Cyclic Structures of Monosaccharides - Video Tutorials & Practice Problems
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concept
Cyclic Structures of Monosaccharides Concept 1
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Video transcript
Here, we're going to say that monosaccharide exist as cyclic hemi ays in aqueous solutions. Now, Cyle takes place when the penultimate alcohol reacts with the C one aldehyde group. If we take a look here in the center, we have the a cyclic monosaccharide in the form of D glucose. And we're talking about the penultimate alcohol. So we're talking about this alcohol here. This alcohol is what's going to help to make this uh o within our structure. Now, from this ac form, we're able to create two possibilities. One where the oh is pointed down when it comes to this hemi acetyl carbon and one where it's pointing up. But how do we know what they are called? Well, this is where the term anim animal comes into play. An animal. These are just epimer produced by cyclist of monosaccharide. Remember an epimer means that we have the same configuration at all chiral centers except for one place. If we were to look, all the chiral centers are the same everywhere except for carbon number one, because that always could either point down or point up that gives us two beers or anim of each other here we're gonna say alpha animal. This is where the where to focus on this. Let's focus on this. This is where the oh group and the C h2o H group are on opposite sides of each other. If we can see the oh group is pointed down here, but the siege to each group is pointed up, they're opposites of each other. So this would be alpha D glucose and then beta anime. This is where anomic oh and C six C h2o H are on the same side. This will be beta. Now remember carbon number one, we can call it the hem ayl carbon, but we can also call it the anomic carbon because it can give us two possibilities. The oh could orient itself down giving us alpha or could orient itself up giving us beta. These would be epimer or ANAs of each other. And so just remember we're going from the non cyclic form of glucose. And once we put it in an aqueous environment, it's gonna coil up to form either this ring or this ring as its dominant forms. So keep that in mind when we're talking about different types of cyclic chem ays that originate from an A cyclic mono monosaccharide.
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example
Cyclic Structures of Monosaccharides Example 1
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Video transcript
Here it says drone Hallworth projection for beta D galactose. All right. So step one is we're going to number the Fisher projection. So 12345 and six and rotate it clockwise to turn it on its side. So now it's 12, 345 and six. Notice that we're starting out with de Galactose, we don't get beta until it's in its cyclic form. Step two. Now is we're going to curl the siege 20 group clockwise, keeping the Carbonel group in the far right corner. So here we're gonna kind of like bend this structure. It's linear right now. We're gonna bend it a little bit curve upon itself so that my C H2O H group and my Carbonel group are in similar positions, similar area with each other. Now, this is gonna be a little bit tricky. We're going to rotate C five. So carbon number five here. So C H2O H faces up, bring oh group close to the carbonel group, right? So basically we're gonna have this oh And this C H2O H moving. So this Coh is gonna move to where siege 20 is. And then see Siege 20 is gonna move here so that we get now S H2O H up here and then the oh rotated over here and here we're gonna close the ring to form the cyclo chemic cy and assign alpha or beta to the atomic oh group. Remember the atomic group is carbon. Number one. It is the carbon. That was the carbonel group here, right? So basically this oh and this carbon are gonna form a bond with each other. This bond here, oxygen ideally wants to only make two bombs. So it's gonna lose its H and we're gonna say that this double bonded O is no longer gonna be double bonded because Carmen can't make five bonds. It's gonna transform into an 08 group. We said here beta which means that the oh group has to be on the same side as AC H2O H group. So oh would be up here oriented up here and this age is still around and have to be down here. So this year will represent our beta de galactose structure.
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Problem
Problem
Draw a Haworth projection for α-D-altrose.
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Problem
Problem
D-ribose is an aldopentose sugar that is found in the DNA. It commonly exists as a five-membered β anomer. Draw D‑ribose in its cyclic hemiacetal form.
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