Alpha/Beta Glucose

[grayfalcon89]

Hi all,

 

I was reviewing my biology from my first year of intro biology at undergrad today, and I came across something that perplexed me.

 

How can one go from glucose to ring-shape and thus, form alpha or beta form? I searched Google and saw the chair conformation to really see why there are alpha and beta conformation, but like, how do you form two C-O bonds to form a ring with 5 carbons and 1 oxygen?

 

Hopefully this is clear! I'm having hard time fathoming how this occurs.

[Mateusz]

Open, chain structure of glucose change in a ring-shape by cyclization. Hydroxyl (-OH) group from C-5 carbon atom attack oxygen atom which is connected with C-1 carbon atom. Then it's possible to create during this cyclization alpha and beta form.

 

Meaby alpha goes to beta by making from ring-shape to chain-shape and than to ring-shape?

[Kaeroll]

Mateusz is correct in saying that the two interconvert via the open chain form. You can follow the formation and decomposition of the ring form using some simple carbonyl chemistry.

[UC]

I'm not sure if you have enough chemistry background to know what I'm talking about when I discuss enantiomers and show electron pushing diagrams, but hopefully you do.

 

Some of the information needed for enantiomers can be found on this wikipedia page and also in the linked thread:

 

http://en.wikipedia.org/wiki/Cahn_Ingold_Prelog_priority_rules

http://www.scienceforums.net/forum/showthread.php?p=493990#post493990

 

I have attached a simplified diagram of the conversion from ring to chain form of glucose. On the right side, I have replaced the carbohydrate tail of the aldehyde group by a simple R for clarity.

 

What you need to understand about the aldehyde group is that it is planar. Therefore, the two sides look different and using the rules on that wiki page can be named as the re and si faces (pronounced "ray" and "psi") When the hydroxyl group attacks, it doesn't discriminate which side of the aldehyde it approaches. The reaction product is tetrahedral. However, the products of the hydroxyl adding to the different sides have different stereochemistry, which leads to alpha and beta dextrose.

 

Now, if the reaction doesn't discriminate which side it attacks, why isn't the ratio of alpha to beta 1:1? Because the reverse reaction occurs readily, and since the alpha form is higher energy (less stable), the reverse reaction will happen more often for it, after which it has a 50:50 chance of becoming either an alpha or a beta. Eventually, you reach an equilibrium corresponding to the observed ratio.

Edited May 25, 2009 by UC

[UC]

What I didn't include in the above reaction diagrams is the movement of some hydrogens around, which occurs via interactions with the solvent, water. You could purify alpha-dextrose and it would last forever if you kept it dry in a bottle on a shelf, but a few hours in water and it'd be back to the mixture of forms.

 

If you want to be more specific, the structure formed there is called a hemiacetal. Google can find you some in-depth reaction mechanisms if you ask it nicely

Edited May 25, 2009 by UC