[victor]

I'm just curious that carbohydrates like maltose is easily soluble in water but the same thing won't happen in cellubiose. The only difference is bond formation where maltose has an alpha glycocidic bond and cellulose has a beta one...
is this kind of bond is the only thing that promote insolubility in water?

[Yggdrasil]

Cellobiose (beta-D-glucopyranosyl-(1->4)-beta-D-glucopyranose) is soluble in water.  Cellulose (a polymer of cellobiose units), however, is insoluble in water.  Molecules of cellulose aggregate in solution because the strands are capable of intermolecular hydrogen bonding.  This creates large fibers with a low surface area to volume ratio.  Since this ratio is low, the cellulose cannot be solvated well enough to disolve (since solvation can occur only at the surface of a molecule).

Amylose (a polymer of maltose units), however, is soluble in water because its structure is such that intermolecular hydrogen bonding is unfavored.  The difference here is in the structures favored by the two types of linkages.  The geometry of the alpha (1->4) linkages in amylose favor an open helical structure (in which intermolecular hydrogen bonding is unfavored) whereas the geometry of the beta (1->4) linkages in cellulose favor a linear structure.  Whereas the linear structure of cellulose is conducive to intermolecular hydrogen bonds, the helical structure of amylose keeps molecules from aggregating.  This maintains a high surface area to volume ratio and allows amylose fibers to remain solvated in water.

The moral of the story: to underdstand the properties of most biological macromolecules, you must consider the overall structure of the molecules.

[victor]

  Thanks for the explaination Boss...I'll save it in my memory (long term one)..

[victor]

Nother question about carbs:
in aqueous equilibrium environment, D-ribose have 4 forms which are: 20% alpha-pyranose, 56% beta-pyranose, 6% alpha-furanose and 18% beta-furanose. PLease draw the Haworth structure for those four.
I get confused when I've to draw the pyranose..it's because that ribose is consisted only by 5 carbon where, one of them in aldehyde group while the other one is achiral and taking form as an alcohol group (the remaining C are chiral). If I've to draw pyranose for D-ribose, do I have to include that achiral C into the ring system?? so, the OH group from achiral C runs addition reaction to the aldehyde group, forming hemiacetal

[Yggdrasil]

Yes, if you are to form a pyranose (six-membered ring), then your ring must consist of the five carbons and the O from the achiral carbon (carbon #5).

[victor]

Ohh...(thank you)₃ you folks so kind (quoted from Numb by Linkin' Park vs JayZ)..

Oh..by the way..anyone have ever try to measure the specific rotation of ribose in aquoeus environment for 37ºC (internal body temperature) ?
Just curions, because the ribose in RNA is taking a furanose structure while in previous statement (in 25ºC) it's said that ribose prefered to occur in pyranose (majority beta-pyranose) form better than furanose.. 
Just wondering about it.. 

[Yggdrasil]

Well, you have to keep in mind that in RNA, the ribose is covalently linked to other ribose molecules, so its in its acetal form, not its hemiacetal form.  While hemiacetals are capable of mutorotation, acetals cannot mutorotate.

[victor]

Ohh...I see...I never pay attention on that aspect.. 
Thanks..
next, D-fructose reduction with NaBH4 results a mixture of D-gulsitol and D-manitol. What can this mixture prove about the configuration of D-fructose, D-manose and D-glucose?
My answer: reduction of D-glucose = D-glusitol; reduction of D-manose = D-manitol. But the only substrate which undergo reduction is only D-fructose...then why there're D-manitol and D-glusitol in the mixture?
or is it like this? fructose has a ketone group in C-2. So, reduction will occur there, resulting alcohol group. then, the alcohol group from C-2 can take place whether in left or right side of the C-2 in Fischer projection. If the OH group take place in the left side in Fischer projection, then it's manitol, if in the right side, glusitol....is it like that?

[Yggdrasil]

fructose has a ketone group in C-2. So, reduction will occur there, resulting alcohol group. then, the alcohol group from C-2 can take place whether in left or right side of the C-2 in Fischer projection. If the OH group take place in the left side in Fischer projection, then it's manitol, if in the right side, glusitol....is it like that?

Exactly correct.  If you want to use some biochem vocabulary, you can say that glucose and mannose are C-2 epimers of eachother (epimers are stereoisomers which vary in the configuration about only one stereocenter).

[victor]

ahh..I see..epimers.. thanks
I'll post again if there's another problem with these sugars.. 

[victor]

Um..actually I got confused whether I have to post it in chemical biology forum or in this Organic chemistry forum. But, since I've posted one about Carbohydrate in this Organic chemistry forum, so I'll continue to post anything related to carbohydrate in this forum....

Um, my question is about qualitative tests on monosaccharides glucose, fructose and arabinose.

1. For Benedict test which is tested on those three samples, I got the results that all of them form a red sediment (which means as positive result). What I wanna ask is, during the formation of red sediment, which is a reduction of Cu²⁺ as CuSO₄ into Cu⁺ as Cu₂O,  meand thst the sugar is oxydized. I got confused about the reaction mechanism,  which in the book is said that this reduction of Cu is related with enol forms of each sugar.
Now, let's see for fructose, it has a ketone group in C-2. What makes me confused is that a ketone group is not as easy as an aldehyde group when it comes to an oxydation (so supposed that ketone sugar wouldn't reduce that Cu), but the fact said a different thing....I think that it's due to formation of enol tautomer of fructose. But, do anyone know about the reaction mechanism? oh, also if you may, please give a link about this....

2. About Seliwanoff and Bial test, actually this is a silly question.. When I was in the Biochemistry lab yesterday, I did the Bial test in acid room (kinda room to place concentrated acid), maybe because Bial reagent is made of Orcynol, conc. HCl and FeCl₃. But, when it comes to do the Seliwanoff test (Seliwanoff reagent is made of Resorcynol (m-dihydroxybenzene) and conc. H₂SO₄ ), I do it not in the acid room (because the Seliwanoff reagent is placed on the desk just like that). Well, considering that both of the reagents contain concentrated acid, how come that Seliwanoff reagent is placeed outside the acid room? Or is it just an error in placing that reagent?

3. Then, for Mollisch reaction. When I did this reaction, I put 3 samples of sugars (glucose, fructose and arabinose) and slowly drops a few drops of alpha-naphtol towards three of them. After that I slowly put conc. H₂SO₄ into the reaction tube. The result it there's a purple ring formed in the middle of the solution, which is said that the acid get reacted with the sugar and formed a furfural which then condensed with alpha-naphtol to form a purple ring as a middle layer. My question is why did the acid won't get react with all the sugar (so it form a three-layer solution) ? and then, do you know what are the substance in that three-layer solution? My guess ia that,  the bottom layer is acid ; middle layer is a furfural-alpha naphtol complex ; and the top layer is the remaining sugar (not in furfural state).

And another favor, could anyone please give me a link about mechanisms in Seliwanoff, Benedict, Bial, Mollisch, and Osazone reaction? and would anyone please tell me the IUPAC name of Orcynol which is used in the Bial reagent?

[Yggdrasil]

These questions could go in either the organic chemistry, biochemistry or chemical biology thread (IMHO chemical biology is essentially the same thing as organic chemistry, just with a more defined purpose).  In the biosciences, there is considerable overlap such that there is not much point to making distinctions between most disciplines anymore.

1.  You are absolutely correct with your explanation.  C-2 ketoses will tautomerize to the same enol which aldoses form, which makes C-2 ketoses reducing sugars.

2.  I'm not familiar with these tests, but I would assume the answer is because HCl is volitile and H₂SO₄ is not.

3.  I'm not familiar with this test, sorry.

However, google or wikipedia may be able to provide you with more information on the mechanisms.

[victor]

Hmm...well, thanks for your answer. Actually I need those explaination for my lab assignment. I've search in google and what I found are methods to do the reaction...not the reaction mechanism...I've also search many biochemistry book and many of them don't explain about the mechanisms. Actually I have many speculations about the mechanisms, but my lab  assistant told me to cite or quote the explaination from biochemistry books..

[victor]

Um...looks like I've made a mistake on posting the previous post.
It's about Seliwanoff and Bial reagents. Well, I'll write the composition of those reagents:

Seliwanoff reagent : known as The resorcynol-HCl test when applied to sugar. It's made of resorcynol (m-dihydroxybenzene) 0.05% in HCl 5N.

Bial reagent : made of orcynol (3,5-dihydroxytoluene) in HCl 5N and also with the presence of FeCl₃.

What I wanna ask is that, both of them contain a strong acid. But why only Bial reagent is placecd in acid room, while Seliwanoff reagent is placed together in ordinary place (I mean there's no securing procedure for Seliwanoff reagent)? Based on the composition, is the placement of these two reagents only a wrong placement? or is that Seliwanoff reagent is not as harmful as Bial reagent? 
I've got this question from my lab assistant. Um, maybe this question sounds silly, but it really makes me wondering about that....