The sugar α-D-Mannose is a sweet-tasting sugar. β-D-Mannose, o...

Question

The sugar α-D-Mannose is a sweet-tasting sugar. β-D-Mannose, on the other hand, tastes bitter. A pure solution of α-D-mannose loses its sweet taste with time as it is converted into the β anomer. Draw the β anomer:

Accepted Answer

Alright. So this practice problem says the sugar alpha d mannose is a sweet tasting sugar. Beta d mannose on the other hand tastes bitter. A pure solution of alpha d mannose will lose its sweet taste with time as it's converted into the beta animer. Draw the beta animer and so really that's all we need to do in this box is draw beta D Mannose and so recall from our previous lesson videos that because mannose has 2 ends in it, that reminds us that mannose is going to be the C2 epimer of glucose and so, what we need to do is recall the structure of glucose and then draw the C2 epimer And so, we know that glucose is going to be an aldohexose. So, it's gonna have an aldehyde group at the very top here, and it's a hexose, so we know it's gonna have a total of 6 carbon atoms. So, we can go ahead and start to draw those. There's 3, 4, 5, 6 carbon atoms, and of course, the last one is gonna be a CH 2 O H group. And then we know that, the pattern of alcohol groups on glucose kinda looks like it's given us the fingers. And so we know that the c 2 hydroxyl will be going to the right, the c 3 hydroxyl will be going to the left, and then of course, the c 4 and c 5 hydroxyls are going to the right. And then everything else we know is gonna be filled in with hydrogens here. And so, this is going to be the structure of glucose, in fact, the structure of D glucose. So if we wanna draw the structure of mannose, again, the two ends tells us that this is gonna be the c 2 epimer. And so carbon 2 is right here. So the only thing we need to do is redraw d glucose' structure and essentially, change this highlighted position. So instead of having the hydroxyl group pointing to the right, we're gonna have the hydroxyl group pointing to the left. And so, we end up getting this structure right here. And so this is going to be the C 2 epimer of glucose, which is going to be mannose or D mannose. And so, of course, if it's telling us, that this is beta D mannose, and beta we know is the configuration of the anomeric carbon, so this is suggesting, the beta itself is suggesting, that this is going to be the cyclic form of beta D Manose. And so, what that means is now we need to convert this linear Fisher projection into the Haworth projection for our final answer. And so, what we need to recall at this point, is the phrase, up, lefting, so, up, lefting and, the other, word, down, right. And so, of course, we know from our previous lesson videos that Beta D Manose is going to form a Pyronose and so, what we can do is draw in the Pyronose, confirmation here so there is our 6 membered ring and, of course we know that there's gonna be a CH2OH group up here and so, numbering this will be our anomeric carbon, so this will be our carbon 1, 2, 3, 4, 5, and 6, and just for keeping the numbers, keeping track of the carbon atoms, we'll also number these carbons, so 1, 2, 3, 4, 5, and 6. And so this is telling us that this is the beta configuration. And so we know that that's gonna tell us the configuration of the anomeric carbon. And so, if it's beta, then we know that the 2 bumps are gonna be on the same side, so that means that the alcohol group here must be on the same side. And so, we expect the alcohol group on the anomeric carbon to be going up in the same direction as the highest numbered carbon. And so, this is going to be the position of the alcohol group on the anomeric carbon, then we can take a look at the C2 carbon and notice that it's going to the left, so we also expect it to be going up on the C2 carbon. The hydroxyl group will be going up, so we can draw hydroxyl group going up here. And then of course taking a look at the C3 hydroxyl group, notice that it's also going to the left, which means that it's also going to be going up. So we can go ahead and draw in the hydroxyl group going up here on the c 3 carbon. And then taking a look at the c 4 hydroxyl group notice that it's going to the right, so we expect it to be going down on the c c4, carbon. So, the c4 alcohol group, we expect to be going down. And so really this is it. This is going to be the beta d mannose anomer that we were looking for. And so, we don't need to fill in the hydrogens here since they are assumed but again, all carbon atoms are gonna have 4 bonds. So, we would expect hydrogens to be coming off but, this is really the simplest answer that we can provide to this answer. And so, that is the end of this practice problem and I'll see you guys in our next video.

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The sugar α-D-Mannose is a sweet-tasting sugar. β-D-Mannose, on the other hand, tastes bitter. A pure solution of α-D-mannose loses its sweet taste with time as it is converted into the β anomer. Draw the β anomer:

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