Draw the following sugar derivatives.
(a) methyl β

Question

Draw the following sugar derivatives.

(a) methyl β-D-glucopyranoside

(b) 2,3,4,6-tetra-O-methyl-D-mannopyranose

(c) 1,3,6-tri-O-methyl-D-fructofuranose

(d) methyl 2,3,4,6-tetra-O-methyl-β-D-galactopyranoside

Accepted Answer

All right. Hi, everyone. So this question is saying to show the structures of the sugar derivatives given below. Now, in this question, we're given four different sugar derivatives to work out here. All of which I'm going to read as we go through each part because I first want to address some of the terminology that we see in these derivative names because recall that the majority of carbohydrates are found as well. Theyre found in the cyclic for right. And they generally exist as either cyclic acetyl or cyclic hemi ays depending on the name used here, we can determine whether it's a five carbon ring or six, five member ring, six member ring, et cetera. Because first, we can differentiate between what's known as a Pinos and a Furano because both Pinos and Finos refer to a cyclic hemal. However, a Pinos oops is the six number ring, whereas a few nose is a five member ring, but both of which are hemi atos. But we can also modify this terminology a bit further depending on whether or not it happens to be an ace top because the term used for sugars in the form of acetyl is a glycoside. So when a pure no sugar is in the acetyl form, right, it's an acetyl instead of a hemi acetyl, it's referred to as Puran oide instead. And similarly, if a Furin nose is converted into a glycoside or converted into an aceto, it's referred to as Furino instead. So with this in mind, I want to go ahead and start with part A which is ethyl alpha D glucan oy because the phrase Perano side at the very end implies that this is going to be a six member ring that is an acetyl. So it's a six member acetyl. And furthermore, the phrase Gluco Peroy indicates that this is the Pran side of glucose. So glucose is our base sugar. And additionally, it also happens to be D glucose because of the phrase D Gluco perianal site. So for part A, the first thing I'm going to do is draw the ring of the glucose. So here I'm going to draw my chair confirmation which should be in the pucker form with carbon one appearing on this bottom right corner here, meaning the oxygen in the ring should be behind it, right. So here is my Pucker chair confirmation. And so the nice thing about D glucose, if you recall is aside from the hydroxy group attached to carbon six outside of the ring, all of the hydroxy groups inside of the ring are going to be in equatorial positions. So that's what I'm going to go ahead and draw first making sure that they're in the equatorial positions except for the hydroxy group on carbon one here because that will depend. So for that, I'm going to refer to the rest of the name for part A here because here part A specifies that our gluc cyanide must be in the alpha configuration. Now, the alpha configuration, if you recall means that your hydroxy group is going to be on the axial position or in other words, pointing downwards, which means that the hydrogen on carbon one should be pointing upwards in the equatorial position. However, this is a Puran oy, which means that it's an acetyl and not a hemi acetyl, which is why the phrase eo at the very beginning comes into play, right? So here because it is ethyl alpha D Gluco Perano side, the hydroxy group on carbon one is going to be replaced with an oxy group instead because that is going to make carbon one here an acetal in the alpha configuration because the oxy group happens to be pointing downwards in the axial position. So now we can go ahead and get started with part B here and part B is 2346 tetra oeld Manop. So this time around it's a pinos, right, which means that we're going to have a six member ring that is a hemi acetal. And so our sugar backbone happens to be D manos based on the phrase D Manop pinos. So once again, it's a six member ring. So I'm going to be drawing a Pucker Chair confirmation and recall that when it comes to substituents, it may be easier to recall how a given sugar relates to the structure of D glucose. So recall that in the case of D manos, the configuration of the hydroxy group is going to change for carbon two only, right. So in the structure of D manos, my hydroxy group is going to be Axio instead of equatorial. However, right, the hydroxy groups on carbons three and four here are going to remain in the equatorial position. And then there's also the hydroxy group outside of the ring on carbon six. So now it's worth mentioning also that the alpha or beta configuration isn't actually specified in this case. So when it comes to the hydroxy group and hydrogen of carbon one specifically, I'm going to be attaching them both with these curve lines, right? Because this indicates that their orientation isn't necessarily specified because it's not mention whether they should be alpha or beta, whether the configuration should be alpha or beta. So one last thing are the substituents on the rings themselves because here we have 2346 tetra ethyl D Manop pinos. So that means that the hydroxy groups on carbons 234 and six are going to be replaced with oxy groups instead of hydroxy groups. So I'll be removing their hydrogens and it's also worth mentioning that the hydroxy group of carbon five is what allows for or what participates in the overall ring structure, right? So that cannot be substituted but all others except the one on carbon one here will be substituted. So they all become oxy groups instead. So next up here is part C which is 136 tri o ethyl D fructose. So part C here is going to involve a five member hem acetyl ring or five member cyclic hem acey based off the structure of D fructose. So for part C here, I'm going to go ahead and draw D fructose first. So here is the structure of D fructose. And what I want to point out here is that carbon one is actually outside of the ring, meaning that carbon two is the hemi acetyl carbon or excuse me, it's the anomic carbon. So once again, the alpha or beta configuration isn't actually specified. So I'm drawing these curved arrows or curved lines to showcase that the hydroxy or the, the hydrogen of the anomic carbon and carbon one are connected to carbon two, but their configuration isn't necessarily specified. So in this case, our substituents involve three epoxy groups on carbons, 13 and six, right? So in this case, the hydroxy groups of carbons 13 and six are each going to be erased and replaced with oxy groups instead. Oops. And here's my last. So that concludes part C and now lets go ahead and conclude here with part D. Part D being ethyl, 2346 tetra ethyl, alpha D galacto Perran. So the phrase alpha D galacto Perano side indicates that this is going to be a six member ring that is an acetyl and is based off the structure of alpha D galactose. So first I will draw alpha D galactose. All right. So here for part D is the structure of alpha D galactose in which the hydroxy group of the anomic carbon is in the axial position and therefore pointing downwards. Now, I want to also mention that the configuration or the structure of D galactose is different from D glucose on carbon or because on carbon four, the hydroxy group is axial instead of equatorial in alpha D galactose. And I also realized that for part C I did not or part B, excuse me, I did not draw hydrogen on carbon five here. So I just wanted to fix that. So now let's look at substituents. We've got ethel 2346 tetra o ethel alpha D galacto pereny. So it is a pranic side, meaning that carbon one is going to be an acetyl with an oxy group. But in addition, the hydroxy groups of carbons 234 and six are also going to be replaced with Foxy groups. So I will erase the hydrogens on my Anno carbon, right? I will erase the hydrogens on the hydroxy groups of each one that I'm going to go ahead and change. And so then I will go ahead and add the epoxy groups to all of them. Last one and there you have it. So here are your final answers or parts A through D. So I know this was a lot, right? But if you stuck around to the end, I thank you so very much for watching and I hope you found this helpful.

Draw the following sugar derivatives.
(a) methyl β-D-glucopyranoside
(b) 2,3,4,6-tetra-O-methyl-D-mannopyranose
(c) 1,3,6-tri-O-methyl-D-fructofuranose
(d) methyl 2,3,4,6-tetra-O-methyl-β-D-galactopyranoside

Key Concepts

Sugar Structure and Anomeric Carbon

Methylation of Sugars

Stereochemistry in Carbohydrates

Watch next