The anomeric carbon it is a stereocenter present in the cyclic structures of carbohydrates (mono or polysaccharides). Being a stereocenter, more precisely an epimer, two diastereoisomers are derived from it, designated by the letters α and β; these are the anomers, and are part of the extensive nomenclature in the world of sugars.
Each anomer, α or β, differs in the position of the OH group of the anomeric carbon relative to the ring; but in both, the anomeric carbon is the same, and it is located in the same place in the molecule. Anomers are cyclic hemiacetals, the product of an intramolecular reaction in the open chain of sugars; whether aldoses (aldehydes) or ketoses (ketones).
The upper image shows the chair conformation for β-D-glucopyranose. As can be seen, it consists of a six-membered ring, including an oxygen atom between carbons 5 and 1; The latter, or rather the former, is the anomeric carbon, which forms two simple bonds with two oxygen atoms.
If you look closely, the OH group attached to carbon 1 is oriented above the hexagonal ring, as does the CH grouptwoOH (carbon 6). This is the β anomer. The α anomer, on the other hand, would differ only in this OH group, which would be located down the ring, just as if it were a trans diastereoisomer..
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It is necessary to go a little deeper into the concept of hemiacetals to better understand and distinguish anomeric carbon. Hemiacetals are the product of a chemical reaction between an alcohol and an aldehyde (aldoses) or a ketone (ketoses).
This reaction can be represented by the following general chemical equation:
ROH + R'CHO => ROCH (OH) R '
As can be seen, an alcohol reacts with an aldehyde to form the hemiacetal. What would happen if both R and R 'belong to the same chain? In that case, you would have a cyclic hemiacetal, and the only possible way that it can be formed is that both functional groups, -OH and -CHO, are present in the molecular structure..
Furthermore, the structure must consist of a flexible chain, and with bonds capable of facilitating the nucleophilic attack of the OH towards the carbonyl carbon of the CHO group. When this happens, the structure closes into a five- or six-membered ring..
An example of the formation of a cyclic hemiacetal for glucose monosaccharide is shown in the image above. It can be seen to consist of an aldose, with an aldehyde group CHO (carbon 1). This is attacked by the OH group of carbon 5, as indicated by the red arrow.
The structure goes from being an open chain (glucose), to a pyranous ring (glucopyranose). At first there may be no relationship between this reaction and the one just explained for hemiacetal; but if you look carefully at the ring, specifically in section C5-O-C1(OH) -Ctwo, it will be appreciated that this corresponds to the expected skeleton for a hemiacetal.
Carbons 5 and 2 come to represent R and R 'of the general equation, respectively. As these are part of the same structure, it is then a cyclic hemiacetal (and the ring is enough to be evident).
Where is the anomeric carbon? In glucose, this is the CHO group, which can undergo nucleophilic attack by OH either from below or above. Depending on the orientation of the attack, two different anomers are formed: α and β, as already mentioned..
Therefore, a first characteristic that this carbon possesses is that in the open chain of sugar it is the one that suffers the nucleophilic attack; that is, it is the CHO group, for the aldoses, or the R grouptwoC = O, for ketoses. However, once the cyclic hemiacetal or ring is formed, this carbon may appear to have disappeared..
It is here where other more specific characteristics are found to locate it in any pyranous or furanose ring of all carbohydrates:
-The anomeric carbon is always to the right or left of the oxygen atom that makes up the ring.
-Even more important, it is linked not only to this oxygen atom, but also to the OH group, coming from CHO or RtwoC = O.
-It is asymmetric, that is, it has four different substituents.
With these four characteristics, it is easy to recognize anomeric carbon by looking at any "sweet structure.".
Above is β-D-fructofuranose, a cyclic hemiacetal with a five-membered ring..
To identify the anomeric carbon, you must first look at the carbons on the left and right side of the oxygen atom that makes up the ring. Then, the one that is linked to the OH group is the anomeric carbon; which in this case, is already circled in red.
This is the β anomer because the OH of the anomeric carbon is above the ring, as is the CH grouptwoOh.
Now, we try to explain what are the anomeric carbons in the structure of sucrose. As can be seen, it consists of two monosaccharides covalently linked by a glycosidic bond, -O-.
The ring on the right is exactly the same one just mentioned: β-D-fructofuranose, only it is “flipped” to the left. The anomeric carbon remains the same for the previous case, and meets all the characteristics that would be expected of it.
On the other hand, the ring on the left is α-D-glucopyranose.
Repeating the same anomeric carbon recognition procedure, looking at the two carbons on the left and right side of the oxygen atom, it is found that the right carbon is the one that is bonded to the OH group; which participates in the glycosidic bond.
Therefore, both anomeric carbons are connected by the -O- bond, and therefore they are enclosed in red circles..
Finally, it is proposed to identify the anomeric carbons of two glucose units in cellulose. Again, the carbons around the oxygen within the ring are observed, and it is found that in the glucose ring on the left the anomeric carbon participates in the glycosidic bond (enclosed in the red circle).
In the glucose ring on the right, however, the anomeric carbon is to the right of oxygen, and is easily identified because it is linked to the oxygen of the glycosidic bond. Thus, both anomeric carbons are fully identified.
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