Glucose hemiacetal form

Glycolysis is a ten-step process that begins with isomerization of glucose from its cyclic hemiacetal form to its open-chain aldehyde form—a reverse nucleophilic addition reaction. The aldehyde then undergoes tautomerixa-tion to yield an enol, which undergoes yet another tautomerization to give the ketone fructose. 

In sucrose, fructose is present as the P anomer. Now, one of these sugars has acted as an alcohol to make a bond to the other sugar. We can look at this in two ways. Either frnctose acts as an alcohol to react with the hemiacetal glucose to form an acetal, or alternatively, glucose is the alcohol that reacts with the hemiketal fructose to form a ketal. In sucrose, the pyranose ring is an acetal, whilst the fnranose ring is a ketal. This all seems rather... 

In aqueous solution, both glucose (hemiacetal) and frnctose (hemiketal) exist as equilibrium mixtures of cyclic and open-chain carbonyl forms. Sucrose, however, is a single stable substance (acetal and ketal), and conversion back to glucose and frnctose requires more rigorous hydrolytic conditions, such as heating with aqueous acid. 

Both six-membered pyranose and five-membered furanose structures are encountered, a particular ring size usually being characteristic for any one sugar. Thus, although glucose has the potential to form both six-membered and five-membered rings, an aqueous solution consists almost completely of the six-membered hemiacetal form five-membered rings... 

We should first consider the open-chain form of glucose 6-phosphate, rather than its pyranose hemiacetal form (see Section 12.2.1). The open-chain aldose has the requirements for enolization, namely a hydrogen a to the aldehyde carbonyl group. Enolization produces in this case an enediol, which can revert to a keto form in two ways, i.e. reforming... 

To explain the above results, glucose is assumed to exist as a cyclic hemiacetal in equilibrium with a small amount of the open-chain aldehyde. In the hemiacetal, C is chiral and two diastereomers (anomers) are possible they are called a- and jS-glucosides. The C of C=0 is always the anomeric position. Cyclic hemiacetals form when an OH and a CHO group in the same molecule can form a five- or six-membered ring that is more stable than the open chain. 

The configuration around the Cj of glucose (i.e. the anomeric C) is not stable and can readily change (mutarotate) from the a- to the / -form and vice versa when the sugar is in solution as a consequence of the fact that the hemiacetal form is in equilibrium with the open chain aldehyde form which can be converted into either of the two isomeric forms (Figure 2.2). 

The protonated amino group, however, has no nucleophilic activity, and does not form a hemiacetal an amino sugar can, therefore, be obtained, as a salt, in an otherwise unfavorable form. For example, 6-amino-6-deoxy-L-sorbose can be isolated as the hydrochloride of a furanose form (22) (presumably a). In alkaline solution, immediate ring-expansion to the pyranose form (23) occurs127 this reaction can be reversed under strongly acidic conditions. 4-Amino-4,6-dideoxy-D-glucose hydrochloride forms a 35 65 mixture of the a- and / -pyranose forms in solution.128... 

The halogen in the acylglycosyl halide is reactive and may be readily displaced, for example, by an alkoxy group on reaction with an alcohol under anhydrous conditions in the presence of a silver or mercury(n) salt. In this case the products are glycosides which are the mixed cyclic acetals related to the cyclic hemiacetal forms of the monosaccharides. In the case of the D-glucose derivative shown below (and of other 1,2-cis acylglycosyl halides) the replace-... 

Figure 122 Glucose open chain (aldehydic form) and cyclic pyranose (hemiacetal) form.

Figure 1.11 The cyclization of the straight chain form of glucose to form the P-hemiacetal cyclic form of the molecule...

The sugar fructose is an isomer of glucose. Like glucose, fructose forms a cyclic hemiacetal, but in this case the ring is five membered rather than six membered. Show the structure for the hemiacetal formed from fructose and show a mechanism for its formation in acidic solution. 

Glycosides are naturally occurring acetals formed from sugars and alcohols. The glycoside salicin, found in willow bark, is formed from glucose and a phenol. Show the structure of the phenol and the steps in the mechanism for the formation of salicin from glucose hemiacetal. 

Two structures for the sugar glucose are shown on page 858. Interconversion of the open-chain and cyclic hemiacetal forms is catalyzed by either acid or base. 

The Cyclic Hemiacetal Form of Glucose Aldoses contain an aldehyde group and several hydroxyl groups. The solid, crystalline form of an aldose is normally a cyclic hemiacetal. In solution, the aldose exists as an equilibrium mixture of the cyclic hemiacetal and the open-chain form. For most sugars, the equilibrium favors the cyclic hemiacetal. 

Glucose exists almost entirely as its cyclic hemiacetal form. 

Draw the cyclic hemiacetal forms of D-mannose and D-galactose both as chair conformations and as Haworth projections. Mannose is the C2 epimer of glucose, and galactose is the C4 epimer of glucose. 

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

In the cyclization of the open-chain form of glucose to form the stable hemiacetal, it may be difficult to work out what has happened. Number the carbon atoms in the open-chain form and put the same numbers on the hemiacetal so that you can see where each carbon atom has gone. Then draw a mechanism for the reaction, glucose open-chain form cyclic form... 

Hemiacetals in sugars are formed in the same way that other hemiacetals are formed—that is. by cyclization of hydroxy aldehydes. Thus, the hemiacetal of glucose is formed by cyclization of an acyclic po/yhydroxy aldehyde (A), as shown in the accompanying equation. This process illustrates two important features. 

Cyclization forms the more stable ring size in a given molecule. The most common monosaccharides, the aldohexoses like glucose, typically form a pyranose ring, so our discussion begins with forming a cyclic hemiacetal from D-glucose. 

We are now set to draw the cyclic hemiacetal formed by nucleophilic attack of the OH group on C5 on the aldehyde carbonyl. Because cyclization creates a new stereogenic center, there are two cyclic forms of D-glucose, an a anomer and a P anomer. All the original stereogenic centers maintain their configuration in both of the products formed. 

In the structures, introduced so far, an O4 rhomb or O5 patterns turned out to be ideal building blocks for carbohydrate-metal complexes. The most important monosaccharide, D-glucose, in any of its hemiacetal forms does not exceed the simple 02 diol pattern ( a consequence is the epimerization reaction of glucose to 8-mannofuranose in the presence of trivalent metal ions). 

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