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Biochemistry

Learn the toughest concepts covered in Biochemistry with step-by-step video tutorials and practice problems by world-class tutors

9. Carbohydrates

Anomer

1
concept

Anomer

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So now that we're familiar with how mono Sacha rides sick lies in this video, we're going to introduce an Immers. So mono sack ride. Civilization also generates Alfa and beta animals. But one in the world are animals. Well, an Immers are literally defined as cyclic sugars that differ on Lee in the configurations of their an americ carbons and recall from our previous lesson videos. We already know that the America carbon is the Onley ring carbon that's attached to two oxygen atoms and the an American carbon can also be defined as the carbon atom that used to be the Carbonnel carbon before civilization. Now we know from our previous lesson videos that upon mono sack ride civilization, the an American carbon is formed and the an American carbon becomes part of either a heavy ass it'll or a Hemi Kitale. But what we have not yet mentioned is that upon mono sacrifice civilization, the animated carbon becomes a Chire ality center, and that means that it's going toe have two possible configurations. The first is the Alfa configuration, and the second is the beta configuration. And so this is what leads to the Alfa and, um er and the beta and um er, depending on the configuration of the an americ carbon. So before we actually define the Alfa and beta animals, let's take a look down below at our image and notice. Over here on the far left, we're starting with a linear or an open change Sugar d glucose, which notice has an alga hide group up here on carbon number one. And it has this highlighted hydroxyl group here on carbon number five. And we've got our muscle man here bending our linear glucose so that these two groups are in close proximity. So notice the hydroxyl group here on C five is in close proximity to the alga hide group here on carbon number one. And we know from our previous awesome videos that the alcohol group is going to act as a nuclear foul and attack this carbon eel group. But this alcohol group can actually attack this carbon eel group in two different ways. And that's what leads to these two different pathways. And these two different results that we see here and so this alcohol group can either attack the carbon eel group from the top side of the carbon deal or it could attack from the bottom side of the carbon. Neil. And so if the alcohol group attacks from the top side, then what we'll get is this animal that we see here, which is the Alfa Animal or Alfa D glue cope. Ira knows. And so what exactly is this Alfa animal? Well, this is gonna be when the an America Carbons hydroxyl group is on the opposite side of its highest numbered carbon. And so when we take a look at this Alfa and, um, er down below, notice that it's an American carbon is this See one carbon here and notice that it's alcohol group is going down below the plane of this ring here. It's pointing in a downwards fashion and notice that the highest numbered carbon atom is carbon number six here, and carbon number six is pointing in an upwards fashion. And so we have carbon number six pointing upwards, and we have the alcohol group of the an American carbon pointing downwards, And so they're pointing in these opposite directions. And so what you can see is that with Alfa here, the hydroxyl groups, uh, is of the animated carbon is reaching down for the ants. And so you can think Alfa is reaching down for the ants, the tiny little ants that air crawling on the floor. And so that's what helps me remember the Alfa Animal. Now, on the other hand, if this hydroxyl group doesn't attack from the top instead it attacks from the bottom. Then what we'll get is this other animal over here. And this is the beta animal or Beta de Luca piranhas in this case. And so, of course, the beta animal is going to be when the an American carbons hydroxyl group instead of being on the opposite side, it's now going to be on the same side of its highest numbered carbon. And so what you'll notice is this time the hydroxyl group is going upwards. It's going up in this direction instead of going down like what it was before. And that means that it's pointing in the same direction is the highest number Carbon carbon number six. And so what you'll see is that carbon number six and the hydroxyl group are both on the same side. And so what helps me remember that the beta confirmation is on the same side is that when you go and write beta, notice that the two bumps of this beta are on the same side as this pole here and so as silly as that might sound that will help you remember that when these two groups are pointing on the same side of the ring that it will be a beta confirmation. And so one thing to point out here is that, uh, these two animals the Alfa animal and the beta animal down below. Uh, these animals are absolutely not mirror images of one another. And so, as close as they might seem, they're not mirror images, which is why we have this broken mirror here to remind you of that. And again, you can see that this alcohol group of the animated carbon is reaching down for the ants. So it's gonna be the Alfa form. And, uh, this, uh, alcohol group of the an American carbon is on the same side as the highest number of carbon, so it's going to be in the beta form. And so it's important to be able to distinguish and recognize these two different animals that can exist for cyclic sugars. And so that concludes our introduction to animals. And we'll be able to get some practice applying these concepts as we move forward in our course, So I'll see you guys in our next video.
2
example

Anomer Example 1

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All right. So here we have an example Problem that's asking which of the following molecules are animals and we've got these five potential answer options down below. Now, recall from our last lesson video that animals are literally cyclic sugars that are identical in every single way except for the configuration of the an American carbon. And so when we took a look at option a comparing molecule one with molecule to, of course, we're gonna be comparing the numeric carbons first. And so recall that the America Carbon is going to be the Onley carbon atom that is bound to two oxygen atoms. And so for sugar one theano Mary carbon is this carbon here and for sugar to the an America carbon is this carbon here? And so notice that the alcohol group on the an American carbon is pointing in the same direction away from the highest numbered carbon. And so what that means is that both of these are reaching down for the ants, and so they're going to both be in the Alfa configuration. But for them to be animals they have toe have opposite configuration. So these two are not going to be animals of each other. So we go and cross off option A. And while we're looking at molecule number one might as well move onto Option C, which is also looking at molecule one and comparing molecule 12 molecule five. And this time, when we look at the animator carbons of both of these sugars, noticed that they're going in opposite directions that sugar one is in the Alfa configuration, as we already indicated. And sugar five alcohol group on the animated carbon is going in the same direction as the highest number. Carbon making a beta configuration. So right s so far. In comparing these two, they seem to be animals. But remember, this must be the Onley way that these two sugars differ from each other for them to be considered animals. But notice that these two sugars differ from each other and mawr than just the configuration that we've already noted. Notice that this carbon over here has its eye drops were going up, whereas this carbon here has its eye drops were going down. So these two carbons differ from each other than mawr than just the configuration of the animated carbon, which means that they are not going to be animals of each other. And we can go and cross off Option C and again, while we're looking at molecule one, might as well move on the option he here, which compares molecule 12 molecule four, and notice that molecule one is a pirate nose with a six member dring, whereas molecule four is a fury knows with a five member drink. So already there's no way that these two can be animals of each other, so we could go ahead and cross off option eat. So now we're between either option B or option D and notice Option d comparing molecule three here to molecule four. When we look at the and America Carbons, notice the animated carbon is here and here notice that they both have the same configuration of beta. And so again, for them to be animals, they have to have opposite configuration. So they have the same configuration. And that means that they're not going to be an Immers, which of course, leaves us with option B molecule to and molecule five and so looking at molecule to and molecule five notice that they're an americ carbons are in opposite configurations that this one is reaching down for the ants. So it's gonna be Alfa, and this one here is reaching up to the same side is the highest number carbon. So same side is gonna be beta. So when we compare the rest of both of these molecules, so the rest of this and the rest of this are identical to each other. And so the only way that these two differ from each other is in the configuration of the animated carbon. And that's what makes them animals. So we can go ahead and indicate that the correct answer for this problem is option B. And that concludes this practice problem. So I'll see you guys in our next video.
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Problem

The ___/___ configuration of a monosaccharide is determined by the ________________ of the chiral carbon furthest from the carbonyl group, while the ___/___ anomers are determined by _______________ of the anomeric carbon.

4
Problem

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

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.

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