n § 24.1 nomenclatuur and structure (page 1235) polyfunctional natural products: carbohydrates...

§ 24.1 Nomenclatuur and Structure (page 1235)

Polyfunctional Natural Products:Carbohydrates

Chapter 24

The two enantiomers of the simplest sugar, glyceraldehyde.

Fig. 24.1

Cx(H2O)y

Hydrated carbon!



Glyceraldehyde is an aldotriose because it has a three-carbon backbone and an aldehyde group.

Fig. 24.2

Problem 24.1* Generalize from the definition of a triose to generate the structures of an aldotetrose and an aldopentose.



Aldo sugars are written with the aldehyde group at the top and the primary alcohol at the bottom. In this scheme, called a Fischer projection, horizontal bonds are taken as coming toward the viewer and vertical bonds as retreating. If the OH adjacent to the primary alcohol at the bottom is on the right, the sugar is a member of the D series. If it is on the left, it is in the L series.

Fig. 24.3



There is no relation between D and L and the sign of optical rotation, (+) and (-).Fig. 24.4



Problem 24.2 Write three-dimensional representations for the following molecules.



Problem 24.3 Write Fischer projections for the molecules in Figure 24.6



The four aldotetroses Fig. 24.8

The four D-aldopentoses Fig. 24.9

22

stereoisomers

both D and L isomers shown

23 stereoisomers

only D-diastereoisomers

shown!



The eight D-aldohexoses Fig. 24.10

24 stereoisomers

only D-diastereoisomers

shown!



D-Fructose, a common ketohexose. Fig. 24.10



Problems 24.4 + 24.5

Problem 24.6Treatment of D-glucose with sodium borohydride (NaBH4) gives D-

glucitol (sorbitol), C6H14O6. Show the structure of D-glucitol and write a brief mechanism for this simple reaction



Although reduction with sodium borohydride, followed by hydrolysis, proceeds normally to give an alcohol, neither NMR or IR reveals large amounts of an aldehyde in the starting material. Fig. 24.12

Why?



Intramolecular hemiacetal formation is analogous to hydration and intermolecular hemiacetal formation. Five- and six-membered ring hemiacetals are easily made, and are often more stable than their open forms. Fig. 24.13



Intramolecular hemiacetal formation is analogous to hydration and intermolecular hemiacetal formation. Five- and six-membered ring hemiacetals are easily made, and are often more stable than their open forms. Fig. 24.13

Example



In an aldohexose, intramolecular hemiacetal formation results in a furanose (five-membered ring) or pyranose (six-membered ring) Fig. 24.14



Fischer projections for D-glucofuranose and D-glucopyranose Fig. 24.5



If two compounds are in equilibrium, irreversible reaction of the minor partner can result in complete conversion into a product. As long as the equilibrium exists, the small amount of the reactive molecule will be replenished as it is used up. Fig. 24.16



Intramolecular hemiacetal formation results in two C(1) stereoisomers called anomers. Fig. 24.17

On the same side as

attacking OH, is -anomer!

On the opposite side as

attacking OH, is -anomer!



The first step in creating a three-dimensional drawing is rotation around the indicated carbon-carbon bond. This motions generates a new Fischer projection. Fig. 24.19



Next, tip the molecule over in clockwise fashion to produce a flat Haworth form Fig. 24.20

Problem 24.9 Draw the flat, Haworth form of the -anomer.



Now let the flat, Haworth form relax to a chair. Don’t forget that there are always two possible chair forms. Fig. 24.21

Problem 24.10* Follow this same procedure for the -anomer.



Problem 24.11 Transform the Fischer projection into a three-dimensional picture of D-mannopyranose

§ 24.2 Reactions of Sugars (page 1246) § 24.2a Mutarotation of Sugars


In water the -anomer is converted into the -anomer and visa versa. Explain!



The - and -anomers can equilibrate through the small amount of the open form present at equilibriumFig. 24.22



Problem 24.12* Write a mechanism for the acid-catalyzed mutarotation of D-glucopyranose (in 3-dimensional structures!)

§ 24.2 Reactions of Sugars (page 1246) § 24.2b Isomerization of Sugars in Base


In base, D-glucose equilibrates with D-mannose and D-fructose, a keto sugar. Fig. 24.23

Explain!



A mechanism of the equilibration of D-Glucose and D-mannose involves formation of an enolate followed by reprotonation Fig. 24.24

Lobry de Bruijn-Alberda van Ekenstein reaction!



Protonation on oxygen generates a double enol, which can lead to D-fructose (or the D-aldohexoses, D-glucose, and D-mannose). Fig. 24.25

§ 24.2 Reactions of Sugars (page 1249) § 24.2c Reduction


Reduction of D-glucose proceeds through the small amount of the open, aldo form present at equilibrium. As the open form is used up, it is regenerated through equilibration with the pyranose form. Fig. 24.26

§ 24.2 Reactions of Sugars (page 1249) § 24.2c Reduction


Problem 24.15* Reduction of D-altrose with sodium borohydride in water gives an optically active molecule, D-altritol. However, the same procedure aplied to D-allose gives an optically inactive, meso hexa-alcohol. Explain.

§ 24.2 Reactions of Sugars (page 1250) § 24.2d Oxidation


Oxidation of an aldohexose with bromine in water gives an aldonic acid in which the end groups are still different. Fig. 24.27

Aldonic acid



Problem 24.16 Write a mechanism for this oxidation. Hint for the first step: What reaction is likely between an aldehyde and water?



Problem 24.17 Examination of the NMR and IR spectra of typical aldonic acids often shows little evidence for the carboxylic acid group. Explain this odd behavior.



Oxidation with nitric acid generates an aldaric acid in which the end groups are both carboxylic acids. Figure 24.28

Aldaric acid

§ 24.2 Reactions of Sugars (page 1251) § 24.2e Ether and Ester formation


Treatment of a sugar with methyl iodide and silver oxide leads to methylation at every free hydroxyl group in the molecule. Figure 24.29

Formation of ethers via

Sn2 process!

Williamson ether

synthesis!!

§ 17.10 Synthesis of Ethers from Alkoxides (pag 845)


Alkoxides can displace halides in an SN2 reaction to make ethers. This reaction is the Williamson ether synthesis. Figure 17.56


Sn2 process!

Williamson ether

synthesis!!

§ 17.10 Synthesis of Ethers from Alkoxides


Alkoxides can displace halides in an SN2 reaction to make ethers. This reaction is the Williamson ether synthesis. Figure 17.56


Sn2 process!

Williamson ether

synthesis!!



A similar process can be carried out in base with a Williamson ether synthesis. Notice in this example that neither the existing ether at C(1) nor the pyranose ring connection is disturbed in the benzylation. Figure 24.30


Sn2 process!

Williamson ether

synthesis!!



?



All the free hydroxyl groups can be esterified with acetic anhydride. Figure 24.30

Formation of

poly esters!



?



Treatment with dilute HCl and alcohol converts only the OH at the anomeric position [C(1)] into an acetal called a glycoside.

Figure 24.32

Acetal function!



Treatment with dilute HCl and alcohol converts only the OH at the anomeric position [C(1)] into an acetal called a glycoside.

Figure 24.32

Glycoside!

Why only methoxylation at the C-1 position?



Although all OH groups can be reversibly protonated, loss of only the anomeric OH leads to a resonance-stabilized cation. Addition of alcohol at this position gives the glycoside. Figure 24.33



Methyl - and methyl -D-glucopyranosides. Figure 24.34



?



Hydrolysis of the fully methylated compounds leads to a hemiacetal in which only the methoxyl group at the anomeric position [C(1)] has been converted into an OH. Figure 24.35



Problem 24.18* Explain carefully why it is only the acetal methoxyl group that is converted into a hydroxyl group.

§ 24.2 Reactions of Sugars (page 1255) § 24.2f Osazone Formation


What product would you expect from the above reaction?



The small amount of free aldehyde present at equilibrium accounts for phenylhydrazone formation at C(1). Figure 24.36



Problem 24.14 Write a mechanism for the above reaction.



The actual reaction is far more complicated!



Osazone formation involves conversion of C(2) as well as C(1) into phenylhydrazones.Figure 24.37



How come?



The C(1) phenylhydrazone is an imine and therefore in equilibrium with an enamine. This enamine is also an enol. Figure 24.38


Reaction of the ketone with phenylhydrazine leads to a new phenylhydrazone that can eliminate aniline to give a new imine. Reaction with a third equivalent of phenylhydrazine leads to the osazone. Figure 24.39


An osazone can be formed from two different aldo sugars that are stereoisomeric at C(2). The stereochemistry at C(2) is destroyed in the reaction.

Figure 24.40


Attention!!!!


§ 24.2 Reactions of Sugars (page 1257) § 24.2g Methods of Lengthening and

Shortening Chains in Carbohydrates


A modern version of the Kiliani-Fischer synthesis generates two new sugars, each one carbon longer than the starting sugar. Figure 24.41

Kiliani-Fischer Synthesis


§ 24.2 Reactions of Sugars (page 1259) § 24.2f Methods of Lengthening and


Problem 24.20 Apply the Kiliani-Fischer synthesis to D-glyceraldehyde. What new sugars are formed? It is not necessary to write mechanisms for the reactions.




Problem 24.21 The following two sugars are produced by Kiliani-Fischer synthesis from an unknown sugar (Fig. 24.42). What is the structure of that unknown sugar?


The Ruff degradation

The Ruff degradation shortens the starting sugar by one carbon. It is the original aldehyde carbon that is lost! Figure 24.43




The Ruff degradation shortens the starting sugar by one carbon. It is the original aldehyde carbon that is lost! Figure 24.43

Mechanism complicated!






The Wohl degradation also shortens sugars by loss of the aldehyde carbon.

Figure 24.43

The Wohl degradation


§ 24.4 The Fischer Determination of the Structures of D-Glucose (and the 15 other Aldohexoses (page 1262)

His starting point:

Arabinose

The natural sugar

arbitrarily assumed to

be the D-enantiomer

Accomplished by Emil Fischer in 1891

The first step in Emil Fischer’s determination of the structure of glucose. The Kiliani-Fischer synthesis applied to D-arabinose leads to D-glucose and D-mannose, which must share the partial structures shown. Figure 24.48



Second step:

Determining the

configuration of C(2) in

arabinose by oxidation

Oxidation of D-arabinose leads to an optically active diacid, which shows that the OH at C(2) in D-arabinose is on the left. Figure 24.49

Only an optically active product formed



Current knowledge about the

structures of D-arabinose,

D-glucose and

D-mannose

What we now know about the structures of D-arabinose, D-glucose, and D-mannose. Only the configuration at C(3) of D-arabinose, which becomes C(4) in D-glucose and D-mannose, is left to be determined. Figure 24.50


Third step:

Determining the


glucose en mannose by

oxidation

Figure 24.51

If true than both

diacids should be optically

active upon

oxidation with nitric

acid!

Assumption: OH on the right!


Third step:

Determining the



oxidation

Figure 24.51

If true than both

diacids should be optically

active upon

oxidation with nitric

acid!


Third step:

Determining the



oxidation

However, if the unknown OH is on the left, one possible diacid is meso, not optically active. Figure 24.52

If true than only one acid

should be optically active!

Assumption: OH on the left!


Third step:

Determining the



oxidation

However, if the unknown OH is on the left, one possible diacid is meso, not optically active. Figure 24.52

If true than only one acid

should be optically active!


Conclusion:

OH on the right!

If the last unknown OH is on the rigtht, two optically active diacdis are produced on oxidation with nitric acid. This result matches the experimental results. Figure 24.51

Experimental Result:

Both sugars gave an optically

active diacid!


Now we know the structure of D-arabinose, L-arabinose, and the two structures (A and B) shared by D-glucose and D-mannose. We do not know which structure belongs to D-glucose and which to D-mannose Figure 24.53



Oxidation with nitric acid renders the ends of a sugar equivalent. Both the aldehyde end and the primary alchohol end are converted into the same group, a carboxylic acid.

Figure 24.54


Two different sugars may

give the same aldaric

acid when oxidized!



Two different sugars may

give the same aldaric

acid when oxidized!

Experimentally: Glucose gives an aldaric acid that is also obtained by oxidation of another sugar named L-Gulose


If D-glucose has the structure A, oxidation of another sugar, L-gulose, can give the same aldaric acid. Note in this case that the L-gulose is drawn with the CH2OH at the top and the CHO at the bottom. Figure 24.55



However, there is no sugar other than B that can give this aldaric acid. As this fact does not match the experimental results, D-glucose must have structure A. Figure 24.56


If glucose would have structure B only glucose would yield this aldaric acid!




However, this is not true!Conclusion: Glucose has structure A!




This outcome also reveals the structure of L-Gulose!


Now we know the structures of these aldohexoses and aldopentoses. Figure 24.57


Structures unraveled sofar!


D-Arabinose and D-ribose give the same osazone, and therefore can differ only at C(2). The structure of D-ribose is therefore known. L-Ribose is simply the mirror image of the D-isomer. Figure 24.58




In a similar way the structures of all the other aldopentoses and aldohexoses were established.

See Figures 24.59, 24.60, 24.61 and 24.62


§ 24.5 Something more: Di- and Polysaccharides (page 1270)

A disaccharide

How can this structure be unraveled?

Examples:

Disaccharides: sucrose, lactose, maltose, Polysaccharide: cellulose



In acid, (+)-lactose is hydrolyzed to D-glucose and D-galactose. Figure 24.63

Hydrolysis shows the individual

monosaccharides!


Hydrolysis in dilute acid means that there must be a glycosidic linkage in (+)-lactose

Figure 24.64


The remaining questions about the structure of (+)-lactose.Figure 24.65



The acid group in lactobionic acid marks the position of the aldehyde in (+)-lactose. As hydrolysis of lactobionic acid gives a gluconic acid (not a galactonic acid), it is glucose that has the free aldehyde in (+)-lactose. Figure 24.66


The acid group in lactobionic acid marks the position of the aldehyde in (+)-lactose. As hydrolysis of lactobionic acid gives a gluconic acid (not a galactonic acid), it is glucose that has the free aldehyde in (+)-lactose. Figure 24.66

We still don’t know which OH is making the connection!!!


The position of the OH used to attach glucose to C(1) of galactose can be determined through a series of methylation and hydrolysis experiments.

Figure 24.67

Methylation!


The position of the OH used to attach glucose to C(1) of galactose can be determined through a series of methylation and hydrolysis experiments.

Figure 24.67

Methylation!

Not methylated


A Fischer projection for (+)-lactose. Figure 24.67


Problem 24.23 Make a good three-dimensional drawing of (+)-lactose



No aldehyde oxidation with Br2/H2O

possible!!!

-> A nonreducing sugar

Sucrose

Suggest a basic structural feature for such a sugar


In nonreducing sugars, there is no free aldehyde group. Attachment must be between both C(1) atoms. Sucrose is an example.

Figure 24.69


Sucrose

No aldehyde oxidation with Br2/H2O

possible!!!

-> A nonreducing sugar


Cellulose and amylose

Figure 24.70


Polysaccharides


§ 24.5 Something more: Di- and Polysaccharides

Problems 24.24; 24.25

§ 24.8 Additional Problems (page 1279)

Problems 24.27; 24.28; 24.29; 24.30; 24.31; 24.32; 24.33; 24.34; 24.37; 24.38; 24.39; 24.40; 24.42; 24.43; 24.44; 24.46; 24.48

n § 24.1 nomenclatuur and structure (page 1235) polyfunctional natural products: carbohydrates...

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