ID #2327

How does glucose go from its straight chain form to ring structure? I don't see the aldehyde group in the ring structure. Is there hydrolysis involved?

Check out the lecture 3 slide that starts with the phrase “Monosaccharides are typically found in lengths of 3, 5, or 6 carbons.” This has the most perfect picture of a ring form of a glucose forming from a chain structure. 
The aldehyde is key because the carbon (number 1 carbon) of the aldehyde group interacts with the hydroxyl group attached to the 5th carbon of the carbohydrate chain. This excludes the 6th carbon from the ring, giving it its distinct 5-C, 6-member ring structure. Because the 1st carbon is now involved in a fourth covalent bond, the double-bonded oxygen from the aldehyde must become a single bond. The oxygen now has two free electrons and must form a bond with a hydrogen to form a hydroxyl group; this is why you do not see a double-bonded oxygen in the ring structure form of glucose. Even if this is not exactly the way it happens, I like to think of the hydrogen that was attached to the hydroxyl group of the 5th carbon hopping over to bind newly single-bonded oxygen of the former aldehyde group. 
Water is not gained or lost in this formation of a covalent bond. If you count the number of oxygens in the chain structure (6) and the ring (6), you see that none are lost. That is a good way to confirm that no waters are gained or lost if you are ever not sure. 

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