In its open-chain form, D-mannose, an aldohexose found in orange peels, has the structure shown here. Coil mannose around and draw it in the cyclic hemiacetal ⍺ and β forms.

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Hi, everyone. Welcome back. Let's look at our next problem. Galactose is a monosaccharide sugar, which can be found in milk. Given the open chain form of D galactose, draw its cyclic hem ay alpha and beta forms. So we have this open chain structure for the galactose, a six carbon sugar. So it's going to be a six ring or six atom ring with one oxygen and five carbons within the ring. How do we draw this? Given the straight changed structure? Well, we start by numbering our carbons, starting with the aldehyde carbon. So over the top, so we number them one through six and then we're going to turn the whole structure on its side. So we're going to flip it sideways clockwise. So I've drawn the backbone as a horizontal line. We have the carbon one on the right side with its carbon carbon and then behind it will be carbon two. So go ahead and put that some numbers there. We want to make sure to keep track of which carbon we're on because we need to keep straight the orientation of the hydroxyl groups. So the hydroxyl groups that were on the left will now be above the line and the hydroxyl groups that were on the right will now be below the line looking at carbon two, that hydroxyl group is on the right. So it's now down below the line. So then we go in and fill in all our other carbons. So we have all six of our carbons, carbons, two and five have hydroxyl groups below the line and carbons three and four have hydroxyl groups above the line. So then in the next step, we're going to see how our ring gets formed by curling this molecule around to see how the ring is going to form. And we do that by starting with carbon six and curving that around up and around over the top. So when we draw that note that we're going to have a hexagonal shaped ring, but it's not closed yet because we need to make some adjustments. So I'm going to draw the top line of the hexagon and then the left side, bottom line and right side. But I'm leaving it still open up on the top right side because that's where we'll form the new bond. And we wanna make sure we have everything lined up properly. So we'd brought carbon six up and around. So we'll write number six up here to keep track and then carbon one, it's now sort of pointing upward that right leg of the not yet closed ring. So we put that in to keep ourselves orientated. So carbon six, we see the CH 20 on the end here, here's carbon five to its left. So carbon five, we know has the hydroxyl group projecting downward. I'm going to leave off the hydrogen just so it's a little less cluttered here. Then carbon four, which is over on the left side has the hydroxyl group projecting upward from the above the ring. Then we're down going around to carbon three. So you can see why the numbering is so important because now we're turning around and making a ring. You just always want to make sure your hydroxyl groups are ending up where they're supposed to be. So carbon three, that hydroxyl group is projecting up and carbon two, that hydroxyl group goes down. Carbon one, we have to remember to add our carbon bond to that oxygen. So now we have the basic orientation of the ring, but recall that we need that oxygen in the ring. We want it in the back right side. I'm also going to add to my ring. This is going to eventually be a Hayworth projection. So I'm going to just make the s the bottom two side legs and the front line, front, bottom line thicker those wedges to show that that's projecting out of the plane of the paper or the screen here. So now how are we going to end up with that oxygen where we want it? Well, we're going to sort of rotate around carbon five call this is a single bond. So there's free rotation around it. It's not a chiral center since there are two hydrogens on it. Recall that the oxygen in the Carbone bond is going to be outside of the ring. That will be part of an a hydroxyl group. Once it gets reduced, the hydroxyl group on carbon five is going to be the interior oxygen. So what we're going to do is rotate around carbon 5, 90 degrees counterclockwise. So the bond to that carbon six will be going up out of the ring and our hydroxyl group will be going to the right, so that oxygen will be then be becoming part of the ring. So we'll draw that rotated form next. So I filled in the rest of the molecule that stays the same focusing here on carbon five. We now have the hydroxyl group projecting to the right and carbon six and that C H2O H projecting up out of the plane of the ring and note that where the carbon six is going to end up depends on whether that hydroxyl group on carbon five was up or down in this case, it was down. So carbon six ends up above the ring. That's important because it's important in determining the alpha and beta forms when we get to our final hemi aid ring. So now I've filled in the parts of the ring that will remain the same as we've already drawn and let's fill in the final completed ring structure. So where there was the hydroxy group, we now have our uh oxygen within the ring that is attached to the carbon carbon and the carbon one. So we'll put a little number there. So we keep track of that. See that carbon one otherwise known as the anomic carbon is the carbon that has the two oxygens bound to it. So it's got the oxygen in the ring. And then we just need to add the oxygen from the carbonell group that's been reduced. So remember that because this is forming a new single bond here with this oxygen that was in a double bond, you can have the oxygen going in either direction. So we could have our oh group. Let's start with a drawing where it's going downward below the plane of the ring. So now I've drawn the other version and here we put the hydroxy group on carbon one projecting upward above the plane of the ring. So there's our two animals there. So put on carbon one and the way we know whether to label them alpha or beta, we compare them to carbon six. That's C H2O H group. If the hydroxy group is on the same side of the ring as the C H2O H, then that is the beta form. So we'll write beta D galactose on our structure with the carbon one hydroxyl group going upward. And then if the hydroxyl group is on the opposite side of the ring from carbon six. That is the alpha form. So that structure is alpha D galactose. Now, that can be a little tricky to remember which is alpha and which is beta. But I sort of remember it by the two vowels. So alpha like A is on the opposite side. A and O are both vowels. So I remember that alpha opposite and then beta, it's like B A consonant like same, so beta, so B and S beta and same side, that's just the way I remember it. So there's our alpha and beta forms and how we determine which is which. So that is our two cyclic hemi Aidy forms of the galactose based on its open chain form. See you in the next video.

In its open-chain form, D-mannose, an aldohexose found in orange peels, has the structure shown here. Coil mannose around and draw it in the cyclic hemiacetal ⍺ and β forms.

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Key Concepts

Aldohexose

Cyclic Hemiacetal Formation

Anomeric Forms