Hemiacetal carbon

Fluoride ion catalyzed reaction of the cyclic nitronate 1 with benzaldehyde provides a cyclic hcmiaeetal 2 in 95% diastereoselectivity18 with tra ,v,tr u.v-relationship of the protons in positions 2, 3 and 4. The relative configuration of the hemiacetal carbon position was not assigned. Upon reaction of the diastereomeric nitronate 3 at 25 °C the 4a-stereocenter is epimerized by fluoride ion such that hemiacetal 2 is once again obtained. In contrast, reaction of 3 at 0°C furnishes the diastereomeric product 4 in 23% yield 8. 

This enzyme [EC 5.1.3.3], also known as mutarotase, catalyzes the epimerization of the hemiacetal carbon atom of aldoses (thus, anomerization). Hence, a-D-glu-cose is reversibly converted to /3-D-glucose. Other sugars can act as substrates (e.g., L-arabinose, D-xylose, D-galac-tose, maltose, and lactose). 

FIGURE 7-6 Formation of the two cyclic forms of D-glucose. Reaction between the aldehyde group at C-l and the hydroxyl group at C-5 forms a hemiacetal linkage, producing either of two stereoisomers, the a and fi anomers, which differ only in the stereochemistry around the hemiacetal carbon. The interconversion of a and fi anomers is called mutarotation. 

The 13C NMR spectrum of phthalic dichloride shows additional signals for a five-membered cyclic pseudo halide, characterized by an sp3 carbon shift of 104 ppm. This shift value is also observed for laevulic acid chloride, which exists in the cyclic form only. Similarly, phthalaldehyde acid (o-formylbenzoic acid) turns out to be hydroxyphthalide, as indicated by a hemiacetal carbon shift of 98.4 ppm. 

Carbon-1 of the second (right hand) glucose unit in maltose is a hemiacetal carbon. Thus, the a and p forms at this carbon atom can equilibrate via the open-chain aldehyde form. Mutarotation is, therefore, possible. Note that carbon-1 of the first (left-hand) glucose unit is an acetal carbon. Its configuration is, therefore, fixed (as... 

Carbon-1 of the glucose unit in lactose is a hemiacetal carbon and will be in equilibrium with the open-chain aldehyde form. Therefore, lactose will be oxidized by Fehling s solution and will mutarotate. 

Each base has a nitrogen atom capable of forming an acetal-like bond to the anomeric (or hemiacetal) carbon of deoxyribose. Recall that formation of an acetal from an alcohol and a hemiacetal involves elimination of water. We can envision that the water molecule is formed from the hydroxyl group of the hemiacetal and a hydrogen atom from the alcohol (or, in this case, the amine). Figure 12.57 illustrates this reaction for a purine base and for a pyrimidine base. 

The a and ft forms of D-glucose differ only at the hemiacetal carbon (C-l the anomeric carbon). 

Let s now address the stereochemistry of the cyclization of D-glucose to a pyranose. Note that carbon 1, the hemiacetal carbon, becomes a new stereocenter when the cyclization occurs. Therefore, two diastereomers of the pyranose, with different configurations at the new stereocenter, are formed when D-glucose cyclizes. Such diastereomers are called anomers. The two anomers for the pyranose form of D-glucose are shown in the following equation ... 

As was the case for cyclohexane derivatives, the chair conformer that has the larger groups equatorial is usually more stable. Therefore, the chair conformers shown in Figure 25.2, which have most or all of the larger substituents equatorial, are more stable than the conformers obtained by ring-flips. The a- and /3-anomers differ only in the stereochemistry of the groups at the hemiacetal carbon. In the a-anomer the hydroxy group on this carbon is axial, and in the /3-anomer it is equatorial. 

Close the ring, and draw the result. Always draw the Haworth projection or chair conformation with the oxygen at the back, right-hand comer, with Cl at the far right. Cl is easily identified because it is the hemiacetal carbon—the only carbon bonded to two oxygens. The hydroxyl group on Cl can be either up or down, as discussed in Section 23-7. 

Draw the chair conformation puckered, as shown in Figure 23-6. The hemiacetal carbon (Cl) is drawn at far right (as the footrest), and the ring oxygen is at the back, right comer. 

The anomers of glucose. The hydroxyl group on the anomeric (hemiacetal) carbon is down (axial) in the a anomer and up (equatorial) in the /3 anomer. The /3 anomer of glucose has all its substituents in equatorial positions. 

A sugar (such as glucosamine) in which a hydroxyl group is replaced by an amino group, (p. 1139) The hemiacetal carbon in the cyclic form of a sugar (the carbonyl carbon in the open-chain form). The anomeric carbon is easily identified because it is the only carbon with two bonds to oxygen atoms, (p. 1112)... 

Cyclization of a hydroxy carbonyl compound always forms a stereogenic center at the hemiacetal carbon, called the anomeric carbon. The two hemiacetals are called anomers. 

Anomers are stereoisomers of a cyclic monosaccharide that differ in the position of the OH group at the hemiacetal carbon. 

Just do what the question says The starting material is numbered from the aldehyde group. We started numbering the product with C6 as that is the only CH2 group and then followed the carbon chain back to Cl which is now the hemiacetal carbon. 

Since a hemiacetal is formed so easily from a carbonyl compound and alcohol, it is not surprising to find that carbohydrates (polyhydroxy derivatives of aldehydes and ketones) frequently exist as cyclic structures in which a hemiacetal linkage is formed intramolccularly. Furthermore, since hemiacetal formation is a reversible process, many carbohydrates exhibit the phenomenon of mutarotation. The liberation of the free aldehyde (V) from the internal hemiacetal of the sugar (IV) destroys the optical activity of the hemiacetal carbon atom (in this case carbon 1), and reformation results in the formation of an equilibrium mixture of two diastereoisomers. 

The NMR spectrum of 26 showed well-dispersed signals of 76 protons, which included seven methyl singlets and two methyl doublets. The 13C NMR spectrum revealed a ketonic carbonyl carbon (5 221.2), two acetal/ hemiacetal carbons (8 108.9,119.8), four oxygenated quaternary carbons, four oxygenated methines, and an oxygenated methylene carbon. In addition to the pyrazine carbons, two sp2 carbon signals resonating at 8 121.2... 

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