5

Draw the expected products of the following reactions. If more than one product is expected, circle the major product Hint: one should have 3 products (9 points, 3 ...

Question

Draw the expected products of the following reactions. If more than one product is expected, circle the major product Hint: one should have 3 products (9 points, 3 points each): CH3 OH HzSO4HzSO4CH3HzSO4OH

Draw the expected products of the following reactions. If more than one product is expected, circle the major product Hint: one should have 3 products (9 points, 3 points each): CH3 OH HzSO4 HzSO4 CH3 HzSO4 OH



Answers

Draw the products formed in the crossed aldol reaction of phenylacetaldehyde (C6H5CH2CHO) with
each compound: (a) $\mathrm{CH}_2(\mathrm{COOEt})_2$; (b) $\mathrm{CH}_2(\mathrm{COCH}_3)_2$; (c) $\mathrm{CH}_3\mathrm{COCH}_2\mathrm{CN}$.

Okay, so rescue the products in the final reactions. The first one is um their action between purple and mean and hbr so the nitrogen only mean has only a pair of electrons that can be used to grab parts on. So therefore it's going to be the base. And hbR whenever you've been acidly hbr hcl or a strong acid like sulfuric acid, it's capable of doing its person. So therefore it's going to be the lowest acid and we'll be able to predict the reaction. So we're gonna get ammonia as well as we have bermuda suspected or ion and this is going to be known as purple ammonium salt. I mean purple ammonium ver mod super mod for burning ammonia for at age three and purple is the three carbons. And the next thing now has a positive charge because again it hurts on and doing a pair of electrons are negative. The next one is with protein and hydrochloric acid. So all Canadians can only undergo radical reactions. If you were to treat beauty in with either Cl two or B. R two and the presence of U. V. Light, we would have radical action and add chlorine to the terminal carbon. But in this case al canes can't react with hydrochloric acid. We could react this with protein and this would produce to iota butane because the key is going to be the nuclear fall and department the acid and we would get a secondary card with that on. But in this case we would have no reaction. The next one is with purple ammonium and the high journey of my arm hc it boss. So that should no longer has a lone pair of electrons. That could be used to grab note for a time. It's therefore, we're not going to have a reaction here in a that I shouldn't had a little pair of electrons that can be used to grab writes on here. We don't have that X one is um be purple ammonium with hydroxide. So this reaction is under basic conditions when we're when we reacted with hydroxide with H three plus, it would be under acidic conditions and hydroxide is going to be the base, ammonia is going to be the acid. It's gonna do any sports on and electrons going to flow back to the nitrogen. So we're going to get purple and meet again and it lost the positive charge because again, a pair of electrons that were negative and hydrogen, I mean the oxygen gained is that person giving us water and no longer has the native charge. Mhm. So the biggest thing here is just identifying which adam has a little pair of electrons. It could be used to grab proton and identifying the acid that's capable of doing the person. This could be hBr, hcl, hydrochloric acid or even a strong acid like sulfuric acid and identify the college base that could be used to attack the carbon cat on in the case of anarchy in reaction, where the intermediate would be the structure and county a beast of hydrochloric acid is just I on, which can it's had the carbon count on and give us the product to iota pew tune. So you really have to follow the flow of electrons and see where they're going.

Okay. This problem is asking us to predict product of these reactions. So for the 1st 1 I noticed that I have my two Carbonell components, right. I have to carbonate molecules in which I have my alpha beta on Saturday system here that's characterized by the president of this Carbonell and on the Alfa Beta position, I had the presence of the soaking. So get Alfa Beta on Saturday system, and then my right side. I have this bay taquito, Esther, and that is my kito. And then on the basic position relative to that key tone, I have the presence of my Esther. So getting bay taquito, Esther. Okay. And that is all going to undergo erection with my socks side, which is going to be a super strong base. Okay, so my super strong base of the stock said is going to go ahead and take off the most aesthetic hydrogen off of these two molecules. So among these two molecules, let's decide which is the most aesthetic hydrogen, And it is actually going to be this hydrogen, right, The awful hydrogen. And that's because we have we actually have a couple different, awful hydrogen to choose from We have this one. For example, we have this one. We have this one, but the most acidic one is the one that is going to be sandwiched between my two carb meals. So this carbon on this one if are too deep throat into this hydrogen, I would have a information of a set of lone pairs. Those lone pairs can go ahead and be d localized into either direction corresponding to the Cardinals. Okay, so that is why that one is most pathetic. And let's go ahead and deponent that. Okay, so I'm actually gonna drop the mechanism for this. I'm just gonna go ahead and take off hydrogen, take off that hydrogen. Meaning I'm gonna have a set of lone pairs on that carpet. So just right here. Okay, so now this set of one Paris can go, had an attack. An electra file. So my electrified in this case is going to be this carbon right here, the beta carbon. And that's because, as we can see here on this alpha beta and cetron system, we have to electrify. Look, sites we have this carbon deal and we have this beta carbon so only strong superstrong nuclear falls. Attack this Carbonell in an alphabet on Saturday system or as a weak to moderate to relatively strong nuclear falls. Attack this beta carbon. Okay, Only the super strong nuclear foes. Such a strong reducing agents like lithium aluminum hydride Renard's Those will attack that carbon. OK, but if we're in alphabet on Saturday system, the only Electra file is going to be the beta carbon. Okay, so with all that in mind, let's go ahead and draw out the product slash intermediate. Okay, so I'm gonna have the same exact molecule appear gonna duplicate that, and I'll place it over here. Getting the connected to this carbon is going to be where this Mollica comes in. So after I took this carbon this carbon with my new profile, I'm gonna have to move the electrons from this carbon carbon bond that double want onto this carbon. Meaning? I'm gonna have to move the electrons from this carbon auction bond onto this auction because the oxygen is considered to be Electra negative. Okay, so with all that in mind, let's draw that out. I'm gonna have the connection to my carbon, which was the beta carbon. We have a connection to my single bonded carbon 100 bond. And over here we have the connection. I see three and connected to that, right? Most carbon. We have a connection to my double bonded carbon, which is single bond into my auction. And then over here, we have the connection to O. C. H three if they should be NH two. Okay, so that is my molecule so far. And now I'll have to do is basically do some touch ups to fix these molecules because this right now is not in the most stable, for I basically have to move the electrons back down from the auction right here. When I move them back down under the single bond to create a double blind, I'm gonna move these set of electrons onto this carbon. So there's extra couple of different ways to show this because after I moved the electrons onto this carbon, I'm just gonna have to protein ate that carbon again. So I'm just gonna actually move the electrons from that talking directly onto an acidic proton such as the I said a proton H o ch 31 methanol. Okay, so I'm gonna move the electrons from that OK, onto that hydrogen, and I'm gonna have to move the electrons onto that oxygen. Okay, so my product is going to be this basically the same thing that I had before, so I'm just going to dedicate the majority of it, takes it just like this. And then instead of having those extra double bonds, I'm gonna have this structure right here in which I got rid of that single bond auction, and now it is a double bond. And then instead of having that came there, I'm just gonna have a single bond with ch two. So that is my structure, and then that looks to be it. Okay, so moving on to this next one. Okay, so for this next one, we have basically a very similar molecule because, as you can see, we have this alphabet unsaturated someone my left. And we also have this, um, very similar to the Bay Taquito, Esther and which we have the presence of this beta Kato nitro. Okay, so let's go ahead and consider that with my base. So I based once again is gonna take off the most acidic hydrogen in this case because we have this alpha beta and sexual system. We're gonna look to this one instead. The basic, you know, my trial. We're gonna take off this proton, okay? Because, as you can see, even though within that trials against a little bit funky in terms of the job geometry, this is basically the same thing is a carbonell. So we can also just drop this like, um, like this if you wanted to, so that would be contrary to what the actual geometry of this molecule is. But that might be a little bit easier to recognize, because again, we have this set of proton in the form of my alfa hydrogen. Okay, so I'm gonna take my pro my base gonna deport at the hydrogen, moving the electrons onto this carbon. So once again, I'm gonna form in Mollica. That looks like this. In which I got rid of that 100 I moved the electrons onto that carbon. Okay, so now I can go ahead and move those electrons onto my beta position of this alphabet on saturated system, just like that. And once again, once I moved the electrons that carbon this carbon, the electrified I'm gonna have to move the electrons away from this double bond. I'm gonna move them onto this single bond, and I'm gonna move the elections from this double bonded oxygen onto the oxygen to my product. Slash intermediate is going to be this in which I have when I drop us a little bigger so I can see all the same screen I have my 63 when I have my carbon auction I c h connected to my well, well, actually, just put my drawn the correct form now. So it's gonna be carbon triple wanted to my next June and then on that carbon, I'm gonna have the connection to the remainder. My compound on the left. So I have a connection to carbon. I have the connection to my C h two ch three, and then I have the connection to my single bonded That's going to actually hydrogen have a connection to my single bonded carbon, where the hydrogen and my double bonded carbon with its If you were the night a charge and then over here I have my O sage three. So I'm running out of room, but I have my o series three. Okay, so with that in mind, let's go ahead and fix this up. So again, I'm gonna move the elections down from this auction onto the single bond to make it double one. I'm gonna move the transfer this cocaine onto anesthetic proton. So, for example, methanol, Ocho. Okay. And that will fix up my molecule to get my correct form, which is going to be this. I'll have my auction connected to my carbon ch three on this side ch ch triple bonded my carbon at least to my nitrogen. So that's my night drag component. Gonna have the connection to on this carbon. I have my carbon hydrogen with my single bond and carbon, which is gonna be ch two. And then I have a connection to my single bonded carbon with my double wanted auction and then oh, ch three on this side And then finally on the left side, we have the connection to it's huge to and see it through. Okay, so that is my final product. And that is going to be my two final products for these two reactions

Okay, This problem is asking us to predict the product of each of these reactions. So first up, I have this Esther direction with my my fox side and a second step of HBO. So notice how we have missed oxide gets oh, connected to a stage three, and then we also have my oh connected to see a three on my reacted as well. Okay, So the reason that we're using the specific base instead of something like starting hydroxide or even Fox said, is because we could have a possibility for trans est verification in which we have a nuclear Phil Kessel substitution and that Oh, say three comes in and attacked my carbon deal. Because this Carbonell is in fact, considered to be Electra Filic gets The reason we're using this specific base is because we could have a situation in which that will save three eventually replaces this Oh, ch three. Okay, so that's the reason why we use while we avoid using something like the fox I because we could have a situation in which we just have oh connected to ch two connected to see his three. If that was the base that we used, but that's kind of an anomaly. The more likely result is going to be an acid base reaction. And the acid base reaction is basically going to be the deep rotation of the most acidic hydrogen. In this case, it's going to be the Alfa hydrogen. So using this based, I'm going to go ahead and deep protein ate that hydrogen moving the electrons onto that carbon. Okay, so the result of that is going to be as follows. And for the remainder of this problem, I'm gonna just draw the skeletal structure just because we do have a lot of carbons that we were working with, and I want to avoid taking too long on this. Okay, so I have 1234 carbons. So 1234 carbons. 234 cups. That's one extra. So 1234 carbons and then have oxygen up here. And then my Oh, siege three over here. Okay. And then again, I d pro nated this carbon, I'm gonna have that lone pairs associated with carbon. Okay, so next up similar to an album conversation or an album additional production, I'm going to have that nuclear file So this nuclear father just created I'm going to have a attack, Another molecule of my starting material. So I'm gonna have another Mullick of this. Esther, I'm going to behave as an elector file to this nuclear file. So something is before I have my molecule. I'm going to use the lone pairs here to behave as a nuclear file to attack this Carbonell get when I attack that Carbonell, I'm gonna have to move the elections up to this oxygen. And that's going to perform this product results in that product. Okay, so I'm gonna have my Esther, and then over here, I'm gonna be connected to my other Esther. I'm gonna form a Tetra Hydro Intermediate in which this carbon correspondents to this carbon and then this carbon corresponds to this carbon. Okay, So connected to that carbon, I have a connection to an oxygen which has a negative charge in it. Okay. And that was my former carbon. You and I have the connection to my with oxide. And then last but not least at the connection to 123 more carbons. So 123 more carbons. Okay, so this is where I'm going to do my eso substitution medically, Phil Kessel substitution? Because right now I have a Tetra Hydro Intermediate, right? This carbon is Tetra federal. I'm going to make it into an SP to hybridize carbon. Okay, I'm gonna do that by go ahead and moving the electrons from this oxygen doctor, that single bond to create a double bond. And in the process of doing that, I'm going to go ahead and make my myth oxide intimate making group. And that's a favorable reaction. Simply because I started out with a Fox said I'm going to end up with Fox said, OK, so it's kind of behaving as a catalyst. Okay, so this is going to be the result. I'm gonna have my oxygen o with my metal group, and then this is going to be connected to it. My auction and then my 123 more carpets. Okay, so that's the result of this butts because I did have to get rid of my meth oxide as my leaving group that is still considered to be basic. And we still we still have this acidic hydrogen. In fact, that's even more acidic than we started off with. So that aesthetic hydrogen is going to get deeper nated by this basic with oxide, it's gonna take off the proton, meaning the electrons are going to go onto this carbon. Okay, so I should end up with this after step one in which I have my Esther. My 23 carbons. Oops. 123 carbons and the connected to that awful carbon have a connection to my Carbonell and then my 123 carpets. But again, I do have those lone pairs on that carbon simply because that's an ascetic proton which I had to dip resonate with My fox said Okay, so the second step, the purpose of HDL, is to go ahead, and protein ate that lone pairs that carbon. So, using a CEO, I'm just gonna go ahead and Protein ate my nuclear fuel. Carbon our basic carbon. Okay, so the end result should be this in what you have. In fact, I'm gonna reorient this to be in a more recognisable form. I'm going to have my company out there way of my three carbons connected to it. So 123 carbons. And then on this carbon. So this carbonara here corresponds to this carbon, I'm gonna have a connection to 12 more carbons case that is going to be my final product. That is my bae Taquito. Esther. Okay, next up, we have this one. So this one is kind of similar. And see how we have the presence of this O. C. H. To save three. We have the same base. So again, we want to avoid trends s certification, which is basically replacing my oxygen group with whatever is right here. But again, that's kind of an anomaly. It's kind of rare, but basically, we have that base. It's gonna go ahead. And D protein eight. My most basic hydrogen, which is this one, My awful hydrogen. I'm going to go ahead and erase one of those. Replace it with that, this one that I'm going to go ahead and D protein. You did it using my if oxide so deep throat need it Moving the electrons onto this carbon, resulting in the formation of this product. I have my carbon compound like this. And then over here I'm gonna be connected to my carbon eel like that on my own and the effort group. And again, I'm having the electrons being moving onto that carbon. Okay. And other. See right here I have the presence of two of these molecules, so I'm gonna go ahead and cracked it with another molecule of my starter material. So if I have another molecule of starting material, when I use that carbonara to go ahead and attack this Carbonell when attacked by cardinal, I'm gonna have to move the electrons up to this oxygen like that. Okay, so this is going to be the result. I'm going to have my Carbonell compound slash Esther and then on this awful carbon That was my former aesthetic proton. I'm going to be connected to my Carbonell. So this card right here that corresponds to this carbon done attack and connected to that carbon have a connection to my auction with negative charge. So that auction with negative charge corresponds to this one and then also connected to that carbon. I have my O with my Ethel group, which is right here, and I also have this group over here, which is connected to one two, and then my I saw a clip. Okay, so that's what I have right now. But again, I can't just end like that. I'm gonna have to get rid of something. In fact, I'm gonna get rid of this group. Okay, so right here is my Tetra Hydro intermediate. I'm gonna move my electrons down from the oxygen onto this single bond to make a double bond relieving myself of that tetra hydro intermediate and getting rid of this. Hey, Fox, that in the process said this is going to be my result of that. Going to have my Esther for here and then attached to this awful carbon. I'm gonna be connected to my carbonell and then my job, You'll group just like that, but again, because we didn't have to get rid of its oxide as my leaving group. That a fox that is still considered to be basic, right, so can go ahead. And D protein ate the most excited hydrogen, which in this case, is the one attached there. Right here. Okay, so it's gonna go ahead and department the hydrogen, moving the electrons onto this carbon. Okay, so after that, I'm gonna have the formation of this product in what you have. Basically the same thing is before, except I just have a suitable impairs on that carbon. Okay, lone pairs right there. And then those own Paris can go ahead and be probated simply by using HCL in a second step. So then I have HD over here. HCL is gonna be the source of protons for that set of lone pairs. So my final product, as I rearrange it, is going to be this in which I have my cardio, my master, and then over here have the connection to Mikey Tone, which is going to be a nice, beautiful attached to it like that. And then on this carbon have the connection to this group right here, which is my isopropyl group. So here's my isopropyl group. Okay, so that is gonna be my final product. That is my Bae Taquito, Esther.

Okay. This problem is asking us to provide the different products in these different reactions with this compound right here. So let's go ahead and try this out. So first up, we noticed that we have basically the same exact reagents as we do on either side. We have L D A and T H E f corrected with D 20 in the second step on that is gonna be on both sides. The difference is the presence of the difference in temperature. Right. We have negative 78 degrees Celsius here, and we have zero degrees Celsius over here. So this one is going to be performed in cold conditions, and this one is going to be considered at least more hot than my other one. This one will be considered hot. Okay, So what is going to be the products for each of these cases in cold conditions? I'm gonna have a kinetic reaction is going to occur, Basically not rapidly, but it's gonna make the most available, deprived nation step. And what I mean by that is that we have these two awful hydrogen on this Carbondale, right? We have enough of hydrogen here. We have an Alfa hydrogen attach the carpet as well. Okay, so the difference is that my l v a my base, my strong base is going to deep rotated. The one that is going to have the most availability slash accessibility, Right, Because this wants to happen as fast as possible. It's not gonna want to take off this proton simply because that proton is to stir quick tender to access. It is not as accessible. Where is this Proton? It's not hysterical. Hindered. So my base is gonna come in and deep rotate deeper, Nate, that hydrogen moving the electrons onto this carbon. Okay, so the result of the happened after that first step is going to be this in which I have my cycle vaccine Mikey tone over here and then on this carbon, I'm gonna have my set of lone pairs, okay? And then the second step of D 20 is basically just pro nation with an isotopic relabeled hydrogen. It's just gonna be the pronation with deuterium. So it's just contak my de of my overall detour. Okay, as it does that, I'm gonna get this as my final product in which I have my key tone. And then over here I have a connection to my deuterium, and then it's gonna be my method group. Okay, so that would be the structure of L V A. T H f in negative 78 degrees Celsius. Flesh cold, colder action with the same exact reagents as my hot reaction. OK, so now let's consider my hot direction and I'll do this part in green. Okay, so in green we have zero degrees Celsius. It's gonna be a little bit harder than my negative 28 degree Celsius. So this means that we have a little bit more chances for thermodynamic stability and by thermodynamic stability, I mean that we're going to want to make the most stable conformers possible because we have the hot conditions. So what I mean by that is that right here, obviously, in the first step in the first reaction, we had this base react with this proton simply because that proton was the most available. I wanted to have the kinetic reaction, but because I had the presence of heat by zero degrees Celsius, that heat is going to facilitate the thermo dynamic product, and that thermodynamic product is going to be the de pronation of the hydrogen that will lead to the most stable conform. Er, So using my base, the same base is before just in the hot air environment. I'm gonna take off this proton because even though that proton is technically more hysterically hindered, I'm gonna form a more stable product. And that is going to be a situation in which the electrons move onto that carbon. So just like this, I'm gonna form. This is my intermediate just like that, in which I have my state of own period on that carpet. Okay, so usually with inorganic chemistry, every time, every time we see a car ban on a tertiary carbon kind of hesitate because in reality we would most likely wanted to be located on a primary right. It would be the opposite of a church, a red carpet. But as faras and items go, if our to move these electrons onto this side of my Carbonell, I can go ahead and d localize that for me in my stable all keen. So, for example, out form this Elke with my intellect, I don't Okay, So, as we can see, this alky informed is actually pretty stable. And that's because it is a concern is considered a tetra substituted. Okay, we have to connections on the set of my team and two connections on that side of my Okay, Okay, so that is a relatively stable cocaine versus If we're to take off this Proton and that product product, we had form this reaction such product. We'd have my in the late I on over here in my o Keane over here. So as we can see, this product is not as stable as this product, and that's simply because of the substitution on the hour came. So the more substitution are, the more substitute this falcon is, the more stable it is. So again, this is going to be a thermal dynamic product in which would make the most stable conformers. And that is gonna be this one right here. Okay, so with that in mind, let's go direct that with in D 20 So DT will is gonna basically protein ate my carbonara. So I'm gonna have the presence of my d o d. This carbon on is going to attack that deuterium moving the electrons onto the OD, eventually making this is my final product key tone. And then deuterium added to that carbon. Okay, so again, we have kinetic product and thermodynamic product.


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