Okay. So, this is um what I'm going to call part 3B um to the slides or the voiceover recordings on amino acids and peptides. And we stopped uh part 3A because my Apple Pencil was running out of juice. So, one of the things I forgot I needed to charge. So anyway, I'm back and this this will all finish this off. So all we need to talk about now is the mechanism for Bach protection um and the uh mechanism for CBZ and Fmock protection and then D protection and we're done. Okay, they're pretty easy. Now the good news is there's nothing new here really. I mean this is just an ASIL this totally just an AIL um transfer reaction. So you'll remember in the chapter on asil groups I mentioned if you take an A asil group with a leaving group Y. So that I should use the right same color. Um so asil group plus a nucleophile and you'll transfer the asil group from the Y group to the nucleophile and a Y group will leave. Okay. And in that chapter, I think we we talked about it a little bit when we're talking about the mechanism for um dcc um amid formation. Um we uh highlighted the asil group in red. Okay. And we used kind of a yellowish green for the leaving group. We should have kept using the red to finish this out because the red group transfers to the nucleophile. We used blue for the nucleophile in that chapter. Okay. So again, we're transferring the asil group to the nucleophile making a new bond between the nucleophile and the asil group and the leaving group leaves. Okay. And we learned that there is a mechanism wherein a nucleophile with an H on it will attack the AIL group. Two in two out. We will make a tetrahedral intermediate. I'm just I think in the interest of clarity maybe I'll just color the leaving group. Okay, that's the leaving group. I don't know. May I don't I don't want to waste your time or bore you to tears, but maybe it'll be useful for you to see that that's the AIL group right there. And we used blue for the incoming nucleophile. Okay. Of course it was. Right. Okay. I think I'll stop using the color because I think you get it. Okay. Um, going to redraw this guy down here. Just took redrrew that in an exam setting. I recommend you do the same. It make it a lot clearer, a lot easier for me to grade, a lot easier to see that you know what's going on. Okay. All right. Once we get to here, this proton will transfer kind of like the beach ball. It'll leave the N and end up on that O. Okay. And I don't care that you show a curved arrow if you want. Okay, you could show something like this, although it doesn't really happen that way. Okay, but if you do show it, you need to show the tail of the arrow starting where the electrons are headed towards the proton and then that bond breaking. Uh, this arrow must break that way because it indicates the electrons in that bond are moving on to the N. Okay, it must be shown that if if you choose and show use arrows to show this. Turns out it doesn't really happen that way because we're going through one, two, three, four member transition state. That's a square transition state two strained. So it doesn't really happen that way. You can show that if you want. All right, we get to here. What we've done is transfer the proton off the end on to the leaving group. I am going to color the lab again so you see what we've accomplished. Okay, we have transferred a proton off of the N. Okay, on to the leaving group. That makes this an even better leaving group. Okay, if that proton does not transfer, this is the better leaving group and we'll go backward. That's what the backward arrow means. It'll just fall right off. So, the productive step involves transferring that proton onto that O to get to here. Okay, I'm going to redraw this guy down here. So, we got color leaving group again. Okay, and we're going to think about how happy this O is with a plus charge on it. The answer is by now, you know, not happy, Bob, right? According to the Incredibles, not happy at all. Company is like a welloiled clock as the saying goes. And so I can't remember what the guy's name is in the Incredibles, but he's a little short little boss guy. The Bob Par throws through the wall. Okay, so this isn't happy. So if the tetrahedral intermediate collapses to where this pair comes down to make a new pie bond and that bond breaks, those come in like that and that bond breaks. We will end up transferring the AIL group which was the red piece. Okay, let me highlight that now. Okay, this is the asil group and it was transferred off of the leaving group. It was attached right here. Okay, it's transferred off of the leaving group on to the nucleophile. This nucleophile. Okay, and here is our leaving group with an H that it picked up off of the N. Okay. Right. So, it's just an ail transfer. You've seen these reactions over and over and over again. You're a pro at these. You already know how. You already know how to do I think 18 of them. I gave you a handout with 18 of them, right? So, piece of cake. That one's not hard at all. Mechanism for B protection. Okay. Erasion to starts with is Bananhydride. And I strongly recommend that you learn that thing structure. Super easy to remember. Turbital oxyarbonil turboloxicarbonil. Okay. Anahhyra boach anhydride. All right. That's the mechanism for boach protection. What's the mechanism for fmok or CBZ protection? The good news is is both of these reagents come asil chlorides. They differ in the nature of the R group. And I'll let you look at this. This is on these uh two. Let me go back. I forgot to tell you what page this is on. Oh, it's on page nine of your peptide handout. Okay. And this other one is on not page nine, but um yeah, it's also page nine. It's also page nine. Reactions 14 to 15. Okay. All right. So, again, we're going to use the same color system. Hope this doesn't bore you too much. You probably get it. There's our nucleophile. We're going to color color our leaving group kind of a fluorescent green kind of isish. There it is again. There it is again. Okay. Comes off there. Blah blah blah. Um, we're going to color the loop red. There it is. There it is. There it is. Our nucleifile has got connected. There it is. There it is. There it is. File blah blah blah. Okay. And we're transferring the a as group which is red. Come on. As it ends up on the blue nucleophile. So it was first attached to the the chlorine. It ends up attached to the incoming domain. We're making an So the amine attacks carbon to in out. we make a tetrahedral intermediate. Now, uh years ago, and this this kind of dates this slide, this this handout's probably at least 15 years uh old. I don't think it's somewhere in 15, but it might be. Okay, I did it by by pencil. Way before iPads were even even a dream. Okay, which pretty that's pretty amazing. And in those days, I used to show the chloride falling off first, which it's arguable it might. Okay, you can't really prove that it does. I've started to teach it this way. Look at 2A. And I'm okay if you you show two or 2A. It'll be easier, I think, if you show 2A because that's how I taught you guys. But earlier generation, I used to teach it this way. And um let me explain why. Uh this is analogous to the step that we saw with the with the bacheloration where the NH transfers to the leaving group. So H+ will transfer to the leaving group and make this an even better leaving group. Okay. If it doesn't transfer, it's it is arguable that that just falls off because the proteinated N is going to be a better leaving group than a CL. Even though Cl minus is a fantastic leaving group, a proteinated N is going to be even better. And so after the nitrogen is added to the carbonial, we get to here. What I'm saying is it is uh highly likely that if this proton does not transfer the Cl, we're just going backward. This the nitrogen is just going to fall right off. It fall right off. I said fall right off. Fall right off. Okay, it falls right off. Then we ain't going nowhere. Okay, we're not going to we're not going to have a a product out of this reaction. Okay, so in order to get a product, I'm redrawing this guy down here. Step 2A. And I'd prefer that you draw the proton transferring the chlorine. That makes us a better leaving group. Now the um protetonated Cl will leave even better it would have before and the N stays on. Okay, these electrons come down kick off HCl. We make HCl away and there's an end point or since this is way is in your handout if you show it this way in an exam setting accept it and say it's okay. Uh and it turns out that that would be really difficult to prove that the proton actually does transfer first, okay, to the chlorine. Uh but it's very logical and more in keeping. It's totally arguable back here. Okay, that if this proton does not transfer off of the end to that O, this is not a better leaving group than that is coming back up here. If that proton does not leave the end and go up on the leaving group, that's not coming off. So in this particular case, it's not arbitrary. That H+ must go up to that O first. Now that's a great leaving group. And it just seemed to be easier to teach that even though chloride you could argue, well, can that just fall off? Maybe. Okay. But it does so while you're still leaving a plus charge on an N. No. this is going to fall off and go back that way first. So, I'd prefer that you show it like this. This is more common, more um similar to the mechanism for putting a bot group on. Either way, at the end of the day, you make an ambit and HCl. Okay? But I'd accept either two and three or 2 A and 3A. Okay? That's that how we put on how we put protecting group on. These reactions are just ail transfers. That's all they are. Let me go back and and circle this. Now, copy it. We spent a whole chapter talking about these. So, we're now all experts in asil transfer. I'm waiting to see it again to show that. That's cool. Okay. All right. Awesome. Let's take a look at mechanisms for cleaving protecting groups. The first one we're going to talk about is cleaving methyl esters or benzil esters. And that is a super simple cleavage reaction is just a soponification. And all you do in a soponification is treat the estester with hydroxide. The hydroxide attacks the carbonal 22. You make this intermediate which is a tetrahedral intermediate. You kick off the aloxxide leaving group and that aloxxide then immediately deproinates the caroxyic acid and you make the caroxilate annion and an alcohol. This is a mechanism we studied um quite a bit in uh the last chapter and was on uh featured in one of our exams. the mechanism for subontification. Again, that is just a subonification. That is all in the world it is. And so this should be like good news. And this is just a review of something you already know. Okay. Okay. We already know. Yay. And we're happy. Look, everybody's happy. So smiley. Okay. All right. Cool. Can I go to the board, please? He said, "That's a reminder to me that if we were in the classroom, I would stand on the desk and then if someone invited me to go to the board, I'm going to show you how you cleave an FOMO group and then also how you cleave a Bach group. First, the FOK. I showed you this earlier um and mentioned that the FOC is cleaved by base and the base can either be trithyl Oh man, I did that. Three eths on the nitrogen. That's a tertiary name. Okay. Or purity. That's the base. This could be pulls off a proton. That bond breaks. We make a double bond here. Kick this out and this pumps up there. What leaves is this. This is just an E2 elimination. It's all in the world. That is E2 elimination. And this is our leaving group. We're deproinating um H and the leaving group to make an alken. Okay. The reason these bases will work is because an annion that builds up here to make the annion at annion is aromatic. And so the CH is quite acidic. It's acidic enough that Namine is basic enough to deproinate it. We get to here and then this spontaneously decomposes to give CO2 and amine. This is the mechanism for that that process. We proteinate the end proteinated ammonium that will come off proton will come off but that's not productive. That's the backer cells. Yeah, comes off and then it goes back on, goes off, go back on, goes off, goes back on. Eventually these electrons will come down because almost like a CO2 activate and make CO2 and okay that is the mechanism for FD protection possible mechanism for modern exams. Okay. All right. Look now at So we have deprotection of the the block. I'll delete that. 66 was not empty. I don't believe why I got deleted. Okay. Um, we may have to have part three C in a minute or maybe I can write it down. Um, hold that thought. So right here check check with Dr. P. I think I've got a version of this set of slides that had that patient um chopped off. This would be page 10 for the peptide handout and that is bo cleavage talked about in class. That may be a reaction that that in the past we have we have omitted. It looks like a pretty long one. So ask me a class. Okay. I the fact there's a blank slide here makes me think that it it it somehow got deleted when I was recharging the pencil. Okay. All right. That is 3B part 3B for now. And check with me about block cleavage.