bpo c chapter 20

7/29/2019 Bpo c Chapter 20

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20CARBOHYDRATES

arbohydrates are a major class of naturally occurring organic com-pounds, which com e by their nam e becau se they usually have, or approximate ,the general formula C,,(H,O),, , , with n equa l to o r g rea te r than three . Amongthe well-known carbohyd rates are var ious sugars, s tarches , and cellulose, a llof which are im portant fo r the mainten ance of life in both plants and animals.Although the s tructures of many carbohydrates appear to be quitecomplex, the chemistry of these substances usually involves only two func-tional groups- ketone o r aldehyde carbonyls and alcohol hydroxyl groups . T hecarbonyl groups normally do not occur as such, but are combined withhydroxy l groups to form hem iacetal or acetal linkages of th e kind discussedin Section 15-4E.A n understanding of stereochem istry is particularly im portant to under-standing the properties of carbohydrates. Configurational and conformationalisomerism play an important role. For this reason, you may wish to reviewC h a p t e r 5 and Sec t ions 12-3 and 19-5.

20-1 CLASSIFICATION AND OCCURRENCEOF CARBOHYDRATESThe s imple sugars , or monosaccharides, are the building blocks of carbo-hydrate chemistry. The y a re polyhydroxy aldehydes o r ketones with f ive , s ix,seven, or eight carbon atoms that are classif ied appropriately as pentoses,hexoses, heptoses, o r octoses, respectively. They can be designated by more


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20-1 Classification and Occurrence of Carbohydratesspecific names, such as aldohexose or ketohexose,to denote the kind of carbonylcompound they represent. For example, an aldopentose is a five-carbon sugarwith an aldehyde carbonyl; a ketohexose is a six-carbon sugar with a ketonecarbon yl:CH2-CH-CH-CH-CHO CH2-CH-CH-CH-C-CH2I I l l I I I I I l lOH OH OH OH OH OH OH OH O OH

aldopentose ketohexoseHowever, it is important to keep in mind that the carbonyl groups of sugarsusually are combined with one of the hydroxyl groups in the same moleculeto form a cyclic hemiacetal or hemiketal. These structures once were writtenas follows, and considerable stretch of the imagination is needed to recognizethat they actually represent oxacycloalkane ring systems:

CHO

an aldohexose a 1,5-cycl ic hem iacetal(open-ch ain form) of an aldohex ose

The saccharides have long and awkward names by the IUPAC system,consequently a highly specialized nomenclature system has been developedfor carbohydrates. Because this system (and those like it for other naturalproducts) is unlikely to be replaced by more systematic names, you will findit necessary to memorize some names and structures. It v~ill elp you to re-member the meaning of names such as aldopentose and ketohexose, and tolearn the names and details of the structures of glucose, fructose, and ribose.For the rest of the carbohydrates, the nonspecialist needs only to rememberthe kind of compounds that they are.

The most abundant five-carbon sugars are L-arabinose, D-ribose,2-deox y-D-ribose,l and D-xylose, which all are aldopentoses. Both the open-chain and cyclic structures of the D-aldoses up to C, are shown in Figure 20-1.

The common six-carbon sugars (hexoses) are D-glucose, D-fructose,D-galactose, and D-mannose. They all are aldohexoses, except D-fructose,which is a ketohexose. The structures of the ketoses up to C, are shown forreference in Figure 20-2. The occurrence and uses of the more importantketoses and aldoses are summarized in Table 20- 1.lDeoxy means lacking a hydroxyl group, and 2-deoxyribose is simply ribose withoutan OH group at the 2-carbon.


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20 Carbohydrates-r L.- o 0 , ,a,,LOOa aa, , 0 0 ,

' 1 1 10 0 0 0 6I 1 I I T0-$I-$I-0-0


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20-1 Classif ication and Occurrence of Carbohydrates

n n i i

1 10 0 6I I I-0-0-0I I


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20 Carbohydrates

CH,OH1,3-dihydroxy-2-propanone(d ihydroxyacetone)

CH,OHIc=oIH-C-OHICH,OH

C=OIH-C-OHIH-C-OHIHO-C-HIH-C-OHI

CH,OH CH,OH CH,OH CH,OHIC=O IC=O c=o IC=OIH-C-OH IHO-C-H IH-C-OH H0-C-HIH-C-OH IH-C-OH HO-C-H HO-C-HH-C-OH IH-C-OH IH-C-OH IH-C-OHIC H 2 0 H ICH,OH ICH20H CH,OH

Figure 20-2 Structure and configuration of the D -ketoses ro m C, to C,.As with the aldoses (Figure 20-I),he cycl ic form is predominantly anoxacyclohexane (pyranose) ring in the free sugar, but the oxacyclopen-tane (furanose) form is shown for fructose because it occurs widely inthis form as the disaccharide, sucrose. Only the a anomers are shown(see Section 20-2B).


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20-1 Classification and Occurrence of CarbohydratesTable 20-1Occurrence, Physical Properties, and Uses of Some Natural Sugars

Sugar M ~ ! [ a ] 20-25"C in H 20 a Occurrence and use

Monosacchar idesTrioses, C3H603D-glyceraldehyde1,3-dihydroxy-2-propanone

Tetroses, C4H804D-erythrosePentoses, C,H,,05L-arabinose

D-ri bose

Hexoses, C6H j20 6D-glucose

D mannoseD -galactose

Heptoses, C7Ht4O7sedoheptulose

Ol igosacchar idesDisacchar idessucroselactosemaltose

Tr isacchar idesraff nose

SY ru P +8.7 Intermediate in carbohy dratebiosynthesis and metabolism.8 1 - As above. '

syrup - 4.5 As above .16 0 +105 Free in heartwood of coniferous trees;widely distr ibuted in combine d formas glycosides and polysaccharides.

87 -23.7 Carbo hydrate com ponen t of nuc leicacids and coenzymes.145 +18.8 Called woo d sugar because it is widelydistr ibuted in comb ined form inpolysaccharides, such as in agriculturalwastes as corn cobs, straw, cottonseedhulls.

+52.7 Free in blood, other body fluids, and inplants; abundant combined aspolysacchar ides.-92.4 Free in fruit juices and honey; com-bined as in sucrose and plantpolysaccharides.+14.6 Comp onent of polysaccharides.t 8 0 . 2 Called brain sugar because it is acomponent of glycoproteins in brainand nerve tissue; also found in oligo-and p olysac charides of plants.

+8.2 Detected in succulent pla nts; an inter-mediate in ca rbohydrate biosynthesisand metabolism.

160-1 86 56 6. 5 Beet sugar and cane sugar.(D-glucose+ D-fructose)202 t 5 2 .6 Mi lk sugar of mammals.(D-galactose + D-glucose)103 +130 Hydrolytic prod uct of starch.(D-glucose+ D-glucose)

(Table continued on fol lowing page.)


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908 20 CarbohydratesTable 20-1 (continued)Occ urrence, Physical Properties, and U ses of Some Natural Sugars

Sugar M P , [ a ]"C in H,Oa Oc curre nce an d usePolysacchar idescel lulose(poly-D-g uco se)

starch(poly-D-g ucose)chi t in(polyethanamido sugar)hemicel luloses

plant gums

Occurs in al l plan ts as a const i tuent ofcel l walls; as structural component ofwoody an d f ibrous plants.As food reserves in animals (glycoge n),and in plants.Found in marine anim als, insects, fungi,and green algae.

aRotation at equil ibr iu m conce ntration of anomers and of pyranose an d furanose forms (Sections20-28 an d 20-2C).

20-2 THE STRUCTURE AND PROPERTIES OF D-GLUCOSE20-2A ConfigurationGlucose is by far the most abundant monosaccharide; it occurs free in fruits,plants, honey, in the blood of animals, and combined in many glycosides,disaccharides, and polysaccharides. The structure and properties of glucosewill be considered in greater detail than those of the other monosaccharides,not only because of its importance, but because much of what can be saidabout glucose also can be said about the other monosaccharides.Glucose is an aldohexose, which means that it is a six-carbon sugar witha terminal aldehyde group, shown by 1CHO 'CHO

I"CHOHII"CIIOH ~ 2 6 - o ~


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20-2 The Structure and Propert ies of D-Gluc oseThe carbons labeled with an asterisk in I are chiral; thus there are 2% orsixteen, possible configurational isomers. All are known- some occur naturallyand the others have been synthesized (see Table 20-1). The problem ofidentifying glucose as a particular one of the sixteen possibilities was solved byEmil Fischer during the latter part of the nineteenth century, for which he wasawarded the Nobel Prize in chemistry in 1902. The configurations he deducedfor each of the chiral carbons, C2-C5, are shown in the projection formula 2.2Although Fischer was aware that natural glucose could be the enantiomer ofStructure 2, his original guess at the nbsol~iteonfiguration proved to be correctand the configuration at C5 is the same as the configuration of the simplest"sugar," D-(+)-glyceraldehyde,3, (see Section 19-5). Therefore natural glucoseis specifically D-glucose:

The complete logic of Fischer's procedures for determination of theconfiguration of glucose is too involved to be explained here in detail; instead,much of it is incorporated into Exercise 20-1. This exercise will give you thesense of one of the finest achievements of organic chemistry, made long beforethe development of the spectroscopic techniques described in Chapter 9. Whatyou will be unable to fully appreciate is the great difficulties of working withcarbohydrates-that is, their considerable solubility in water, instability tostrong oxidizing agents and acidic or basic reagents, reluctance to crystallize,and their tendency to decompose rather than give sharp melting points. For-tunately for Fischer, many different pentoses and hexoses already wereavailable from the efforts of earlier investigators, and the principles of opticalisomerism were well understood as the result of the work of van't Woff.

Two of the key ideas used by Fischer can be illustrated best with aldo-tetroses because they have only two chiral carbons and far fewer possiblestructures to consider. Writing the four possibilities as the aldehyde ratherthan hemiacetal structures, we have 4-7.Of these, 4 and 5 constitute a pairof enantiomers, as do 6 and 9. These pairs can be identified by careful oxida-tion of the terminal groups to give the corresponding tartaric (2,3-dihydroxy-

W e will re ly heavi ly o n project ion form ulas in th is chap ter a s a way of represent ingthe configurations of carboh ydrate s . If you ar e unsu re of theii- meaning, we suggesttkat you review Sect ions 5-3C and 5-4. I t i s especial ly imp or tant to be able to t rans la teproject ion formulas in to models and saw horse drawings, as shown in Figures 5-12and 5-13.


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20 Carbohydratesbutanedioic) acids. Oxidation of both 4 and 5 gives meso-tartaric acid, whereasoxidation of 6 and 7 gives, respectively, (+) and (-) tartaric acids:

CHO CHO CHO CHOI I IH-C-OH HO-C-H

IHO-C-H H-C-OHI I IH-C-OH HO-C-H H-C-OH IWO-C-H

H-c-OHIH-C-OHIC 0 2 Hmeso-tartar ic a c id

HO-6-H H-&--OHIH-C-OH IHO-C-HIC 0 2 H ICO2HDg-(+) tartaric acid L,-(-) tartaric acid

It should be clear from this that the configurations of 6 and 7 are establishedby being related to the respective chiral tartaric acids. However, we have noway of telling which of the tetroses represented by 4 and 5 is D and which isL, because, on oxidation, they give the same achiral tartaric acid. What we needto do is relate one or the other of the chiral carbons of these tetroses to thecorresponding carbon of either 6 or 7. One way that this can be done is by theWohl degradation, whereby the chain length is reduced by one carbon by re-moving the aldehyde carbon:CHO CH=NOH CN1 1 1 -HCN

NH20H > CHOHHOH (CH3CQ)2Q CHOH- HOI -H 20 I IAs applied to 4, 5, 6, and 7, the Wohl degradation forms enantiomers ofglyceraldehyde:

CHO- - - - - - - - CHOI CHO - - - - -- --H-C-OH I IHO-C-MI - -C-OH +-H-C-OH IW-C-OHI C H 2 0 HCH2OH IC H 20H


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20-2 The Structure and Properties of D-GlucoseC H O- - - - - - - - C H OC H O - - - - - - - -HO-C-H H-C-OH

-HO-C-H +-HO-C-H I IHO-C-H

Here we see that 4 and 6 give the same enantiomer, D-glyceraldehyde, andtherefore have the same configuration of their highest-numbered asymmetriccarbon. In contrast, 5 and 7 give L-glyceraldehyde and thus must be similarlyrelated. By this kind of procedure, the configurations of 4 to 7 can be estab-lished unequivocally, although, as you might imagine, it is far easier to do onpaper than in the laboratory.Knowing the configurations of the tetroses aids in establishing the con-figurations of the pentoses. Thus 4, by Kiliani-Fischer cyanohydrin synthesis?can be converted to a mixture of two aldopentoses, 8 and 9, by adding a carbonat the aldehyde end of the molecule. The configurations of the two carbons atthe lower end of the starting material remain unchanged, but two diastereo-meric aldopentoses are formed in the syntheses because a new chiral centeris created:

C H O C H OIHO-C-H IH-C-OHIIH-C-OH 1 IH-C-OH H-C-OH

Products 8 and 9 present a new configurational problem, but a lessdifficult one than before, because the configurations of two of the three chiral

3The steps of the Kiliani-Fischer synthesis are:0CN CO,H CHOCHO

II I I

HCN CHOH H,o, H@ CHOH -H,O Na-Hg CHOHCHOHI IOH@' CHOH I CHOH IIpH ' CHOHCHOHI CHOH ICHOH ICHO ICHOHI I I I


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20 Carbohydratescenters already are known. Controlled oxidation of 8 and 9 will give diferentdiastereomeric 2,3,4-trihydroxypentanedioic acids, 110 and 11, respectively:

C H O C 0 2 H C 0 2 H C HOIHO-C-H IHO-C-H IH-C-OH Iw-e-owIH-C-OH IH-C-OH IH-e-OH < H-L-OHIH-C-OH IH-C-OH IH-C-OH IH-C-OH

C H 2 0 H C 0 2 H CO2I-n C H 2 0H8 10 d l 9

Of these, 11 is achiral (meso), whereas 10 is chiral. Therefore, by simplydetermining which oxidation product is optically active, and hence chiral, wecan assign the configurations of 8 and 9. Direct comparison of these syntheticaldopentoses with the naturally occurring compounds then could be used asproof of the structure of natural aldopentoses. By this reasoning 8 turns outto be D-arabinose and 9 is D-ribose.

Some of the key reactions in carbohydrate chemistry involve oxidation of al-doses to carboxylic acids. You will encounter some of these if you work Exer-cise 20- I . There is a simple nomenclature system for these acids. In abbreviatednotation, the products of oxidation at C1, C6, or both are called:

HOCH2- (CHOH),,-CHOaldose

J \HOCH2- (C HO H) ,-C02H H02C-(CHOH),,-CHO

aldonic acid alduronic acid\ JHOZC- (CHOH),,-C02H

aldaric acid

The carboxylic acids derived from glucose are therefore gluconic acid, glu-curonic acid, and glucaric acid.

Exercise 20-1 The lo gic necessary to solve this pro blem essential ly is that used byFischer in his classic work w hich established the configurat ions of glucose , arabinose,and mannose.a. The project ion formulas for al l the theoret ical ly pos sible n -aldo pe ntos es, HOCH,(CHOH),CHO, are shown in Figure 20-1. One of the D -ald op en tos es is the naturallyoccurr ing D -arabinose, which is enantiomeric with the more abundant L-arabinose.Oxidat ion of n-arabinose with nitr ic acid gives an optical ly act ive 2,3,4-tr ihydroxy-pentanedioic acid. Which of the n-aldopentoses could be n-arabinose?


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20-2B Hemiacetal Formation. Anomers of Glucoseb. D-A rabin ose is converted by the K i l iani-Fischer synthesis3 to D-g luc os e andD-m annos e. What do these transformations tel l about the relationship between theconfigurations of mannose and g lucose?c. Oxidation of D -g lucose and D -mannose gives the 2,3,4,5-tetrahydroxyhexanedioicacid s, glucaric and mannaric ac ids, respectively. Both are op tical ly active. What arethe configurations of the D - and L-arabinoses?d. D -Glucaric ac id c an form two different y-monolactones, whereas D -mannaric acidcan form only one y-m ono lacton e. What are the configurations of D-glucose a nd D-mannose?Exercise 20-2 a. Deduce possible configurations of natural galactose from thefollowing observations. Give your reasoning. (1) D -Galactose gives a pentose by oneWohl degradation. On nitr ic acid oxidation this pentose gives an optical ly active2,3,4-tr ihydroxypentanedioic acid. (2) The pentose by a second Wohl degradationfol lowed by nitr ic acid oxidation gives D-tartar ic acid.b. Write reasonab le mech anisms for the reactions in volve d in the Wohl degra dation .

20-2B Hem iacetal Formation. Anomers of GlucoseAlthough glucose has some of the properties expected of an aldehyde, it lacksothers. For example, it forms certain carbonyl derivatives (e.g., oxime andcyanohydrin), and can be reduced to the hexahydroxyhexane (sorbitol), andoxidized with bromine to gluconic acid (a monocarboxylic acid). (With nitricacid, oxidation proceeds further to give the dicarboxylic acid, D-glucaric acid.)

C H 2 0 H C H O C 0 2 H C 0 2 HIH-C-OH IH-C-OH IH-C-OH IH-C-OHHO-C-H IHO-C-H HO-C-H HO--C-H

I Na-HgH-C-OH < H-C-OH Br2 > IH-C-OH ~ ~ 0 3 ,-L-OHIH-C-OHIC H 2 0 HIH-C-OH IH-C--OH IH-C-OHIC H 2 0 H C H 2 0 H C 0 2 H

D -g lucose D-g luconic ac id D -g lucar ic ac idsorbi to lGlucose also will reduce Fehling's solution [Cu(II) --+ Cu(I)] and Tollen'sreagent [Ag(I)-+Ag(O)] and, for this reason, is classified as a reducing sugar?41n general, reducing sugars are hemiacetals or hemiketals and the nonreducing sugarsare acetals or ketals. The difference is between the presence of the structural elements

I IC-0-C-0-H and C-0-C-0-C.I I


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914 20 CarbohydratesHowever, it fails to give a hydrogen sulfite addition compound and, althoughit will react with amines (RNH, ) , the products are not the expected Schiff's

\bases of the type ,C=NR. Furthermore, glucose forms two diflerent mono-methyl derivatives (called methyl a-D-glucoside and methyl P-D-glucoside)under conditions that normally convert an aldehyde to a dimethyl acetal:

methyl a- and p-Dglucoside

All of these reactions can be explained on the basis that the carbonylgroup is not free in glucose but is combined with one of the hydroxyl groups,which turns out to be the one at C5, to form a hemiacetal, 12 or 13. Why aretwo hemiacetals possible? Because a new asymmetric center is created at C1by hemiacetal formation, and this leads to diastereomeric forms of D-glucosecalled a-D-glucose and P-D-glucose. In general, carbohydrate stereoisomersthat differ only in configuration at the hemiacetal carbon are called anomers:,nomeric carbon\anomersof IH O - 3 ~ - ~glucose IH - 4 ~ - - ~ ~ HO-C-HH-C-OH

I6 ~ ~ 2 ~ ~

ICH,OHa-D -glucose p-D g ucose

12 13

Although formulas 12 and 13 show the configurations at each of thechiral centers, they do not provide the crucial information for understandingthe properties of glucose with respect to the arrangement o f the a to ms in space.Conversion of a projection formula such as 12 or 13 to a three-dimensionalrepresentation is not at all a trivial task. We have indicated the procedure fordoing it before (Sections 5-3C and 5-5) and, if you wish practice, there are


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20-2B Hem iacetal Formation. Anomers of G lucose 915

examples in Exercise 20-3. The result of these procedures applied to 1 2 and13 are the so-called I-Iaworth projection formulas, 14 and 15,and the sawhorseconformations, 16 and 17:

You should be able to satisfy yourself that the configuration at C 5 isthe same in both Fischer and Haworth representations. This amounts to askingif 1 8 and 19 represent the same configurations:

They do, but if you have trouble visualizing this, it will be very helpful to use aball-and-stick model to see that 1 8 and 19 are different representations of thesame configuration. If you do not have models, remember that if transpositionof any three groups converts one projection into the other, the formulas areidentical. Thus 1 8 and 19 have the same configuration because 1 8 becomes19 by transposition of C4 with CW20H, hen C4 with H.


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20 CarbohydratesX-ray studies of crystalline a- and P-D-glucose show that these mole-

cules have their atoms arranged in space as correspond to 16 and 17. This iswhat we would expect from our studies of cyclohexane conformations (Sec-tions 12-3A to 12-3D), because for the P form, all of the substituents on theoxacyclohexane ring are in equatorial positions, and for the a form, all exceptthe hydroxyl at the anomeric carbon (Cl ) are equatorial.

Exercise 20-3 Determine for each of the following sets of structures whether theycorrespond to the same stereoisomer. The left structure in each ex am ple is a Fischerprojection formula. Models wil l be helpful!

HO-C-HIPCH, -------]CH,OH

H-C-OH

and

IHO-C-H HOI andH-C-OH CH,OHI

c. HO-c-HIHO-C-H- IHO-C-HIHO-C-HI

Hand H


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20-2C Mutarotation of the Anomeric Forms of Glucose20-2C Mu tarotation of the Anom eric Forms of Glu cos eAlthough the crystall ine forms of a - an d p-D-glucose ar e quite stable , in solu-t ion each form slowly chang es into an equilibrium mixture of both. Th e processcan be observ ed a s a decrease in the optical rotat ion of the a ano m er (+112")o r an inc rease fo r the p anomer (t-18.7") to the equilibrium value of 52.5".T h e phenomenon is known a s mutarotation and commonly i s observed forreducing sugars. Both acids and bases catalyze mutarotat ion; the mechanism,Equ ation 20- 1, is vir tually the sam e as described for acid- and base-catalyzedhemiacetal and hemiketal equil ibria of a ldehyde s and ketones (see Sect ion15-4E):

p-Dg ucose aldeh yde form

A t equ i lib rium, abou t 64 % of the P anom er and 3 6% of the a anornerare present . T h e am oun t of the f ree a ldehyde form a t equil ibrium is very small(about 0.024 mole percent in neutral solut ion). Prepond erance of the P a n o m e ris a t t r ibuted t o th e fact th at glucose ex ists in solut ion in th e chair conformationwith the large -CH ,OH group equatorial . In this conformation, the hydroxylsubst i tuent a t C1 is equatoria l in the /3 anomer and axial in the a a n o m e r ;hence the ,G anomer is sl ight ly more stable . When glucose is in the a ldehydeform, the hydroxyl a t C 4 a lso could add to the a ldehyde carbonyl to prod uce ahemiacetal with a f ive-membered r ing. This does not occur to a signif icantdegree with glucose because the hemiacetal with the six-membered r ing andmany equatoria l groups is more stable . With other sugars, mixtures of f ive-and s ix-membered hemiace tal o r ke ta l r ings and the ir respec t ive anom ers a reproduced in water solut ion.Carbon- 13 nm r spec t ra (Sect ion 9- 10L ) provide an especially powerfultool for studying th e anom eric forms of sugars. Fo r example, with glucose the


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20 Carbohydrates

100 9 80 7 60ppm from TMS

Figure 20-3 Proton-decoupled 13C nmr spectrum at 15.1 MHz of glucosein water solution, showing the peaks of both the a! and p anomers. (As inFigure 9-50 the relative peak heights do not correspond to relativeamounts of a! and p forms present.)

resonances of C 1, C3, and C5 of the a anomer are seen in Figure 20-3 to beshifted substantially upfield relative to those of the P anomer becabse of theaxial substituent effect (Section 12-3D).

Exercise 20-4 Draw the chair conformation of P-D -gluc os e with a l l of the su bsti t-uent groups axia l . Explain how hydrogen bonding may compl icate the usual con-siderations of steric hindrance in assessing the stabil i ty of this conformation relativeto the form with all substituent groups equatorial.

20-2D Aldosev etose RearrangementsIn the presence of dilute base, D-glucose rearranges to a mixture containingthe anomers of D-glucose (- 64%), D-fructose (- 3 I%), and D-mannose (3%).This interconversion undoubtedly involves enolization of the hexoses by way


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20-2D Aldose Ketose Rearrangementsof a common enediol intermediate according to Equation 20-2:

D -glucose(partial structure)

T L H Z O

CHOHIC-OH + HOenediol

I +HO-C-H

The rearrangement is of interest because the corresponding enzymatic inter-conversion of aldoses and ketoses is an important part of the biosynthetic,photosynthetic, and metabolic pathways, as we shall see in Section 20-9. Al-though the biochemical rearrangement also may proceed by way of enediolintermediates, it is highly stereospecific and yields only one of two possiblestereoisomeric aldoses. For example, glucose, but not mannose, can be enzy-matically interconverted with fructose as the 6-phosphate ester derivative:

CH=O CH-OH CH20 HI I 11CH-OH - -OH " C=OI I I

aldose enediol ketose(glucose Bphosphate) (fructose 6-phosphate)


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20 Carbohydrates20-3 CONVENTIONS FOR INDICATING RING SIZEAND ANOMER CONFIGURATIONSOF MONOSACCHARIDESThe oxide ring is six-membered in some sugars and five-membered in others,and it is helpful to use names that indicate the ring size. The five- and six-membered oxide rings bear a formal relationship to oxa-2,5-cyclohexadieneand oxa-2,4-cyclopentadiene that commonly are known as pyran and furan,respectively:

For this reason, the names furanose and pyralnose have been coined to denotefive- and six-membered rings in cyclic sugars. The two forms of glucose areappropriately identified by the names a-D-glucopyranose and P-D-glucopyra-nose. Likewise, L-arabinose, D-xylose, D-galactose, and D-mannose occurnaturally as pyranoses, but D-ribose (in combined form) and D-fructose occuras furanoses (see Figures 20- 1 and 20-2).There is an important question as to which one of the two anomericforms of a sugar should be designated as cr and which one as P. The conven-tion is simple; the a anomer is the one that has the s ame configuration of theOH at the anomeric carbon as the carbon which determines the configurationof the sugar itself:

D sugara anomer D sugarp anomer L sugara anomer L sugarp anomer

Exercise 20-5 Mak e a sawhorse d raw ing of what you believ e to be the favored con-formations of a- and @-D ribopyranose and of a- and @-D idopyranose.


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20-4 Derivatives of Glucose 92120-4 DERIVATIVES OF GLUCOSE20-4A Determ ination of the Oxid e Ring SizeAlthough we now have powerful spectroscopic methods available to determinethe s izes of the oxide rings formed by the s imple monosaccharides, the way inwhich this was done chemically for glucose highlights the difference in reac-t ivi ty betw een eth er an d alcohol functions.The acid-catalyzed methylation of glucose with methanol to give twodist inct glucosides, methyl a-D-glucoside and methyl p-D-glucoside, cor-responds to displacement of the hemiaceta l hydroxyl by m ethoxyl to form anacetal (see left s ide of F igure 20-4).T h e remaining fou r hydroxyl grou ps can be m ethylated in basic solutionby dimethyl sulfate or by methyl iodide and si lver oxide in N,N-dimethyl-methanam ide, H C O N (CH ,) ,, solution. Hydrolys is of e ither of these penta-methyl glucose derivatives with aqueous acid affects only the acetal l inkagean d leads to a tetramethylated glucose, 20, as shown in Figure 20-4.T h e pyra nose ring s tru cture of D-glucose originally was established byH irst , in 192 6, by convert ing D-glucose to a tetra-0-meth yl-~-glucose andshowing that this substance actually was 2,3,4,6-tetra-0-methyl-D-glucose,20.T h e key fea ture of 20 is the fact that all but the tw o carbo ns involved inhemiacetal formation are protected from oxidation by being substi tuted with0-methyl groups in place of hydroxy groups . The larges t f ragment i sola tedfrom oxidation of Hirst 's tetra-0-methyl-D-glucose was a tr imethoxypen-tanedioic acid, 21, and beca use the two carboxyl carbons must h ave been theon es originally involved in ring forma tion, the oxide ring m ust be be twee nC 1 an d C 5 :

Re age nts th at specifically oxidize vicinal glycols [e.g., WaIO,, Pb-(O,CCH, ) ,, and N aBiO,; Section 16-9 A] ar e quite helpful in determiningthe cyclic s tructures of sugars . With periodate, the numbers of moles of oxi-dant consumed and the moles of methanoic ac id and methanal produced are


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20 Carbohydrates

methyla - D -g lucosidem p 166"

methylp-Dg lucos idem p 105"

OCH,

Figure 20-4 Haworth pro ject ion formulas showing the format ion a ndreact ions of 0-m ethy l de r ivat ives of glucose. The notation (H,OH) in 20means that the anomeric conf igurat ion is unspeci f ied.

different for each type of ring structure. The cleavage reactions that normallyare observed follow:R RICHOH I104' , CHO

- -- - - - - - +CH,OH HCHO

RICHOHRICHO

- - - - - - - -I +CHOH 21040 > HCOzH----I---- +CHOH CHO

I

I ICHoH 10~0 CHO_ _ _ _ I _ _ _ _ > +CHOHI CHOI


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20-4B React ions with Amines a nd Hydrazines; Osazone Formation 923

Exercise 20-6 From the following information deduce the ring structures of thesugars. Give your reasoning.Sugar Mo les of lo,@ Mo les of HC02H Mo les of H2C0(as methyl g lycoside) consumed formed formedmethyl a -D -mannoside 2methyl a-D -ri boside 1a m ethyl g lycoside of

an aldohexose 3a m ethyl g lycoside of

an aldohexose 2

20-48 Reactions with Amines and Hydrazines;Osazone FormationA s we s ta ted previously , g lucose forms som e, but no t a l l, of the com mo ncarbonyl derivatives. The amount of free aldehyde present in solution is sosmall that it is not surprising that no hydrogen sulfite derivative forms. Withamines, the p rodu ct is not a Schiff 's base but a glucosylamine of cyclic s truc-ture analogous to the hemiacetal s tructure of glucose, Equation 20-3. TheSchiff 's base is likely to be a n intermediate that rapidly cyclizes to the gluco-sylamine:

Haldehyde form ofD -g lucose Sc h f f 's base

H OHanomers of N-phenyl-D-g lucosy lamine


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20 CarbohydratesThe reaction of glucose with an excess of phenylhydrazine (phenyl-

diazane) is particularly noteworthy because two phenylhydrazine moleculesare incorporated into one of glucose. Subsequent to the expected phenyl-hydrazone formation, and in a manner that is not entirely clear, the -CHOH-group adjacent to the original aldehyde function is oxidized to a carbonylgroup, which then consumes more phenylhydrazine to form a crystallinederivative called an osazone, or specifically glucose phernylssazone:CHO CHzNNIIC6W5 CH=N NHC6 H5"I C,H5NHNH2 ICHOH > CHOH - C=NNHC6H5I I Iglucose phenyl hydrazoneof glucose g lucosephenylosazone

The sugar osazones usually are crystalline and are useful for charac-terization and identification of sugars. Fischer employed them in his workthat established the configuration of the sugars. The kind of information thatcan be obtained is illustrated by the following example:

I(ewow4 C6H5NHNH, C=NNHC6H5 C,H5NHNH, C=oexcess (GHOH), excess (CHOH),CH,OH IC H 2 0 H C H 2 0 H

glucose andmannose phenylosazone fromglucose, mannoseand fructosefructose

Because the s ame phenylosazone arises from glucose, mannose, and fructose,the configurations of C3, C4, and C5 must be the same for all three sugars.

Exercise 20-7 D -Arabinose and D - r ibose g ive the same phenylosazone. D -Riboseis reduced to the opt ica l ly inact ive 1,2,3,4,5-pentanepentol, r ib i to l , D-Arab inose canbe deg raded by the Ruf f method, whic h involves the fo l lowing react ions:

The tetrose, D -erythrose, so obta ined can be oxid ized with n i t ric a cid to meso-tar tar icacid. Show how th is in format ion can be organized to establ ish the conf igurat ions ofD -arabinose, D - r i bose, r ib i to l , and D -erythrose.


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20-5 Glycos ides 92520-5 GLYCOSIDESAlthough abundant quanti t ies of glucose and fructose are found in the freestate, they and less common sugars occur widely in plants and animals com-bined w ith various hydroxy com pound s. Th e bonding is through oxygen tothe carbonyl carbon , as in the a - a nd P-methylg lucosides d iscussed in Sect ion20-4A, to g ive aceta l o r keta l s t ructures . These substances are somet imessimply called glycosides, but i t is desirable t o spe cify that the bonding is throughoxyge n by using the name 0-glycos ide. H ydrolysis of an 0-glycoside givesthe sugar and the hydroxy compound, ca l led the aglyeone component of theglycoside.A specific exam ple is glucovanill in , w hich can b e isolated from th e greenfruit pods of vanilla, a climbing orchid cultivated in several tropical countries.Hydrolysis gives glucose and the aglycone, vani l l in , which is the principalingredient of vanilla flavoring. A s the vanilla po ds m ature , a natural hydrolysisreact ion procee ds t o the exten t that th e pod s may be covered with small crys-tals of vanillin.

sugar part ag lycone par tg lucovan i l l in( p isomer)

vanil l in

T h e configurat ions of glycosides ar e designated by th e sam e conve ntionused for the sugar anomers. Thus i f a glycoside of a D sugar has the D con-figurat ion at the anom eric ca rbo n, i t is designated a s th e a-D-glycoside, andif it has the L configuration it is called the p-D-glycoside (see Section 20-3).If the sugar involved in glycoside formation is glucose, the derivative is aglucoside; if fructose, a frucdoside; if galactose, a galactoside, and s o on. Whenthe hydroxy compou nd, o r ag lycone, is anothe r sugar , then the g lycoside is adisaccharide, an d if th e sugar is already a disaccharide, th e glycoside is a tri-saccharide, and so on.Am ong the natura l p roducts that occur a s g lycosides (most comm onlya s P-D-glucosides) are m any plant pigments ( the anthocya nins) , the f lavorings


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20 Carbohydratesvanillin and amygdalin, and many steroids (such as the cardiac glycosides andand saponins). The structures of some of these substances will be discussedin later chapters.

Not all glycosides are 0-glycosides. A group of N-glycosides of specialbiological importance are derived from heterocyclic nitrogen bases andD-ribose and 2-deoxy-D-ribose. They commonly are known as nucleosides, ormore specifically, as ribonucleosides and deoxyribonucleosides; the N-glycosidelinkage is always P :

r i bonuc leoside(part ial structure) deoxyr i bonucleoside(part ial structure)

The N-glycoside of D-ribose and the nitrogen heterocycle, adenine, isadenosine:

p-D - r ibose adenine

adenosine(a nucleoside)

A nucleotide is a phosphate ester of a nucleoside. The hydroxyl groupat the C5 position of the pentose is the most common site of esterification. Thenucleotides of adenosine are ATP, ADP, and AMP (Section 15-5F).

A dinucleotide is a combination of two nucleosides through a commonphosphate ester link. Familiar examples are NADO, NADH, FAD, and


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20-6 Disaccharides; General Types and Propert ies 927FADH, (Section 15-6C). Polynucleotides are polymers of nucleosides linkedthrough phosphate ester bonds. Polynucleotides also are called nucleic acids(RNA and DNA) and are the genetic material of cells, as will be discussed inChapter 25.

Exercise 20-8 Work out a mechanism for the acid-induced hydrolysis of N-glyco-sides . Pay special attention as to where a proton c an be add ed to be most effective inassisting the reaction. Wo uld you ex pec t that adenosine wo uld hydrolyze more, orless, readily than N-methyl-a-r ibosylamine? Give your reasoning.

20-6A General Types a nd PropertiesCombinations of two or more of the simple sugars through glycoside linkagesgive substances known as polysaccharides. They also are called oligosaccha-rides if made from two to ten sugar units. The simplest oligosaccharides aredisaccharides made of two molecules of simple sugars that can be the same ordifferent. There are two ways in which the simple sugars can be joined with0-glycoside links, and it probably is easiest to visualize these as shown in the"stripped-down" formulas, 22 and 23:

acetal

(.-carbon

Ihydroxy H O)-Icarbon22

.hemiacetalcarbon

carbon23

You should look at 22 and 23 carefully to be sure that you recognize thedifference between them."n 22, sugar A is acting as a simple hydroxy com-pound, the aglycone of the sugar G to which it is linked by an 0-glycoside

"For now, we will ignore the possibility of different anomers of the disaccharide ortheir component sugars.


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20 Carbohydratesl inkage.6 Hydrolysis of 22 at th e glycoside link th en will proce ed a s follows:

aglycone22

Disacchar ides such as 22 are like glucose in being reducing sugars (Section20-2B), because Component A has the hemiacetal grouping that is openedeasi ly to the aldehyde form in the mildly alkaline condit ions used for theTollen 's and Fehling 's solution oxidations. Because ther e is a f e e hemiacetalgroup, reducing sugars also form osazones and they mutarotate (Sections20-4B and 20-2C).Disacchar ides of type 23 ar e different in tha t eac h sugar, G and G', isact ing as both a glycoside sugar and as an aglycone. The l inkage betweenI Ithem is tha t of a doub le-barreled aceta1,-0-C-0-C-0-, an d ther e is

no hemiacetal grouping in the molecule. Therefore these are nonveducingsugars as fa r a s th e s tandard tes ts go . However , hydrolys is of the 0-glycos idelinkages of 23 do es generate reducing sugars with hemiacetal carbo ns:

T h e manner in which sugars are linked together to form oligosaccharides was eluci-dated by W . N. Haworth, who received the Nobel Prize in chemistry in 1937 for thisand other contributions to research on the structures and reactions of carbohydrates.


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20-6 Disacchar ides; G eneral Types and P ropert ies 929In general, we find that the nonreducing disaccharides give none of the car-bony1 reactions observed for glucose, such as mutarotation and osazoneformation, except when the conditions are sufficiently acidic to hydrolyze theacetal linkage.

Among the more important disaccharides are sucrose, 24, maltose, 25,cellobiose, 26, and lactose, 27:

sucrose,24

maltose,25

ce l lob iose,26

OW lactose,27

Sucrose and lactose occur widely as the free sugars, lactose in the milk ofmammals, and sucrose in fruit and plants (especially in sugar cane and sugarbeet). Maltose is the product of enzymatic hydrolysis of starch, and cellobioseis a product of hydrolysis of cellulose.


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20 CarbohydratesTo fully establish the structure of a disaccharide, we must determine (1)

the identity of the component monosaccharides; (2) the type of ring junction,furanose or pyranose, in each monosaccharide, as it exists in the disaccharide;(3) the positions that link one monosaccharide with the other; and (4) theanomeric configuration (a or P) of this linkage.

Hydrolysis of disaccharides with enzymes is very helpful in establish-ing anomeric configurations, because enzymes are highly specific catalysts forhydrolysis of the different types of glycoside linkages. For instance, a-D-glucosidase (maltase) catalyzes hydrolysis of a-D-glycosides more rapidlythan of p-D-glycosides. The enzyme emulsin (found in bitter almonds) in con-trast shows a strong preference for P-D-glycosides over a-D-glycosides. Yeastinvertase catalyzes hydrolysis of P-D-fructosides.

Exercise 20-9 a. Which of the disaccharides, 24 hrough 27, ould you expect tobe reducing sugars?b. Determine the configuration of each of the anomeric carbons in 24 hrough 27 aseither a or p.c. Determine which m onosac charides (neg lect the anomeric forms) wi l l be prod uce don hydrolysis of 24 hrough 27. e sure you s pecify the c onfigurations as D or L.

20-66 Structure of SucroseWe know that sucrose consists of the two monosaccharides, glucose andfructose, because hydrolysis with acids or enzymes gives equal amounts ofeach hexose. Further, sucrose is not a reducing sugar, it forms no phenylosa-zone derivative, and it does not mutarotate. Therefore the anomeric carbonsof both glucose and fructose must be linked through an oxygen bridge insucrose. Thus sucrose is a glycosyl fructoside or, equally, a fructosyl glucoside.

Because sucrose is hydrolyzed by enzymes that specifically assisthydrolysis of both a glycosides (such as yeast a-glucosidase) and P-fructosides(such as invertase), it is inferred that the glucose residue is present as ana glucoside and the fructose residue as a P fuuctoside. If so, the remaininguncertainty in the structure of sucrose is the size of the rings in the glucose andfructose residues.

The size of the sugar rings in sucrose has been determined by the reac-tions shown in Figure 20-5. Methylation of sucrose with dimethyl sulfate inbasic solution followed by hydrolysis of the octamethyl derivative gives2,3 4,6-tetra-0-methyl-D-glucopyranose (Section 20-4) and a tetra-o-methyl-D-fructose. This establishes the glucose residue in sucrose as a glucopyranose.The fructose residue must be a fructofuranose because periodate oxidation ofsucrose consumes three moles of periodate, whereby one mole of methanoicacid and one mole of a tetraaldehyde are formed. On bromine oxidation


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20-68 Structure of Sucrose

sucrose tetraaldehyde

C 0 2 H C H 2 0 H C0,H(H,OH) 4- t e t r a - 0 - m e t hy l - ~fructose I I I(not characterized at the H-C-OH + CHO C=O + H--C-OHt ime this experiment I I 1C H2 0H C 0 2 H C 0 2 H Iwas ca rr ied out) CH,OHOCH, D -glycer ic glyoxyl ic hydroxy- D -glycer ic2,3,4,6-tetra-0-methyl- ac id ac id pyruvic ac idD -g lucose ac idFigure 20-5 Summary of reactions used to establ ish the r ing structureof sucrose

followed by acid hydrolysis, the tetraaldehyde gives 3-hydroxy-2-oxopropanoicacid (hydroxypyruvic acid, HOCH2COC02H), oxoethanoic acid (glyoxylicacid, OCHC0 2H), nd D-glyceric acid (HOCH2CHOHC02H). ucrose there-fore has structure 24, and this structure was confirmed by synthesis (R .Lemieux in 1953).

Exercise 20-10 Draw Haworth and conformational structures for each of the followingdisacchar ides:a. 6-0 -p -D-g ~ C ~ p y r a n ~ ~ y l - p - D - g l ~ C ~ p y r a n ~ ~ eb. 4 - O - p - ~ - g a l a C t 0 p y r a n 0 ~ y l - a - ~ - g l ~ ~ 0 p y r a n 0 ~ ec. 4-0-p-D-xylopyranosyl-p-L-arabinopyranosed. 6-0-a-D-galactopyranosyl-p-D-fructofuranose


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20 CarbohydratesExercise 20-11 Show how the structure of maltose can be deduced from thefol lowing:(1) The sugar is hydrolyzed by yeast a-D -gluco sida se to D-glu cos e.(2) Maltose mutarotates and forms a phenylosazone.(3) Methylation with dimethyl sulfate in basic solution fol lowed by acid hydrolysisgives 2,3,4,6-tetra-0-methyl-D -glucopyranose and 2,3,6-tr i -0-methyl-D-glucose.(4) Bromine o xidation of m altose fol lowed by methylation and hydrolysis gives2,3,4,6-tetra-0-methyl-D-glucopyranose and a tetramethyl-D-gluconic acid, whichreadily forms a y-lactone.

Exercise 20-12 Cellobiose differs from maltose only in its behavior to enzymatichydrolysis. It is hydrolyzed by yeast P-D-g ucosidase. What is its structure?

Exercise 20-13 Show how the structure of lactose may be deduced from thefol lowing:(1) The sugar is hyd rolyze d by /I-D-galactosidase to a mixture of equal p arts ofD-glucose and D-galactose.(2) Lactose mutarotates and forms a phenylosazone.(3) Bromine oxidation of lactose fol lowed by hydrolysis gives D-g lucon ic ac id andD-galactose.(4) Methylation and hydro lysis of lacto se gives a tetra-0-methyl -D-galactose and2,3,6-t r i -0-methyl -D-glucose.The same galactose derivative can be obtained from themethylation and hydrolysis of D -galactopyranose.(5 ) Bromine oxidation of lactose followed by methylation and hydrolysis yieldstetra-0-methyl- l ,4-gluconolactone and the same galactose derivative as in (4).

20-7 POLYSACCHAR IDES20-7A Cel lu loseThe fibrous tissue in the cell walls of plants contains the polysaccharidecellulose, which consists of long chains of glucose units, each of which is con-nected by a p-glucoside link to the C4 hydroxyl of another glucose as in the


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20-7 Polysaccharidesdisaccharide cellobiose (i.e., P- 1,4):

cel lu lo se st ructure(the hydroxyls on the r ings are omit ted for clar ity)

Indeed, enzymatic hydrolysis of cellulose leads to cellobiose. The molecularweight of cellulose varies with the source but is usually high. Cotton celluloseappears to have about 3000 glucose units per molecule.

The natural fibers obtained from cotton, wood, flax, hemp, and jute all arecellulose fibers and serve as raw materials for the textile and paper industries.In addition to its use as a natural fiber and in those industries that depend onwood as a construction material, cellulose is used to make cellulose acetate(for making rayon acetate yarn, photographic film, and cellulose acetate butyrateplastics), nitric acid esters (gun cotton and celluloid7), and cellulose xanthate(for making viscose rayon fibers). The process by which viscose rayon is manu-factured involves converting wood pulp or cotton linters into cellulose xanthateby reaction with carbon disulfide and sodium hydroxide:

The length of the chains of the cellulose decreases about 300 monomer unitsin this process. At this point, the cellulose is regenerated in the form of finefilaments by forcing the xanthate solution through a spinneret into an acid bath:

7Celluloid, one of the first plastics, is partially nitrated cellulose (known as pyroxylin)plasticized with camphor.


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20 CarbohydratesA few animals (especially ruminants and termites) are able to metabolize

cellulose, but even these animals depend on appropriate microorganisms intheir intestinal tracts to hydrolyze the P- 1,4 links; other animals, including man,cannot utilize cellulose as food because they lack the necessary hydrolyticenzymes. However, such enzymes are distributed widely in nature. In fact,deterioration of cellulose materials- extiles, paper, and wood- y enzymaticdegradation (such as by dry rot) is an economic problem that is not yet ade-quately solved. Efforts to turn this to advantage through enzymatic hydrolysisof cellulose to glucose for practical food production have not been very success-ful (see Section 25- 12).

20-7B Starch and Related Com poundsA second very widely distributed polysaccharide is starch, which is stored inthe seeds, roots, and fibers of plants as a food reserve- a potential source ofglucose. The chemical composition of starch varies with the source, but inany one starch there are two structurally different polysaccharides. Both con-sist entirely of glucose units, but one is a linear structure (arnylose) and theother is a branched structure (amylopectin).

The amylose form of starch consists of repeating 1,4-glucopyranoselinks as in cellulose, but unlike cellulose the linkage is a rather than P (i.e.,a-1,4):

mal tose unit > I

amy lose

Hydrolysis by the enzyme diastase leads to maltose.


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20-76 Starch and Related CompoundsIn amylopectin, amylose chains are joined by a-1,6 inkages:

amylopect in

Animals also store glucose in the form of starchlike substances called glycogens.These substances resemble amylopectin more than amylose, in that they arebranched chains of glucose units with a-1,4- and a-1,6-glucoside links.

Starch is used in paper manufacture and in the textile and food industries.Fermentation of grain starches is an important source of ethanol. Hydrolysisof starch catalyzed by hydrochloric acid results in a syrupy mixture of glucose,maltose, and higher-molecular-weight saccharides. This mixture is calleddextrin and is marketed as corn syrup. The hydrolysis does not proceed all theway to glucose because the a-1,6 glucosidic link at the branch point is not easilyhydrolyzed. Enzymes also catalyze hydrolysis of starch, but the enzymea amylase is specific for a- 1,4 links and, like acid-catalyzed hydrolysis, gives amixture of glucose, maltose, and polysaccharides (dextrin). The enzyme a-1,6-glucosidase can hydrolyze the a-1,6 links at the branch points and, when usedin conjunction with a amylase, completes the hydrolysis of starch to glucose.

A very interesting group of polysaccharides isolated from cornstarch hydroly-sates are known as cyclodextrins. One of these compounds, cyclohexaamylose,is a large doughnut-shaped molecule with a central cavity that literally canengulf a small, relatively nonpolar organic molecule and hold it in water solu-tion, similar to a micelle (Section 18-2F). As with micelles, unusual reactivityis exhibited by the bound molecules. An example is the change in the ortho-para


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20 Carbohydratesratio in electrophilic substitution of methoxybenzene by hypochlorous acid,HOCI, in the presence and absence of cyclohexaamylose:

Apparently the cyclohexaamylose wraps around the methoxybenzene in sucha way as to protect the ortho carbons from attack by HOCl but to leave the paracarbon exposed. I t is this kind of specificity that we need to generate in reactionsbefore we can claim to have synthetic reactions under control.

Exercise 20-14 Explain how the /3-D-glucoside units of cel lulose pro duc e a polymerwith a stronger, more co m pa ct ph ysical structure than the a-D-g luco se units of starch.Models wi l l be helpful.

20-7C Other Important PolysaccharidesMany polysaccharides besides starch and cellulose are important componentsof animal tissues, or play a vital role in biochemical processes. One example ischitin, a celluloselike material that is the structural component of the hard shellsof insects and crustaceans. The difference between chitin and cellulose is thatinstead of being a polymer of glucose, chitin is a polymer of 2-deoxy-2-N-ethanamidoglucose (N-acetyl-p-D-glucosamine):

Heparin is a very important and complex polysaccharide derivative thatoccurs in intestinal walls and has a major use as a blood anticoagulant, espe-cially in connection with artificial kidney therapy. Heparin also has shown great


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20-7C Other Important Polysaccharides 937promise in the treatment of patients with extensive burns, by promoting bloodcirculation to burn-damaged tissue. The structure of heparin can not be definedprecisely because its composition depends on the source of supply. The majorcomponents of the polysaccharide chain are D-glucuronic acid, L-iduronic acid,and the same 2-deoxy-2-aminoglucose (D-glucosamine) that is a constituent ofchitin (although in heparin it occurs as the a anomer):

H HHO

HH H OH OH H OH

D-glucuronic acid L-iduronic acrd 2-deoxy-2-amino-D -glucose(p anomer) (a anomer) (a anomer)

The general construction of heparin involves the linkage of the anomericcarbons of one of the components with the 4-hydroxyl of another. A key featureof the heparin structure is the presence of sulfate groups that occur as hydrogensulfate esters (Section 15-5B) and as sulfamido groups, -NHSO,H, on the2-deoxy-2-amino-D-glucose units in the chain. Hydrogen sulfate groups also arelocated on the 2-hydroxyls of the L-iduronic acid units of the chain. In additionthere are N-ethanoyl groups attached to some of the 2-deoxy-2-amino-D-glucosenitrogens that are not connected to -SO,H.

Heparin is clearly an extraordinarily complex substance with many highlypolar groups, and its mode of action as an anticoagulant is not clear. At present,because of increases in the use of artificial kidney machines, heparin is in rathershort supply.

Among the plant polysaccharides are the pectins, which are used as jellingagents in the making of preserves and jellies from fruit. Also important are thealginates from seaweeds and gums from trees, which are used as stabilizers andemulsifiers in the food, pharmaceutical, cosmetic, and textile industries. Thepectins principally are polysaccharides of the methyl ester of D-galacturonicacid, whereas the alginates are polysaccharides made up of varying proportionsof D-mannuronic acid and L-guluronic acid. The plant gums are similar materials.

p-D -galacturonic acid a-L-guluronic acid p-D mannuronic acid

There are other polysaccharides besides cellulose in the cell walls of plants.These are called hemicelluloses, but the name is misleading because they areunrelated to cellulose. Those that are made of pentose units (mainly xylose) aremost abundant. They accumulate as wastes in the processing of agricultural


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938 20 Carbohydratesproducts , an d o n t reatm ent w i th acids they yield a com pound of cons iderablecommercial impor tance, oxa-2,4-cyclopentadiene-2-carbaldehydefurfural):

hemicellulosesorpentosans Hz0 > H H oxa-2,4-cyclopentadiene-2-carba ldehydeH O H (furfural)

Exercise 20-15 Write Fischer projections, Haworth projections, and sawhorse con-formational drawing s for the following:a. a-D-glucuronic acid b. p-D- iduronic acid C. a - ~ - g u l u r o n i c c id

20-8 VITAMIN CT h e "ant iscorbutic" facto r of f resh f ru i ts , which prevents th e development ofthe typ ical symptom s of scurvy in hum ans , is a carbohydrate der ivat ive knowna s vitamin C o r ascorbic acid. This subs tance is no t a carboxyl ic acid , bu t alactone, and o we s i ts ac id ic proper t ies (and ea se of oxidation) to the presenceof an enediol grouping. I t belongs to the L ser ies by the glyceraldehydeconvent ion:

I \C=O L-ascorbic ac idH-C, /1 OHO-C-H

M ost an imals a re ab le to synthes ize vi tamin C in their livers but, in thecourse of evolution, ma n has lost this capacity .

Exercise 20-16 Explain how you could account for the fact that ascorbic acid ismost stable in the enediol form rather than having its C3 and C2 carbons arrangedeither as -C(=O)-CH (0H)- or as -CH(OH)-C(=O)-.


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20-9 Formation of Carbohydrates by Photosynthesis20-9 FORMATION OF CARBOHYDRATESBY PHOTOSYNTHESIS

Carbohydrates are formed in green plants by photosynthesis, which is thechemical combination, or "fixation," of carbon dioxide and water by utilizationof energy from the absorption of visible light. The overall result is the red~lct ionof carbon dioxide to carbohydrate and the formation of oxygen:

lightxCO, + x H 2 0- C . H 2 0 ) ,+ x 0 2green plants carbohydrateIf the carbohydrate formed is cellulose, then the reaction in effect is the reverseof the burning of wood, and obviously requires considerable energy input.

Because of its vital character to life as we know it, photosynthesis has beeninvestigated intensively and the general features of the process are now ratherwell understood. The principal deficiencies in our knowledge include just howthe light absorbed by the plants is converted to chemical energy and the detailsof how the many complex enzyme-induced reactions involved take place.

The ingredients in green plants that carry on the work of photosynthesisare contained in highly organized, membrane-covered units called chloroplasts.The specific substances that absorb the light are the plant pigments, chlorophylla and chlorophyll 6, whose structures are shown in Figure 20-6. These highlyconjugated substances are very efficient light absorbers, and the energy sogained is used in two separate processes, which are represented diagrammat-ically in Figure 20-7.

CH, CO,CH,OICO,R '

R = CH3 (chlorophyl l a ) ; CHO (chlorophyl l b)

Figure 20-6 The structure of chlorophyll a and chlorophyl l b, showingcis-trans relationships of the substituents


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20 Carbohydrates

Figure 20-7 Simpl i f ied representat ion of the photoreact ions in photo-synthesis. The o xidation of water is l inked to the red uction of NADPB byan electron-transport ch ain (da sh ed l ine) that is co up led to ATP forma-t ion (photophosp horylat ion) .

One photoprocess reduces nicotinntnide adenine ditzucleotide phosphate(NAD PO) to N A D P H . Th ese d inuc leo tides , shown below, d iffe r f rom N A D Band N A D H (Sect ion 15-6C) in having a phosphate group a t C2 of one of ther ibose uni ts . T he oxidized form, NAD PO, behaves l ike N AD O and receivesthe equivalent of H:O at C 4 of the n icotinamide ring to form N A D P H :


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20-9 Formation of Carbohydrates by Photosynthesis 946

NADPH

The other important photoreaction is oxidation of water to oxygen by thereaction:

The oxygen formed clearly comes from H,O and not from CO,, because photo-synthesis in the presence of water labeled with 180 roduces oxygen labeledwith 180 , whereas carbon dioxide labeled with 180does not give oxygenlabeled with 180.Notice that the oxidation of the water produces two electrons,and that the formation of NADPH from NADPOrequires wo electrons. Thesereactions occur at different locations within the chloroplasts and in the processof transferring electrons from the water oxidation site to the NADPOreductionsite, adenosine diphosphate (ADP) is converted to adenosine triphosphate(ATP; see Section 15-5F for discussion of the importance of such phosphoryla-tions). Thus electron transport between the two photoprocesses is coupled tophosphorylation. This process is called photophosphorylation (Figure 20-7).

The end result of the photochemical part of photosynthesis is the formationof O,, NAD PH , and ATP. Much of the oxygen is released to the atmosphere,but the NADP H and ATP are utilized in a series of dark reactions that achievethe reduction of carbon dioxide to the level of a carbohydrate (fructose). Abalanced equation is

The cycle of reactions that converts carbdh dioxide to carbohydrates is calledthe Calvin cycle, after M. Calvin, who received the Nobel Prize in chemistryin 1961 for his work on determining the path of carbon in photosynthesis.

Carbon enters the cycle as carbon dioxide. The key reaction by which theCO, is "fixed" involves enzymatic carboxylation of a pentose, D-ribulose1,5-pho~phate.~

8All of the reactions we will be discussing are mediated by enzymes, and we will omithenceforth explicit mention of this fact. But it should not be forgotten that these areall enzyme-induced processes, for which we have few, if any, laboratory reagents toduplicate on the particular compounds involved.


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20 CarbohydratesCH,OPQ,~@ CH,OPO,~@IC = O I@o,c--C-OHI __ I _H-C-OH CO2 > C=O H20,IH-C-OH IH-C-OH

A subsequent hydrolytic cleavage of the C2-C3 bond of the carboxylationproduct (this amounts to a reverse Claisen condensation; Section 18-8B)yields two molecules of D-3-phosphoglycerate."

In subsequent steps, ATP is utilized to phosphorylate the carboxylate groupof 3-phosphoglycerate to create 1,3-diphosphoglycerate (a mixed anhydrideof glyceric and phosphoric acids). This substance then is reduced by NADPHto glyceraldehyde 3-phosphate:

0 ~ 0 , " OCO,@ ATP ADP O\C/I NADPH NADP@H-C-OHI uCHO

H-C-OHICH,OPO,~@D -glyceraldehyde3-p hosphate

Two glyceraldehyde 3-phosphates are utilized to build the six-carbon chainof fructose by an aldol condensation (C, + C,+ ,), but the donor nucleo-phile in this reaction is the phosphate ester of dihydroxypropanone, which isan isomer of glyceraldehyde 3-phosphate. Rearrangement of the C, aldose to

W e will henceforth, in equations, designate the various acids we encounter as thephosphate and the carboxylate anions, although this is hardly reasonable at the pHvalues normal in living cells. Glyceric and phosphoric acids are only partially ionizedat pH 7-8. However, it would be equally unrealistic to represent the acids as beingwholly undissociated.


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20-9 Formation of Carbohydrates by P hotosynthesisthe C3 ketose (of the type described in Section 20-2D) therefore precedes thealdol addition. (For a discussion of the mechanism of the enzymatic aldol reac-tion, see Section 17-3F.) The fructose 1,6-diphosphate formed is then hydro-lyzed to fructose 6-phosphate:

C H , O P O ~ ' ~1-3-d hvd roxv-2-propanone phosphate

From what we have described thus far, only one atom of carbon has beenadded from the atmosphere, and although we have reached fructose, five pre-viously reduced carbons were consumed in the process. Thus the plant has toget back a five-carbon sugar from a six-carbon sugar to perpetuate the cycle.Rather than split off one carbon and use that as a building block to constructother sugars, an amazing series of transformations is carried on that can besummarized by the following equations:

transketolasec, + C3 > c4 + C5aldolaseC, + C3-7trans ketolaseC7 + C3 > C5 + c5

These reactions have several features in common. They all involve phosphateesters of aldoses or ketoses, and they resemble aldol or reverse-aldol con-densations. Their mechanisms will not be considered here, but are discussedmore explicitly in Sections 20-IOA, 20-IOB, and 25-10. Their summation isC, + 3C3---+C5, which means that fructose 6-phosphate as the C, componentreacts with a total of three C, units (two glyceraldehyde 3-phosphates and onedihydroxypropanone phosphate) to give, ultimately, three ribulose 5-phos-phates. Although the sequence may seem complex, it avoids building up pentoseor hexose chains one carbon at a time from one-carbon intermediates.

The Calvin cycle is completed by the phosphorylation of D-ribulose 5-phos-phate with ATP. The resulting D-ribulose 1,5-diphosphate then is used to startthe cycle again by combining with carbon dioxide. There is one sixth morefructose generated per cycle than is used to reform the ribulose 1,5-diphosphate.This fructose is used to build other carbohydrates, notably glucose, starch,and cellulose.


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944 20 Carbohydrates

Exercise 20-17* Write mec hanism s, supported by analogy in so far as p ossible, forthe carboxylation and cleavage reactions of Equation 20-4 as you would expect themto occur in the absence of an enzyme. Both reactions can be reasonably expected tobe induced by OH@, nd it may be helpful to review the prope rt ies of enols d esc ribedin Section 17-1.

20-10 THE GENERATION OF ENERGYFROM CARBOHYDRATE METABOLISMWe will consider here the reverse process of photosynthesis, namely how car-bohydrates, especially glucose, are converted to energy by being broken downinto carbon dioxide and water.

A general summary of the several stages involved is shown in Figure 20-8.Initially, the storage fuels or foodstuffs (fats, carbohydrates, and proteins) arehydrolyzed into smaller components (fatty acids and glycerol, glucose and othersimple sugars, and amino acids). In the next stage, these simple fuels are de-graded further to two-carbon fragments that are delivered as the CH,C=Ogroup (ethanoyl, or acetyl) in the form of the thioester of coenzyme A,CH,COSCoA. The structure of this compound and the manner in which fattyacids are degraded has been considered in Section 18-8F, and amino acidmetabolism is discussed briefly in Section 25-5C. This section is concernedmainly with the pathway by which glucose is metabolized by the process knownas glycolysis.

In the conversion of glucose to CH,COSCoA, two carbons are oxidized tocarbon dioxide with consumption of the equivalent of two oxygen molecules:

For further oxidation to occur, the CH,COSCoA must enter the next stage ofImetabolism, whereby the CH,C=O group is converted to CO, and H2 0 .This

stage is known variously as the citric acid cycle, the tricarboxylic acid cycle, orthe Krebs cycle, in honor of H. A. Krebs (Nobel Prize, 1953), who first rec-ognized its cyclic nature in 1937. We can write an equation for the process asif it involved oxygen:

Notice that combination of the reactions of Equations 20-5 and 20-6, glycolysisplus the citric acid cycle, oxidizes glucose completely to CO, and H20:

But, as you will see, none of the steps uses molecular oxygen directly. Hence


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20-10 The Generation of Energy from Carbohydrate MetabolismCarbohydrates Fats Proteins

Figure 20-8 Perspective of the metabolic scheme whereby carbohy-drates, fats, and proteins in foodstuffs are oxidized to CO,, sh ow ing thelink between glyco lysis, the citr ic aci d cycle, and oxidative p hosphory-lation

there must be a stage in metabolism whereby molecular oxygen is linked toproduction of oxidizing agents that are consumed in glycolysis and in the citricacid cycle.

The coupling of oxygen into the metabolism of carbohydrates is an extremelycomplex process involving transport of the oxygen to the cells by an oxygencarrier such as hemoglobin, myoglobin, or hemocyanin. This is followed by aseries of reactions, among whichNADH is converted to NAD@with associatedformation of three moles of ATP from three moles of ADP and inorganic

STAGE l

STAGE ll


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20 Carbohydratesphosphate. Another electron-carrier is flavin adenine dinucleotide (FAD;Section 15-6C), which is reduced to FADH, with an associated production oftwo moles of ATP from two moles of ADP. These processes are known asoxidative phosphorylations and can be expressed by the equations:

3ADP 3ATPNADH + $0, + H@ NAD@ + H,O

2ADP 2ATPFADH, +7i0, FAD + H,O

Oxidative phosphorylation resembles photophosphorylation, discussed in Sec-tion 20-9, in that electron transport in photosynthesis also is coupled with ATPformation.

By suitably juggling Equations 20-7 through 20-9, we find that the metabolicoxidation of one mole of glucose is achieved by ten moles of NAD@and twomoles of FAD:

6C02+ 10 NA DH + 2FADH2+ 10H@ (20-10)The overall result is production of 36 moles of ATP from ADP and phos-phate per mole of glucose oxidized to CO, and H,O. Of these, 34 ATPs areproduced according to Equation 20-10 and, as we shall see, two more comefrom glycolysis.

20-10A GlycolysisGlycolysis is the sequence of steps that converts glucose into two C, fragmentswith the production of ATP. The C, product of interest here is 2-oxopropanoate(pyruvate):

ADP ATP 0CGH1,OG I~ C H , C C O , ~glucose pyruvateThere are features in this conversion that closely resemble the dark reactionsof photosynthesis, which build a C, chain (fructose) from C:, chains (Section

20-9). For example, the reactants are either phosphate esters or mixed anhy-drides, and the phosphorylating agent is ATP:ROH +ATP- O-P0,20+ ADP + H@


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20-10A GlycolysisFurthermore, rearrangements occur that interconvert an aldose and ketose,

O H 0I I IRCHCHO RCCH20Hand the cleavage of a C, chain into two C, chains is achieved by a reverse aldoicondensation:

Also, oxidation of an aldehyde to an acid is accomplished with [email protected] a related reaction in photosynthesis (Section 20-9) that accomplishes thereduction of an acid to an aldehyde and is specific for NADPH, not NADH:

glycolysisRCHO + NAD@+ H 2 0- C02@ NADH + 2H@RCHO + NADP@+H 2 0 < photosynthesis RC02@+NADPH + 2H@The detailed sequence in glycolysis is summarized in Figure 20-9 and each ofthe steps is identified more specifically in the ensuing discussion.

First, glucose is phosphorylated to glucose 6-phosphate with ATP. Thenan aldose ketose rearrangement converts glucose 6-phosphate intofructose 6-phosphate. A second phosphorylation with ATP gives fructose1,6-diphosphate:

ATP ADPOH HO

glucose g ucose-6-phosphateATP ADP

HHO HO

fructose 6-phosphate fructose 1,6-diphosphate

At this stage the enzyme aldolase catalyzes the aldol cleavage of fructose1,6-diphosphate. One product is glyceraldehyde 3-phosphate and the other is


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20 CarbohydratesGlucose

Glucose 6-phosphate

Fructose 6-phosphate

Fructose 1,6-diphosphateII1,3-Dihydroxypropanone phosphate Figure 20-9 The glycolyt ic sequence. [Thedashed arrows in the reverse direction indi-+ cate the steps in the synthesis of gluc ose f rom

Glyceraldehyde 3-phosphateG AD@ pyruvate (glyconeogenesis) that dif fer fromthose in glyco lysis. ]1,3-Diphosphog ycerate:;;

.L2-Phosphog lycerate

IPhosphoenolpyruvate

I ATP!IsADPPyruvate

1,3-dihydroxypropanone phosphate. Another ketose e ldose equilibriumconverts the propanone into the glyceraldehyde derivative:

2 " 0 3 ~ 0 CHO CH,OPO~"aldolase I IH-C-OH + C=OI

C H , O P O , ~ ~I

CHZOHfructose1,6-diphosphate(ketose form)

glyceraldehyde 1,3-d hydroxypropanone3-phosphate phosphatetriose phosphate I

isomerase


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20-10A Glycolysis 949The next step oxidizes glyceraldehyde 3-phosphate with NAD Bin the presenceof phosphate with the formation of 1,3-diphosphoglycerate:

IH-C-OH + NAD@ + ~ ~ 0 4 ~ 'I

~ ~ ~ 0 ~ 0 3 ~ 'glyceraldehyde3-phosphate

IH-C-OH + NADH + H@ICH,OPO,~@The mixed anhydride of phosphoric acid and glyceric acid then is used toconvert ADP to ATP and form 3-phosphoglycerate. Thereafter the sequencediffers from that in photosynthesis. The next few steps accomplish the forma-tion of pyruvate by transfer of the phosphoryl group from C3 to C2 followedby dehydration to phosphoenolpyruvate. Phosphoenolpyruvate is an effectivephosphorylating agent that converts ADP to ATP and forms pyruvate:

1,3-diphosphog ycerate 3-phosphoglycerate

cop cop co,'I I ADP ATP 1H-C-OPO,~@ -Hzo> C - O P 0 3 2 ~ A,=OICH,OH IICH2 ICH:,2-phosphoglycerate phosphoenol pyruvate pyruvateThe net reaction at this point produces more ATP than is consumed in thephosphorylation of glucose and fructose (see Exercise 20-20).What happens thereafter depends on the organism. With yeast and certainother microorganisms, pyruvate is decarboxylated and reduced to ethanol. Theend result of glycolysis in this instance is fermentation. In higher organisms,pyruvate can be stored temporarily as a reduction product (lactate) or it can beoxidized further to give CH,COSCoA and CO,. The CH,COSCoA then enters


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20 Carbohydratesthe citric acid cycle to be oxidized to CO, and H,O, as discussed in the nextsection:

N A D HCH,CHO T-----+ CH,CH,OH fermentationNAD@0 /

pyruvate

citric acid cycle;1CoASH

Exercise 20-18* From the discussion in Section 15-5F, it should be clear that thereaction of an a lcohol phosphate with ADP to give ATP, RO P 03 20 +ADP- TP +ROH, is not l ikely to have a favorable equi l ibr ium constant. Explain why one mightexpect the fol lowing reaction to be m ore energetically favorable.

c o pI6 - 0 ~ 0 , ~ ~AD P- =o + ATPIICH2 ICH,Exercise 20-19* The heat of co mbustion of glucose (s) to C 02 (g ) and H 20 (/) s 670kcal m ole- ', whereas that of 2-oxopropanoic acid(/) is 280 kcal mole-'. Ne glectin gthe heats of solution of the com poun ds in water, estimate the energy of glucose (aq) +0,+ CH3COC02H(aq)+ 2H20(I) .Exercise 20-20* The following interconversion is catalyz ed b y the enzyme triosephosphate isomerase:C H 2 0 H C H O1 tr iose phosphate IC=O f ' H-C-OHI i some raseCH,OPO,~@ ICH,OPO,~@Explain how you might use bond energies to estimate whether the equi l ibr ium con-stant, K, for this reaction w ould be greater, or less, than unity.Exercise 20-21* Assuming that one molecule of glucose is oxidized to two moleculesof 2-oxopropanoic ac id (pyruv ic ac id) , how ma ny moles of ATP are formed from ADPin the overal l reaction by the sequence of steps given in Figure 20-9?


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20-10B The Citr ic Acid (Krebs) Cycle20-108 The Citr ic A cid (Krebs) Cy cle

Glycolysis to the pyruvate or lactate stage liberates heat, which can help keepthe organism warm and produce ATP from ADP for future conversion intoenergy. However, glycolysis does not directly involve oxygen and does notliberate CO,, as we might expect from the overall process of the metabolicconversion of glucose to carbon dioxide and water (Equation 20- 10). Theliberation of CO, occurs subsequent to pyruvate formation in a process calledvariously, the citric acid cycle, the Krebs cycle, or the tricarboxylic acid(TCA) cycle.

The initial step, which is not really part of the cycle, is conversion ofpyruvate tc -thanoyl CoA (acetyl CoA):

ICH,COCO,@ + HSCoA + NAD@- H,CSCoA + CO, + NADHThis reaction looks simple but actually occurs in four discrete steps that involvea complex of enzymes having a molecular weight of about 4,500,000. We shallpass over this interesting and rather well-studied reaction as we describe thecitric acid cycle. A simplified representation of the citric acid cycle is shownin Figure 20-10, and it will help to refer to this diagram as each of the steps init are discussed in more detail.

To achieve the oxidation of acetyl CoA on a continuing basis, intermediatesconsumed in certain steps must be regenerated in others. Thus we have asituation similar to that in the Calvin cycle (Section 20-9), whereby the firststage of the cycle achieves the desired reaction (CO, formation) and the secondstage is designed to regenerate intermediates necessary to perpetuate the cycle.

The entry point is the reaction between acetyl CoA and a four-carbon unit,2-oxobutanedioic acid. An aldol-type addition of the CH,CO group to this C,keto acid extends the chain to a branched C, acid (as citric acid):

co,@ II CH,o=c II CH,COSCoA H O - C - ~ ~ ~ @+ H,O I + HSCoAy-32 CH,I2-oxobutaned ioate(oxaloacetate) 3-hydroxy-3-carboxypentaned oate(citrate)Dehydration-rehydration of citrate converts it to isocitrate:

CH, CH ,Ico,@ IC O , ~2-hydroxy-3-carboxypentanedioate(isocitrate)


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START HERE

.cC, acetyl CoA

20 Carbohydrates

citrate

N A D H

NAD@hydroxybutaned oate C,"'(malate)

t\H . 0 4

t rans-butened oate C,"(fumarate)FAD

Figure 20-10 The c i t r ic ac id cyc le

From here, oxidation of the hydroxyl function with NADe gives a keto acid,which loses C 0 2 eadily (Section 18-4) and affords 2-oxopentanedioate:

c o p c o pI co,@IHO-C-H O=C IO=C

2-oxopentaned ioate(2-ketog utarate)


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20-10B The Citr ic Acid (Krebs) CycleWe now have a C, keto acid that can be oxidized in the same way as the C,keto acid, pyruvic acid, to give a butanedioyl CoA:

I + C 0 2+ NADHICH,Ico2@CH,

I&o,@butaned ioyl CoA(succinyl C o A )

Two molecules of CO, now have been produced and the remaining part of thecitric acid cycle is concerned with regeneration of the CoA for forming acetylCoA from 2-oxopropanoate, and also with regenerating the 2-oxobutanedioate,which is the precursor of citrate. The steps involved are

c H 2 H 20 ,GDP CH, FAD-HSCoA, GTP ' I -FADH,'CH2 ?"2

CH,I

The hydrolysis of the acyl CoA in the first step is used for energy storage byconversion of guanosine diphosphate (GDP) to guanosine triphosphate (GTP):

guanosine diphosphate(GDP)


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20Carbohydrates

HO OHguanosine tr iphosphate(GTP)

The hydration of the tmizs-butenedioate (Section 10-3G)and the final oxida-tion reaction (Section 15-6C) have been discussed previously.

Exercise 20-22* The react ion ADP + RC-SR' + ATP + R C 0 2 H+ HSR'0IIis subs tantial ly more favorable than the correspo nding react ion with R-C-OR. On

the b asis of the valence-bon d treatment, explain why this should b e so.Exercise 20-23* Citr ic ac id is prochiral. Nonetheless, if one were to introduce acetyl

0IICo A labeled with l4 C (radioact ive carbon) at the ca rboxyl group, CH3-14C-SCoA,into the citr ic ac id cyc le, the 2-oxopentanedioate ac id (2-ketoglutarate) formed in thefourth step of the cycle would have a l l of the 14C in the carb oxy late gro up farthestaway from the ketone carb ony l group . For some years, this result was us ed to arguethat citr ic a cid i tself co uld not be an interm ediate in the formation of 2-oxopentane-dioate. Review Section 19-8 and explain how, in stereospeci f ic enzyme-inducedreact ions, citr ic ac id cou ld be an intermediate in the formation of 2-oxopentanedioateeven if the I 4 C wou ld not appear equally in both carboxyl ic carbons of the product.Exercise 20-24* What analogy can you draw from react ions studied in previouschapters to the cleavage RC OCH2CO SCoA+HSCoA- RCOSCoA +CH3COSCoA?What reagents would you expect to cause this reaction to occur in water solution?Exercise 20-2 5* A first step in unrav elling the mec han ism of the m etab olism of fattyacid s was ma de in 1904 by F. Knoop, who found that dogs m etabolized 4-phenyl-butanoic acid to phenylethanoic ac id an d 3-phenylpropionic ac id to benzoic acid.What does this pattern of results indicate about the mechanism of degradation offatty acid s? G ive your reasoning.Exercise 20-26* A very strong man can l i ft 225 kg (500 Ib) 2 m eters (6.5 f t). Mu scleaction gets its energy from the reaction ATP + H 2 0- DP + H2P04@, processwith a AGO of -7 kcal .


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20-10C Alternative Routes in Carbohydrate Metabolisma. Assum ing 50% efficienc y in the use of the hydroly sis free energy, how many gramsof ATP (M W 507) would have to be hydrolyzed to achieve this lift ing of the weight?(One kg raised one meter requires 2.3 cat of energy.)b. How m any gram s of gluco se wou ld have to be ox idiz ed to CO, and water to re-ple nish the ATP used in Part a on the b asis of a 40% conversion of the energy of com-bustion to ATP? (AGO for com bustion of glucose is -686 kcal.)

20-10C Alternative Routes in Carbohydrate MetabolismThere is an alternative route, called the pen tos e phos pha te pa thway , by whichglucose enters the glycolytic sequence to pyruvate. This route achieves theoxidative decarboxylation of glucose to give ribose, as the 5-phosphate ester.The essential steps are

NADP@ NADPHglucose 6-phosphateA>-phosphogluconateHz0 NADP@NADPH , CO,

ribose 5-phosphate c-- ibulose 5-phosphate

One purpose of this oxidative route is to generate NADPH, which is the re-ducing agent required by the organism for biosynthesis. The other purpose isto produce ribose, which is needed for the biosynthesis of ATP, CoA, NADG,FAD, RNA, DNA, and so on. However, the demand for NADPH is higherthan the demand for ribose, so there must be a way of channelling the excessribose back into the metabolic cycle. This is accomplished by the conversionof ribose into glycolysis intermediates, fructose 1,6-diphosphate and glycer-aldehyde 3-phosphate (see Figure 20-1 1). The reactions that accomplish thisare very similar to those of the Calvin cycle (Section 20-9), only in reverse.They may be summarized as

transketolasec5 + c5 > c,+ c7

transaldolasec7 - C3 > c4 + c6

transketolasec5 + c4 > C3 + C6

The net result is that three pentoses are converted into two molecules offructose and one of glyceraldehyde (3C5- C6+ C,).The relationship of the pentose-phosphate pathway to glycolysis is shown inFigure 20-1 1. The steps involved in the pentose shunt are readily reversible,but there are several steps in glycolysis that are not. These are the phosphory-lation steps (see Figure 20-9). Yet, there has to be a return route from pyruvateto glucose. This route is called gluconeogenesis and, in animals, takes place in


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20 Carbohydrates

co2NADH

Lactate pyr:vate

~ c e t ) ; lCoA

Figure 20-11 Schemst ic representat ion of metabol ism of glucose byway of g lycolys is and c i t r i c ac id cyc le, and the pentose phosphate,lactate, and glucon eogen esis l inks.

the liver. We shall not discuss the steps in gluconeogenesis except to indicateagain that they are not all the reverse of glycolysis. For comparison, the stepsthat differ are indicated in Figure 20-9 by dashed lines.Why is lactate formed from pyruvate in the metabolism of glucose? Pyruvate+ N A D H + H e- actate + NAD@ s a dead-end path, but it does furnishthe NAD@ needed for glycolysis in active muscle. This route for formingN A D e is important, because in circumstances of physical exertion, the rate ofproduction of NAD@ rom oxidative phosphorylation may be slower than thedemand for NAD@ , in which case a temporary supply is available from thepyruvate+ actate reduction. The lactate so formed builds up in muscle tissueunder conditions of physical exertion and is apt to cause muscles to "cramp."Th e excess lactate so formed ultimately is removed by being converted back topyruvate by oxidation with NAD@ .

The beauty of the metabolic cycle through pyruvate, shown in summary inFigure 20- 11 , is the way it can be tapped at various points according to whetherthe organism requires ATP (from glycolysis), NADH (from pentose shunt), orNAD B (from the lactate siding).


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Supplementary ExercisesAdditional ReadingW. W. Pigman and D. Horton (Eds.), The C arbohydrates, Chem istry a n d Biochem -istry, Academic Press, New York, 1972.L. Stryer, Biochemistry, W. H. Freeman and Company, San Francisco, 1975, Chap-ters 11-1 3.

Supplementary Exercises20-27 A naturally occ urring op tica lly ac tive pentose (C5H,,)O5) redu ces Tollen'sreagent and forms a tetraethanoate with ethanoic anhydride. I t gives an optical lyinac t ive phenylosazone. Write al l the po ssib le structures for this pentose that are inac co rd with each of the experimental observat ions.20-28 A hexos e, C6Hl,06, wh ich we s hall call X-ose, on redu ction with so diumamalgam gives pure D-sorbitol, and upon treatment with phenylhydrazine gives anosazone different from that of D -gluc os e. Write a project ion formula for X-ose andequations for its reactions.20-29 Com poun d A, C5Hlo04, is opt ic al ly act ive, forms a diethanoate ester withethanoic anhydride, but does not give a silver mirror with Agm(NH3),. When treatedwith d i lute acid, A y ields m ethanol and B , C4H 8O4. is opt ical ly act ive, reducesAgB(NH3),, and forms a triethanoate ester with ethanoic anhydride. On reduction, Bgive s op tica lly inac tive C, C4H,,0,. M il d oxid ation of B give s D , a carboxyl ic acid,C 4H 80 5. reatment of the am ide of D with di lute sodium hypochlorite solut ion gives(+)-glyceraldehyde (C3H60,). (For a desc ription of this rea ction see Section 23-12E.)Use these facts to derive structures and stereoch em ical configurat ions for A, B , C,and D . Write equations for all the reactions involved.20-30 Draw Haworth- and conformation-type formulas for each of the following:a, methyl 2,3,4,6-0-tetramethyl-cx-n-glucopyranosideb. p-D-arabinofuranosyl a-L-arabinofuranoside6. L-sucrose20-31 Sugars condense with anhydrous 2-propanone in the presence of an acidcatalyst to form cycl ic ketals known as isopropyl dene derivat ives:


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20 CarbohydratesThe react ion of D-g luco se with 2-propano ne and an ac id catalyst prod uces a mono-and a di isoprop yl idene der ivat ive. Ac id hydrolysis of the di isop ropyl iden e der ivat ivegives the monoisopropylidene compound. 0-Methylat ion of the diketal derivat ive(Section 20-4A) fol lowed by hydrolysis of the ketal groups forms 3-O-methyl-D-glucos e. 0-M ethy lat ion of the monoketal derivat ive fol lowed by hyd rolysis of the ketalfunct ion forms a tr i-0-me thyl-D-glucos e. This tr i-0-m ethy l-D-g luco se when O-meth-ylated forms an isomer of penta-0-methyl-D-glucopyranose. This isomer whensubjected to hydrolysis in di lute acid yields an isomer of 2,3,4,6-tetra-0-methyl-D-glucop yranos e (20, Figure 20-4). Write structures for these cy cl ic ketals which agreewith the experimental evidence. Give your reasoning. (Review Sections 20-2C and20-4A.)20-32 Co mp lete the fol lowing seque nce of reactions, writ ing structures for al l theproducts, A-I.

(1 mole)a. a-D -g ucofuranose HCI > A (see Exercise 20-31)

heate . E + F - G + H- 2 0

20-33 Write a mec hanism for the interconversion of an aldohexose an d a ketohexosethat is catalyzed by h ydroxide ion. What produ cts wo uld you expect starting withD-glucose?20-3 4 The gl yc os ide am yg da lin (C2,H,,011N) is hy dro lyz ed with the a id of theenzyme em ulsin (but not with the enzym e m altase) to give D- gluc os e, HCN , andbenzenecarbaldehyde. 0-Methylat ion of amygdal in, fo l lowed by acid hydrolysis,g ives 2,3,4,6-tetra-0-methyl-D-glucose and 2,3,4-t r i -0-methyl-D-glucose. Write astructure for amygdalin that f its with these observations.

bpo c chapter 20

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