Draw the ⍺ and β anomer of D-talose in pyranose ring form:

Question

Structural representation of D-talose in pyranose form, showing ⍺ and β anomers with hydroxyl groups.

Accepted Answer

Hi, everybody. Let's take a look at our next problem. It says provide the structure of both animals of the monosaccharide shown below in pino ring form. And our monosaccharide is D gulose, which we see as an aldose. So it has that aldehyde as carbon one. And then if we number our carbons going down the straight chain form, we see we have six carbons total ending in that ch 20 And then in terms of our internal hydroxyl groups, carbons 23 and five have the hydroxyl groups on the right and carbon four has it on the left. We're going to make a pinos ring. So that means a six membered ring. But recall that our anomic carbon is one of the things in the ring. So there will only be five carbons in the ring. So carbon six will be left out. Now, if you don't remember how to transfer our straight chain structure to a ring structure, um we can sort of do it in a step wise fashion by first turning the structure 90 degrees to the right. So that would end up with a horizontal line to be the backbone. Carbon one is going to be on the right side with its aldehyde group and then carbon six with its C H2O H, it's on the left side and then we draw little vertical lines to be our internal carbons. So number them 234 and five. And then all the oh groups on the right side are now pointing down. So 23 and five are pointing down and four is pointing up. So that's pretty straightforward. We could do that in the ring. But sometimes it gets a little tricky is remembering the orientation of that C H2O H group with carbon six when you form the ring. So drawing it like this and going this way is a nice way to remember that. So this is kind of like, you know, think of drawing a caterpillar with that carbonel group as its head and now imagine its tail carbon six sort of coming up and curving over its head. So now we'll draw this as if it's starting to form a ring. So we don't have a ring yet, but we draw sort of this top line. It will be the top of the ring and C H2O H carbon six is on the right side of that line. And then we know this is carbon five on the left side of that line, carbon five has an oh group going down. Draw that in there, carbon four is going to be the left side of what will be a hexagon. Carbon four is an O age group going up. Then we will be at carbon three, which is an O age group going down, draw the bottom line of the hexagon. That's carbon two, which has another oh group going down. And then we go up to carbon one, our anomic carbon, which again, we still have a carbonel group here. We haven't formed our ring yet. Now, the key is we need the hydroxyl group on carbon five to react with our carbon carbon. So to do that, we need to rotate the bonds around carbon five to end up with our oh group on the right. So that means we have to rotate these bonds 90 degrees to the left. That will mean that our C H2O H group with carbon six will be pointing up to put our oxygen from our hydroxyl group on the right. So if you can't remember how to position that carbon six, drawing it in this way, will help you, you have to imagine this rotation, see where that group ends up. And you can see that it will point up. So when I draw my final ring structure, so now it's a hexagon with oxygen in that ring at the top, right? And my anomic carbon on the right side. So that's carbon one. Now, I've got a hexagon here and again, carbon 20 group going down carbon 30 going down, carbon four, going up and then on carbon five, I have that ch 20 group. So now what are my two an animal? Well, I recall that hydroxyl group now there on my animatic carbon can point either up or down, that was a, you know, that was a double bond carbon. So those that oxygen could have attacked the carbon carbon from either top or bottom. So I get my two an animal. And how do I remember? Which is which? Well, if I put that hydroxyl group on my anomic carbon pointing up, it is on the same side of the ring as it C H2O H group. And when it's on the same side, I remember it as same is a cons, it starts with S which is a consonant like BB for beta. So this is the beta anim. So beta D Golos and then we can draw a ring. So I just drew that ring again. And now if I put that hydroxyl group on my anomic carbon, putting down, it's on the opposite side of the ring from that C H2O H and I think opposite starts with O, it's a vowel like a for alpha. So this is alpha de goos. So official terminology we can say here that if my Anno meric O group is, is CIS to the C H2O H group outside the ring, then that's my beta an animal. And if my Ant America age group is trans to that C H2O H group outside the ring, then it's the alpha de goos. But again, I like to remember same, it's equals beta because they're both consonants and opposite equals alpha since they're both vowels. So that's just my little shorthand. But again, we had to draw both an animal of the de Golos and we have them here. If our oh group is pointing up on the same side of the ring as our carbon six, that is the beta anor, our oh group is pointing down on the opposite side of the ring of our carbon six. That is alpha de goos. See you in the next video.

Draw the ⍺ and β anomer of D-talose in pyranose ring form:

Verified video answer for a similar problem:

Key Concepts

Anomers

Pyranose Ring Structure

D-Configuration