Chirality of a carbohydrate

[tiny529]

Okay... Stereochemistry has never been my strong point, so seeing all of this chiral stuff in the carbohydrates chapter is kind of freaking me out. My book is confusing the heck out of me. Is there any trick to figuring out optically active carbons in a carbohydrate ring?

 

 

Only 4 problems left, BTW... Out of 80 problems covering 8 chapters, 3 of which I haven't even done yet. The chirality is a multiple guess problem, so if worst comes to worst, I'll just... guess. I've got 3 "guess the product/intermediate" problems left but I don't have it in me to try them tonight. Tomorrow is another day! I'm going to bed.

Edited March 18, 2011 by tiny529

[hypervalent_iodine]

Are you given the structure of a carbohydrate and asked to determine stereochemistry in terms of ® and (S) or are you also being asked to determine if it is a D or L sugar and what the name of it is, etc.?

[tiny529]

Are you given the structure of a carbohydrate and asked to determine stereochemistry in terms of ® and (S) or are you also being asked to determine if it is a D or L sugar and what the name of it is, etc.?

 

None of the above... =) We are given 4 different unnamed structures and are told to mark which ones have optically active carbons. They are all in chair form and show the -OH groups as either axial or equatorial. I haven't learned any of this material and it seems rather foreign. The other chapters that I haven't done yet at least still made sense, even if it was new material. This carbohydrate stuff is really confusing. =(

[Horza2002]

To see if a carbon centre is optically active, you only need to know one thing....does it have four different subtituents. If it does, it is chiral, if doesn't then its achiral.

 

For example, bromo-chloro-flouro methane (CHBrClF) is chiral because there are four different grous bound to the central carbon. However chloforom (CHCl₃) isn't because it only have two different groups around it. If you don't need to assign the R/S notation to the stereocentres, then you don't need to know anything about chair/boat conformations.

[Dan_Ny]

If you don't need to assign the R/S notation to the stereocentres, then you don't need to know anything about chair/boat conformations.

Ring conformations have no influence at all on the configuration of the stereocenters, so, even if you have to assign the R/S notation, the conformation does not matter. Now, I don't know how your structures look like, but what horza said, four different substituents on a carbon, applies also for carbons in a chain or a ring (like in sugars). So, look at you structure - it should have one oxygen in the ring and a CH2OH group attached to the ring next to that oxygen next to it and looks perhaps a bit like this picture:

Well, does carbon 1 have two substituents that are the same? So, ist it optically active?

Now look at the carbons in the ring. How many different substituents do they (all) have? Careful, it is sometimes not enough just to look at the next atom.

(And here is a bit more about the CIP rules and optical activity: http://tigger.uic.ed...xt/chapter5.htm )

Edited March 18, 2011 by Dan_Ny

[Horza2002]

Dan_Ny, I know that the conformation has no influence on the R/S assignment; it is only the configuration that does. I asked that because if you have to assign the R/S centres, then you need to be able to draw the carbon with the lowest priority group facing away from you. This is much easier to do if you know the relationship between axial and equatorial substituents. That is why i said that.

 

This is the most common/simplest for of stereochemistry, there are other forms (like axial chirality for example). HOwever, if you are simply looking at sugars, then you will only need to know the stereogenic centre one. Even if you are not going to assign the R/S notation to each stereogenic centre, it is often simpler to use the CIP rules to see if they are indeed stereognic centres.

[Dan_Ny]

Dan_Ny, I know that the conformation has no influence on the R/S assignment; it is only the configuration that does.

 

Well, I know that you know, Horza2002. Just giving you a hard time. Yeah I understand and you are rigt, knowing the conformation can make it easier to see the priorities on a ring carbon.

 

Well I definitely forgot everything about axial and planar chirality but this would be the ideal time to look it up again. And what was that stuff with the re and si face of trigonal or prochiral molecules? Hm, I should know that...

 

The structure was actually for tiny529, I don't know if it is any help, though.

Edited March 18, 2011 by Dan_Ny

[Horza2002]

Lol thanks for that Dan_Ny just what I need

 

Axial chirality arises from allene (two double bonds next to each other, C=C=C). This arises because the two double bonds are orthogonal to each other because of the arrangement of the orbitals.

 

The Re and Si face is for prochiral molecules; these are compounds that are not chiral but if you add another group, then it will become chiral. The Re face is the face that attack leads to the R enantiomer where as the Si face leads to the S enantiomer.

 

I've attached a file that shows them in diagrams...I often find drawing/seeing chirality problems written down, it makes it soooo much easier than doing it in your heads.

[tiny529]

Ring conformations have no influence at all on the configuration of the stereocenters, so, even if you have to assign the R/S notation, the conformation does not matter. Now, I don't know how your structures look like, but what horza said, four different substituents on a carbon, applies also for carbons in a chain or a ring (like in sugars). So, look at you structure - it should have one oxygen in the ring and a CH2OH group attached to the ring next to that oxygen next to it and looks perhaps a bit like this picture:

Well, does carbon 1 have two substituents that are the same? So, ist it optically active?

Now look at the carbons in the ring. How many different substituents do they (all) have? Careful, it is sometimes not enough just to look at the next atom.

(And here is a bit more about the CIP rules and optical activity: http://tigger.uic.ed...xt/chapter5.htm )

 

Yeah, that's exactly what my pictures look like. I've got 4 different ones and I have to say which ones are chiral. I understand that to be chiral, the carbon needs to have 4 different substituents, but the way I see it, all of the carbons except for C1 have 4 different substituents. =\ But all 4 isn't one of the multiple choice choices.

 

I forgot to note that it's the product of the carbohydrate after being treated with HNO3 that needs to be chiral.

[Horza2002]

So first of all, you will need to work out what you get if you treat each of them with nitric acid. There are 6 carbons in that structure that Dan-Ny provided and 5 of them are chiral (as you said, carbon 1 isn't chiral).

[tiny529]

Ugh... My brain is fried on this stuff. Since this was a multiple guess, I just guessed until I guessed right. I'll worry about understanding it when we get to that chapter. We're still in the middle of learning about carboxyls... Not even close to carbohydrates yet.

 

I did end up completing them all, amazingly enough. I just hope my prof doesn't decide to spring even more problems on us since my trial period is up in just a couple of days. Good thing I'm not taking a lab... I can't go ballistic and blow up the chemistry department if he does add more problems...

[Horza2002]

If you want any of us to expalin you the correct anwsers, then post the structures and the questions and we'll explain it you. I'm sure myself, Hypervalent and Dan_Ny will give you the help you need.

[Dan_Ny]

Lol thanks for that Dan_Ny just what I need

 

Axial chirality arises from (...)

I've attached a file that shows them in diagrams...I often find drawing/seeing chirality problems written down, it makes it soooo much easier than doing it in your heads.

 

You are great, thanks a lot, horza2002! Actually, the re si thing is much easier than I thought ;-) I just looked up the third kind of chirality, planar chirality (because I finally wanted to get that in my head, too). According to wikipedia (http://en.wikipedia....lanar_chirality)

 

"This term is used in chemistry contexts, e.g., for a chiral molecule lacking an asymmetric carbon atom, but possessing two non-coplanar rings that are each dissymmetric and which cannot easily rotate about the chemical bond connecting them: 2,2'-dimethylbiphenyl is perhaps the simplest example of this case. Planar chirality is also exhibited by molecules like (E)-cyclooctene, some di- or poly-substituted metallocenes, and certain monosubstituted paracyclophanes. Nature rarely provides planar chiral molecules; cavicularin being an exception."

 

(...)

 

To assign the configuration of a planar chiral molecule, begin by selecting the pilot atom, which is the highest priority of the atoms that are not in the plane, but are directly attached to an atom in the plane. Next, assign the priority of three adjacent in-plane atoms, starting with the atom attached to the pilot atom as priority 1, and preferentially assigning in order of highest priority if there is a choice. When viewed from the side of the pilot atom, if the three atoms form a clockwise direction when followed in order of priority, the molecule is assigned as R, otherwise it is assigned as S.

 

 

wow. I could not have been putting this better myself

 

 

tiny - there are two things that might happen if you treat your carbohydrate with HNO₃ (also depending on the concentration of your nitric acid). At very low concentrations and as an aqueous solution, the sugar will (acid catalyzed) eliquibrate and exist partly in it's open chain form (and, to be precise, in the hydrate form of the aldehyde, if i'm correct). Now, look at the picture below. How many chiral carbons do you have in the open sugar form? Does the number differ from the closed form?

 

The second possibility (which I like because it shows how rude and unfriendly these inorganic acids are ), is, of course, that it will rip your sugar apart. HNO₃ is a strongly oxidizing acid and I don't know what it's gonna do, but you won't recognize you sugar after it. Or your stereocenters.

 

 

Ah and like horza2002 said, in the closed form it's five, not four stereocenters.

Edited March 19, 2011 by Dan_Ny

[Horza2002]

Just to pouint out, there are a few other types of chirality that exist. One of them is helical chirality; which way the helix is cooled up. The most common form of this is DNA. In almost all cells, DNA exists it the B-DNA form. Among other difference, the helix is wound up "right handed"...there also exists another form of DNA called Z-DNA in which the coil is wound up "left-handed".

 

Yes, sugar stereochemsitry is very complicated because one of the stereocentres only exists in the cyclic form. Incidently, the aldehyde in the straight chain form is prochiral. Therefore the two faces of the aldheyde (top and bottom) will be Re and Si....which enatiomer you get depends on which side the hydroxy group attacks the carbonyl. In solution, there is actually a mixture of the two epimers at that centre.

 

As Dan_Ny has said, for this problem, using nitric acid is kinda like going after a fly with a bazooka. It is very strong and in practise would tare the sugar to pieces, you'll get a mixture of elimination, epoxidation, polymerisation and racemisation reactions all going.