Provide the mechanism for the cyclic hemiacetal formation of t...

Question

Provide the mechanism for the cyclic hemiacetal formation of the following hydroxycarbonyl.

Mechanism for cyclic hemiacetal formation from hydroxycarbonyl with H3O+ catalyst.

Accepted Answer

Alright guys, so before we can practice drawing the cyclization mechanism of a sugar, why don't we practice drawing the hemiacetal hemiacetal mechanism on a much more simple molecule that just has the important pieces? It has the carbonyl, in this case a ketone, and it has the Oh on the end, but it doesn't have any of the OHs in the middle, which is just going to make this easier to draw. So it says provide the mechanism for the cyclic hemiacetal formation of the following hydroxyl carbonyl, meaning guys this is the same exact mechanism you learned in the acetal section of your carbonyls chapter, okay? So nothing I'm doing here is going to be new. This is a complete application of hemiacetals. So what's the first step of this mechanism guys? You guys remember? Remember that it's always protonation, right? So I'm going to go ahead and just start off with the molecule I've been given and I'm going to protonate it. Cool. Let's go ahead and redraw it underneath with the new proton. Now it has a positive and that Oh is still there. What's the next step? Do you guys remember? We practice this so much guys. It's going to be a resonance structure. So let's go ahead and draw that resonance structure. And remember that the way that the resin oops remember that the way that the resonance structure forms is that your carbonyl can shift electrons up to the O. Meaning that now this Oh is still here but I'm going to have a positive charge on the carbon instead of on the O. Of course I'm going to put this in brackets to show that this is a resonance structure. And the whole reason that I like drawing that positive charge is because it gives me a very easy target for my nucleophile to show where it's going to nucleophilically add. Now usually I would take my first equivalent of alcohol from out here and I would attack and make the beginning of my hemiacetal. But remember guys, there's no alcohol here. This is not a regular hemiacetal, this is a cyclic hemiacetal where the only source of alcohol comes from the molecule itself. So that means that this Oh actually would come in and that's what does the attacking. Okay, it's the O. Now if we want to figure out how big the new ring is going to be because this is intramolecular ring forming, I have to let's number from nucleophile to electrophile, so it'd be 1, 2, 3, 4, 5, and 6. So I know I'm going to get a 6 membered ring. Okay? So this all happens reversibly, so I'm just going to put double headed arrow, equilibrium arrows. And now let's draw what this new ring would look like. Now there are a lot of ways that you could draw this ring. I'm just going to go ahead and draw a 6 membered ring like this where the o came in attack from here and this is the carbon 6. So like this is 6, 5, 4, 3, 21. Now guys this is not normally the way that you number sugars or anything, I'm not this is not a sugar. I'm just trying to name my ring so it makes sense with what you're seeing here. Here. Does that make sense? Is that fine? How I kept 6 in the same position and then I just made a ring around it, okay? So now that we have those atoms numbered, we have to add our other groups to it, kind of plug and chug, right? So what other groups do we have? Well for sure 1 is going to have an H, so let's put a positive. For sure 6 is going to have a methyl and it's going to have an Oh. Cool? Awesome. So are we done? No. We need to deprotonate. So let's go ahead and use our conjugate base of water to come in here and pull off that H and give us let's just put more well I'll put them down here reverse oval arrows give us our final cyclic hemiacetal that now looks like this and like this. Cool? So see how this is a cyclic hemiacetal? Why? Because this carbon is attached to 2 R groups, RR. And it's attached to Oh and it's attached to OR. So that's a hemiacetal and this is stable. Now guys we're almost done but I just want to point something out. Is there a chiral center on this molecule? There wasn't on the original one but now there is. Isn't this chiral? Because it has 4 different groups on it. So I randomly drew the Oh going up, but it could have also gone down right? So let's go ahead and just draw that we would have another enant- not enant- well yeah it's an enantiomer but another stereoisomer which is that the same molecule could have occurred, but it's with the Oh facing down and with the methyl facing up. Up. Cool? So do you guys see how you would get a cyclic acetal but with 2 stereo isomers. Oh my god, I can't spell. Okay. And this is very important because this is going to be very crucial once we actually start cyclizing sugars, those stereo isomers are going to be important. Cool guys, so hopefully before we talk about cyclizing actual sugars you guys are now more familiar, more refreshed on what the Hemiacetal mechanism is. Alright, cool. So let's move on to the next video.

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Provide the mechanism for the cyclic hemiacetal formation of the following hydroxycarbonyl.

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Monosaccharides - Forming Cyclic Hemiacetals practice set