D-Fructose can form a six-membered cyclic hemiacetal as well as the more prevalent five-membered cyclic form. Draw the α isomer of D-fructose in the six-membered ring.

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Hello everyone. And welcome back. The next problem says Talos is a monosaccharide that forms a six membered membered cyclic hemal known as its Pyin nose form. Draw the alpha anmer of D Talos in its py py form. So first, we need to know what de Taus looks like and we'll start with its straight chain structure. So go ahead and make sure you have looked that up or know that. So we've got this straight change structure here and we will start by numbering the carbon. So we can talk about the orientation of the hydroxyl groups because it's going to be important to keep that straight. So the first that aldehyde carbon is always labeled number one. And then as we go down, we see, carbon two has the hydroxy group on the left, carbon three, on the left, carbon four, on the left, carbon five has its hydroxyl group on the right. And then carbon six is that C H2O H at the bottom of the chain. So the first step in drawing its cyclic structure is to rotate this whole straight charu structure clockwise to the right. So now I draw my backbone as a horizontal line, I put carbon one on the right hand side with its carbon group. I'm going to leave the hydrogens off this sideways drawing just to make it a little less cluttered. Actually, I'll put them on, I'll let, I think about it. We'll leave them off in the next step. So then when I go back to carbon two, it's now to the left of carbon one. And if we call carbon two, its hydroxyl group is on the left. So when we rotate it counter clockwise, excuse me, when we rotate it clockwise, then that puts the hydroxyl group above the line above that backbone with the hydrogen below. So let's just fill in our other carbons. I can see then carbon three, that hydroxyl group is above carbon four. The hydroxyl group is above carbon five. Though the hydroxyl group is below the line and then we've got carbon six C H2O H over on the left. Now to start to think about the ring formation, we start starting with carbon six. We sort of curl that carbon six up and around to be the ring. So the way that works travel arrow. So we have a top line from the ring and that's going to be carbon six on the left side of that line, we're not going to close the ring yet. So we draw the left side of the ring and then the bottom of the ring and then just the um one at the bottom, right side of the ring. So it hasn't been joined into a ring yet. And we're ending with carbon one on the right side here. So number our carbons to keep them straight, we got our six carbons. Carbon one has the carbon bond to the oxygen there. Carbon six is that C h2o H. So now let's fill in our hydroxyl groups in the proper locations. So again, this is why this numbering is so important, especially because we're now going around in a circle and we've got some on top of the other rather than all in a straight line. So probably to be safe, we have to keep in mind if we start up at the top, right, which sort of feels the best, that's carbon six. So then we're going the other way, we're going backwards in numbering. So just be careful and make sure you're matching your numbers, which is why I draw them in both places. So then carbon five, the hydroxyl group is below the ring and here's where I'll drop the hydrogens. Otherwise it's going to get too cluttered here. So carbon five, the hydroxyl group is going down, carbon four, it is going up, carbon three, it's going up and carbon two also going up and now is where we need to recall that the carbonyl oxygen will be reduced to become a hydroxyl group and be outside of the ring, the oxygen that's going to be internal to the ring is going to be to highlight this in pink, the oxygen from the hydroxyl group on carbon five. So we need to rotate around carbon five to place that oxygen so that it will be come within the ring. So we're essentially going to rotate, recall we can do that. We've got a single, single bonds here. So there's free rotation and this is not a chiral center. So we can rotate everything 90 degrees counterclockwise, rotate that bond to carbon six. So it's facing upward and rotate the hydroxyl group so that it would be pointing to the right. So now let's draw after we rotate that. So I added in, I had forgotten to add in um in my previous drawing. This is going to be a Hayworth projection when we get this um ring drawn. So I've thickened into wedges, the lower part of that ring indicating that's projecting out of the plane of the screen. So just keep that in mind. So now I've redrawn everything that remains the same, leaving that top line empty, so I can fill it in the right way. So we rotated 90 degrees counterclockwise around carbon five. So now I have that top bar of the ring now has the hydroxyl group and the carbon six C H2O H is now pointing upward. So we'll label carbon six and five. So we can see that carbon one is going to be the, you know, form the ham acetyl because that's going to have bonds then to two oxygens, once we close the ring, now, when that oxygen gets reduced to become a hydroxyl group, it was in a double bond, it had trigonal planar geometry. Now it's going to be in a single bond, there will be tetrahedral geometry around that carbon one. So that hydroxyl group can have two different positions. It can either be going up or down above or below the plane of the ring. So those are our alpha or beta forms. So let's draw the ring and then figure out our question is only asking us for the alpha animal. So we're going to leave off that last hydroxyl group and then think about which is the alpha form. So we'll draw our closed ring. So I've got now the oxygen where it should be when we draw these cyclic structures should always be in that top right corner of the ring. And here we have carbon one on the right side where it always goes. So we just have to figure out which way is our oh group pointing on carbon one to be the alpha animal. Well, the way I always remember it is alpha, which is like a, so like a bowel means opposite. And that refers to the position of carbon six. So I put C six position and beta, we think of as B A consonant like S in front of same. And again, referring to the C six position. So alpha, opposite two vowels, beta, same two consonants. So we need the alpha ener. So it should be on the opposite side of the ring from C six. So remember that C six is projecting up out of the ring that C H2o H group. So the hydroxyl group must be on the opposite side and therefore must project down from carbon one below the plane of the ring. So there we have it, we've drawn our alpha anmer of D Talos in its pinos form the six membered cyclic he acetyl. And that means putting the hydroxyl group on carbon one down below the plane of the ring in the next video.

D-Fructose can form a six-membered cyclic hemiacetal as well as the more prevalent five-membered cyclic form. Draw the α isomer of D-fructose in the six-membered ring.

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Key Concepts

Cyclic Hemiacetal Formation

Anomeric Carbon

Furanose vs. Pyranose Forms