Glucose cyclic hemiacetal formation

There are two different cyclic hemiacetals because the carbonyl carbon of the open-chain aldehyde becomes anew asymmetric center in the cyclic hemiacetal. If the OH group bonded to the new asymmetric center is down (trans to the primary alcohol group at C-5), then the hemiacetal is a-o-glucose if the OH group is up (cis to the primary alcohol group at C-5), then the hemiacetal is jS-o-glucose. The mechanism for cyclic hemiacetal formation is the same as the mechanism for hemiacetal formation between individual aldehyde and alcohol molecules (Section 17.12). 

If the carbonyl and the hydroxyl group are in the same molecule, an intramolecular nucleophilic addition can take place, leading to the formation of a cyclic hemiacetal. Five- and six-membered cyclic hemiacetals are relatively strain-free and particularly stable, and many carbohydrates therefore exist in an equilibrium between open-chain and cyclic forms. Glucose, for instance, exists in aqueous solution primarily in the six-membered, pyranose form resulting from intramolecular nucleophilic addition of the -OH group at C5 to the Cl carbonyl group (Figure 25.4). The name pyranose is derived from pyran, the name of the unsaturated six-membered cyclic ether. 

Figure 9.5 Cyclic, hemiacetal structures of D-glucose. The reaction between an alcohol and aldehyde group within an aldohexose results in the formation of a hemiacetal. The only stable ring structures are five- or six-membered. Ketohexoses and pentoses also exist as ring structures due to similar internal reactions.

A vast range of natural sugars exempMly these cyclic addition products. A typical sugar exists predominantly in the form of a hemiacetal or hemiketal in solution, although this is an equilibrium reaction, and the open chain carbonyl form is always present to a small extent (<1%). The formation of a six-membered cyclic hemiacetal from glucose is achieved by attack of the C-5 hydroxyl onto the protonated carbonyl (conjugate acid). 

On the other hand, if ammonolysis first occurred at carbon atoms 4 or 5, the formation of a cyclic hemiacetal would be favored, and the possibility of obtaining a diacetamide would be correspondingly decreased. Such an explanation accounts for the formation of iV-acetyl-D-gluco-furanosylamine reported by Hockett and Chandler. Here, the first product of the reaction is a D-glucofuranose, which then condenses with acetamide. A similar explanation accounts for the results obtained by Brigl, Miihlschleger and Schinle with 2-thioethyl-3,4,5,6-tetrabenzoyl-oZde%do-D-glucose. 

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. 

In this series of equilibrium steps, the hemiacetal ring of a-glucose opens to yield the free aldehyde. Bond rotation is followed by formation of the cyclic hemiacetal of 6-glucose. The reaction is catalyzed by both acid and base. 

The structural lability of the cyclic hemiacetals is a consequence of their oxo-cyclo tautomerism, which allows a rapid interconversion of the (2R)-configurated enantiomer into the (2S)-enantiomer and vice versa via the ring-opened oxo form (Fig. (4)). An analogous and well investigated process is the formation of an equilibrium mixture of a-D-glucose and p-D-glucose on dissolution of the pure a- or P-isomer in water by mutarotation. 

In the presence of oxygen, intermediate 10 can undergo oxidation and give rise to the formation of the aminoketo compound 11 and subsequently an acid, the so-called Strecker acid [36], The formation of Strecker acid, however, is not possible if dicarbonyls such as 4 are present in form of their stable cyclic hemiacetals, as shown by Hofmann and co-workers [36]. The same group revealed also that Strecker aldehydes can be formed via an oxidative degradation of Amadori compounds, as shown for the Amadori compound of phenylalanine and glucose [37],... 

Monosaccharides can have both cyclic and straight chain structures. Cyclomonosaccharides are formed when straight-chain monosaccharides are dissolved in water. In this case, the number of asymmetrical carbon atoms increases to five. This process is known as hemiacetal formation. Let us examine the formation of the cyclic structure in the glucose molecule. 

Hemiacetal formation is fundamental to the chemistry of carbohydrates (see Section 11.1). Glucose, for example, contains an aldehyde and several alcohol groups. The reaction of the aldehyde with one of the alcohols leads to the formation of a cyclic hemiacetal (even without acid catalysis) in an intramolecular reaction. 

There are four chiral centers in the openchiral center at the carbon involved in hemiacetal formation, giving a total of five chiral centers in the cyclic form. 

Many of the stereoelectronic effects in the list above govern reactivity, but the next section will deal with how stereoelectronic effects affect structure—and in particular conformation. Some of the most important saturated oxygen heterocycles are the sugars. Glucose is a cyclic hemiacetal—a pentasubstituted tetrahydropyran if you like—whose major conformation in solution is shown below. About two-thirds of glucose in solution exists as this stereoisomer, but hemiacetal formation and cleavage is rapid, and this is in equilibrium with a further one-third that carries the hemiacetal hydroxyl group axial (<1% is in the open-chain form). 

Finally, in this vein, it is important to note that the carbohydrates (Chapter 11) composing significant amounts of our biosphere commonly exist as hemiacetals and hemiketals, and thus the cyclic forms of glucose (a pair of diasteromeric pyrans) (Table 8.6, item 13) predominate over the open form, which lies in equilibrium between them (Scheme 8.53). Indeed, shorn of its elaborate extra functionality, the cyclization simply represents the same sort of hemiacetal formation seen with methanol (CH3OH) and benzenecarbaldehyde (benzaldehyde, CeHsCHO) (Table 8.6, item 11) shown above (Scheme 8.51). 

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