Draw the Haworth structures for α- and ß-D-glucose.

Question

Accepted Answer

Welcome back everyone. Our next question says, what are the Hayworth structures for the alpha and beta isomers of D gulose? So recall that the Hayworth structures are the structures that involve a ring. I'll just draw a little sketch here where you show the top three lines of, for instance, here I draw it hexagonal ring, six membered ring with single lines. And then you sort of do these solid wedges for the front showing that this ring is projecting out of the plane of the screen. So those are, that's a Hayworth structure. And let's recall how we draw those for sugars. So our sugar is D gulose. So we start with a Fisher projection that straight chain form so we can look up uh de goose. So that's going to give us a moment and I'll put that structure up. We have a six carbon sugar. It's an Aldo hexose. So if we number our sugars, I've drawn the fisher projection here. So carbon one is our aldehyde group. And then if we go down a number up to carbon six carbons, two and three have the hydroxyl group on the right. Carbon four has it on the left carbon five on the right, and then carbon six is not a chiral carbon. So we have our Fisher projection. And if you remember right away how to translate this into a ring, go ahead. If you need a little more of a um leave the process described in more detail. Let's walk through that. So we remember we start by rotating the whole structure 90 degrees clockwise. So we'll draw a straight line and put our carbon one on the right. Carbon six on the left and then our four lines here uh uh perpendicular. So then all of our uh groups that were on the right because we've rotated, it are now down below the line. All our groups that were on the left are above the line. So filling and remembering that this is carbon one on the right and carbon six on the left. So we want to keep track and make sure we're putting things in the correct order. So filling in on the top. So that's our groups on the left. So starting with carbon two, we have hhohh and then underneath the line, starting with carbon two. Oohoh. And again, if you remember how to, you know, you can just remember how to draw the ring by remembering that the groups on the right will be below the plane of the ring, the groups on the left above the plane of the ring. But if you can't remember that off the top of your head, this can help you to visualize it. And then the next step in visualizing that ring, we imagine this carbon six group kind of coming around to form the ring showing us how that hydroxyl group and the aldehyde group will react. So if I draw it before it's joined, so now I have carbon six C h2o H on the top, right, with its bond to carbon five, then I will draw it's not a ring yet, but I'm drawing it sort of curve, the backbone sort of curved around ending in my aldehyde group. So I've drawn the double blond to the O there. And if I number it to keep track, that's carbon one. So the ring is not closed yet, but I can see how my groups are going to go here. So again, carbon two, that's and I'm going to, before I draw those on to remind myself how this is going, I'm going to shade in those wedges on the bottom of my ring to remind myself that this is sticking out of the plane of the screen. So carbon two, then that oh group is going down. Carbon three, it goes down, carbon four goes up, carbon five goes down and then recall that we're going to need to rotate around carbon five because we need our hydroxyl group on carbon five to react with the aldehyde group. So we're going to rotate so that B carbon six, its C H2O H group will then be going straight up and then the hydroxyl group on carbon five is sticking out to the right where it can react to close the ring. So that's kind of why even though it's a little tedious, I like to draw this just to remind myself of where everything ends up. But if you can remember all that, that's great. You can draw that a little faster. So now our final ring, we have an oxygen on the top right of the ring, fill in my ring here it's now closed, go on my solid wedges, then we're going to fill in. We have carbon six sticking up C H2O H and then fill in our other groups. Oh Going up on carbon four, down on carbon three, down on carbon two, we need to draw the alpha and beta isomers. And this refers to which way the hydroxyl group on our anomic carbon or carbon one is going. So first, I'm just going to duplicate this drawing and then we'll discuss which way it goes for the two forms. So one second here, so that hydroxyl group can either go up above the plane of the ring or down below. And we use alpha and beta to describe its orientation with respect to that this carbon group in carbon six. So let's draw first the version where the hydroxyl group is going up. So when it does this, both the hydroxyl group, the on the anomic carbon. And this last carbon group here are both above the plane of the ring. They're on the same side of the ring. So I remember same starts of the consonant just like the letter B. So this is the beta form with that hydroxyl group on the same side. And then with the hydroxyl group going down below the plane of the ring. So now the we'll go ahead and highlight that in pink just to remember that these two groups that we're referring to and how they relate to each other. So now that hydroxyl group and the carbon six are on opposite sides of the ring and opposite starts with a vowel like the letter A. So this is the alpha version. So these are our two alpha and beta isomers or an animal. So we'll go ahead and put a box around that, that is our answer. We had to draw the Hayworth structures. So we've got those sort of three dimensional rings and we see that the alpha and beta versions are the with the two directions that that hydroxyl group can be above or below the plane of the ring. See you in the next video.

Draw the Haworth structures for α- and ß-D-glucose.

Verified video answer for a similar problem:

Key Concepts

Haworth Projection

Anomeric Carbon

α- and ß-D-Glucose