Answer the following questions regarding the following cyclic ...

Question

Answer the following questions regarding the following cyclic monosaccharide shown below:

Cyclic monosaccharide showing hemiacetal carbon for biochemistry chapter on anomers.

A) Clearly label the hemiacetal carbon.

B) The monosaccharide is a(n) _________ anomer.
a) Alpha (α).
b) Beta (β).

C) Draw the opposite anomer.

D) Draw the α stereoisomer that differs in the arrangement of substituents at C2.

Accepted Answer

Alright. So here we have a 4 part practice problem broken up into part a, part b, part c, and part d way down below. And it says, answer the following questions regarding the following cyclic monosaccharide shown below which is right here. And so part A says, to clearly label the hemiacetal carbon and so what we need to realize here is that the hemiacetal carbon is just going to be the anomeric carbon itself. And so, of course, we know that the anomeric carbon is gonna be the only carbon atom that's bound to 2 oxygen atoms. And so taking a look at this cyclic sugar over here, notice that the anomeric carbon is right here and it's bound to the ring oxygen and the oxygen atom of the hydroxyl group. And so if we want to clearly label this atom, we can go ahead and put an arrow here like this and just label it the hemiacetal or, in other words, the anomeric, carbon. And so we can move on to part b now, which says that the monosaccharide is a blank anomer and the answer options are either alpha or beta anomer. And so when we want to consider the configuration of the anomer, of course, we're going to look at the anomeric carbon. And so notice that the anomeric carbon has its hydroxyl group pointing upwards. And when it's pointing upwards in the same direction as the highest numbered carbon, So if we were to number these carbons, of course, we would want the anomeric carbon to have the lowest possible number. So this would be carbon, this carbon would be carbon number 1, carbon 2, carbon 3, carbon 4, carbon 5, and then of course, carbon number 6 would be this one up here. So now we can see that the hydroxyl group is pointing upwards in the same direction as the highest numbered carbon. And so this means, that it's going to have the beta configuration. Since again, the bumps here of the beta are on the same side just like the hydroxyl and the highest number carbon are in the same side. So, what we can say is that the answer here is going to be B beta is the correct answer for part B. And so now, moving on to part C, it wants us to draw the opposite anomer and so if we want to draw the opposite anomer, the opposite anomer is only going to differ in the configuration of the anomeric carbon. So, that means that we just need to draw the same exact structure that we see here And so when we do that, we get this structure here. And again, we only wanna change the configuration of the anomeric carbon. So we're only gonna change the direction that the alcohol group is pointing in. Instead of having the alcohol group pointing upwards in the same direction of the highest number carbon, we want it to be going downwards and reaching towards the ants and so we can go ahead and put the hydroxyl group down below and put the up here. So, now, we have this is going downwards and the highest numbered carbon is going upwards. So, that makes this the alpha anomer and that's it for part C and so moving on to part D, notice that it wants us to draw the alpha stereo isomer that differs in the arrangement of the substituents at C2 which is the C2 carbon And so if we want the alpha stereo isomer, that means that we want the one that we just drew. So, all we need to do is, copy the same exact structure that we have here and we can paste it here and so, we want to change the substituents at carbon number 2, the C2 carbon and so, recall that the anomeric carbon here is carbon number 1 and so the C2 carbon is going to be this carbon right here. So, that means that we want to change the arrangement here. So, instead of having the hydroxyl group going up on the C2, we want the hydroxyl group going down. So we can go ahead and put the hydroxyl group, here, put it in green, hydroxyl group going down here, and of course the hydrogen atom we can have going up. And so really this is the only thing that we needed to do was change this, position here at C 2 and that is the answer for part d. And so that concludes our 4 part practice problem and I'll see you guys in our next video.

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