Question

BIOCHEMISTRY/ORGANIC CHEMISTRY

which carbon is easily oxidized on molecule 1 (bottom) and molecule 2 (top)? why?

НО НО но 5 он

Answer 1

Molecule1:

Ii is pentose sugar is five carbon cyclic furanose form, also known as ribose from the nucleic acid and derived by cyclization of C4-hydroxy group with the aldehyde (CHO) group at C1 position of five membered monosaccharides.

Thus the C1 carbon in the pentose is most oxidized carbon atom, the example of a pentose sugar are ribose (C₅H₁₀O₅ ) and deoxyribose (C₅H₁₀O₄)

Molecule 2:

It is pyranose form of the glucose monosaccharides having six carbon atom. The hydroxyl group at C5 position of glucose, forms a hemiacetal with the aldehyde at C1 position which resulting glucopyranosegisomer. In biological system the series of enzyme-catalysed reactions like glucose is oxidized to form carbon dioxide and water yielding energy in the form of ATP. The Glucose (C₆H₁₂O₆) and fructose (C₆H₁₂O₆) having six carbons and so are called hexose sugars.

Therefor the most oxidized carbon in this molecule is C1 carbon. Thus, the CHO group at C1 is highly oxidized than the CH₂OH group at C6 position