conforn ations and Cyclic Forms of Sugars;Mutaroh typically with either five-or sixmembered ring.open-chain structure intoa picture of the cyclic hemiacetal is tricky:Here's a step-by-step way to do it. 1.Write in wedges and dotted lines and (for the carbonyl group. CHO CHO OH OH HO- -CHO OH OH OH H OH H- OH -OH H.OH CH.OH D-Glucose is shown step-by-step in the next paragraph.) By convention,the p cedure for L-series sugars is modified such that the top of the structure is rotated down d tside.That allows the andstructures tookemage when placed Ho B-D-Glucopyranos B-L-Glucopyranose
24-2 and 24-3. Conformations and Cyclic Forms of Sugars; Mutarotation Because sugars contain alcohols and carbonyl groups, they can (and usually do) form cyclic hemiacetals, typically with either five- or six-membered rings. Translating an open-chain structure into a picture of the cyclic hemiacetal is tricky: Here’s a step-by-step way to do it. 1. Write in wedges and dotted lines and (for a D-series sugar) lay the structure on its side, with the top moved down to the right. Then, locate the OH group that you will use to form the cyclic hemiacetal with the carbonyl group. 2. Rotate around the COC bond to the right of the OH group you picked out until the OH is horizontal and pointing away from you. (Make a model!) Then wrap this left-hand end of the chain behind the plane of the paper to put the OH near to the carbonyl carbon. Finally, make the hemiacetal bond. (This sequence is shown step-by-step in the next paragraph.) By convention, the procedure for L-series sugars is modified such that the top of the structure is rotated down to the left when it is laid on its side. That allows the L and D structures to look like mirror images when placed side-by-side. Mirror -D-Glucopyranose CH2OH O OH OH OH HO -L-Glucopyranose CH2OH O HO HO OH HO Will form a five-membered ring (a furanose) Will form a six-membered ring (a pyranose) HOCH2 CHO H H OH H OH OH H OH D-Glucose CH2OH CHO OH HO H H OH H OH H CH2OH CHO OH HO H H OH H OH H 424 • Chapter 24 CARBOHYDRATES: POLYFUNCTIONAL COMPOUNDS IN NATURE 1559T_ch24_423-438 10/20/05 4:54 AM Page 424
1559T_ch24_423-43810/20/054:54 AM Page425 EQA Keys to the Chapter·425 For a furanose For a pyranose HOCH2 --CHO HOCH2 --CHO H--C---OH HOCHH OHH HO- --CHO HO- --CHO 日0n日0 CH2OH HO-C-H 9 CHO CHO 人 HOCH CH,OH CH2OH OH DOH OH OH OH OH HO B-D-Glucof a-D-Glu B-D-Glu If you like.you can derive the cyclic structure of an L-sugar by doing the D-sugar first and then writing its ether with just traces of open chain and furanose structures.The interconversion of B and anomers is called mutarotation. 24-4 tho old on istry,a couple of new reagents are introduced.These mainly allow selective reactions to be carried out on the
If you like, you can derive the cyclic structure of an L-sugar by doing the D-sugar first and then writing its mirror image. Remember, sugars in solution typically exist as equilibrium mixtures of open chain plus cyclic hemiacetal structures. For glucose, this equilibrium mixture contains 63.6% -pyranose, 36.4% -pyranose, together with just traces of open chain and furanose structures. The interconversion of and anomers is called mutarotation. 24-4 though 24-8. Polyfunctional Chemistry of Sugars Although most of the reactions in these sections are old ones, typical of either alcohol or aldehyde/ketone chemistry, a couple of new reagents are introduced. These mainly allow selective reactions to be carried out on the For a furanose For a pyranose OH OH -D-Glucofuranose -D-Glucopyranose OH O CH2OH HO H CH2OH C H OH HO CH2OH C Rotate HO Wrap OH around back Wrap OH around back Close ring Close ring Rotate H OH CH2OH H H HO OH OH OH -D-Glucofuranose Anomers Anomers OH O OH OH OH O -D-Glucopyranose CH2OH HO OH OH OH O HOCH2 CHO H H OH H OH OH H OH HOCH2 CHO H H OH H OH OH H OH HO CHO OH H H H OH HOCH2 CHO H OH H H OHH OH CHO OH H H OH CHO OH H H OH HOCH2 HO H HOCH2 HO H Keys to the Chapter • 425 1559T_ch24_423-438 10/20/05 4:54 AM Page 425
1559reh24_423-43810/20/054:54 AM Page426 426 chapter 24 CARBOHYDRATES:POLYFUNCTIONAL COMPOUNDS IN NATURE reactions in thes for their mprtance in pactic of chmisry.the shorten sugar chains and the use of some very clever logic dealing with the consequences of the stereo h of the sted fo sence of optical a s in Nature material just completed.Mother Nature has developed a co oup of one sugar forms an ace etal by reac with glycoside.s ina previous secti).Again,the determination of unknown structures represents a common typ aretwoimportant earures that distinguish sugarsconta taining sugars (1)undergo mutarotation and (2)are readily oxidized by mild oxidants like Ag.The latter fea is the basis for the Tollens's test forom of the sugars ese are examples of "nonred Solutions to Problems 30.You get CHO This compound is the mirror image (enantiomer)of p-lyxose(Figure 24-1). Therefore,this sugar is L-lyxose,a diastereomer of D-ribose H -OH H —OH HO- -H CH-OH 31.(a)D-Aldopentose (Note:Only one stereocenter!):(b)L-aldohexose:(e)p-ketoheptose 32. CHO CHO HO -H O H HO- H OH HO- -H HO_ HO- CH-OH L.-Ribove L-Glucos
multifunctional sugar molecule. Examples are Br2 in H2O, which oxidizes only the aldehyde group of an aldose to CO2H, and HNO3, which oxidizes both the end carbons of an aldose, forming a dicarboxylic acid. The reactions in these sections have been chosen for their importance in practical aspects of sugar chemistry, the material in the latter half of this chapter. 24-9 and 24-10. Step-by-Step Buildup and Degradation of Sugars Determining sugar structures was a major effort involving the development of reaction sequences to lengthen or shorten sugar chains and the use of some very clever logic dealing with the consequences of the stereochemistry of sugars and sugar derivatives. The “Fischer proof” illustrates the main techniques used. Note that the process repeatedly makes use of the synthesis of dicarboxylic acids, which are tested for the presence or absence of optical activity. An optically inactive diacid is assumed to be a meso compound, containing a plane of symmetry, and this kind of information is used to narrow down possible structures for unknown sugars. 24-11 and 24-12. Complex Sugars in Nature These sections present simple extensions of the material just completed. Mother Nature has developed a convenient method for linking sugar molecules together: An alcohol group of one sugar forms an acetal by reactions with the hemiacetal group of another. This connection, called a glycoside linkage, is just a fancier version of a simple acetal formed by reaction of a sugar with an ordinary alcohol, like methanol (to form a “methyl glycoside,” as in a previous section). Again, the determination of unknown structures represents a common type of problem. There are two important features that distinguish sugars containing free hemiacetal groups from those lacking them. In solution, hemiacetals are always in equilibrium with aldehydes. So, hemiacetal-containing sugars (1) undergo mutarotation and (2) are readily oxidized by mild oxidants like Ag. The latter feature is the basis for the Tollens’s test for “reducing” (that is, oxidizable) sugars. Some of the sugars in this section contain acetal but not hemiacetal groups. Find them! These are examples of “nonreducing” sugars. Solutions to Problems 30. You get This compound is the mirror image (enantiomer) of D-lyxose (Figure 24-1). Therefore, this sugar is L-lyxose, a diastereomer of D-ribose. 31. (a) D-Aldopentose (Note: Only one stereocenter!); (b) L-aldohexose; (c) D-ketoheptose 32. L-Glucose Systematic name: (2S,3R,4S,5S)-2,3,4,5,6- Pentahydroxyhexanal CH2OH CHO HO H H OH HO H HO H CH2OH CHO HO H HO H HO H L-Ribose Systematic name: (2S,3S,4S)-2,3,4,5- Tetrahydroxypentanal CH2OH CHO OH HO H H OH H 426 • Chapter 24 CARBOHYDRATES: POLYFUNCTIONAL COMPOUNDS IN NATURE 1559T_ch24_423-438 10/20/05 4:54 AM Page 426
34.Make models if you peed n CHO (b) atoH neu CHO d 35.See in the and fe)are u-sueat
33. I know it sounds like an awful thing to make you do, but you might just have to review Sections 5-5 and 5-6 for this one. The Study Guide text for these sections may also help. (a) L-Glyceraldehyde; (b) D-erythrulose; (c) just D-glucose (upside down!); (d) L-xylose; (e) D-threose 34. Make models if you need to. (a) (b) (c) (d) 35. See procedure in the Study Guide text for this chapter. Careful—(b) and (c) are L-sugars. (a) (b) (c) OH OH HOCH2 HOCH2 -Furanose O HO HO OH OH -Furanose -Pyranose O HO HO CH2OH HO OH OH OH O CH2OH H OH HO OH HO OH -Furanose O OH HO OH -Furanose O CH2OH L-Psicose H H H HO O HO HO CH2OH D-Idose CH2OH CHO HO H H OH HO H H OH CH2OH D-Psicose H H H OH O OH OH CH2OH D-Altrose CH2OH CHO HO H H OH H OH H OH Solutions to Problems • 427 1559T_ch24_423-438 10/20/05 4:54 AM Page 427
1559Tch24423-43810/20/054:54 AM Page428 EQA 428.Chapter 24 CARBOHYDRATES:POLYFUNCTIONAL COMPOUNDS IN NATURE HOCH, CH,OH OH CH2OH OH (d) OH OH OH HO OH OH CH,OH OH CH2OH o-Furanose B a-Pyranose B HOCH> CH2OH HO -H CH2OH OH CH,OH OH (e) OH HO OH HO { HO OH CH>OH IOH CH>OH a-Furanose a-Pyranose & 36.No.They are all hemiacetals and therefore are capable of readily interconverting their o and B anomers. HO CH,OH 00H 一0 37.(a) 一OH (b)HO OH OH OH HO ⊕ OH HOCH2 CH,OH ⊕ HO个 HO 07 OH (c) HO OH (d) ∠OH HO (d)is an unusual case where the CH,OH is forced to be axial to allow all four OH's to be equatorial. 38.Base-catalyzed enolization allows the interconversion to take place.But notice that the product is not an ordinary enol.It has hydroxy groups on both carbons of the double bond:It is an enediol.Therefore. when it tautomerizes it has the option of losing a proton from either of two hydroxy groups,giving either the original ketone or the isomeric aldehyde. CH2OH CHOH HC-OH HC-OH H H20 =0 C-Q: C-OH -OH Ketose Enolate Enediol HC-0 Hc-O HC- 6: H2O OH HC-OH- H←一 OH Aldose Other enolate
(d) (e) 36. No. They are all hemiacetals and therefore are capable of readily interconverting their and anomers. 37. (a) (b) (c) (d) (d) is an unusual case where the CH2OH is forced to be axial to allow all four OH’s to be equatorial. 38. Base-catalyzed enolization allows the interconversion to take place. But notice that the product is not an ordinary enol. It has hydroxy groups on both carbons of the double bond: It is an enediol. Therefore, when it tautomerizes it has the option of losing a proton from either of two hydroxy groups, giving either the original ketone or the isomeric aldehyde. Aldose Other enolate H2O OH H2O OH OH HC O HC OH HC O C OH HC O C O O CH2OH C Ketose Enolate Enediol H2O OH H2O OH CHOH C HC C OH OH HC OH O O C HO O HO OH OH CH2OH HOCH2 O HO HO OH OH CH2OH O HO HO OH OH O HO OH OH OH OH -Furanose -Pyranose OH O HO HO OH HO O CH2OH CH2OH OH CH2OH OH CH2OH HOCH2 HO H OH CH2OH OH OH OH HOCH2 CH2OH -Furanose O -Pyranose CH2OH HO OH OH OH O CH2OH OH CH2OH OH 428 • Chapter 24 CARBOHYDRATES: POLYFUNCTIONAL COMPOUNDS IN NATURE 1559T_ch24_423-438 10/20/05 4:54 AM Page 428
1559T_ch24_423-43811/8/0510:14Pa9e429 ⊕ EQA Solutions to Problems.429 (a)CH2OH H,C=0 C=0 →C02 that is.2 formaldehyde I CO, CH2OH H,C=0 (b)HC=O HCO2H HC-OH HCO,H HC 一OH HCO,H that is.4 formic acid 1 acetaldehyde HC -OH HC-OH HC=0 CH3 CH (e)CH2OH H,C=0 (HC -OH4一4HC0H that is,4 formic acid 2 formaldehyde H2C=0 40.(a)0 COOH COOH (iii) HO- -H HO- H HO -H H -OH H -OH H -OH COOH CH-OH D-Threonic acid D-Tartaric acid D-Threitol COOH CH=NNHCHs (b)(i) COOH OH H C-NNHC Hs o HO- -H HO- H H -OH H 一OH H —OH CH-OH COOH D-Xvlonic acid D-Xylaric acid
39. Each carbon atom in the starting sugar is alongside its product after HIO4 cleavage. The number of hydrogens on each carbon remains the same before and after cleavage. (a) (b) (c) 40. (a) (i) (ii) (iii) (iv) (b) (i) (ii) COOH COOH D-Xylaric acid OH HO H H OH H D-Xylonic acid CH2OH COOH OH HO H H OH H D-Threose phenylosazone* CH2OH NNHC6H5 CH NNHC6H5 H OH C D-Threitol CH2OH CH2OH HO H H OH D-Tartaric acid COOH COOH HO H H OH D-Threonic acid COOH CH2OH HO H H OH (HC OH)4 CH2OH H2C O H2C O that is, 4 formic acid 2 formaldehyde CH2OH 4 HCO2H OH HC O HC HC HC OH OH HC OH CH3 CH3 HCO2H HCO2H HCO2H HCO2H HC O that is, 4 formic acid 1 acetaldehyde O CH2OH CH2OH C H O 2C CO2 H O 2C that is, 2 formaldehyde 1 CO2 Solutions to Problems • 429 1559T_ch24_423-438 11/8/05 10:14 Page 429
1559Tch24423-43811/10/052:22Page430 EQA 430 chapter 24 CARBOHYDRATES:POLYFUNCTIONAL COMPOUNDS IN NATURE ( CH2OH () CH-NNHC.Hs (c)( COOH -OH C=NNHC Hs HO- -H HO- -H HO- 一H -OH HO- CH2OH -OH CH-OH D-Xylitol D-Galactonic aci COOH CH2OH CH-NNHC.Hs H H 一0H C=NNHCoHs HO- H HO- 一H HO- 一H HO- HO- H- 一OH H -OH H -OH COOH CH2OH CH2OH D-Galactaricacid p-Galactitol *Same as D-erythrose phenylosazone 41.(a)D-gulose (Figure 24-1):(b)L-allose(all OH's on left side) 42.(a)Arabinose and lyxose.Ribitol and xylitol are meso compounds (b) CH-OH CH-OH CH2OH =0 H OH HO- -H 、HO H -OH *H0 -OH CHOH H- H -OH -OH OH -0H FCH.OH M. A new stereocenter is generated at C2.so two diastereomeric alditols are produced.In contrast. esimpler because no new stereocenter will be generated:therefore HOCH2 Also with an acetal,not a hemiacetal,at Cl
(iii) (iv) (c) (i) (ii) (iii) (iv) *Same as D-erythrose phenylosazone. **Same as D-lyxose phenylosazone. ***Same as D-talose phenylosazone. 41. (a) D-gulose (Figure 24-1); (b) L-allose (all OH’s on left side) 42. (a) Arabinose and lyxose. Ribitol and xylitol are meso compounds. (b) A new stereocenter is generated at C2, so two diastereomeric alditols are produced. In contrast, reduction of any aldose will be simpler because no new stereocenter will be generated; therefore, only a single product can form. 43. (a) and (d), because they still possess hemiacetal functionality. In (b) and (c), the OH at C1 in glucose has become OCH3, and the molecule is now an acetal, incapable of mutarotation. (e) is Also with an acetal, not a hemiacetal, at C1. HOCH2 CH3 CH3 HO O HO O O D-Fructose CH2OH CH2OH O HO H H OH H OH NaBH4 CH3OH D-Glucitol CH2OH CH2OH HO H H OH H OH H OH D-Mannitol CH2OH CH2OH HO H HO H H OH H OH D-Galactose phenylosazone*** CH2OH HO H H OH NNHC6H5 CH NNHC6H5 HO H C D-Galactitol CH2OH CH2OH OH HO H HO H H OH H D-Galactaric acid COOH COOH OH HO H HO H H OH H D-Galactonic acid CH2OH COOH OH HO H HO H H OH H D-Xylose phenylosazone** CH2OH HO H H OH NNHC6H5 CH NNHC6H5 C D-Xylitol CH2OH CH2OH OH HO H H OH H 430 • Chapter 24 CARBOHYDRATES: POLYFUNCTIONAL COMPOUNDS IN NATURE 1559T_ch24_423-438 11/10/05 2:22 Page 430
1559T_ch24_423-43811/8/0510:14Page431 ⊕ EQA Solutions to Problems.431 44.(a)The and acid via a stabilized carbocation. 土Cm 8--Cm (b)The oxygen at Cl in this case is an acetal oxygen,not a simple methyl ether.As in (a).mild (e)Four methyl glycosides are possible (refer to the structures for fructofuranose and fructopyranose in Section 24-2) HOCH2 CH2OH HOCH2 OCH3 KI HO HO HO 1 OCH, HO B CH.OH HO HO VOCH HO CH,OH H HO ann 45.Arabinose(as a B-pyranose)forms a double acetal (Section 24-8).So does ribose.because in its a-pyranose form all four hydroxy groups are cis. H00 CH;COCH.H HO OHO CH; a-D-Ribopyrano H.CH
44. (a) The oxygen at C1 of an aldopyranose is a hemiacetal oxygen, not a simple alcohol oxygen. It can therefore be methylated the same way a hemiacetal can be converted to an acetal—with methanol and acid via a stabilized carbocation. (b) The oxygen at C1 in this case is an acetal oxygen, not a simple methyl ether. As in (a), mild aqueous acid is sufficient for hydrolysis as a result of the same stabilized carbocation intermediate shown above (the mechanism is just the reverse of the one shown). (c) Four methyl glycosides are possible (refer to the structures for fructofuranose and fructopyranose in Section 24-2). 45. Arabinose (as a -pyranose) forms a double acetal (Section 24-8). So does ribose, because in its -pyranose form all four hydroxy groups are cis. HO O HO OHOH CH3 CH3 CH3 O O CH3 O O O CH3COCH3, H -D-Ribopyranose HO HO HO O CH2OH OCH3 HO HO HO O CH2OH OCH3 -D-fructopyranoside Methyl O HO -D-fructofuranoside Methyl HO OCH3 CH2OH O HO HO HOCH2 HOCH2 OCH3 CH2OH O O CH3OH O OCH3 O OCH3 H H O OH H2O O OH2 Solutions to Problems • 431 1559T_ch24_423-438 11/8/05 10:14 Page 431
1559r_ch24_423-43810/20/054:54 M Page432 432 chapter 24 CARBOHYDRATES:POLYFUNCTIONAL COMPOUNDS IN NATURE HOHO HO OF -0 a-D-Xylopyrano CH3 CH H:C CH OH 0 B-D-Lyxopyranose 46.(i)The sugar has seven carbons and is a ketose.because HIOa treatr produces a mole of CO2 (ii)Because the sugar forms the same osazone as an aldose,its ketone must be at C2.So far.therefore.you have the following partial structure: CH2OH C=0 CHOH CHOH Stereochemistry unknown HOH」 -OH CH2OH (iii)and (v)tell you that CHO CHOH CHO CHOH CHOH CHOH CHOH CHOH CHOH H-C-OH H-C-OH
The best xylose and lyxose can do is have only one pair of adjacent cis hydroxy groups, so these readily form only monoacetals. 46. (i) The sugar has seven carbons and is a ketose, because HIO4 treatment produces a mole of CO2 (see a similar reaction of D-fructose, Section 24-5). The sugar has two CH2OH groups (leading to two formaldehydes) and four CHOH groups (leading to 4 mol of formic acid). (ii) Because the sugar forms the same osazone as an aldose, its ketone must be at C2. So far, therefore, you have the following partial structure: (iii) and (v) tell you that CHO CH2OH Aldoheptose A CHOH CHOH CHOH CHOH C OH Ruff Ruff H CH2OH Aldohexose B CHOH CHO CHOH CHOH H C OH CH2OH CH2OH C O CHOH Stereochemistry unknown CHOH CHOH H C OH D-sugar O HO HO OH OH CH3COCH3, H -D-Lyxopyranose O HO HO O O CH3 H3C HO O HO HO OH HO O HO O O CH3 CH3 CH3COCH3, H -D-Xylopyranose 432 • Chapter 24 CARBOHYDRATES: POLYFUNCTIONAL COMPOUNDS IN NATURE 1559T_ch24_423-438 10/20/05 4:54 AM Page 432