For the given hemiacetal form of a monosaccharide, all the carbon atoms has to be numbered, the anomeric carbon atom has to be identified and whether it is a α o r β anomeric form has to be given Concept introduction: Monosaccharides forms cyclic hemiacetals, the oxygen atom in the – O H group of second last carbon (from bottom) attacks the carbonyl group of the same compound. The carbon atom in carbonyl group of monosaccharides is called anomeric carbon. Haworth projection is used to represent sugars in their cyclic hemiacetal forms. In Haworth projection, α -indicates, the – O H group of anomeric carbon C 1 and CH 2 OH substituent on the C 5 carbon is trans to each other. β -indicates, the – O H group of anomeric carbon C 2 and CH 2 OH substituent on the C 5 carbon is cis to each other.

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Chapter 20.4, Problem 20.11P

Interpretation Introduction

Interpretation:

For the given hemiacetal form of a monosaccharide, all the carbon atoms has to be numbered, the anomeric carbon atom has to be identified and whether it is a α or β anomeric form has to be given

Concept introduction:

Monosaccharides forms cyclic hemiacetals, the oxygen atom in the –OH group of second last carbon (from bottom) attacks the carbonyl group of the same compound.

The carbon atom in carbonyl group of monosaccharides is called anomeric carbon.

Haworth projection is used to represent sugars in their cyclic hemiacetal forms.

In Haworth projection,

α-indicates, the –OH group of anomeric carbon C1 and CH2OH substituent on the C5 carbon is trans to each other. β-indicates, the –OH group of anomeric carbon C2 and CH2OH substituent on the C5 carbon is cis to each other.

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