Look at the open-chain form of D-mannose and draw the two glycosidic products that you expect to obtain by reacting D-mannose with methanol.

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Hi, everybody. Let's look at our next problem. It says, what are the two glycolytic products formed by the acid catalyzed reaction of D galactose with ethanol. And we're given a straight chain structure of D galactose here. Well, first, let's think about what kind of reaction will occur. How do you get a gly acidic bond? Well, this occurs, we do have acid catalysis and this will occur when we have a hemi aid or Hema kyl group on one molecule that reacts with a hydroxyl group on another molecule. So our two molecules are sugar and ethanol. So ethanol, of course is only two carbon. So ch three C h2o H in an alcohol group. So that must be the alcohol group. So we need in this case, he or hemi Aidy group because our sugar here is an aldehyde. Well, what's the hem acetal form of a sugar? That would be the cyclic form. So let's draw the cyclic form of de galactose. So recall how we do that. We need to number our carbons in our straight line chain. So we start at the top here with that aldehyde group. That's carbon one. And then just go down the chain 2345 and six. And we flip this whole structure 90 degrees clockwise. So we draw that, as we recall, we'll draw a straight line. Now, when we make this cyclic form, recall that we'll have an alpha and beta form depending on where our extra cyclic oxygen is with respect to the plane of the ring. So I'm going to draw my cyclic form in the middle, so we can draw the two forms um on either side. So first, I'm going to draw the straight chain rotated up here. So we'll have our carbon carbon. That's carbon one on the right. And then we have 1234 carbons here that I, so I draw four lines. And then at the end, I have carbon six as AC H2O H group. So go ahead and number those to keep track of where everything is. So now we just need to fill in all our hydroxyl groups. Recall that anything on the right will now be below the line. So I'm going to go ahead and starting with carbon two hydroxyl group below the line hydrogen on carbon three hydrogen, on carbon four hydroxyl and carbon five. Now, I'll fill in above the line. Carbon two, of course, hydrogen, carbon three hydroxyl carbon four hydroxyl carbon five hydrogen. So I've got everybody in the right spot. Now, we just need to sort of virtually bring around carbon six to sort of curve the whole thing around and draw how our ring will start to form. So with a straight line on top, there is carbon six with its CH 20, we're not going to connect our ring just yet. So here's carbon five, I'm going to just put those two numbers on here. So we're kind of keeping ourselves oriented and then go down to the left, that's carbon four, the leftmost carbon. And now we're making the bottom of the ring going down to the right, the bottom line and then going up into the right. That's where carbon one will be. So our carbonel group is there and then recall that we sort of shade in these bottom parts of the ring to show that this is projecting out of the plane of the screen towards us. And now I'm just going to number my carbons to keep track of where everybody is. Oops, this is why it's important because I'm numbering them wrong. So I started 65. I need to go backwards 4321. Carbon one is the aldehyde carbon and now fill in those hydroxyl groups where they go so on carbon five. And see here again, this is where it gets a little confusing because carbon five is over to the left on my straight line structure. But carbon five is, you know, we're now curved up. So carbon five has a hydroxyl group below the plane of the ring. Carbon four, the hydroxyl group goes up, carbon three up and carbon two down. So everybody's on the ring where they're supposed to be. Now, we need to look at our two forms of the ring that will form alpha and beta. So we'll have our Cyle organization and I'll draw it on either side here. The reversible reaction recall that we're going to need to flip. So I'm going to just erase and redraw to save a little space. We want to make sure that the hydroxyl group on carbon five is in a place to react and be within the ring. So we rotate around carbon five. And if oh is going to go into the position where the carbon six was that rotates carbon six up, so we'll erase those two groups to rotate around carbon five. So now we have hydroxyl group out to the right and carbon six here above the plane of the ring. So label that as six being a little crowded there. And so we can now draw the ring because we can see how this will react. And we end up with are rings that draw the top line, oops, not in red and blue, draw the top line. I'm drawing on the right first, we'll make this our um first ring. So we know that we have that oxygen on the top right of the ring as it should be. And so we have the top line oxygen go down. Now there's a carbon, that was our carbon carbon. So that's carbon one. And we can see it's our anomic carbon, it has a bond in oxygen in the ring and the carbon oxygen has now become a hydroxyl group. Now, here's where we have our two forms, that hydroxyl group can go up. So here we'll draw it up above the plane of the ring four. And now I'm going to go all the way over to the left just to start the other ring. So oxygen going to carbon one and that hydroxyl group on carbon one can go down. So that's our two forms I will add in some numbers. So we're keeping track anomic carbon is now carbon one. We've got our oxygen, we have carbon five on the top, we've already added in carbon two, just putting those numbers in there just to keep us oriented as we draw. So now I'm on to carbon two. I will finish my ring, just draw that ring structure in there and shade that in. And now we just have to add in our hydroxyl group and our carbon six. So moving on to carbon two, that hydroxyl group goes down, carbon three, up, carbon four up and then attached to carbon five, we have carbon six going up as a CH 20 group. So on this structure, we have R carbon six and our hydroxyl group, the aoc cyclic oxygen that's bonded to our anomic carbon are on the same side of the brain. So I always sort of think of it as same starts with a consonant B is a consonant. So same equals beta. So this is beta D galactose here and then that's going to make our other ring. So let's go over to our other ring here. Fill in the rest of the ring shade in the front sides. V in our hydroxyl groups. 08 going down on carbon two, up on carbon three, up on carbon four in carbon six. That's the H2O H group going up on carbon five. And we see that now our oh group and our carbon six are on opposite sides, opposite starts with o of bowel as does A for alpha. So this is alpha, the galactus. Well, those are cyclical sugars but our question recalls recall that we're asking for what are the two glycolic products. So we recall that's when that Hammy acetyl group is going to react with the hydroxyl group in our ethanol. So we just need to draw those two products. So I have drawn the parts of our two rings that are not reacting with ethanol just so you don't have to watch me draw that all out called the ethanol has those two carbons and then the oxygen group. So it will react with my hydroxyl groups on my anomic carbon. So on alpha de galactose over here, it's reacting with that oh group on carbon one which goes down So let's draw that line down. But now we have this oxygen and then two carbons, the rest of the ethanol molecule. So I will draw a little two carbon chain there. That's the gly acidic bond. And then, so that's going to be just going to scroll up a little bit our beta form. So we would call this compound at all beta d now called galacto side since it's no longer Jes galactus. And then we recall that our other product will be when our hydroxyl group on carbon one went up above the plane of the ring. So draw that bomb going up, we have oxygen and then our two carbon chain. So this will be ethyl alpha D galacto side. So here are two products based on the alpha and beta form of the ring. When they react with ethanol to form a glyco cytic bond, we end up with our two products of ethyl beta de galactoside and ethyl alpha de galactic side where our eth ethyl chain has been added on has reacted with the oxygen outside the ring that's on our anomic carbon. See you in the next video.

Look at the open-chain form of D-mannose and draw the two glycosidic products that you expect to obtain by reacting D-mannose with methanol.

Verified video answer for a similar problem:

Key Concepts

Glycosidic Bond Formation

D-Mannose Structure

Methanol as a Reactant