Draw the product of the following 1→4 condensation and name the glycosidic bond:
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Question

Draw the product of the following 1→4 condensation and name the glycosidic bond:

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Accepted Answer

Hi, everyone. Let's take a look at our next problem. It says draw the product and name the type of gly acidic bond formed in the following 1 to 6. So not the number two, but one arrow six condensation reaction. And we have two mono sac rides here, both six carbon molecules with a six membered ring. So one of the carbons outside of the ring and we know our linkage is from carbon one on our first mono sac ride to carbon six on the second. So let's start with our first monomer. All of these two are identical. So we don't need to keep them straight and find carbon one, which by definition will be the anomic carbon. So the one with bonds to two oxygen. And so that's going to be this one on the right side, which is typically how these are drawn. That's carbon one on our first. And then given that, that's carbon one. When we look at our second molecule starting with carbon one, an americ carbon, we just go around the ring 234. So we're on to carbon five and now we have to go outside of the ring for carbon six. So carbon six is this carbon outside of the ring? So that's a little, you know, different looking compared to often what we see in these diess. So there's our uh I'll just highlight the carbon there, there's RC six. So let's draw the product of the condensation reaction. Well, to name the bond, we have to know by the name it alpha or beta. If it's a bond between the oxygens, the hydroxyl groups on two anomic carbons, then you have to name them both but only one anomic carbon is involved. So let's look where the hydroxyl group is on this anomic carbon. It's below the plane of the ring. And we compare it to the location of the C H2O H group. On the other side of that ring, oxygen, these two groups are on opposite sides of the ring. So I went to right opposite. And the way I remember it is opposite, starts with a vowel like A for alpha. So when these two are on opposite sides of the ring, then the position of this a group is alpha. So our gly acidic bond that will be formed is going to be called alpha alpha and then open parentheses. And then one and then the arrow to the right six glyco bond alpha, 1 to 6 Glyco bond. So that's going to be our first answer, name the bond and then let's draw the product. So we'll draw our first ring it switch other pen line across the top oxygen on top. Right. And then we draw our hexagon. I'm going to go sort of backwards around the rain because I'll need to form the bond with that hydroxyl group on carbon one. So starting with the oxygen in the ring, I'm going back to C five where I'll add my C H2O H group, that's carbon six on carbon four. I have a hydroxyl group going up above the plane of the ring. Carbon three, another one going up, carbon two, that hydroxyl group goes down. And then I've worked my way around to carbon one. So I saw that, that hydroxyl group went down on the original molecule. It's an alpha molecule. So in this case, it will also go down. Now we often draw the glyco cytic bonds or most of them draw with our bonds sort of going out diagonally to the linking oxygen and then on to the next molecule. But in this case, it's not linking to a carbon within the ring. It's linking this external carbon. So we can just draw this going straight down and then we'll draw a dag in a line from that oxygen to the C H2O H that was carbon six. So we lose the hydroxyl group there or we lose a water molecule. So we draw water coming off here. H2O. So we lose one hydrogen and 10 group and we're left with this oxygen and this ether linkage. So this oxygen is linked to a carbon with two hydrogen CH two and then we go down. So there's the CH two group, that little junction or that vertex in my bottom line there just gonna roll up a bit since I have to go down for my next part, like a little more room for the diagram. And then that is carbon five. So we've drawn, we've got carbon six as that ch two within that bond. And then carbon five is within the ring, draw the ring oxygen or hexagon, which for some reason, I drew kind of stretched out there and then draw all my hydroxyl groups. So again, now, carbon one is over here on the right. It's got hydroxyl pointing down carbon two pointing down carbon three going up above the plane of the ring and carbon four down. So we've drawn our product here circle that surround it and our gly acidic bond, our alpha 1 to 6 gly glyco bond. Is that bond in the middle there? Linking the two monomers? See you in the next video.

Draw the product of the following 1→4 condensation and name the glycosidic bond:
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Verified Solution

Key Concepts

1→4 Condensation Reaction

Glycosidic Bond

Product Structure