Question

Choose the correct answer from the alternatives given.
Glucose ($C_6{H}_{12}{O}_6$) can exist as both an open chain form and a dosed ring form Earlier, it was thought that glucose occurs in the closed ring form What possible difference between these forms would give scientists a clue that the open chain form was not present

• A. Only open chain forms can undergo condensation, which does not occur with glucose
• B. An open chain presents ends with functional groups (in this case aldehyde), and glucose failed to undergo typical aldehyde reactions, a phenomenon that could be explained by having no end functional group in a ring structure
• C. Because glucose is solid at room temperature, it must have saturated hydrocarbon chains
• D. Glucose could not be denatured so it must be a tight chain

Answer 1

The correct option is B An open chain presents ends with functional groups (in this case aldehyde), and glucose failed to undergo typical aldehyde reactions, a phenomenon that could be explained by having no end functional group in a ring structure

Glucose (C${}_6$H${}_{12}$O${}_6$) is the essential constituent of human blood and is a simple sugar component of carbohydrates. It exists in both the open chain structure and closed cyclic ring form. The open chain structure has a straight chain of six carbon atoms with the presence of an aldehyde group at its end but it fails to undergo aldehydic reactions. This is because the glucose ring is not very stable and can be easily broken by strong reagents like HCN to give the intermediate aldehyde form which reacts with them just like an aldehyde. But weak reagents like NH${}_3$ are unable to open the chain and cannot react with it. This explains the inability of glucose to form aldehyde and ammonia. So, the correct option is "an open chain present ends with functional groups (in this case aldehyde), and glucose failed to undergo typical aldehyde reactions, a phenomenon that could be explained by having no end functional group in a ring structure".