a. In an aqueous solution, d-glucose exists in equilibrium wit...

Question

a. In an aqueous solution, d-glucose exists in equilibrium with two six-membered ring compounds. Draw the structures of these compounds.

Accepted Answer

All right. Hello everyone. So this question says that D galactose predominantly exist as six membered ring compounds in an aqueous solution, draw the structures of these compounds. And here we have the structure of D galactose on the left side of the street. All right. So first and foremost, recall that a six carbon cyclic sugar is referred to as a pin nose. But to be more specific, a py nose ring refers to a six member ring in which five of the atoms in the ring are carbons and one of them is oxygen. So in order for cy ization to take place the hydroxy group of carbon number five in the hexose, right in the parrot chain is going to attack the carbon carbon. In other words, position one, this means that in the cyclic form of our sugar carpet, number one is now going to be sp three hybridized and it's going to be Cairo. And so because of that newfound stereo chemistry, because of the fact that carbon one is going to be P three hybridized in the structure of the ring from there, we can go ahead and talk about animals. Now, an differ from each other in the positioning of that hydroxy group of carbon. Number one, because recall that after cyc conversation, the carbon oxygen on position one is converted into a hydroxy group. And so just to go ahead and clarify the reason why the question is mentioning multiple different six member ring compounds is because de galactose is going to form two distinct anomous. This is because the hydroxy group of carbon number one can be either above the ring or below the ring. And so that's going to differentiate between both animals here. So before we go ahead and talk about those enemas and how to go ahead and position the oxygen or the hydroxy group of carbon. Number one, let's start with the general procedure. How do you draw a pyre? No ring? Well, first lets go ahead and start by drawing a six member ring in which five of the atoms are carbon and one of them is oxygen. Now here, a ring structure is also known as a Hayworth projection. And so I am going to add some shading to our first three carbons here just to showcase three dimensionality. But recall also that in a conventional Hayworth projection of carbon number one, that is the anomic carbon is going to be depicted on the far right side. And so we can go ahead and finish labeling all of our carbons going in a clockwise direction. So now that we have our base, now that we have our ring. Let's talk about substituents. Now, in the structure of a pure nose ring, carbon number six of the parent chain is going to be treated as a substituent branching off of carbon five in ad sugar, specifically the substituent on carbon number five is going to be above the ring. So I'm going to go ahead and draw precisely that. So for this question, we're going to go ahead and leave carbon one until the end. I want to go ahead and talk about our other stereo centers that is carbons 23 and four. When it comes to those positions, converting a Fisher projection into a Hayworth projection. That ring structure. The idea is to imagine taking the Fisher projection and lying it down on its right side. This means that hydroxy groups to the left point upwards and hydroxy groups to the right point downwards. So looking at carbon number two, for example, in the fissure projection given for the galactose, its hydroxy group is pointing downwards or excuse me is pointing to the right in the fisher projection, which means that it points downwards in the Hayworth projection in the pyre nose ring. On the other hand, for carbon number three, that hydroxy group is pointing to the left, which means that it should point upwards in the Hayworth projection. And the same is true for carbon number four or the hydroxy group on carbon number four. And so here for both an animal in other words, both six membered ring compounds that exist from D galactose, all of these positions are going to stay the same. And so the difference between them and the reason why they're called animas is going to be the positioning of the hydroxy group on carbon number two. So now let's talk about alpha versus beta, an animal, the alpha ener just to go ahead and start us off is going to have the hydroxy group of carbon number one, pointing in the opposite direction relative to the substituent on carbon number five. So in both an animal here, the substituent of carbon number five is pointing above the ring. And so the alpha anmer is going to have the hydroxy group. Carbon number one is going to have its hydroxy group pointing downwards so that these two substituents are on opposite sides or trans to each other. This means that the beta anmer as those two substituents this to each other. So if the substituent on carbon number five is pointing upwards, so too will the hydroxy group of carbon number one and there you have it here, we have both an animal. And so with that being said, thank you so very much for watching. And I hope you found this helpful

a. In an aqueous solution, d-glucose exists in equilibrium with two six-membered ring compounds. Draw the structures of these compounds.

Key Concepts

Anomeric Carbon

Haworth Projection

Mutarotation

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