TopScore orgo question..

[prydA]

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In topscore, test 2, o-chem question #86..

Why is sucrose non-reducing? Doesn't the fructose ring in it have a free hemiacetal group? i know lactose is considered a reducing sugar.. what about it makes it a reducing one when compared to sucrose? i know the basic definition is that there is an aldehyde group it can open up to, but i dont understand it in this case.

And why is fructose considered a ketohexafuranoside? did they mean the sucrose instead? i understand the furan part since it has 5 carbons, but what does the hexa- prefix have to do with anything?

thanks for any help!

 

[Orgodox]

In topscore, test 2, o-chem question #86..

Why is sucrose non-reducing? Doesn't the fructose ring in it have a free hemiacetal group? i know lactose is considered a reducing sugar.. what about it makes it a reducing one when compared to sucrose? i know the basic definition is that there is an aldehyde group it can open up to, but i dont understand it in this case.

In sucrose unlike most polysaccharides, the glycosidic bond is formed between the reducing ends of both glucose and fructose, and not between the reducing end of one and the nonreducing end of the other. The effect of this inhibits further bonding to other saccharide units. But the main thing is that even though fructose alone did have the hemiacetal group when it is in sucrose if does not

To be a reducing sugar you would look for the hemiacetal group. This is not present in sucrose.