Trying to understand “reducing sugars” - is the C1 in the ring considered “more oxidized” than C1 in the aldehyde? Even though both C’s are *technically* bound twice to O and once to H?

Answer 1

The term reducing sugar actually refers to the open/linear form of the sugar. So what we're referring to is presence of a ketone or aldehyde. Thinking back to organic chemistry, we know that when we react an alcohol with a carbonyl we get a hemiacetal/hemiketal. That is what happens in cyclization of the sugar. So, if we want to go backwards through cyclization to open the ring and make a reactive carbonyl, we need a hemiacetal/hemiketal, meaning we need an alcohol sitting on C1. And if we react that alcohol with something like another sugar to form a glycosidic bond, we no longer have that hemiacetal/hemiketal and can't open the ring to form an aldehyde/ketone.