Monosaccharides - Reduction (Alditols)

Concept: Concept

8m
Video Transcript

so it turns out that one of the important ways that monosaccharides can react is through reduction and that's what we're going to explore right now. So, guys monosaccharides are essentially polyols, meaning multi alcohols with carbonyls. So, what that means is that they can undergo a series of oxidation and reduction reactions. So, guys we're not done with the reactions like we have a lot of reactions coming up because monosaccharides have so many functional groups that are reactive that they're just so many different places they could react and the one I want to talk about right now is called monosaccharides reduction, which yields a compound called alditols, okay? So, reduction of monosaccharides produces polyols known as alcohols or sugar alcohols, okay? Now, the term sugar alcohol is misleading because you might think that it's either a sugar or it's an alcohol it's neither because remember that to be a sugar you need to have at least one ihd and it turns out that sugar alcohols have been reduced all the way down to just a series of alcohols with no carbonyls. So, there are no longer considered sugars, they're their own molecule called either sugar alcohols or alditols, cool? Now, alditols are used industrially as sugar substitutes, food thickeners and medicinally as laxatives usually, okay? So, you might have heard of these molecules before, if you look at the back of like your food contents, if you like kind of read like what your food is made out of, a lot of times you're going to see molecules like sorbitol, mannitol and what these are these are sugar alcohols that they're reduced carbohydrates that are digested by the body very similarly to alcohol but, I'm sorry, very similarly to monosaccharides, but they usually have less caloric value they're not as sweet they're good at like thickening foods and stuff like that. So, guys the reducing agent that you really need to know is nabh4, nabh4 is one of the most common reducing agents that's used. Remember, that nabh4 sodium borohydrate is a weak reducing agent. So, it's perfectly able to reduce an aldehyde to an alcohol and that's what we see here, what we see is that if I were to take d-mannose and reduce it with sort of a burro hydrate and then use h3o plus to quench the negative charge and to add a proton basically the second part is just the protonation step, what we're going to get is a molecule that looks just like d-mannose except you now have an alcohol in place of the carbonyl. So, that's a sugar alcohol, let's look at the glucose so the glucose, remember, what the difference between mannose and glucose there epimers, right? Just at the c2 position. So, if I were to go ahead and reduce d-glucose I would get a very, very similar sugar alcohol to mannitol. Notice that they look almost exactly the same, again my carbonyl gets reduced, the only difference is that this OH is facing in different directions on each of these, okay? Now, the reason is because the c2 position, the chirality isn't changed reduction only occurred here and here. So, that means that the stereochemistry of your original sugar is conserved after your reduce, okay? So, you can reduce glucose to get sorbitol, you can reduce mannose to get mannitol. Now, what about the strain situation where maybe you want equal amounts of both, if you wanted that to happen, what you could do is you could actually reduce d-fructose because whereas reduction of aldoses produces one product as I just showed you those are both aldoses glucose and mannose, reduction of ketosis forms two products due to c2 racemizations, let me show you. So, what you have going on here is that this is the second carbon, right? The ketosis on the second carbon, and the way the mechanism works, you're basically donating an h minus and you can donate that h minus from either side, you could either donate it from this side, where you put the alcohol to the right or you could donate it from this side, which would make alcohol go to the left, so it turns out that if you're reducing a ketose with nabh4 you're trying to get a mixture of both products. So, in this case, let's look at my fructose, fructose. Remember, the bottom three look exactly the same as glucose, they look exactly the same as mannose. Notice the only difference is that instead of having another OH there have a ketone, that ketone can turn into an alcohol facing right or left, it gets racemized, which is why the reduction of fructose is going to lead to two different sugar alcohols or two different alcohols, okay? Make sense? Now guys, the mechanism for this follows the same exact mechanisms that we use in the reduction chapter, when we first learned how to reduce with lithium aluminum hydride and sodium borohydrate but let me just go over it just so that you guys will remember, and I know some of you guys are gonna ask. So, guys remember that sodium borohydride is a good hydride donor and it specifically it's a nucleophilic hydride donor. So, I have B, H, H, h, H negative and then there's just a sodium hanging around with a positive charge, those are ionically associated, okay? And remember that these reducing agents are a good source of nucleophilic h minus, it's different than, if I just use NaH like, for example, by just use NaH, this is a base, this is not a good nucleophile, okay? But, if I use nabh4 it turns it into a nucleophile, okay? So, the way this would work is that it's nucleophilic addition, nabh4 acts as a hydride donor and attacks the carbonyl versus be a nucleophilic addition and then protonation occurs. So, my h comes in attacks the partial positive of my carbonyl, kicks up electrons to the O, what I get is my tetrahydro intermediate with a negative here and with an h here, and then protonation occurs where I have whatever protonating agent it could just be water, that comes in and add the proton. So, what I'm going to get is an alcohol plus an h and I looks behind me. So, I'm just going to move up. So, guys, this is the general mechanism of reduction and this is how it works, it works on the straight chain form of the of the sugar. So, in this case once you have the straight chain form you can add the h and you can turn it into alcohol, cool? Awesome guys, so that is a little bit about alditols and, or sugar alcohols, let's move on to the next video.

Problem: Determine the structure of the alditol formed when b-D-xylofuranose is treated with NaBH4 and then water. Explain how NaBH4 can reduce the hemiacetal group of the furanose. 

6m