Predict the product(s) for the following reaction. Provide the...

Question

Predict the product(s) for the following reaction. Provide the mechanism of the imine hydrolysis step if required.

Reaction scheme for β-D-xylofuranose with HCN and hydrogenation steps.

Accepted Answer

Predict the products for the following reaction and provide the mechanism of the amine hydrolysis step if required. So let's go ahead and look at what we're given and figure out what we're dealing with. So it looks like I have a cyclic sugar and I'm reacting that with HCN and then a hydrogenation that includes H2, palladium, but then barium sulfate as well. So this is indicative of a poison catalyst that would happen in a Kiliani Fischer synthesis. So just go ahead and write that down. And I'm going to go ahead and just call it Kiliani Fischer chain lengthening to remind us that this is a reaction that is going to try to add a new carbon to my sugar. So however many carbons I have in this sugar right now, it looks like 5. 1, 2, 3, 4, 5. By the end of this whole process, I should have a 6 carbon sugar. Cool, so guys the very first step of this reaction is that I need to convert my cyclic sugar into a straight chain. Because cyclic sugars actually can't react with HCN at all, you need an aldehyde present. So what we want to do is convert this to the Fischer projection, the straight chain, and just you know this happens readily in any kind of acid or base in an aqueous solution. HCN has acid in it, so we would say this would easily happen. So let's go ahead and draw out what the carbon chain, the Fischer projection should look like here. And let's go ahead and list out our positions. So I have position 1, 2, 3, 4, and 5. Cool. Awesome guys. So now we just have to start adding in everything. We know that positions, let's just use a thinner one. 234 get these lines across them. 1 and 5 are the ends so so they don't. Cool. So let's go to carbon 1 first of all. Carbon 1 here matches carbon 1 here and we know that that's an aldehyde. That one's easy, it's going to be an aldehyde and we we know that because it's the anomeric carbon. Okay, notice that this is the anomeric carbon. Right now I have the Oh going up. If the Oh had been going down it makes the same thing. Remember that the whole reason that the anomeric carbon can either be an alpha or beta is because it used to be trigonal planar and then it gets attacked. So basically what I'm saying is that first one is just an aldehyde. That one's easy. Okay? Cool. So let's go to carbon 2. Carbon 2, we're going to use the downright rule. Downright says that if the Oh is facing down, you draw it on the right for the Fischer projection. So we're going to draw it over here. For 3, I'm going to use down right, it's going up so it's left. Now for 4, 4 is actually the stereo descriptor for this 5 carbon chain. So I can't use down right anymore. What I would use is the rule that says that if this is going up it's it's a D sugar and if it's going down it's an L sugar. Okay, now according to my name it already said that it was a D sugar, but my stereo descriptor confirms that the CH2OH is going up, meaning that it's a D sugar meaning that the 4th position should go to right. Cool? So perfect and then finally 5 is just the CH2OH. Cool, so that was some good practice learning how to turn reminding ourselves how to turn cyclic sugars into Fischer Projections, one of the most important skills for this chapter. Cool. So now that is our sugar and now we're ready to do our reaction. So guys the first step, it doesn't actually ask for the mechanism of the first step, but we know that the mechanism is going to be a cyanohydrin formation. So imagine that you have CN negative and H positive because HCN will dissociate in solution and the mechanism will look like this. The negative attacks the positive, kicks up the electrons to the o, and then it gets protonated. So our product at the end here of this first step should be the same sugar. It's going to take a little while. Plus my cyanohydrin. So it's going to be Oh c triple bond N, and H. And just so you guys know, I drew it going to the right, but really that should be a squiggly line because I don't know what direction it's going to go. At the end I'll put a squiggly line. Cool, so that's the end of step 1. Step 2 is that I need to reduce, okay? So step 2 I'm going to reduce with my poison catalyst. So I'm going to use my H2, palladium, and barium sulfate. And what that is going to give me is the same thing. Now notice I have 4. So I have Oh Oh Oh CH2 H H H H. But now up here what I have is CH NH. So now I have an imine. Bueno. Awesome. So now I have my imine and now we're going to have to hydrolyze it with water and acid. We already have acid from the HCN, but guys it says that we can't just hydrolyze it in one step or without a mechanism, we have to draw the whole mechanism. So this is the part that's going to take a little while because we have to draw the whole imine hydrolysis mechanism and if you guys need a reminder of this mechanism, I did teach it in the imine section of carbonyl chemistry, but let's go over it here. Okay, so let's go ahead and say that we are hydrolyzing this with H h3o plus which is what you would get when you combine water with any acid or with any slight amount of acid you'll get h3o plus. Okay. So in Amine Hydrolysis just with any acetylase mechanism, we're going to go ahead and protonate the nitrogen first. So let's go ahead and do that And I should be drawing all these as reversible arrows, not the other ones, but this one for sure. So now let's go ahead. What's probably going to be most annoying about this mechanism is that we have to draw the sugar every time and I'm probably I'm actually gonna cheat because for the sake of your time I don't want to have to draw this thing over and over So I'm going to go ahead and copy this so that we can use it every time in the mechanism. Okay. And you can just pause if you're falling behind. Cool. So that means that now I'm going to get CH double bond. Wow guys I had too many H's in my first one. So guys I'm sorry this is why you need a pencil or that's why I need an eraser. So this is actually just supposed to be NH not NH2. NH2 is what happens after. But remember, neutral nitrogen should have 3 bonds, so it was actually drawn incorrectly before. But now that we added the extra H, it should be n h two positive. Does that make more sense? I'm sorry guys. I was drawing quickly and I messed up. Okay? But that is the right answer now. Perfect. So now we have that and now we have water and remember we want to add water to that carbon, so that we can eventually make a carbonyl. So this mechanism goes like this, pushes up the electrons to the nitrogen. So now what do I have? I have an intermediate structure that looks like this. Carbon with water with a positive charge because now it has an extra bond. With a nitrogen that's just NH2 that's neutral now and with hydrogen. Cool? So now guys we're going to do a proton transfer. Remember we're trying to get rid of the nitrogen and we're trying to make the O into a carbonyl. So the N is going to grab one of the H's and give electrons to this H. So now this is going to give me carbon NH3 positive HOH. Cool. And maybe you guys can predict what's the next step. We need to get rid of the ammonium at this point and make a carbonyl. So I what I can do well I'll just keep that towards the end. What I can do is I can use the electrons on the O to make a double bond and then to kick off the ammonia. So now I'm going to do this. Paste. Now what I have is a newly formed carbonyl that's still protonated so it has a positive charge but we're almost there, nOH. And now we're ready just to deprotonate and you can deprotonate with whatever. You could deprotonate with your water. So let's go ahead and do that. Let's deprotonate with our water to reform the catalytic acid which was H3O plus And I know that I already used the water but we're going to assume there's another equivalent of water around that could be used. And there you go, that's our aldehyde. So guys sometimes imine hydrolysis, the idea that an N can turn into an n double bond can turn into an o double bond seems magical to students sometimes. Like how did that happen? This is the exact mechanism that you would follow follow to hydrolyze the amine into the carbonyl. Cool? By the way one more thing, I didn't actually include the arrows for the CN adding to the CN over here, but basically you should know that this is also reversible. So this is a mechanism that looks like this. Forwards arrow, little backwards arrow. So even though it's going to favor the Cyanohydrin because the reaction conditions are going to be created in such a way to favor the Cyano Hydrogen. This is a reversible reaction, so you could theoretically use base to pull off this H and kick off the CN. So now if you wanted to, you could go in the backwards direction. That's all I'm saying. But for right now, we're going to go in the forwards direction because that's what we want for our end product. So now guys, let's just go ahead and also add the stereochemistry which which I'm going to use as squiggly lines to indicate that we don't actually know where that Oh ended up. It could be on the right, it could be on the left, you could also draw both the one on the right and the one on the left and say just mixture of diastereomers. They're not enantiomers because there's a lot of other chiral centers, but you could say mixture of epimer or mixture of diastereomers and that would be fine. Or you could just draw it like this, which implies you have a mixture of those epimers. Awesome guys, so this is a really long problem but I hope that you understand the Kiliani Fischer Synthesis better now. Let's go ahead and move on to the next video.

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Predict the product(s) for the following reaction. Provide the mechanism of the imine hydrolysis step if required.

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