Monosaccharides - Wohl Degradation

Hey guys. Now, we're going to discuss a reaction that is essentially the opposite of Kiliani-Fischer chain lengthening and that is the wall degradation, which is a chain shortening reaction, let's go ahead and look into it. So, guys as I mentioned with Kiliani-Fischer aldose aldehydes are susceptible to the same exact carbonyl reactions that we would have learned in our aldehydes and ketones section and scientists realized a long time ago that they could strategically take advantage of that aldehyde group to do the same types of reactions that we would have done in carbonyl chemistry but for purposes that are specific to sugars, okay? So, what did I tell you guys about cyanohydrins you can use cyanohydrins to add carbons in Kiliani-Fischer, right? But I also told you that cyanohydrins are reversibly added so it is possible to actually take them off after you put them on. So, opposite to Kiliani-Fischer in this type of reaction, which is called wall degradation the aldehydes are transformed into cyanohydrin to then be reversibly removed in base, so the idea is that we're making cyanohydrins again but for the opposite purpose, this time we want to put them on so that we can take them off and remove a carbon, okay? Now, the chain is going to be shortened by one carbon at every cycle of wall degradation and theoretically you could keep shortening the chain until you run out of alcohol groups. So, just like you Kiliani-Fischer you could just keep shortening that chain, okay? Now, one big difference between Kiliani-Fischer and wall degradation is that because this is a change shortening reaction and not lengthening you're not going to create multiple epimers, in fact, the wall degradation is always going to result in a single epimer product, so notice that there are no squiggly lines here, we know the exact stereochemistry of every alcohol because all the stereochemistry is retained, which I'll show you in a second, okay? Now, the reason this happens is because in, I'm sorry, in wall degradation the c2 stereocenter is lost at every cycle. Now, what am I talking about? I'm talking about second carbon, this guy here, okay. Notice that right now that is a in OH that's facing towards the right direction so that is in R configuration, right? But, what we're going to do is we're actually going to at the end of this reaction turn that carbon into an aldehyde. So, it's still going to be chiral after the fact after done the wall degradation? No, it used to be chiral and now it's going to be a chiral. So, what that means is that it's possible for multiple sugars to produce the same product because, for example, if I reacted to the all degradation with an OH that was faced towards the right on the c2 position and then I also did a wall degradation with another sugar that was the same exact sugar except that the OH is towards the left, so it's at the opposite c2 epimer, if you reacted with both of those epimers you get the same exact product because regardless of which one you started with it's going to turn into an aldehyde, okay? Now, I know that I've been just talking a lot.

Now, I'm going to go into the mechanism but we are going to practice this concept later. So, don't worry, we're going to come back and I'm going to show you guys how multiple sugars could lead to the same product using the wall degradation. So, what I want to do now, is really just go into the reagents cuz I've been talking too much. So, first of all, let's start with d-glucose, d-glucose the very first step is that we're going to turn it into an amine derivative. Now, what we're going to do here guys is we're going to take advantage of the fact that in carbonyl chemistry, what did you learn about ketones and aldehydes? that if you react them with primary amines, let's say NH2R, right? So, it's a primary amine, what are you going to get as a product? this is a review, in review of the section of carbonyl chemistry called amines you could search for it in your search bar if you want, what you're going to find is that this becomes a double bond N, and a single bond R, okay? And, this is what we called an amine, right? But then we also talked about in that section how if you have a Z group instead of the R group. So, instead of n h R imagine that it's NH2Z, meaning it stands for something electronegative something like OH, okay? then the product instead of being, I'm sorry, instead of being NR it's going to be NZ, okay? And guys that Z is going to be the hydroxyl group in this case. So, specifically the primary amine derivatives that we're using is called hydroxylamine and when you react hydroxylamine having an OH with a aldehyde, what you're going to wind up getting is an amine derivative, so this is called an amine derivatives and specifically the amine derivative is called an oxime, why is it called an oxime? that's the specific name that we give to the amine derivative of hydroxylamine and all this is reviewed in your amine derivative section of clutch. So, if you want to just type that in to go over these reactions again that's fine, but I'm just saying this is all you need to know that you go from, the first step is you go from your saccharide and then you make it oxime, which is really just a type of amine, great. Now, we have our oxime, this is all review, the next step is that we actually want to rearrange that oxime or that amine derivative into a cyanohydrin. Now, this reaction is a reaction that I don't expect you to know as well because it's so much, it's a much less emphasized reaction in organic chemistry and it's called the Beckmann rearrangement. Now, guys the Beckmann rearrangement is not unique to the wall degradation, there are other reactions in organic chemistry that use the Beckmann rearrangement but it's just one that we don't focus on as much as we do amine. So, I'm not going to make you, hold you accountable to know this reaction very well, in fact, what I want you to do more than anything is just memorize what happens in this step and what happens and this step is that we use an anhydride, this is the formula for anhydride. Remember, anhydride, we're going to use anhydride to dehydrate that oxime and rearrange it into a nitrile or a cyanohydrin, okay? So, this is basically a rearrangement that I'm not going to hold you accountable for this mechanism, your professor won't either, not in this section, you don't need to draw the whole rearrangement you just need to know that that oxime is going to rearrange to a cyanohydrin, okay? Now, here's the, cool? Part, we learned that cyanohydrin add reversibly so that means that it's possible to also kick them off if you use base and that's going to be our third step, our third step is going to be you to use a strong base usually methoxide and what we're going to do is we're going to reform the carbonyl and kick off the cyanohydrin, which will eliminate one carbon, let's see what the mechanism would be, it would be, negative grabs the H from z2 then those electrons from the O and the J turn into a carbonyl. So, now I'm going to get a double bond O and then I'm going to kick out my CN, okay? And guys I put here this is an alpha elimination, not because I want you to memorize it but because I want you know, this is an elimination reaction, technically, what we're doing is we're getting rid of two single bonds and we're making a double bond, which is the definition of elimination you get rid of two single ones you make one double one but it's not the same as beta elimination, which would have been in Orgo one. Remember, you had like E2 and stuff like that, this is not that, it's just a different type of elimination called alpha elimination because it all happens on the same carbon on that alpha carbon everything takes place, and anyway guys, we end up getting and this is really the most important part of this whole thing is you end up getting a monosaccharide that now instead of having a hydroxyl group as a C2 position that C2 position became an aldehyde and we lost HCN, okay? So, basically, we lost HCN and we got this in return, okay? So, now this specific monosaccharide would be called d-arabinose why? Because notice that the chirality is all of these chiral centers has to be retained, so this is going to be the same exact monosaccharide as the one we started with except that we chopped off the very top part, okay? So, that's the wall degradation does that make more sense now guys? So, basically wall degradation is all about using this series of reagents to chop off the very top carbon and what that means is that we're going to get a stereo specific product, okay? Awesome guys, let's go ahead and move on to the next video.