Acetals & Ketals --> Aldehydes & Ketones

[FROGGBUSTER]

So acetals and ketals are not reducing sugars because they cannot be converted back to the open-chain aldehyde and ketone form, and hemi-acetals and hemi-ketals ARE reducing sugars because they can be converted back. I think these are the reasons Chad gave.

But acetals and ketals CAN be converted back to the open-chain form if you throw in H30+ and heat the solution.

So what gives? Shouldn't they be reducing then?

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[mh0000]

Look at the mechanism for the addition of an alcohol to an aldehyde. The steps are in equilibrium and are reversible. You will need an acidic environment to shift the equilibrium back towards the hemiacetal and free aldehyde (open chain). The reason why acetals are not reducing is because the oxygen on the anomeric carbon is bonded to another sugar to form a glycosidic bond. YOu need an acidic environment to hydrolyze this bond. This of course occurs in vitro. biological systems operate differently.

 

[FROGGBUSTER]

Look at the mechanism for the addition of an alcohol to an aldehyde. The steps are in equilibrium and are reversible. You will need an acidic environment to shift the equilibrium back towards the hemiacetal and free aldehyde (open chain). The reason why acetals are not reducing is because the oxygen on the anomeric carbon is bonded to another sugar to form a glycosidic bond. YOu need an acidic environment to hydrolyze this bond. This of course occurs in vitro. biological systems operate differently.

Ah, so basically you're saying, oxidation-reduction reactions can't take place in acidic environments correct? I guess I should have realized that, lol.

Thanks.

 

[mh0000]

Ah, so basically you're saying, oxidation-reduction reactions can't take place in acidic environments correct? I guess I should have realized that, lol.

Thanks.

No I'm not saying that at all. I'm saying ethers are not oxidized to carboxylic acids directly. In order for a sugar to be a reducing sugar, it needs to be in the open linear chain, because it is only in this form that its aldehyde exists. It is this aldehyde that becomes oxidized. When the sugar forms a ring, it forms a hemiacetal (which is not a aldehyde). The reason why people say hemiacetals are reducing is only because it is equilibrium with the open chain. So it is only indirectly capable of being oxidized. When this hemiacetal forms a glycosidic bond to another sugar to create a disaccharide or polymer, it forms an acetal bond. Thus in order for this monomer (the original monosaccharide) to be oxidized (ie a reducing sugar), it needs to revert back to its chain form. Therefore it needs to break this glycosidic bond (the acetal) via a catalyzed hydrolysis (such as an aqueous acid). Once free, this sugar can equilibriate back between its ring and open chain form.

Also, some oxidizing agents need an acidic environment to oxidize an aldehyde to a carboxylic acid (such as the chromates)

 

[FROGGBUSTER]

No I'm not saying that at all. I'm saying ethers are not oxidized to carboxylic acids directly. In order for a sugar to be a reducing sugar, it needs to be in the open linear chain, because it is only in this form that its aldehyde exists. It is this aldehyde that becomes oxidized. When the sugar forms a ring, it forms a hemiacetal (which is not a aldehyde). The reason why people say hemiacetals are reducing is only because it is equilibrium with the open chain. So it is only indirectly capable of being oxidized. When this hemiacetal forms a glycosidic bond to another sugar to create a disaccharide or polymer, it forms an acetal bond. Thus in order for this monomer (the origianl monosaccharide) to be oxidized (ie a reducing sugar), it needs to revert back to its chain form. Therefore it needs to break this glycosidic bond (the acetal). Once free, this sugar can equilibriate back between its ring and open chain form.

Okay I get that acetals and ketals are incapable of being directly oxidized and need to be converted back into the open chain form first. But if you throw enough acid and heat into the solution, the polymer should be hydrolyzed and eventually turn back into the aldehyde or ketone. Once it is in this aldehyde or ketone form, theoretically it could be oxidized (or so I thought!).

However, I found a site that says this:

In contrast, acetal forms (glycosides) are not reducing sugars, since with base present, the acetal linkage is stable and is not converted to the aldehyde or hemiacetal. The outcome is that in a reducing sugar the anomeric carbon is in an aldehyde or hemiacetal. In a non-reducing sugar, the anomeric carbon is in an acetal.

Source: http://chemistry2.csudh.edu/rpendarvis/dipolysacch.html


This is how I'm interpreting it: in the presence of acid, yes the acetal and ketal can be converted back to aldehyde and ketone, but you can only oxidize the ensuing aldehyde and ketone under basic conditions. You can't carry out oxidation under acidic conditions.


Am I interpreting this incorrectly?

 

[mh0000]

Okay I get that acetals and ketals are incapable of being directly oxidized and need to be converted back into the open chain form first. But if you throw enough acid and heat into the solution, the polymer should be hydrolyzed and eventually turn back into the aldehyde or ketone. Once it is in this aldehyde or ketone form, theoretically it could be oxidized (or so I thought!).

However, I found a site that says this:

Source: http://chemistry2.csudh.edu/rpendarvis/dipolysacch.html


This is how I'm interpreting it: in the presence of acid, yes the acetal and ketal can be converted back to aldehyde and ketone, but you can only oxidize the ensuing aldehyde and ketone under basic conditions. You can't carry out oxidation under acidic conditions.


Am I interpreting this incorrectly?

No you're reading it correctly. Hydrolyze the acetal/glycosidic bond with acid. Then in a second step, neutralize the pH with some base. Under neutral conditions, the nucelophilic attack by an alcohol is not favorable, so it won't reform the hemiacetal/acetal that readily. Now one change the pH to whatever it needs to be for the given reagent to test for reducing sugars.

 

[FROGGBUSTER]

No you're reading it correctly. Hydrolyze the acetal/glycosidic bond with acid. Then in a second step, neutralize the pH with some base. Under neutral conditions, the nucelophilic attack by an alcohol is not favorable, so it won't reform the hemiacetal/acetal that readily. Now one change the pH to whatever it needs to be for the given reagent to test for reducing sugars.

So when people say acetals and ketals are not reducing sugars, it really just means under either acidic or basic conditions correct? Because if someone really wanted to, he could do those series of steps that you just said to go from acetal --> aldehyde or ketal --> ketone, and then oxidize it. Is this right?

 

[mh0000]

So when people say acetals and ketals are not reducing sugars, it really just means under either acidic or basic conditions correct? Because if someone really wanted to, he could do those series of steps that you just said to go from acetal --> aldehyde or ketal --> ketone, and then oxidize it. Is this right?

Yes. It would erroneous to say that acetals/ketals are inherently reducing sugars. They are not. But they can be turned into one via acid hydrolysis. However, just keep in mind that because a sugar has an acetal link, doesn't mean that it isn't a reducing sugar. Most disaccharides are reducing sugars. The difference is that the oxidation of the aldehyde is not happening where the glycosidic link is, but rather on the sugar with a "free" anomeric carbon. Sucrose is not a reducing disaccharide. Look at the structure to determine why.

 

[FROGGBUSTER]

Yes. It would erroneous to say that acetals/ketals are inherently reducing sugars. They are not. But they can be turned into one via acid hydrolysis. However, just keep in mind that because a sugar has an acetal link, doesn't mean that it isn't a reducing sugar. Most disaccharides are reducing sugars. The difference is that the oxidation of the aldehyde is not happening where the glycosidic link is, but rather on the sugar with a "free" anomeric carbon. Sucrose is not a reducing disaccharide. Look at the structure to determine why.

Alright thanks man.

 

[LetsGo2DSchool]

Good **** mh0000, I bookmarked this page. And thanks Frog for creating this thread.