Question

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Please explain with detail

Assign the anomeric centres of the following disaccharide. OH OH он O HO OH 2 HO OH 1 OH Select one or more: a. 1-beta O b. 1-alpha O c. 2-alpha d. 2-beta

Answer 1

Anomeric center of a saccharide (aldose or ketose): It is an epimer at the hemiacetal/acetal carbon in a cyclic saccharide. The anomeric carbon is the carbon derived from the carbonyl carbon ( ketone or aldehyde functional group) of the open-chain form of the carbohydrate molecule.

An epimer is basically a stereoisomer that differs in configuration at a particular stereocenter.

The anomeric carbon appears at C-1 of an aldose and C-2 of a ketose

In saccharides when the -OH group present at the anomeric carbon is projected below the plane of the cyclic system then the configuration is assigned as alpha (α) and beta (β) is used when the projection is above the plane of the system.

Now in the given disaccharide the C-1 of the left side ring (chair form) is projected upward in equatorial direction . Hence C-1 will have a beta (β) configuration.

But in the right side chair form then the anomeric carbon -OH group is pointing downward making it a alpha (α) configuration.

The resulting disaccharide is formed by the linkage between C-1 and C-4 of the two rings.

Hence the final answer is 1-beta(β) , 2-alpha (α)

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You can find the general mechanism of ring formation of saccharides in any book.

Thank you !