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7. Rank the following compounds in order of increasing acidity. Place the number rank (1= lowest acidity; 4 = highest acidity) in the blank below the structure (4 p...

Question

7. Rank the following compounds in order of increasing acidity. Place the number rank (1= lowest acidity; 4 = highest acidity) in the blank below the structure (4 points)OHCH3CHzOH(CHa)3C-0-H CF3CHzOH

7. Rank the following compounds in order of increasing acidity. Place the number rank (1= lowest acidity; 4 = highest acidity) in the blank below the structure (4 points) OH CH3CHzOH (CHa)3C-0-H CF3CHzOH



Answers

Rank the following compounds in order of increasing acidity.

Acidity is essentially just a measure of how readily a molecule will give up a proton. So I've drawn here, the three molecules were asked to rank in order of decreasing acidity and then some possible carbo cat ions that form when we get the donation of a proton. Now remember that the donation of a proton can technically occur any position, but some positions will be more energetically favorable than others. And in this position we actually get a conjugated pi system here and here in these two molecules and we can form some residents structures which lead to stability. So in this molecule we can actually form this resonant structure. And here we actually have two possible resonance structures. We have this one right here where we push the arrow in this direction. Can we get this conjugated system or we have this resonant structure where we push the era from that direction and we get this conjugated Hi system. So what does this mean? Well, in this species, right here, this will be the most acidic. This will be most acidic number one because we have the three resident structures from this conjugated ah Hi system. Once we have the cat, a cat, excuse me, carbon county on this will be the most stable with the most electron de localization. And since this has one resident structure, this this can I on. That formed will be the set will be the second most stable species. And when we have no conjugated pI system, we only have one resident structure. And so that will actually be the least stable carbon can I on and therefore the least acidic proton at this position. So in terms of decreasing acidity, um it's going to be this one first, this one second and this one third. Um So there you have it.

This is the answer to Chapter 19. Problem number 47 Fromthe Smith Organic chemistry textbook. Ah, And in this problem were given these three molecules and were asked to order them from least two most acidic. Um, and then we're asked to give an explanation as to why. Ah, and so this, uh, middle molecule here is going to be least acidic. Ah, and as far as why it will be least acidic. Um, it's because ah, the car back silly gas. It is not substituted directly on the aromatic. Right? The car back Selic acid is separated from the ring by one carbon. And so Ah, this car back. So, like, acid is not much different than any regular alcohol carb. Oxalic acid. Um, we don't really get are the stabilizing effects of the aromatic ring. Um okay, so that's, uh that's the least acidic s o the intermediate, uh, acidity molecule is going to be this one. So inter immediate acidity. Um, and again, uh, with the car box, Silly gas. It's separated from the ring by one carbon. We're not really going to get the same strong affect that we would if the car about Selic acid were substituted directly onto the aromatic ring. But we do have the nature group on the ring, so there's gonna be some inductive electron drawing effect. Ah, and so this molecule will be somewhat more acidic than in the middle molecule here. The least acidic molecule. Oops. Say at least acidic. Um, and then so that the first molecule is going to be the most acidic. Um, and it's going to be the most acidic, Uh, four Two reasons. Really. So the car box silly acid here is substituted directly onto the ring. And so we have a ll of, uh, the the possibilities of drawing resident structures where we can, um uh Well, okay, so I'll actually just just draw this resident structure. So, um, the electron withdrawing effect of the para nitro group is going to increase the acidity here of the other thing and let me move these over. Okay, um is without drawing every every single resident structure that we could draw here. There is a resident structure. Once this this is the protein ated. So let me go ahead. And de Protein ate this. So it'll look like this negative charge there. Um, so there is a resident structure that we could draw where this molecule is going to look like this. So we would obviously have to move through some some residents structures in between. Uh, but there is a potential residents structure. Ah, where we have our car box Selic acid here while our core box late. I on rather here. Eso like that. Um And then we've We've moved the positive charge from our ah nature group. Well, eat. Okay. You know what? Um, sorry. Ah. Trying to be as as un confusing as possible. Um, and also trying to draw as little as possible. So Okay, so the nature group the night your group looks like this, right? And so this is how we conceive of a nitro group where one of the oxygen's has two bombs into lone pairs and no charge. And the other oxygen has one bond and three lone pairs in the native charge. And then the nitrogen has ah, positive charge on it. So this is what, um this is what our r d protein ated? Uh, Karl Marx allee gas. It would look like and so we could draw a resident structure for this where So we can draw a resident structure where both of these oxygen's have negative charges are the nitrogen still has a positive charge. This oxygen still has as negative charge from being Cipro nated. Uh, but we could, um, move, move electrons like this so these could go here thes could go here, and this bond could break. And so if we do that, we end up with a positive charge on this carbon. And so now, now we have positive charge in a negative charge that are closer together, and so they're gonna help to stabilize one another. Um, And so for, for all of those reasons, that makes this first molecule here by far the most acidic. Ah, and that's the answer to Chapter 19. Problem number 47.

We want to wreck the following acids in increasing acidity. So remember the lower peak a the stronger the acid so lower PK is equal to stronger acid. So let's just go ahead and label these 1 to 4. So what is the lowest peak? A. That we have are actually, since they say increasing, let's start with the highest peak A cause. I would be the weakest acid and will it looks like that 19.3. So first going to have acid tone and then the next highest is going to be this 9.9 for fiddle. Then it's the PK of nine for the 24 Penton die own. And then lastly, we have our acetic acid at about 4.76 so that would be the ordering in increasing acidity strength.

This is the answer to Chapter 19. Problem number 17 Fromthe Smith Organic chemistry. Textbook on this problem gives us two sets of molecules on and asks us to rank them from least acidic to most acidic. Um and so I'm going to to do that with numbers 12 and three. Eso won is going to be least acidic. Ah, and three will be most acidic. Ah, and to obviously will be intermediate on dso in both of these sets molecules, What we need to realize is that these are benzo gases, derivatives. So we're talking about an aromatic ring with the carb oxalic acid. Um, immediately substituted onto that ring. Um, and then, ah, each of these Siri's is basically comparing, um uh, electron withdrawing in electron donating groups. Eso it. Looking at Siri's A. We're comparing just Ben's oic acid to para claure Ben's oic acid into para methyl benzo Gassid. And so the metal group is going to be electron donating, and so that's going to decrease the acidity. Here s o ah, the metal benzo Gassid is going to be the least acidic. Uh, the just men's oig acid with no ah Paris substitue in't is going to be the intermediate molecule here and the para claure Ben's oic acid where the chlorine group withdraws. Electron density from this aromatic ring is going to be the most acidic here. And so really, uh, in Sirius B, we're making, um, not the exact same comparison, but essentially the same comparison. So we're comparing electron donating a and and withdrawing groups. Um, and so Ah, the meth Oxy group is going to be donating, so that's going to be the least acidic. Ah, the metal group is going to be slightly donating s Oh, that's going to be intermediate on. And then the key tone eyes going to be electron withdrawing. And so that's gonna be the most acidic. Um, yeah. And so Thio, answer this question. We really just need to remember the rules that the textbook lays out ah, for affecting the acidity, the way this substitutes on an aromatic rain affect the acidity of a benzo Gassid. And that's the answer to Chapter 19. Problem number 17


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