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(6) Drw the product formed following the cnamine hydrolysis below . Then show capiss tjuS pushing mechanism for its formation:I0'...

Question

(6) Drw the product formed following the cnamine hydrolysis below . Then show capiss tjuS pushing mechanism for its formation:I0'

(6) Drw the product formed following the cnamine hydrolysis below . Then show capiss tjuS pushing mechanism for its formation: I0'



Answers

Show the mechanism for the acid-catalyzed formation of $23 \mathrm{c}$ starting with the product obtained from its hydrolysis.

Okay. So rest assured mechanism for the hydraulics of Macedo using an acid. And to settle is formed from the addition of two alcohols too a galahad. So we have the addition of two meth oxy groups and we're to treat this with With H 3-plus 2 some acid. It's going to marry draw with us. So one of these auctions is going to do you appreciate the hydrogen and we'll get the formation of a better leaving group. So bonds can form here and it's gonna expel the one group. Next step is for a water molecule to attack at the carbon carbon. Breaking the five armed and screw the addition of O. H. Two positive charge. So this is gonna be is he pregnant about another water molecule serving alcohol and meth oxy group. And the meth oxy group is going to do you appreciate the acid again. So turning itself into a better leaving group. And as a result the oxygen on O. H. Is going to use a pair of electrons to form a bond and expel the oh CH three Go down here. And the last step is for a water molecules to deeper in eight the O. H. Thank you, Mr Kayser. We get our products the key tune from the Seattle and the regeneration of the acid your surplus.

Our organization. Open door, right. This little acidic condition, this is the end goal is to make support meaningful. All right things. Yeah. Everyone move. Yes. And when you're already respond, Yeah, it is a contributor. That dialogue, How much of this product can be explained when they visit all the protein National One more regular pretty door. Yeah, Yeah, yeah. Mm. Yes, Yes. This is attracted by an order. The idol brought it home after generalization. Which use warnings. Hardly that we want fully and positive here. It's mhm. Proton transfer occurred 20. Preparation his phones or it's two plus double 10 mhm. Under this condition. Good living. Look, Water. The Utah. Did you even know snapping or club the minister problems. So this is gonna can you off nationality Done? Yeah.

In this problem, I'm writing the reaction. So just look at it carefully. CS three CH 2 C double one oh Ch two, Ch 2, CS three in by Denzel. Any BH four, it will change into see http CH two, see it all minus CH two, CH two. See it to the and when it is reacted with the tour, it will change into Ch three, Ch 2 C h o D, Ch two, Ch 2 CS three. So option B. So option B is got it and said. But this problem.

This is the answer to chapter 20 to problem number 65 Fromthe Smith Organic chemistry textbook. Ah, and this problem asks. Well, this problem tells us that acid catalyzed hydraulics ists of this molecule forms compound A, which shows a strong peeking. It's IRA at 17 70 Wave number. Ah, and then were given some anymore Spectra as well. Ah, and were asked to draw the structure for a and draw a stepwise mechanism that accounts for its formation. Okay, so, um, actually, move the anymore. Piece over. All right, um, so, first of all, we're given a formula. See six each 10 02 s. So the first thing we can do is calculate the HD I, um or the unsaturated number or the degrees of on saturation. Um, and so that's gonna be two times the number of carbons plus to minus the number of hydrogen Sze s o. We get 14 minus tennis for a love that over too. So our HD I equals to us. We have two rings to double bonds or a ring and a double bond. Okay, um, so next, looking at the I r. Uh, the only thing we get from the i r. I's a peek at 17. 70 on. So this suggests a carbon oxygen double bond. Okay, on. And certainly that could explain one of our degrees oven saturation in this molecule. Okay, so now, looking at the anymore, we have a 1.27 p p. M. Single it with an integration of six. This to me suggests to methyl groups. Ah, and obviously they be equivalent. Since they're giving just the one signal, there s so then we have a 2.12 ppm triplet with an integration of two and a 4.26 p. P. M. Triplet, also with an integration of two cities, are both triplets, and they both integrate to two. Ah. And so, um, I think that these air to methylene groups on I believe that they are splitting one another. So two methylene group's splitting one another. Um would look like this. We have a carbon Neil somewhere in this molecule. We also have a second oxygen to account for. So I think that we have ah, and Esther here. Um, and so one way that we can put this all together, um, to get our six carbons, 10 hydrogen tze and two Oxygen's accounted for. Ah, and in a way that's gonna make sense with our anymore would be if we have a black tone, Uh, and so it's gonna be a lack tone with two metal groups here, Um, and so obviously ah, the signal for our metal groups are gonna be the methyl groups. Um, and then our first triple it with an integration of two is gonna be the two protons on this carbon. Um, and then our one that is further down field is going to be, uh, the methylene group that is attached to an oxygen. And that's why it's so be shielded. Okay, s So there's molecule a, um, And now, as far as a mechanism that accounts for its formation, that's gonna look like this. So the first thing that's gonna happen is going to be that a molecule of water is going to add to this carbon of the night trial. So after that first step, uh, we will be right here. Um and so the oxygen eyes only gonna have one lone pair, and it's gonna have a positive charge. Ah, and our nitrogen is now gonna have to own pears, and so it is gonna have a negative charge. Um, and to resolve those charges, um, and make everything have the number of bonds that it wants. Ah, we can just call this form Formerly a proton transfer. Okay. And so the result of that, it's going to be this molecule. Okay, um, and next thing that's going to happen here is that this nitrogen is going to get protein ated. And so to do that, we'll use some of the acid that we have in this molecule are some of the asset that we have in this reaction? Rather. Sorry. So this nitrogen will get protein ated like this. Ah, and the result of that will be this molecule. Ah. And so now the nitrogen is gonna have no lone pairs, and it's gonna have a positive charge. Um, and actually, let me change the way that I drew this eso what can have next, uh, is simultaneously, um this hydrogen on the alcohol can be grabbed by a water molecule. Ah, and at the same time, one of the carbon nitrogen bonds can break. And so the result of this is going to be the formation of a meid. Okay, s So we have a name? I'd now, uh, the next thing that's gonna happen, um, is going to be pro nation of Ah, the oxygen of our A meid. Ah, and so again, we do have acid in this reaction. Ah, and so the acid can carry that out. Yeah, okay. And so the oxygen now has Ah, one lone pair and a positive charge on it. Um, and of course, when a carbon you'll oxygen is pro donated like that, um, it makes the carbon carbon more reactive. And so that is going to allow this alcohol on the other side of the molecule to come and attack. Ah, that carbon. Uh, these electrons from the carbon oxygen double bond will revert to the oxygen. And so now we have formed a ring, which is which is good, right? Because we said the product molecule a was ring. Um, so it's good that we've started to form this ring. Um, Okay, so we have our tetra. He'd roll intermediate here. This oxygen has one lone pair and a positive charge. Okay, um and so the next thing that's going to happen eyes deep pro nation of that oxygen. So again, water can pick up a proton there. Okay, Um, all right. And so from here, we're going to pro Nate the mean, which is going to make it into a good leaving girl. I guess I should have written a little smaller, cause it looks like I am about to run out of space, so I'm probably going to have to finish this problem on 1/3 page. But so, anyway, our nitrogen has been pro donated. Eso um Well, maybe I can now I can't squeeze it all in. All right. Um, our nitrogen has been pro donated. So what's going to happen now is, um, the lone pair from this oxygen is going to add in to reform the carbon. Neil. Ah, and simultaneously the nitrogen will leave with those electrons from that carbon nitrogen bond. And so I continued on the third page here, um, that will get us to this intermediate and the very last thing that needs to happen. Eyes deep Throat nation here. Um, and that could be accomplished by water yet again. So you see the utility of water in these systems, they can accept a proton on after it's accepted a proton. And it's the age 30 plus it can donate a proton eso It serves almost as a buffer in some of these organic re agents. Ah, some of these organic reactions rather Okay. Ah. And so this is now our final product. Ah, a So again here on page three is the final step, which is just a deep throat nation. Um, you're on Page two is the rest of the stepwise mechanism guy. And then here on page one is the formula and spectral information that we used to deduce the structure of a, um, which happily agrees with what we got when we worked it out. Mechanistic Lee. And that's the answer to chapter 20 to problem number 65.


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