Trehalose, a disaccharide found in the blood of insects, has the following structure. What simple sugars would you obtain on hydrolysis of trehalose? (Hint: Rotate one of the rings in your head or redraw it rotated.)

Question

Trehalose, a disaccharide found in the blood of insects, has the following structure. What simple sugars would you obtain on hydrolysis of trehalose? (Hint: Rotate one of the rings in your head or redraw it rotated.)

Accepted Answer

And everybody, let's take a look at our next problem. It says sucrose, a common disaccharide found in plants can undergo hydrolysis. What monosaccharide would you obtain from this reaction? We have our structure of sucrose. We have a six membered ring or pinos bound to a five membered ring or finos with a glyco cytic bond between them. So when we have a bond undergoing hydrolysis, it means we're adding water to break the bond. So we have an oxygen in between the two of these. So we'll add h2o and in the end, both of the carbons that were linked by this bond will have a hydroxyl group on there where the bond was. So now we need to think about what are the two sugars that make up sucrose. Sucrose is made of glucose and fructose. So glucose is our pinos ring and fructose is our buro ring. So let's think about what they'll look like once we break that bond by hydrolysis. So sucrose here, we will draw sorry glucose, we will draw our top line the oxygen. So draw our hexagonal ring here. This is a Hayworth projection. So we thicken the three bottom lines to show them projecting out of the plane of the paper. And looking at our glucose over here, we can see this goes with convention in terms of how we would draw it. We have our oxygen ring on the top back, right, our anomic carbon will be the one with bonds to two oxygens. So it would be the one participating in that Glyco bond. And so the anomic carbons on the right now where we had an oxygen bond that the bond of oxygen was going downward below the plane of the ring. So our hydroxyl group will also be pointing downward below the plane of the ring. Now, for ease of referring to them, let's go ahead and number these carbons. So the anomic carbon in this Pinos ring will be carbon one. Let me just go around 23, 45 and then recalls glucose has six carbons and carbon six. Is that one that's projecting up in AC H2O H group? So there is carbon six. So let's continue filling in all our hydroxyl groups. We added the one on carbon one, the one on carbon two goes down, the one on carbon three goes up on carbon four goes down. So the last thing we need to do is determine whether we're talking about an alpha or a beta Anno or here. So we look at our anomic carbon, look for the exocytic oxygen. So that's going to be the hydroxyl group. It is pointing down below the plane of the ring. Now trace over to the oxygen in the ring. Look at the carbon on its other sides of carbon five. Then look for the C H2O H group coming from that. So in this case, carbon six that is going up above the ring are these two groups, the oh group and the C H2O H group, are they on the same side of the ring or opposite, they're on opposite sides of the ring. So that means this is the alpha version, think of two vowels, O for opposite A for alpha, they're on the same side of the ring. That would be the beta aimer think two consonants, same S beta B, both consonants. So they're on opposite sides of the ring. So this is going to be alpha D glucose plus, we have a molecule fructose. Now first, I'm going to draw underneath, I will kind of directly copy how the structure is drawn in our sucrose molecule. And then we'll think about the convention for drawing a fo by itself. So the oxygen is at the top. And then on the right side, we have AC H2O H projecting down then a hydroxyl group going up, then a hydroxyl group going down now or two hour carbon that was in the glyco bond. It also will have a hydroxy going down since the bond went down. And then on that same carbon, we have AC H2O H group projecting up. So we do have to rotate this around because by convention, when we draw Furin noses, we draw them oxygen on top anomic carbon. On the right, in this case, where's our anomic carbon? Well, it's on the left side, it's the one that was participating in the Glyco bond. So interestingly, we number this on like a pin nose ring and the purose ring, the anomic carbon is carbon two carbon one is projecting above carbon two. And then we go around our ring. Carbon three, carbon four, carbon five and then carbon six is projecting down below the ring bonded to carbon five. So let's rotate this around. So we have it with our anomic carbon on the right. So carbon three downward, carbon four going up, carbon five has carbon six going downward. So then again to determine alpha and beta, we start with our anomic carbon and look at the exocytic oxygen. So highlight that. And again, this gets a little tricky because you have another carbon there. It's picking up. So you have two C H2O H groups outside the ring. So that can be a little confusing, but we're not using the same one here we go to because that will automatically be opposite, go to the oxygen in the ring then to the other side. So looking at carbon five and that C H2O H group. So carbon six, I'll write that in there six and say, are they on the same or opposite sides of the ring. In this case, they're on the same side. So this is beta D fructose or the other molecule. So there you go drawn in the conventional way, our two products are monosaccharide. We obtained from the hydrolysis of sucrose are alpha D glucose and beta D fructose. See you in the next video.

Trehalose, a disaccharide found in the blood of insects, has the following structure. What simple sugars would you obtain on hydrolysis of trehalose? (Hint: Rotate one of the rings in your head or redraw it rotated.)

Verified video answer for a similar problem:

Key Concepts

Trehalose Structure

Hydrolysis Reaction

Monosaccharides