Glucose acyclic aldehyde

In general, a six-membered pyranose form is preferred over a five-membered furanose form because of the lower ring strain, and these cyclic forms are very much favoured over the acyclic aldehyde or ketofte forms. As can be seen in Table 1.3, at equilibrium, the anomeric ratios of pyranoses differ considerably between aldoses. These observations are a direct consequence of differences in anomeric and steric effects between monosaccharides. The amount of the pyranose and furanose present in aqueous solution varies considerably for the different monosaccharides. Some sugars, such as D-glucose, have undetectable amounts of furanose according H-NMR spectroscopic measurements whereas others, such as D-altrose, have 30% furanose content under identical conditions. 

The substrate specificity of glucose-phosphate isomerase illuminates additional stereochemical subtleties of the isomerase reaction (JOO). In the aldose to ketose direction, the enzyme potentially operates on an equilibrium mixture of substrate forms composed of two cyclic hemiacetals (the a- and /3-anomers, of glucose 6-phosphate) and trace quantities of the acyclic aldehyde form lEq. (17)] ... 

The equilibrium existing between products of addition of H-X nucleophiles to the carbonyl group (C=0) and the reactants from which the products form has been utilized widely in carbohydrate chemistry (Chapter 11). As already pointed out (Scheme 8.53), the cyclic hemiacetal (pyranose) form of glucose is in equilibrium with the acyclic aldehydic form. However, reasonably, the hemiacetal as either the a- or 3-glucopyranose can, in an alcohol medium, be converted to an acetal. 

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. 

To translate the acyclic form of glucose into a cyclic hemiacetal, we must draw the hydroxy aldehyde in a way that suggests the position of the atoms in the new ring, and then draw the ring. By convention the O atom in the new pyranose ring is drawn in the upper right-hand comer of the six-membered ring. 

Glucose is an equilibrium mixture of cyclic forms (hemiacetals containing a tetrahydropyran), and a small concentration of acyclic polyhydroxyaldehyde, which is responsible for many of the observed chemical reactions. This illustrates the inherent stability of chair conformers of saturated six-membered systems. The propensity for cyclisation is a general one 5-hydroxy-aldehydes, -ketones and -acids all easily form six-membered oxygen-containing rings - lactols and lactones respectively. 

Infrared spectroscopy has been used to show that the acyclic form of this sugar is not present to any appreciable extent because it equilibrates to give different cyclic forms. The acyclic form of the glucose has an aldehyde group present and thus should obviously produce a significant carbonyl absorption in its IR spectrum. As the spectrum does not show this absorption, it is clear that the acyclic form cannot be present... 

The monosaccharide D-glucose, whose chemistry is representative of all aldoses containing four or more carbon atoms, exists predominantly in the two pyranosc forms 4 and 5. These are six-membered hemiacetals formed by the reversible cyclization of the acyclic polyhydroxy aldehyde 3 (Eq. 23.1). In the cyclic forms 4 and 5, the ring carbon that is derived from the carbonyl group is referred to as the anomeric carbon atom. The specific rotation, [a] (Sec. 7.5), of a-D-(+)-glucose (4) is +112 whereas that of the -anomer 5 is +19°. When crystals of either pure 4 or pure 5 are dissolved in water, the [a]p changes to an equilibrium value of +52.7°. This process is termed mutarotation. At equilibrium in water, the a- and p-forms are present in the ratio of 36 64 only about 0.03% of D-glucose is in the acyclic form 3. 

In an aqueous solution, the open-chain form of o-glucose is in equilibrium with the two cyclic hemiacetals. Because formation of the cyclic hemiacetals proceeds nearly to completion (unlike formation of acyclic hemiacetals), very little glucose is in the open-chain form (about 0.02%). Even so, the sugar still undergoes the reactions discussed in previous sections (oxidation, reduction, imine formation, etc.) because the reagents react with the small amount of open-chain aldehyde that is present. As the open-chain compound reacts, the equilibrium shifts to produce more open-chain aldehyde, which can then undergo reaction. Eventually, all the glucose molecules react by way of the open-chain form. 

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