Created by Ryan Scott Patton.
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- at3:35, I still don't understand why you chose that Oxygen as the nucleophie...(7 votes)
- The Answer is Stereochemistry. There is a chance that other OH's will attack, but they would be less stable and the major product will be a 6-sided ring. Because...We know that a intramolecular ring is more "comfortable" when the bond angles are larger (this is due to steric strain of the molecules). Well since we know that, we also know that if the OH on C3 did the nucleophilic attach you would have a 4-sided molecule. Lets compare the angles between a 4 and 6-sided ring. 4-sided rings have a bond angle of 90 degrees (Higher strain), and a 6-sided ring has a bond angle of 120 degrees (Lower Strain). Thus a six-sided ring is more favored.(15 votes)
- I thought that cis was alpha and trans were beta.(1 vote)
- would the same principle be applied if it was a D sugar instead of an L sugar? at5:37, will the Haworth diagram be similar? What about the fischer?(2 votes)
- In trying to draw sucrose as a haworth drawing my OHgroups on C3, C4, and C6 are on opposite sides from the correct answer. I drew the linear structure correctly and used the "uplefting, downrighting" method to draw sucrose so don't understand how I keep drawing it incorrectly. Please help, maybe there'ssomething I am not understanding.(1 vote)
- Are you labeling your carbons correctly on your Fischer vs Haworth? I used to draw my hydroxyls the wrong way too until I realized that I was numbering my Haworth projection the wrong way! He shows the numbering at6:30. Good luck!(2 votes)
- at4:53I don't understand why there is a HOH off the anomeric carbon where did that come from?(1 vote)
- I don't understand how you went from the Haworth diagram to the chair confirmation. Can you explain how you know where to put the substituents in terms of axial and equiatorial?(1 vote)
- when i dissolve glucose in water it is gonna be occurring in alpha and beta configurations only or alpha and beta and open chain form..?(1 vote)
- I was wondering if the 4th c would be the one attacking in a Furanose ? because this way we will make a 5 member ring and it is not as stable as a 6 member ring! Thank you(1 vote)
- So, around8:22Ryan mentions the alpha and beta configurations.
Is it correct to visualize the top hydrogen and double bonded oxygen as rotating around the C1 atom? And depending on which side they happen to be on when the nucleophilic attack happens, an alpha or beta configuration will form?(1 vote)
- When converting the Haworth Projection to the Chair Conformation, how do you know whether the OH is in the axial or equatorial? For example, in B-D-Allopyranose the C3 carbon is in the axial position but the B-D glucopyranose the C3 carbon is in the equatorial position. Is there a certain trick to memorizing them or do you just have to look at the steric hindering?(1 vote)
- Alrightie, so we've been speaking so far about carbohydrates as chains of carbon atoms. And these are chains of carbon atoms that feature an aldehyde or a ketone functional group. And that falls into this general kinda one-to-two-to-one ratio of carbon, hydrogen, and oxygen. And of course I'll keep using glucose as an example. Now I've also used the term polyhydroxylated to refer to the numerous hydroxyl groups that are in these carbohydrates. And really I bring all of this verbiage back up to hopefully spark your ability to see that carbohydrates have all the makings of an internal, or intramolecular, I guess, reaction between the carbonyl carbon here, and one of the hydroxyl groups. Because essentially what we have is carbonyl and alcohol chemical reactions kinda just waiting to happen. What happens when an alcohol nucleophile attacks an aldehyde or a ketone? Well, if there's an excess of alcohol, we end up with a product that is either an acetal, or a ketal. But what happens if there's only one nucleophilic attack by an alcohol, if we just have one alcohol? And that's gonna be the case in the ring-closing, intramolecular reaction we have going on here. Well in that case, we end up with a hemiacetal, or a hemiketal. And really that terminology is just kind of a review of acetal and ketal chemical reactions that would fall under, I guess if you're looking in an organic chemistry book, aldehyde or ketone reactions, probably in the carbonyl section. So let's show how this process is happening in the context of our glucose, over here. First, I'm gonna highlight the particular hydroxyl oxygen that's gonna act as the nucleophile. So, we'll make that pink. And after being deprotonated, so after losing this proton, this oxygen is gonna have an extra set of electrons right here. And those electrons are gonna target that carbonyl carbon. So I'll draw the carbonyl carbon in green. And remember that the carbonyl carbon has a partial positive charge on it. It has a partial positive charge, because a lot of the electron density in this double bond is being hogged by this oxygen, so the oxygen has a partially negative charge, and the carbonyl carbon is partially positive. And that makes it a perfect target for the nucleophile that's been created in the deprotonation process of this oxygen. And so after the oxygen's electrons attack this carbonyl carbon, what's gonna happen is the electrons from this double bond are kinda gonna kick back up to the oxygen up here, and eventually they're gonna attract another proton, and will form another hydroxyl group out of some of the electrons from that bond. Now you might be asking, and it's a perfectly valid question, why it's this particular oxygen, the one I highlighted, that's acting as the nucleophile. And you're gonna see as soon as I get the product drawn, that we've formed a six-member ring, so it really has to do with product stability. And if you remember, the basis for the formation of the ring in the first place, was the increased stability over the straight carbon chain. So it makes sense that we're gonna form the most stable ring that we can. Now when we end of with a six-membered carbohydrate ring, such as the case with glucose here, we call the product a pyranose. The -ose again, as the suffix for sugar. And the pyra- part, to indicate that this ring is a sugar with six carbons. And then if the carbohydrate ring is a five-carbon ring, we call it a furanose, which is a bit easier for me to remember, because furanose and five both start with the letter F. So that's kinda the memory jogger for me. And maybe a good example for that would be ribose, with its five-carbon chain, but I'll kinda stop there, because almost every ring-forming carbohydrate that I can think of with biological implications, at least, forms either a five- or a six-membered ring. So pyranoses and furanoses. So just by convention, you can see that I have placed the O in this corner up here, and that places the formerly carbonyl carbon down here right below it. And it's actually no longer the carbonyl carbon, but it's still significant because it's the only carbon here that is bonded to two oxygen atoms, the highlighted oxygen and it's bonded to another hydroxyl group, as well. So I'll keep a distinguishing color. And we also distinguish its name now as the anomeric carbon. So that's the anomeric carbon. And then we can go ahead and fill in the rest of the substituents in the diagram. So, a hydroxyl group and another, and another. And we call this diagram a Haworth diagram. So, Haworth diagram. And the Haworth diagram doesn't show us the actual configuration of the ring, because in reality, six-membered rings are gonna show up in a more-stable chair shape. But it is beneficial in telling us which substituents are above or below the ring. So to keep this convention straight in my mind, I remember the phrase, "Downright uplefting." So, "Downright uplefting." Kind of a play on, I guess, the phrase, "That's downright uplifting." But, "Downright uplefting," because as I fill in the substituents, those on the right side of the Fisher diagram will point down, and those on the left side of the Fischer diagram are gonna point up. So we can actually see that that one's up, and we'll make sure that this numbers off right. This one's up as well, And maybe we'll start numbering with one, two, three, four, five, six. And we can do that over here. This would be one, two, three, four, five, six. So our three carbon in the Haworth diagram is pointed up, and our three carbon on the Fischer diagram has its substituent on the left. So down, right, up, left. And as we get to the last carbon group, which kinda forms this tail down here, I remember that if it's a D-sugar, that group is gonna point up in the Haworth projection. So this is a D-sugar. And you can see in the Haworth projection that this last carbon points up as well. And really this is gonna be the case for a lot of sugar chemistry that you deal with, because again, we're entomatically programmed to digest D-sugars, so we often end up with this last group pointing up. Now the last thing I want to show you is the chair confirmation, so the chair confirmation. And that's because this is the kind of diagram that's gonna give us a sense for the actual configuration of D-glucose. But it really does just follow suit with the Haworth projection as far as the substituents being above or below the ring. So let me just kinda keep filling in the substituents here. I'll number them off, again, just so you can kind of see that there's some consistency here. So we've got one, two, three, four, five, six. And again this three-carbon right here is the only one with the hydroxyl group pointing up. And I guess I better change the color of our one-carbon, to keep that consistent, as well. Now I didn't indicate the position of the anomeric carbon's hydroxyl group yet, because I think it makes more sense to show it in this diagram. Remember that the original nucleophilic attack by the oxygen way back over here, that could've created two different products: one with an R configuration about the anomeric carbon, and the other with a S configuration. So that last hydroxyl group can actually be in two different positions. One one hand, the hydroxyl group would be cis to the last carbon in the equatorial position. So it'd be cis to this last carbon over here. And it's in a equatorial position. And we call this the beta anomer. Then on the other hand, I guess, it can be trans to that last carbon group, which would place it in the axial position down here. So I guess it could be down here in the axial position, and we call it the alpha anomer when the hydroxyl group is in the axial. And I kinda remember that a little bit easier: alpha for the axial position of that substituent. And I guess I've also heard that if fishies are down in the sea and birds are up in there air. So if that helps you keep them straight, you might be able to use that also. Now you gotta remember that what caused this ring to close in the first place, was some amount of acid or base. And the amount of acid and base in water is actually kinda capable of doing that. Because that's what facilitated this ring-closing process in the first place. And in water, the ring can actually open and close spontaneously. And when it opens up, the C1 and C2 bond right here can actually rotate, and when it closes again you can form either the alpha or the beta product. So this thing is constantly opening and closing to form the two different products. And we call that process, where it opens, and rotates, and closes again, mutarotation. So this thing is mutarotating in the water at all times. So mutarotation. And the outcome is that we end up with both configurations, the beta and the alpha, in the equilibrium concentration. So for glucose, that's gonna be about 36 percent alpha, and about 64 percent beta. And the reason that the alpha configuration is less favored in equilibrium for glucose, is because the transpositioning of the hydroxyl group creates some stearic hindrance. But this is pretty individualized for different sugars. So, I guess the most general rule, I suppose, that you could apply to all cyclic sugars, would be to say that the beta anomer, again anomer, that the beta anomer is the one with the anomeric oxygen, in the cis position with respect to the last carbon.