If a sugar is nonreducing, does that mean it doesn't ionize in water?

Question

I'm thinking of sucrose, which I learned is nonreducing, what I take to mean it doesn't accept/take electrons from the surrounding solution. Does this also mean that it doesn't ionize and thus has a Van't Hoff factor ( i ) of 1?

Answer 1

I believe the confusion lies with the term Reducing sugar. From my perspective sucrose is not an ionic compound, it is covalently bonded. I believe that this should indicate to you that it does not "ionize" or the term that I was taught "dissociate" since it has nothing to dissociate from.

To quote wikipedia:

"The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved, and the concentration of a substance as calculated from its mass."

Since it does not "dissociate," I would say that the answer is no, it does not "ionize," and therefore has a Van't Hoff factor of 1 when mixed with pure water.

Like anything with electrons, they can be lost. This takes place in our bodies through more complicated reactions than I understand completely. I believe however, according to wikipedia, that sucrose has a Glycosidic bond which when broken, will separate sucrose into smaller sugars, if this takes place in your solution then I believe it will have a Van't Hoff factor > 1.

Answer 2

There's more than one way to skin a cat accept an electron. Actually, in the case of a reducing sugar, it donates an electron. When it does so, it is oxidised. But that doesn't mean that it ionizes in water!

When you keep a reducing sugar in the presence of an oxidising agent, the electron is transferred over to the agent, reducing it. The sugar may form an ion, or it may form a carboxylic acid (depending on the medium). The electron is fuelling the reduction half reaction of the oxidising agent we have used. If we do get a carboxylic acid, there is a chance for it to ionize(again, depends upon the medium) and we may get a Van 't Hoff factor $>1$ {*}. But remember that this is in the presence of an oxidising agent. If we just dunk a sugar in water, it won't be able to reduce anything, and will be chemically inert, giving a Van 't Hoff factor of 1.

^{*Also, the reaction with the oxidising agent may provide additional ions, increasing the Van 't Hoff factor for the oxidising agent}