D-Fructose can also form a six-membered ring. Draw the β anomer of D-fructose in the six-membered ring form.

Question

Accepted Answer

Hi, everybody. Let's look at our next problem. It says draw the beta anmer of D tao in its six membered cyclic form. So we need to look at what the straight chain or fish or projection as of the sugar. So we need to look up taos and we would see that it is a keto hexose. So six carbons with a ketone rather than an aldehyde. So that means we have a ketone group of carbon two. So we'll draw our vertical line. We know we have just AC H2O H A carbon one since it has a ketone, not an aldehyde and then six carbons, our carbon six will also be AC h2o H and then we have four internal carbons, but C two is the ketone group. So draw that double bond to an O at the carbon two position and then lines for carbon 34 and five. And then we need to know for Taga tose that at C three and C four, we have hydroxyl groups on the left. So put those in and then since it's ad tagged dose, we know that of course, our C five hydroxy group must be on the right, then I'm just going to add our numbers for our carbons here to help us keep track and add in our hydrogens. So we have our Fisher projection. Now, we need to think about how the ring will form. Well, this one can be a bit tricky because we're used to seeing internal hydroxyl groups react with our anomic carbon. But in this case, if it was the hydroxyl group on C five, that would only give us a four membered ring and or excuse me, a five membered ring. And we're told we need a six membered ring. But if I'm starting with C two and I need a six membered ring, then it has to be the oxygen or the hydroxyl group on carbon six that reacts that's different than we're often used to seeing. So that oxygen on carbon six is what's going to react with our carbon two. So let's draw our six membered ring and then think about where our groups are gonna go to make a beta animal. So we draw the top bar of our ring with oxygen on the top, right? As is the convention draw the rest of the oops, rest of the Hexagon there. And then we know we put our anomic carbon on the right and a little red dot For that, on the right side, our anomic carbon in this case is going to be C two rather than C one. So we need to be careful about that. Now, see one, I'm not going to put on here yet because of course, I have to figure out which way it's pointing in the alpha and beta forms because I need the beta anma. So leave C two and C one alone for right now. But we'll go around and look at our other carbons. So starting with C two as of my anomic carbon moving clockwise, then I have C three and that oh group is on the left. So I know that must be pointing up. So there is C three of the hydroxyl group pointing up C four also on the left. So also pointing up and then C five goes to the right. So that is pointing down. Now again, this is where it looks a little weird to us given the sugars that we're often used to seeing here C six within the ring. Well, it has no hydroxyl group. Its hydroxyl group has reacted with C two. So C six just has two hydrogens on it. So that definitely looks weird compared to the sugars you're used to seeing. OK. Well, here comes the issue. I'm used to referring to C this position SC five. I'm used to referring to C six to decide how my alpha or beta is pointing because we often think of, you know, if it's, if it's uh trans, if it's opposite to that carbon six C H2O H group, then it's alpha. If it's CIS or on the same side, then it's beta, but we don't have ach two H group sticking up out of the ring in this case. So then we need to remember the sort of foundational rule to get to alpha and beta, which is a bit tricky because it feels backwards to us compared to the way we usually think about alpha and beta. But what we do is we look at the highest numbered chiral carbon. So in that case, that's C five. So right C five and which way is the hydroxyl group, you know which side is it on in the straight chain form, we have to specify that because many of our sugars like glucose, that oxygen is going to be within the ring when you make the cyclic form. So in this case, which side is it, it's on the right side, obviously, it's a deform. So that's on the right now, let's look at our anomic carbon. So anomic carbon, if the hydroxyl group is in the same, on the same side as the hydroxyl group, on our highest number, croc carbon C five, that's going to be the alpha form. So again, that is confusing. It's opposite from the way we often think about. But remember that we're comparing hydroxyl groups and then on the on the anomic carbon, if the hydroxyl group is on the opposite side, the left side or up in the ring, that will be the beta form So we know we need the beta animal. So that means the hydroxyl group on C two must go up to be on the opposite side as the hydroxyl group on C five. So I'm actually going to highlight those groups in blue. Actually, I don't like blue. That's a little too dark and pink to keep in mind they're on the opposite side. Said the hydroxyl group on the opposite side to make them beta. Again, it's because we're comparing hydroxyl groups, not the hydroxyl group with that C H2O H. So we only kind of think about it this way, we have a kind of unusual ring as we do here. So now that we've placed our hydroxyl group on C two, my little two in there, now we can place carbon one. So then on pointing down from C two, we have C H2O H that is carbon one. So we've got our final answer here, the beta AER of D Tao in its six membered cyclic form. So we have the oxygen top right and then going clockwise. C two has a hydroxyl group pointing up and AC H2O H which is carbon one pointing down then carbon three with the hydroxyl group pointing up carbon four with the hydroxyl group pointing up does carbon five with a hydroxyl group pointing down carbon six with no hydroxyl groups joined back to the oxygen. So there is our final diagram of our beta form. See you in the next video.

D-Fructose can also form a six-membered ring. Draw the β anomer of D-fructose in the six-membered ring form.

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Key Concepts

Anomer

Cyclization of Sugars

Haworth Projection