Reducing vs nonreducing sugars

[arc5005]

I don't understand how to tell the difference between reducing and nonreducing sugars when looking at the glycosidic bonds between different molecules. I also seem to struggle with being able to recognize where the anomeric carbon is. Maybe this is the problem? Can anyone help me out please?

here is a question, i continue to struggle with:

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Raffinose can be described as a:


A. reducing sugar
B. nonreducing sugar
C. disaccharide
D. glycoprotein

B) nonreducing sugar
As shown by the molecular structure, raffinose is clearly not a disaccharide. It is considered to be olgiosaccharide (or even a polysaccharide). Since there are no amino acids attached to and of the three sugar residues, it cannot be a glycoprotein. This means that raffinose is either a reducing sugar or a non-reducing sugar.
Sugars having anomeric carbon atoms that have not formed glycosides (containing an acetal linkage) are called reducing sugars. Recall that the cycli and linear forms of both aldoses and ketoses are readily interconverted. The aldehyde function in galactose and glucose can easily be oxidized by an oxidizing agent like Benedict's reagent (a solution of copper(II) sulfate and sodium citrate in aq base) to the corresponding carboxylic acid (see below)

Fructose is an alpha-hydroxy ketone and alpha-hydroxy ketones are easily oxidized to the diketone by Benedict's reagent. The Benedict's reagent, which is a blue solution, is reduced to a red precipitate. If the aldehyde and ketone functional groups remain tied up in a glycosidic bond (as shown in the structure of raffinose), then they cannot react with the Benedict's reagent.

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[aldol16]

The anomeric carbon will always have the following shape: R-O-C-OH. In other words, it will be directly attached to an oxygen atom within the ring and also an alcohol. You can imagine collapsing the alcohol and kicking out the ring oxygen to re-form the aldehyde. In short, if you can re-form the aldehyde, it is a reducing sugar because that carbon is exposed to the redox processes of the cell. But if you can't, then that carbon is not redox-active and is a non-reducing sugar.

 

[JustinM88]

The anomeric carbon will always have the following shape: R-O-C-OH. In other words, it will be directly attached to an oxygen atom within the ring and also an alcohol. You can imagine collapsing the alcohol and kicking out the ring oxygen to re-form the aldehyde. In short, if you can re-form the aldehyde, it is a reducing sugar because that carbon is exposed to the redox processes of the cell. But if you can't, then that carbon is not redox-active and is a non-reducing sugar.

Not sure if my thinking is correct, so I'd like your input on this...
As I'm sure we've all said many times before, Reducing agents get oxidized.
(Reducing agents = Reducing sugars.)
Reducing sugars can be oxidized @ the anomeric carbon, after said anomeric carbon (C1) (the Hemiacetal) undergoes its ring opening into an Aldehyde like so:

Aldehydes can be oxidized.
Reducing agents are oxidized.
Reducing sugars can be oxidized.
Reducing sugars contain free Aldehydes!









Sooooo...
Raffinose cannot get oxidized at its anomeric carbons, and thus it is not a Reducing Agent (aka Reducing Sugar):

 

[Lawpy]

Not sure if my thinking is an oversimplification, so I'd like your input on this...
Reducing agents (aka Reducing Sugars) get Oxidized.
Reducing sugars can get oxidized, like this (@ the hemiacetal anomeric carbon):













Sooooo...
Raffinose cannot get oxidized at its anomeric carbons, and thus it is not a Reducing Agent (aka Reducing Sugar):

This sounds right. Reducing sugars have hemiacetal groups, which in noncyclic form contain a free aldehyde group that’s oxidized into a carboxyl group. Nonreducing sugars don’t have any hemiacetal groups, which means there are no free aldehyde groups that can be oxidized (remember in normal circumstances, ketones can’t be oxidized further). Since raffinose doesn’t have any hemiacetal group, it is a nonreducing sugar.

More information: Reducing Sugars — Master Organic Chemistry

And:

 

[JustinM88]

This sounds right. Reducing sugars have hemiacetal groups, which in noncyclic form contain a free aldehyde group that’s oxidized into a carboxyl group. Nonreducing sugars don’t have any hemiacetal groups, which means there are no free aldehyde groups that can be oxidized (remember in normal circumstances, ketones can’t be oxidized further). Since raffinose doesn’t have any hemiacetal group, it is a nonreducing sugar.

More information: Reducing Sugars — Master Organic Chemistry

And:

A question just crossed my mind. So when looking at that free Fructose's anomeric carbon, when the ring opens up, it's got a Ketone which can't be oxidized. How is it considered a Reducing Sugar then (reducing agents get oxidized, right?)? Is the hemiacetal group on that same Fructose being turned into a ketone upon ring opening considered "oxidation" even though that C still has 2 bonds to O either way?

 

[Lawpy]

A question just crossed my mind. So when looking at that free Fructose's anomeric carbon, when the ring opens up, it's got a Ketone which can't be oxidized. How is it considered a Reducing Sugar then (reducing agents get oxidized, right?)? Is the hemiacetal group on that same Fructose being turned into a ketone upon ring opening considered "oxidation" even though that C still has 2 bonds to O either way?

This is because the reagents used are basic, and under basic conditions, fructose isomerizes into glucose and mannose. For more information, see:

Why does fructose reduce Tollen's reagent and Fehling's solution?

Lobry de Bruyn–van Ekenstein transformation - Wikipedia

And:

 

[JustinM88]

This is because the reagents used are basic, and under basic conditions, fructose isomerizes into glucose and mannose. For more information, see:

Why does fructose reduce Tollen's reagent and Fehling's solution?

Lobry de Bruyn–van Ekenstein transformation - Wikipedia

And:

Ah, totally forgot about the Ketone Aldehyde equilibrium thing