PM1B - Autumn Functional Group Chemistry Flashcards

Question

IHow is the strength of an acid determined, and discuss in reference to pka what makes a stronger and weaker acid

Answer 1

The strength of an acid is indicated by its pKa value: The smaller the pKa, the stronger the acid. The larger the pKa, the weaker the acid. pKa informs us about the equilibrium between the charged species (H+ and A-) and the uncharged species (HA):

Answer 2

what the lewid acid is doing is accepting a pair of electrons from the clorine. This makes the clorine atom in the middle electron deficient as in this case its got its electrons to itself, however when it donates and when lewis acid polarising and accepts that electron pair, the chlorine in the middle is now sharing its electrons with the lewis acid, and now its electron deficient. Now the chlorine that’s electron difficiant, so it takes the other chlorines electrons and make that electron deficient. So the chlorine at the end becomes electron defficiant, and what this is saying is, if we have ALCL3 to the lewis acid, with chlorine then we have a source of halogen plus (CL+) But in a exam if we just say CL2 means I have CL+ that is sufficient.

Answer 3

we understand that the aromatic ring, reacted with E+, in this case E+ is actually C+. The electrons in this ring will react with this C+. So this is like an addition step so far where we have added our electrophile to our aromatic ring, and we then get our carbocation here. Now we have our CL- here from what was ALCL4, which becomes ALCL3 (Back to our lewis acid?), And the CL – can grab that proton, that allows the carbon-hydrogen electrons, to form the new carbon-carbon bond, this neutralises the positive charge and reform our aromatic ring. So to summerise, the first bit is very similar to our alkenes, the new bit is saying, once we get to our different situation where we have got the hydrogen, ( we had the hydrogen in our alkene version, but there wasn’t a driving force for the alkene to reform) the driving force in this example is we reform the aromatic ring, that is energetically favourable. The other driving force is, the positive charge is delocalised and spread out across the ring,

Answer 4

The FEBr3 is a form of lewis acid, its going to do the same thing the ALCL3 did, it will accept the elctron pair, and the bromime will become BR+. Once we have BR plus we can do our reaction as we did with the alkene but we start it with a benzene ring. So the electron come out the double bond down the PI bond, react with the bromine which is going to be electron deificiant, that allows the bromine-bromine bond to break which will try and neutralise the charge over here to make this a more stable structure. This means we have added the electrophile to our aromatic ring. We may or may not be asked about resonance structures but in this example we will. So we have to think how to move the positive charge, we can use the electrons in this bond to do that/ So the electrons in this bond can form a double bond here, that will neautralise the positive charge, so the carbon lacking electrons will get electrons from this pair here, which will move the positive charge to the bottom. We can then draw another resonance structure showing the electrons moving out of the double bond here and neutralise that charge again. So what that means is, the carbon is laxcking in one electron, its given them away, by moving the electrons in this bond down to form a new bond here, that carbon gets its electrons back and is sharing the electrons in that bond. So we end up with 3 resonance structures in which the positive charge is moved to 3 different positions. That’s the resonance that shows how a charge is stabilised, but what we then do is, reform the aromatic ring. So we know is we have our BR floating around, what will happen is that will remove the hydrogen, the electrons in the carbon-hydrogen bond will form a new carbon-carbon double bond which will give us our product where we have substituted the hydrogen for the bromine., therefore forming an electrophilic aromatic substitution.

Answer 5

How we add something other then a halogen. This is adding a carbon which spreads our chain out so we can use other functional groups and we can build our drug molecules. Reaction mechanism same as previous slide, all we remember is once we see lewis acid again, and this time we have R which has some carbon atom which has an hydrogen attached to it. We put that in the presence of our lewis acid, this will give R+. Chlroine does the work where it reacts with the lewis acid, In this case the clorine doesn’t want to have that positibe charge, its given its electrons and its unhappy. So it will immidiantly take the electrons from the R-CL, that will give us R+ as our electrophile. Continuing from R plus – the it’s the same thing where the electrons come out the double bond, and it reacts with our electrophile. Then the electrophile adds to one of the carbons, (doesn’t matter which one). We then form a carbocation, then finally the hydrogen is used to reform that double bond. In this case the added extra is, we have our cl- which will help remove our hydrogen. (bottom example). We then end up with an alcohol benzene, we can then add this carbon chain to our aromatic ring. So essentially we can build our molecule up.

Answer 6

We have to recognise here that, for example if we have a primary carbocation here (CH3), its almost SN2 like, for example the electrons are coming out here reacting with this carbon that’s electron deficient because its got this great leaving group here. We can still do these reactions with primary carbocations, as the carbocation is not on its own as its attached to something different so we look at it as a whole. Normally primary carbocations are not stable but its different as we have our lewis acid in there. Middle one less likely as we have a bigger aromatic ring, so access is limited, so its more likely SN1 as the carbocation has to break off. If we see an example where the R-CL produces a primary carbocation. We may think its not possible as its not stable, However In this example we have a lewis acid in there it helps stabalise the carbo-cation, and it happens as a Sn2 reaction. If we have something that would form a stable carbocation its more likely SN1.

Answer 7

In this example and reaction we can perhaps perform something like an SN1 like reaction by making this a good leaving group, this would then lead to forming a tertiary carbocation, this would react with our neucleophile. But if we can add in our aromatic ring, our aromatic ring can act as a neucleophile. So we can do an SN! Followed by an aromatic substitution.

Answer 8

Do to this we need to have harsh conditions that can generate a reactive electrophile. So we take nitric acid, and react it with a strong acid, often something like sulfuric acid,, we will see nitric acid and sulfuric acid together. What this does is we protonate the nitric acid, this is hard as the acid actually wants to give a proton, so we have to put in something that’s more acidic, to provide a proton, for the nitric acid to get protonated with, that’s why it has to be a strong acid. What this does it forms this water leaving group and the nitrate can use its electrons that its got in this oxygen to kick out the water and form NO2+. All we need to understand is if we see nitrate acid and sulfuric acid together, in an aromatic reaction, what we are actually forming is a source of NO2+.

Answer 9

whats important is we recognise we have no2 plus, hopefully then we can see if we have an electrophile. The nitrogen is positively charged, the nitrogen doesn’t want to be positively charged, it wants to get some electrons back from somewhere. Its positively charged because the oxygen is more electronegative, so its taken the electrons. So its need to react with something that’s not an oxygen, that way it can get more of a share of electrons in that new bond. So we do the same reaction. So we begin by, the electrons coming out the double bond, react with the electrophile. For the nitrogen to accept that, the nitrogen has to do something as it cant have more then 4 bonds, so thenitrogen oxygen double bond breakes, that sends the negative charge and electrons to the oxygen ( which is fine because its electronegative and happy to take electrons). (2nd pic) then we end up in the situation where we have our nitrogen at the top added to our carbon, then we have our carbocation again. The only extra bit here is recognising we have to do something with this nitrogen oxygen bond. Showing resonance structures (bottom), we can show movement of electrons, we can take that charge we can move it to stabalise it, finally we have the hydrogen, but we don’t want it anymore, we can take the hydrogen away, use the electrons in the carbon-hydrogen bond, to reform our aromatic ring. And finally we have added our nitrate group to our aromatic ring (top right). It a substitution as the hydrogen has left and the nitrate has been added.

Answer 10

So here's an example where we've got our benzene ring. when we add nitric acid and another strong acid like sulfuric acid, it gives us a source of NO2+. The combination of nitric acid and sulfuric acid gives you a source of NO2 plus. So that's our electrophile. So then we just go through the process. We say we take the electrons out of the aromatic ring, we form a new carbon-nitrogen bond, the nitrogen can't have more than four bonds. So we need to allow space for that extra bond to form. And we do that by breaking one of the nitrogen oxygen double bonds. That leaves us with this structure here. We recognize that there are these resonance structures. So we can move that charge around the ring. And then finally, we recognize that when we form an aromatic ring, we lose the proton, we use those electrons to reform the aromatic ring. you can choose whichever double bond you want. It doesn't have to be this one. I'm just using the same one each time so you can understand that it's the same no matter what the electrophile is. But you can use any of these bonds.

Answer 11

So we've got our electrophile, its undergone that first reaction. Imagine if we have another substituent on it. So we're just going to call that x in this example. If that x sits on this carbon here, that ends up with a positive charge. So in fact, the carbocation, if our substituent is sat next to that carbocation, it's going to have an impact. Because what we know is that carbocation is saying this is an area of electron deficiency. So we have something which withdraws electrons away from that. It's going to make that carbon even more electron deficient. So that's going to stabilize that carbocation. So that will have an impact on what happens in the reaction. On the other hand, if we have something which can donate electrons into that, that’s going to stabilise the carbocation. We've seen other examples where if you stabilize the carbocation, that often helps to drive the reaction forward. The SN1 reaction. It's only going to work if you have a stable carbocation. We looked at actually with even just with a simple aromatic system, the fact that you can move that charge around the ring stabilizes the carbocation. And that makes it more favorable. Things that stabilize the carbocation generally are going to be good for your reaction. Thinks it's Destabilize a carbocation are not gonna be good.

Answer 12

The effect of the substituents will all depend on whether it can influence the carbocation. If it can't influence a carbocation is not going to have any effect at all. It depends on what kind of substituent you've got. Whether it's electron withdrawing, electron donating. It's also going to matter where on the aromatic ring your substituent is. Something we haven't talked about yet, is there are different positions on that ring. What we mean by that is we've done our first part of the reaction. We've added our electrophile across this bond and we've formed the carbocation. from this carbon (electrophile) that we've added it. We can name the different carbons that are now in there. So what we mean by ortho and para is the one directly next to the substituent. So at the moment, the one that got the carbocation on, that is considered the ortho position. The carbon up here, there's also next to the electrophile would also be considered an ortho position. The next one along. That's now next to this carbocation that is considered the Meta position. This other carbon involved in this double bond here, this would also be considered the Meta position. The final one is the one that's directly opposite would be considered the para position. And that's important because the ortho and the para positions are electron deficient We're just saying, let's do the same reaction we've done, We add our electrophile with the electrons out of the double bond at the aromatic ring, we end up with a carbocation. So this carbocation at the moment is in the ortho position. So that's telling us that carbon is electron deficient. If we draw a resonance structure, we move that positive charge around to here that positive charge now sits on the power position. So that's telling us in our resonance structure that the ortho position and the para position both have electron deficiency. If we move it around, again, it goes back up to here, which is also the ortho position. Not once have we drawn a resonance structure where the charge ends up in the Meta position. And the naming that we're using is about looking at where the electrophile is adding. And then saying this is ortho, meta or para, It's not about taking the substituent and saying where it's adding. So if we know the ortho and para positions are the ones that are electron deficient in our resonance structures. What that means is if we put a substituent in the position, it's going to have the biggest effect because the meta position could have a substituent, but because that positive charge never sits on the meta carbon, then anything in that position won't have an effect on whether you can stabilize or destabilize this charge. And if you have an electron withdrawing group, it will destabilize. An electron donating group will stabilize.

Answer 13

There were two ways that we can do this. Up to now, we've often talked about inductive effects. If you think about hyperconjugation when we're talking about stabilizing the carbocation, that's often through the sigma bonds. We can do the same thing here. There are 2 things to things, they are inductive and resonance. Inductive effects is about pushing electron density through the Sigma bonds. Here's an example. We start with toluene, We add our electrophile. And you can now see that these two are ortho to each other, the electrophile and the CH3 group are ortho to one another. If we then do the resonance structures, which we know, we can go from the ortho here to the para here. So the ortho here, we can see that the substituent here is going to have an effect on that carboncation Because the carbocation sits on the carbon that the substituent is directly attached to. And with a methyl group, just like with the carbocation with hyperconjugation. There's this inductive effect. What that means is they can push electrons into that positive charge through the sigma bond. So that's something that's favorable. Now, if we imagine something like a nitro group. So a nitrile group is a classic electron withdrawing group. If you're not sure about whether it's electron donating, electron withdrawing. Towards the end, you'll see there's a lovely little table that says these groups are electron donating, groups are electron withdrawing. These groups are a bit of both. something like a nitro group, which is very electron withdrawing. And we can understand why it's electron withdrawing. The nitrogen itself has got a positive charge. Nitrogen is more electronegative than carbon. The nitrogen is already electron-deficient because of the oxygens. So any spare electrons, it's going to pull them towards itself. So having this situation where you've got the two positive charges next to one another, that's gonna be very unfavorable. So this is where these kind of electron withdrawing groups, will draw the electrons away from the carbocation that would de-stabilize a carbocation. That's not gonna be very favourable. If you have an electron donating group, you're going to have a more favorable reaction that's in the ortho position.

Answer 14

Exactly the same thing happens in the para position. So this time we're asked. can we use a different double bond? You absolutely can. So here we're saying, here's our group up here. If we add the electrophile at the bottom here, these are now para to one another. so whichever way you count it ortho, meta, para or if you go from the electrophile, ortho, Meta and para. So they are para. The relationship is a para to each other. Once again, if you draw the resonance structure, your electrophile is added here. In this bond. This carbon here has got the positive charge. If you use the electrons in this bond to neutralize that, to move it around, it ends up once again sitting on the carbon that the methyl groups attached to, or your substituent is attached to. And so once again, we have this inductive effect that can push electrons into that positive charge and stabilize it. If you have electron withdrawing group that will draw the electrons out and destabilize. So we can see with the ortho and para, the charge sits on the carbon that the substituent is attached. What about the meta position. Got exactly the same starting material. In this case, we're now using a different double bonds when adding our electrophile. And again, if you count, whichever way you do it, we go ortho, meta from the methyl group to the electrophile. Or if you started the electrophile, you go ortho Meta. So the relationship between those two they are meta to one another. Now we have done the reaction, the charge states here. If you do the resonance structures, you can move it across to this carbon here. Or you can move it down to the carbon at the bottom. What you'll notice is in the three resonance structures, not once does that positive charge end up on the carbon that the substituent is attached, so it never sits where the methyl group is. Therefore, the fact that that methyl group can do these inductive effects doesn't matter because there isn't anything for it to stabilize. Because this positive charge doesn't actually land directly on that carbon, the methyl groups attached to it. So what that's telling us is the ortho and para are the ones. When you see a group is ortho or para, it will matter if it's electron donating. It will be favorable. If it's electron withdrawing, it will be unfavorable. With the meta If it's electron donating, nothing will happen. But more importantly, if it's electron withdrawing, nothing much will happen. Won't provide unfavorable interaction. What we'll see what that means is that with ortho and para substituents, if you have an electron donating substituent, That's going to be beneficial to have in the ortho and para position. So if we have a substituent on your ring. You electrophile is going to add either ortho or para to that substituent because that will provide this stabilizing interaction. From here. If you have an electron withdrawing group in the ortho and para position, that'll be de-stabilizing. So it'd be very unfavorable. So if you try to add an electron withdrawing group, it's not going to add to the ortho or para position because that's not going to be very favorable, It's going to add to the Meta position, so the electron withdrawing group adds to the Meta position. There is no destabilization from that electron withdrawing group.

Answer 15

The kind of inductive that gives us that signal. But we can also look at what's called a resonance stabilization or resonance effect, which is in theory exactly the same. So all of what we've said still stands. We add our electrophile. If it adds to the ortho position ( diagram) our charges are currently here. We can move that around until it sits in this position where the substituent is. The resonance effect is about, there's a lone pair of electrons on the substituents. So in this case it's an, it's an NH2 group. Nitrogen has a lone pair of electrons. We have a positive charge here, and we know that that positive charge is electron-deficient. It's a bit like when we're doing addition of bromine to an alkene. The bromine have lots of electron density. the carbon right next to it was lacking electrons so the bromine donated them to form a three membered ring with the bromine in it. What we're saying here, we've got extra electrons here and we are lacking electrons here. We can actually donate that pair of electrons. What that means is we've got our standard three resonance structures where we share this charge around the three carbon atoms. But if you have a group that's able to do this kind of resonance stabalisation, resonance donation, we get a fourth resonance structure. What that is saying is our charge now is not just spread over three atoms, which we know is favorable, start with, but its also spread over a fourth atom. So that means it's even more stabilizing as its not just stabilized at the three atoms we spread over four.

Answer 16

Again, here we have our amine, we add our electrophile. We end up with our charge here. The NH2 group and the electrophile on meta to another. So ortho Meta. You move the charge around in your resonance structures. And again, there is no resonance structure that we can draw. That puts the positive charge next to that nitrogen. So although The nitrogen has this lone pair of electrons that it's ready to say, Look, you can use this pair of electrons to help stabilize that charge, because the positive charge never ends up here, it cant do that stabilization as efficiently. Then the other one will be the para position. Again, you add the electrophile, so this NH2 group and this electrophile are para to one another, we get our charge here again. We move that charge around the ring using our resonance structures. What we end up with at some point it will end up on the substituent. And once again, we can do that stabilization. That lone pair of electrons can be donated to that carbocation. And it provides another resonance structure. So another form of stabilization. Electron donating groups in those ortho and para positions can make a difference. They can stabilize it. So if you've got a substituent and you're trying to add an electrophile to your ring, we know that there's gonna be a carbocation formed because that's part of the reaction mechanism. So what it's saying is that carbocation will at some point end up on the ortho or para carbons because substituent is on there that can donate electron density in, that's going to be positive, That's going to be favorable. Therefore, if your ring has an electron donating group on it, your electrophile is going to add to either the ortho or the para position to that substituent. If your trying to add an electron withdrawing group. Well, it's not going to want to add to the ortho or para position because that's going to lead to this destabilization and very unfavorable conditions. So an electron withdrawing group will want to add to the meta position because at that point, it's a neutral effect. It doesn't get this stabilization or destabilization effect. So an electron withdrawing group will add to the meta.

Answer 17

The other thing is to recognize that you have a substituent on your ring, you will actually change the reactivity of the ring as well. So an example being, Here's another example of an electron donating group. You've got lone pair of electrons on the oxygen that can donate into the ring. If they donate Into the ring then we can draw a resonance structure where we formed a positive charge up here, and a negative charge down here. This is not, it's not a formal charge, but what it's saying is we're recognizing that the electron density can flow through this series of Pi or p orbitals and pi bonds. And it can reside here. We can then move that pair of electrons again through the ring. Each time we're doing that, we're adding an extra bond to this carbon. The carbon here has got four bonds. Remember there's a hydrogen in there, and it can't have four bonds, so it needs to break a bond. We break the double bond. That would put the electrons down here. Again, you can do the same thing. You can use those electrons to move around the ring. What this is saying is these four resonance structures, the electron density in the ring can sit on the ortho, not the meta. They can also sit on the para position. What that tells us is, if you donate electrons into the ring, that electron density will predominantly land on the ortho and para position, which means they have a slightly higher electron density. We know that in our electrophilic aromatic substitution, the ring is acting as a nucleophile. So the ring is the thing that's donating the electrons. So wherever we have higher electron density, that's the bit that's going to act as our nucleophile. So in that case, these ortho and para positions when we have a substituent on the can donate the electrons in, they are the ones that have more electron density. So they are the ones that are going to react with the electrophile. They are the ones that are more reactive.

Answer 18

Same thing happens with electron withdrawing groups, but rather than activating it, they deactivate the ring. Again, here's a classic example of a electron withdrawing group. We have this nitrogen here. Nitrogen at the moment is having to share all its electrons with two oxygens. They are far more electronegative than the nitrogen. So all of the electrons the nitrogen has are being drawn away into these oxygens. So the nitrogen, although it will still retain a positive charge. This is a more stable environment to the nitrogen because it's sharing some electrons with the oxygen, but actually it's sharing two pairs of electrons with the carbon. So in this bond here, kind of got more of a share of one set of electrons, but here it's got more of a share of two electrons. So what that means is it can take the electrons out of the ring, helps stabilize the nitrogen. Ultimately leads to the carbon here being electron deficient. And then we draw our resonance structures. We show, just like we have done many times before. We can show that charge being shared around the ring (3rd diagram) from the ortho to the para and back to the ortho position. And what that ultimately means is when you have an electron withdrawing group on your ring the ortho and the para position are electron deficient. It's the same picture as before. But when it's electron donating, these are more electron rich when it's electron withdrawing these are electron poor. So they are going to be less reactive. Once you have that electron withdrawing group on your ring. And you tried to add another electrophile. You're trying to add another substituent onto this ring. Well, the ortho and the para position aren't going to be reactive because they are more electron deficient. So if you're trying to do that, the meta is the one that still has its normal electron density. It's a little less just in general, but this is where the electrons are more likely to be. This is where it's going to react with another electrophile. We've got the directing effect that tells us where it's going from the point of view of once you add your second electrophile, there is the stabilizing effect or de-stabilizing effect of the charge that's formed. But we also can explain the directing effects via which carbon is reactive if we have an electron donating group on it, the ortho and para, more reactive. If we have an electron withdrawing group on it, the Meta position is more reactive.

Answer 19

many we get in a situation where we have a substituent that seems if it can do both electron withdrawing and electron donating, how do we decide what effect it has? And it really comes down to which one is stronger. So if we take the example that we've been looking at for electron donating with this NH2 group. So if the NH2 group is the substituent on the ring, the ring will react with the electrophile. In ortho position here, If it reacts with the ortho position, the positive charge ends up here. So the NH2 group can have an effect. What effect can it have? It can donate the electrons in to that bond. Resonance stabilize the charge and bring it up to the nitrogen. So We can form this resonance structure. Because the other thing that nitrogen can do is, the nitrogen is more electronegative. So the nitrogen and that nitrogen carbon bond is going to want to draw the electrons out of the ring. So this is already electron-deficient. It's attached to a nitrogen that's going to pull the electrons towards itself that are there anyway. So this inductive effect would be destabilising and this resonance effect would be stabilizing. Which one do we think wins out? What comes down to the strength of it. If you think the inductive effect is about just manipulating the share of the electrons. It's about saying normally it's 5050, but perhaps in a carbon-nitrogen bond, the nitrogen has got 80 percent but Carbs only got 20%. Here. We're talking about, I'm going to put in a whole two more electrons into this system. I'm going to stabilize this charge by throwing two extra electrons. So that's going to have a much bigger effect, adding those two electrons as opposed to a small change in the share of electrons here. The nitrogen might still have 80 per cent of the electrons in this Sigma bond. But the carbon now has a share of two electrons that it didn't have before. That share of two electrons is going to be much more powerful than the loss of the electrons in this bond. The resonance effect is stronger. And that's generally the resonance effect is going to be strong because it's about using two electrons that we weren't in the system to start with, then the inductive effect. So this amino group actually is a very electron donating group and a very activating group.

Answer 20

I guess do the opposite. They are very electronegative. And they also really don't like giving up their electrons or lone pair of electrons. In this case, the chlorine has a very, very strong inductive effect. It withdraws electrons away from that ring an awful lot. But actually, it's very, very reluctant to donate its lone pair of electrons in to the bond. It's very, very rare that you see a carbon-chlorine double bonds. Because the chlorine halogens just they don't like giving up their electrons. So in the case of the halogens, they are. Electron withdrawing. So they deactivate the ring. However, because there is this ability to weakly release electrons, again, when it comes to the directing effect, they actually direct ortho and para. The halogens are one of those where when you see a halogen, you have to just think again. It deactivates the ring, but it does actually direct ortho and para. So although it has this electron withdrawing effect, it also has a weak electron donating effect, but because it's a resonance effect, when it comes to that kind of directing, it does actually help. And so it directs ortho and power.

Answer 21

So electron donating or electron releasing groups, these ones are very strong ones and you can see why they're strong. That single atoms or effectively attached and what this line means is this is where the aromatic ring is off this line. So the nitrogen oxygens are attached directly to the aromatic ring. They have lone pairs of electrons. They can do resonance stabilization and there's nothing else that could take those electrons away. They're really helpful, you can do that. Then the examples we just looked at, things like the amides and the esters. They have the lone pair of electrons that can do that donating effect. But it's not as strong because there's these competing resonance effects where the lone pair actually can be withdrawn by the oxygen. So they do still donate, but not as strongly as these ones at the top. And then you have some weak ones. So these are the ones that can only do inductive. They have no resonance stabilization, they are purely inductive effects. They do stabilized, but not as much. All of these electron releasing groups are ortho para directors, but all those reasons we've said, they can push electrons in and stabilize that positive charge on the ortho and the para position. You then have the electron withdrawing groups. So the nitrate one is the classic one that we've talked about. The Amine, you think, well, we've got an amine up here. So why is this one not so good? The reason is this one is protonated, positively charged. It's got three other groups on it plus the aromatic ring, because it's used its lone pair of electrons in those other bonds, they are not available for the resonance stabilization, so the effect for this, it's going to be inductive a bit like this nitrogen here again, the lone pair of electrons are being used to bond to these oxygens. So the only thing it can do is withdraw electrons. And it will do that very strongly because it itself is lacking electrons. Then you've got things like CF three, CCl three. And again, that's because that carbon is electron deficient. It's attached to three halogens. Halogens are pulling electrons inductively away from the carbon. So the carbon is electron deficient in its own right. So if it's next to another carbon that's electron deficient, there is no sharing of electrons, there is nothing you can do. The electrons are withdrawn. Then you have again some moderate ones. And this will be down to the fact that again, these carbons are all delta positive They've all got something else in them. And these ones here, predominantly these carbonyls we will see in the next, next week and the week after why that carbon is electron deficient. But it is. So these are again electron-deficient areas of low electron density. And if their neck to another carbon with low electron density, that's really unfavorable. And then the final ones that always throw the spanner in the works, these are halogens. The halogens generally withdraw electrons inductively, so they deactivate your ring. But because they do have. A bit of resonance stabilization. They actually direct ortho and para. You have to remember three classes. Theres the standard electron releasing groups, the standard electron withdrawing groups. And then there's a halogen sat on their own which do a bit of both. They deactivate but they also donate.

Answer 22

Basically the table that you've just looked at, we should be able to predict what product of these reactions should be. We've got substituents attached to our rings. We're adding in some groups. Where are those groups going to end up, are they going to end up in the ortho position, The meta or the para position. Nh2 is electron donating. And we know that electron donating will direct ortho and para. So we would expect it to either go here or here. Now, which one does it choose? what we need to remember as well is an NH2 group. Not only is it electron donating to direct ortho and para is also activating. Once you add the chloro group, it's still reactive. So the answer is it adds to both. If you put ortho or para, I'd be very happy. Okay. That's what I'm expecting. But equally recognizing that it can add to all three, because here this is the ortho position. Here's the para position. This is another ortho position. It could add to one of those first And then the second one could add in a third one. The other ones,You've got a choice to make. Down here. We've got an electron donating group, an electron withdrawing group. On, here. We have an electron donating group and another electron donating group. So they're going to have conflicting systems. So the electron donating group here with direct ortho and para. But of course it can't direct para because the nitrate groups there. So this donating group would only be able be ortho. But what about the nitrate group? Because the nitrile group will also have an effect. So the nitrile group is electron withdrawing so it would direct meta. So the meta position would be here. This is directly ortho. This CHthree would donate ortho but not para. So one would direct to this bond, one more direct to this area. But the resonance effect is stronger. Remember, this resonance effect would make the ortho for this more reactive. And so you end up with them adding across here. This one would add to the ortho down the bottom, And this one would also add to the ortho down here because the OH group has the strongest donating effect.

Answer 23

The simplest carbonyl now we can come up with is formaldehyde. The structure format of our Formaldehyde is made up of three atoms. Has a carbon in the middle, double bonded to an oxygen, and then there are two hydrogens attached. So we can understand that because the carbon is covalently bonded, its got three Sigma bonds, meaning Sigma bonds to three other atoms, which means it's used S orbital and 2 of its p orbital to make up those signals bonds. That's why it's SP2 hybridized. And then what's leftover is the p orbital? It's that p orbital which can then overlap and share electrons with the oxygen's p orbitals that gives rise to this dipole. So that's why it's SP2 hybridized. The other thing is trigonal planar. There's no stereochemical outcome of this particular reaction. there are much more complex systems and we can design it that way. But what we're going to talk about, we don't have to worry about what stereo isomer for me because it's Flat and planar, We can attack either side of the carbon oxygen double bond. So there is no stereochemistry.

Answer 24

Oxygen is more electronegative than carbon. So what that means, because your carbon is double bonded to an oxygen, the oxygen is more electronegative. It means the electrons in that bond in both, affecting both the double bond and the single bond that we've drawn towards the oxygen. So the carbon you can almost consider as being positively charged and the oxygen has been negative charge. Double bond is generally weaker than a single bond. And that's because the electron overlap isn't as strong. In a single bond we have orbitals, big overlap in the contact area, The electrons are shared in that orbital, and that makes it a really strong bond. In the double bonds or in these pi bonds, you've got the p orbitals and there's a small amount of overlap, but it's not overlapping as much. And so the electrons are a bit more, they're always more diffuse in that space. Means that if they've got a choice of where to go to be near to the carbon or the oxygen. They can gravitate a more one way or the other. All about carbon being electron deficient. If it's electron deficient, it mean if we add a nucleophile in our system, the nucleophile can attack. we need to recognize that because of the oxygen is actually very electron rich. It's got its own electrons. But it also has this greatest share of electrons in the double bond, carbon-oxygen double bond. It's electron rich. That means itself can act as a nucleophile and react with an electrophilic species.

Answer 25

we've got the carbon is electron deficient, we've got the oxygen's electron rich. It means the carbonyl can undergo nucleophilic and electrophilic addition. We have our nucleophile our electric electron deficient carbon and we have our electron rich oxygen. So this is electron deficient, we have got the nucleophile that has got this extra electron is negatively charged. It's going to send those electrons to form a new bond between the nucleophile and that carbon. The carbon, in order to receive those electrons, it has to make space for that. So how's it going to do that? Well, because the electrons in this carbon oxygen double bond, are already been taken away from the carbon anyway, we said that the oxygen's electronegative, it's drawing the electrons away. So the carbon will give its electrons up to the oxygen because it can get its electron from somewhere else, this nucleophiles and provide them with those electrons, which means the oxygen can take the electrons away. So what you end up with is you add your nucleophile, you break the carbon oxygen double bond. So we break this carbon-oxygen double bond and we get this negative charge. The other thing we can do is an electrophilic addition. So we can take our standard electrophile with its positive charge, it's lacking electrons. We've got this oxygen that has all these electrons plus it's drawn electrons away from the carbon. We've got plenty of electrons go around. They can use one of its lone pair to attack the electrophile. Donate those two electrons to form a new bond between the oxygen and whatever electrophile you have. Well that will do, it will create positively charged oxygen because the oxygen has given away its electrons. One of them, it keeps hold of in that bond. But one of them effectively, it's given away to the electrophile, so it now has a plus two charge. But what we can say is we recognize as being carbon oxygen double bond. The electrons are predominantly towards that oxygen anyway. So we can draw this resonance structure where we say, let's have the positive charge on the carbon rather then the oxygen, because we know oxygen doesn't like to give up electrons. its electronegative, it will be much happier to be neutral species or negatively charged species than a positively charged species. And what we're also saying, this carbon cation, is stabilized because you can spread it out across by the carbon and the oxygen. But in reality, it would be the carbonyl that is most electron deficient

Answer 26

we're likely to have a negatively charged nucleophile i.e OH minus the classic example of a base sodium hydroxide, you've got a negatively charged nucleophile. So that's going to do exactly what we've just seen before. It reacts with this electron deficient carbon. We then break that carbon-oxygen double bonds. We form this tetrahedral species. So we've gone from SP2 to SP3 hybridized. We've got our oxygen with a negative charge. At this point, we use sodium hydroxide that's usually aqueous, is in water. So we have water molecule floating around and this oxygen negative charge can grab a proton from water, extract a proton, break the bond between the hydrogen and the oxygen. What that does is it makes this OH minus an OH. You can also see that we end up with OH minus, which is our base and our nucleophile. It goes back to the start guys through the system again. That's basic conditions.

Answer 27

So we've got lots of electron density with this oxygen. A proton which has no electrons, really wants to get an electron from somewhere. The oxygen says fine. You can have some of my electrons for now. And we end up with this protonation that leads to the conjugate acid. What that means is that this can lose a proton and get back to where it starts. So once we've got this, if you remember, when we protonated our oxygen with it being electrophile, that ultimately did is we said, of course we recognize that this positive charge is on the oxygen as we have drawn it. But most of that is electron deficient, it actually sits on the carbon. So this carbon now is more electron deficient because of that positive charge than the carbon was on the starting material. So now we can take something like a neutral nucleophile e.g methanol or ethanol, which is uncharged, okay? But it has a lone pair of electrons on this oxygen that can be used as a nucleophile. Now we are much more reactive electrophile here, much more electron deficient carbon. We've got this lone pair of electrons and oxygen are in our nucleophile. We can now do that. Attack the nucleophile onto the acid, conjugate acid. We break the carbon-oxygen double bonds. Okay? And we end up again with our tetrahedral SP3 hybridized carbon. This time the oxygen now is negative, is neutral, no charge. The hydrogen that it got from here on it. The nucleophile now positive charge but like with the original example where we've got the water extract the proton, we need to recognise we can lose that proton quite easily. And that will give us this tetrahedral When we added the nucleophile and the OH group.

Answer 28

two things looking at a carbonyl. Either get charged nucleophile with an uncharged carbonyl. So that's our basic conditions. Charged nucleophile, attacking our electron deficient carbon, we break the carbon oxygen double bond, we form our sp3 tetrahedral intermediate. We then grab a proton to get rid of that negative charge on the oxygen. And we get our product where we we added the nucleophile to our carbonyl. The other example is where we take an uncharged nucleophile and we need to react it with a carbonyl and activate it.. we activate it by protonating it. We add some acid to the System. The Oh group, grabs, the proton, gets charged and this charge with this oxygen polarises the carbon-oxygen double bond even further. What it means, it makes that carbon even more electron deficient and therefore a weaker nucleophile can react with it, it will be more attractive to that. Then we have our uncharged nucleophile attacks. We get again our tetrahedral intermediate nucleophile is still protonated now, we need to lose that proton to form our product. And what you can see we end up with additionl across the double bond. First example the nucleophilic attack first followed by the electrophilic attack. The other example The electrophilic attack first followed by the nucleophile. The only difference is in this top one, we have to do another bit of reaction to get to the product. Whereas in this one, we needed to make this reactive mean to activate it before this happened. Once we have this then we have to lose a proton to get to the product.

Answer 29

Aldehydes are more reactive then ketones due to the carbon atom being more electrophilic. In a ketone, a ketone is going to have two carbon groups attached to your carbonyl carbon, lots of carbons. This is delta positive. And what we said is that if the delta positive nature of the carbon that drives the reactivity, the more electron deficient the carbon is, the more reactive it is. Anything that can stabilize it will mean for carbonyl group is less reactive. Anything that can destabilise when we make that electron deficiency worse will make your carbonyl are more reactive . So in this case, we're not looking at something that's making it worse. We're looking at where in the ketone we can stabilize that. ketone provides more stability to that delta positive carbon because it pushes more electron density, makes it less delta positive and less reactive.The aldehyde can only do half the amount of stabilization. So it's more reactive.

Answer 30

So we start off by saying, here is our carbonyl. Okay, we're going to use an alcohol, we're going to use ethanol, methanolor whatever it is. It's uncharged. By adding the acid to it, we're going to activate our carbonyl. So we're following that second pathway is saying, a weaker nucleophile requires us to make a more reactive carbonyl. We do the same process. We protonate the oxygen (1st pic ) That gives us this positively charged oxygen, which we know actually means that the carbonyl carbon is more electron deficient. Then we can take our uncharged nucleophile. So this is an alcohol group. So the R prime can be any. So you can see methanol, you can see ethanol. They're pretty much the two that we use most of the time. Now that we've got our reactive carbonyl center, the lone pair of electron from the oxygen can react now with that, at this point, they wouldn't have been reactive enough. The lone pair wouldn't have a reason for doing that reaction. But here, far more reactive carbon center, the reaction can occur. So the electrons come and attack this carbonyl carbon. And as we said, to allow that to happen, to accept that electron pair, the carbon needs to make space for it and the way it make space for it is by saying, well, let's give up the electrons in that carbon-oxygen double bond. There will already be taken by the oxygen anyway. So it's not much effort to say, yeah, you take those electrons away. I'm going to get my electrons from this alcohol. So we break the carbon-oxygen double bond. 3rd pic ---Then we can compare that as saying, imagine it's a positively charged carbon.We know in reality it's not formally there is this resonance structure. So these two are basically saying the same thing. Just one showing. The charge on the oxygen, one showing with the charge on the carbon, they both end up with the same product. So here you can see we've got our alcohol added to the bottom. We formed our tetrahedral intermediate, just like we did with any of the other general nucleophiles. And at this point, we need to get rid of that charge. So we can recognize we've got acid in the system, therefore, we can lose a proton. It's quitestraightforward it as acid there, that means protons are happy to exist on their own. So we can take the electrons out of this oxygen-hydrogen bond, produce H plus, but we neutralize this oxygen. (Far right bottom) So we generate what's called a hemiacetal and we regenerate our acid. Now, Bottom most diagram - if we've done this on a ketone example, this is the aldehyde, aldehyde or ketone. It would be a hemiketal. Exactly the same pathway. Is exactly the same. Attacked here. Adds our alcohol at that point. And then instead of having OR here and OH here, we've got another R Group that's just applying that exact same mechanism. Electrophilic bit first, then the nucleophilic bit.

Answer 31

We can do it with the base. So we can say what would happen with a basis. We would take our aldehyde. We would use a base that would deprotonate our alcohol. So what we mean by that is we would have to add a base. That would mean that the OH bond would break, that hydrogen would be lost as a proton. So we need to use a base. It's going to be strong enough to take the proton away from the ethanol. We're talking about strong base to do that. But what that ultimately do under basic conditions where you'll beleft with an 0 minus. So it could be an ethoxy, ETO minus methoxy. Any O minus would be the two we use. But we would end up with this negatively charged oxygen. And so then you look at that first idea that says, sometimes it's charged nucleophile. If it's a charged nucleophile, it can react straightaway. If it's not, we need to protonate the carbonyl group. But in this case, charged nucleophile, uncharged carbonyl. It's a simple nucleophilic attack first to this carbon, which is electron deficient. For the carbon to accept those electrons, it needs to break a bond. So it's going to break the carbon-oxygen double bond. I willkeep saying that just to really hammer the point home. Hopefully this will be the year that no one will, will get it wrong and no one will break the carbon and oxygen that they'll always break down first, but we'll see. So we end up, we add the OR group, we form our tetrahedral carbon. The oxygen now has a negative charge. We've got solvent around. In this case, we can actually use the alcohol as the solvent. But you could have water as well. But you would end up extracting the proton, which has the effect of regenerating our base, can go back into the system. But we protonate. We grabbed hydrogen onto this OH or this O minus to give us our hemiacetal. So the product is the same. It's just the order in which we do things is different

Answer 32

so we can take our hemiacetal and we can add some acid to it. We can then protonate what was effectively our alcohol again. We end up back here (2d pic ). You can see the equilibrium arrow is now at this point. This means that by protonating it, this becomes a better leaving group. As it leaves it produce the alcohol that's a very stable molecule. So it can leave as a nice stableleaving group. We end up with our positive carbon, which we know actually is in resonance with this positive charge oxygen (3rd 4h pic ). And that point, we wouldn't show this normally. We would show re, formation of the carbon oxygen double bond and removal of the leaving group at the same time. So that's, that's an easy way to unjust showing you, just to remind you that what we're really saying is that this carbon is electron deficient, so it needs to get the electrons from somewhere. And at that point, we lose the proton. We regenerate our acid, and we go back to our starting material, our aldehyde (upmost right). So what I've shown you there is the first one I showed you. You can take the acids, you can protonate your carbonyl. You can then react with the alcohol and produce your hemiacetal. In this case, we're saying we can take the hemiacetal, we can protonate it, is we can do that electrophilic bit first. We can then lose a leaving group and then gain a proton. So it's just the reverse

Answer 33

Again. Can you do the same thing with the base? Yes, you can. In this case, the place will deprotonate first. Forming our negatively charged oxygen. Then we can re-form the carbon-oxygen double bond, and we can have this leave as a leaving group. Now, the alkoxy ani needs to be a stable leaving group. So again, by havingunder basic conditions we can help that to happen. But the whole point of this is to show that it's under equilibrium, so things are going backwards and forwards. The forward reaction is going to be the reaction of the carbonyl. The reverse reaction is gonna be reforming that carbonyl. Now actually the equilibrium is often favorites the carbonyl compound, it doesn't favor the reaction of the alcohol with the carbonyl. So that product, that hemiacetal, its going to be not very stable. So the way we can improve it is we can try to stabilize the hemiacetal to make it happen

Answer 34

Here's an example of an aldehyde. Got an alcohol in there, and we've got an acid. So hopefully that should lead you to say, oh, this must be an acid catalyzed reaction.You've got the classic things that you've got HCL - . The moment we see HCL See how you've got to think it's acidic conditions. It's astrong acid. That's going to mean that we've got protons floating around. We also know that with this, we need a strong nucleophile is not going to work for methanolby itself isn't a strong nucleophile. It can be nucleophilic, but it is going to need some help. So that should lead us to say, right? What we do here is we need toprotonate this oxygen. So we use a lone pair of electrons, we form a new bond with the proton. And that activates our carbonyl. That gives us that really electron deficient carbon atom. They're ready to react. We then bring in our methanol)3d part ). Methanol oxygen has a lone pair of electrons that can be nucleophilic. We've got an electrophile here, lots of lacking electron density. We've got the electrons here. So we do that reaction (arrow) We try and form a new bond between the oxygen and the carbonyl carbon. To accept those electrons, the carbon needs to makespace for them. It makes space for them by breaking the carbon oxygen double bonds.Has an added advantage of that positive charge on the oxygen, You can take those electrons and formally neutralize that charge and become neutral. We then form this neutral OH, up here. Our carbonyl has gone from SP2, SP3 hybridized. We've added our alcohol here. Now the alcohol oxygen is protonated. It's got a positive charge again, the oxygen really doesn't want that. So it finds a way to remove that. And it says, well, we know that there's protons in the system. Therefore, we can take the electrons out of this oxygen-hydrogen bond, give them to the oxygen that neutralizes that charge. And the proton leaves as H plus. And we end up with our hemiacetal

Answer 35

Other one you can do if you want to us e carbon is via a Grignard reagents. . if you see things like Magnesium, Ether , mg BR is a Grignard reagent, effectively is a source of R- . bit like the sodium borohydride is H minus. If you see R MGBR, it's really a source of C minus. But again, it doesn't really exist on its own. It's part of this metal complex, So then in this case we say we recognize that R M g, x, whatever that might be, is the source of R minus. So what that means is the electrons in the R and magnesium bonds are the nucleophile. they can react with the carbonyl carbon to make space, we break the carbon-oxygen double bond. So the reaction mechanism is exactly the same as all the other ones. It's just this time we've got the R- only thing you have to remember that slightly different here is that R minus actually comes from the R MG what we actually formed because the magnesium is going to have positive charge, you form this ionic interaction. But all we do is add an acid And you end up again with an alcohol. With a hydride, we add a hydrogen to it to provide an alcohol. Here, the Carbon. depending on if we start with.

Answer 36

We can also exactly the same reaction but with cyanide. So you can take something like H and C . Hydrogen Cyanide wont help us here. But Although it can be acidic, it's got a relatively high PKA, not acidic enough to provide a proton for The acid catalyzed reaction to what it actually does is it provides a source of CN minus. If you see something like HCN, it can form CN minus. And again, it's exactly the same reaction. Charged nucleophile definitely do the nucleophilic part. First. We have the nucleophile attack. We make space for that bybreaking the carbon oxygen double bonds. We form our tetrahedral intermediate, or added the cyanide here. Then we have got our proton. So we can do the electrophilic part. So we can grab that proton from the HCN and finish with cyanohydrin. using cyanide, it's not really the safest thing to use. So actually what we'd normally use something like sodium or potassium version of it. Which means when you come to this , you'd be adding a bit of water. So it's really just being aware that if you see a cyanide pay to C C triple bond N, it's giving you a source of CN minus. And you can then feed that into your standard general mechanism. You might see it as a sodium or potassium.

Answer 37

Here's a ketone. We've got a grinyard reagent here. So that's going to give you a source of ME minus you've got some acid in there as well. So you just need to apply that standard reaction mechanism, starting with a charge, very strong nucleophile, reacting with your carbonyl. In this case, it will be producing an alcohol So we've got our ketone and grinyard reagent . I'm what we recognized that this bond between the Methyl group and the magnesium is where the electrons are there available for bonding. That forms our nucleophile which attacks the electron deficient carbon. To make space we break the carbon-oxygen double bonds. So the electrons move This way. That gives us our tetrahedral intermediate. we've added another CH3 group that we have this kind of ionic interaction with the – charged oxygen and the positive charged magnesium. We add our acid and what's going to happen there is that negative charge is going to pick up the acid . And then we've produced our tertiary alcohol.

Answer 38

Nitrogen is always the one. That becomes a little bit more tricky. So why does it get tricky where we start with r, we're going to need an aldehyde. So this example, and we've got an amine. The amine has got that lone pair of electrons. It could use to react with the carbonyl carbon. Remember the aldehyde is more reactive than ketones. In order to come to accept that it does the same thing of breaking the carbon oxygen double bond. We end up with our tetrahedral species. And here the oxygen has got a negative charge because the electrons in this carbon oxygen double bond have now been taken by the oxygen, which gives it an extra electron. But nitrogen has used its lone pair of electrons to form a new bond. So it's now, it keeps holding one of them, but the other one is given to the carbon to replace one of the oxygens. So the nitrogen now has a positive charge. And this is where we start to do a proton transfer. So lovely word. We can just write proton transfer or think that minus H plus, plus H plus. What we're saying here is, we have something thats negatively charged and positively charged. This positive charge has got hydrogens attached to it. We know all of this equilibrium. We know that what can happen is those, those charges can move. And it's hydrogens can move, so the hydrogen can leave . It might've taken by water, or by another molecule i.e Amine . It might be taken by another molecule of our intermediate here. But ultimately the hydrogen leaves the nitrogen and transfer to the oxygen. So by doingthat, we go from having this two charges, to having two neutral atoms. What happens next? Because if we know that we can transfer protons, we know that we can protonate oxygen, and can protonate nitrogen. So if we protonate the oxygen, and if you remember going back to the acetal example, where we protonated the OH group. We thought water is a good leaving group. If we do that, we're going to lose water rather than just any obvious tetrahedral species. The other thing here is just to point out that in order for this to happen, we needed acid. I think we have too much acid and we have a really strong acid. What actually happens is Amine gets protonated to start with. So we justhave to be careful. So sometimes you'll see in these kind of questions, it says its pH is controlled. So it will tell you this is pH four to six. What that's telling you, there's not too much, just enough so that the reaction can proceed, but not too much that we can protonate this nitrogen. So what then happens is instead of just staying at this point with the amino alcohol with a bit of acid,we can protonate this oxygen. that makes it a good leaving group. Then the nitrogen lone pair we can now form instead of a carbon-oxygen double bond, we can form a carbon nitrogen double bond. Do that, we need to lose the water. What we end up with is a iminium cation. Which when we protonate it forms imine.

Answer 39

But in this particular case, because you've got the oxygen and the nitrogen, we have choices. So we could protonate the nitrogen again and do the reverse reactions do. Orwe can protonate the oxygen to form the imine. If we protonate the nitrogen, we re-form the carbon-oxygen double bond, and we would kick out the amine We go back to where we started. If we protonate the oxygen, we then form a carbon-nitrogen double bond and we kick out water as our LG. We end up with this protonated imine cation.So which one actually happens. We think oxygen is more electronegative than nitrogen. So if you have a choice between protonated oxygen or nitrogen, nine times out of ten, you're going to protonate the nitrogen . Because the oxygen really doesn't want to give up its electrons. It doesn't want to have that positive charge that will beore unstable Then the nitrogen containing positive charge. But in this case, what we're actually looking is the product that actually gets formed at the end. So the product that gets formed at the end here the oxygen is positively charged and that's reasonably unfavorable. Here, the nitrogen positive charge, and that's more favorable.

Answer 40

we've got this protonate nitrogen with two hydrogens. That means the hydrogens are available to proton transfer. We can transfer one of those over tothe oxygen. Keep one on the nitrogen. We've now got two neutral species which are a bit happier. But we can say, well, if we can do one more proton transfer, we can do more than that. We've got a small amount of acid, PH Six, Therefore, we can protonate this oxygen here. Okay, So we've protonated this oxygen. We've now got this really nice water molecule leaving group. So therefore, let's make use of that. So the nitrogen again use its lone pair to form this carbon-nitrogen double bonds. In order for the carbon to accept that I needed to do something, needs to make space so it breakes carbon-oxygen bond. it makes sense because the oxygen is positively charged. Again, it's going to want to get as many electrons as it can It's going to be pulling the electrons in this single sigma bond towards itself to the carbon would be electron deficient. If it accepts the electrons from the nitrogen is going to have more of a sharing these electrons in it. So it makes sense for carbon to let those electrons go. We end up with water as a nice LG. That produces our iminium cation. This point we can lose that proton. Or what we showed you if we added sodium cyanoborohydride, right? So we are adding our source of Hminus. So again, I've drawn it as Hminus, but we recognize it doesn't exist. As H- its own, it's part of that complex metal hydride. But for the purpose of this reaction mechanism. I'mshowing you this H minus. We've then got our carbonyl carbon. Nitrogen is positively charged. That means that this carbon is going to be more electron deficient because nitrogen going to pull the electrons away. Therefore, the H- is going to attack the carbonyl again to make space for that then We're going to break the carbon-oxygen double bond. That adds or hydrogen in and what we've ended up with is nitrogen, neutralized positive charge. And we've ended up with Amine, started with the ketone

Answer 41

So we're going to start by looking at what I'm going to call a substitution reaction. So when we look to those nucleophilic reactions last week,with the simple carbonyls, we saw that for all of them they will reverse. So it waspossible to reform our carbonyl. Okay. So we had our nucleophile, this case it's our charged nucleophile to attack our carbonyl carbon. In order for that to happen, the carbon-oxygen double bond breaks, the carbon needs to make space for the electrons coming in. We can show that, you know, that can be reversible, that we can re-form the carbon-oxygen double bond and kick out the nucleophile. And often we can pick up an electrophile. Electrophile can do this. So there was lots of reversible processes. So what we're going to look at now is what happens if there's already the leaving group. So this is where the nucleophile that's atacked effectively acts as a LG. . what if there is a LG already there. (L here is just a generic LG). . And what we can see here that the same reaction occurs, the neutrophil, again, generic nucleophile attacks the electron deficient carbon, the back carbon to accept those electrons, it means to break the carbon-oxygen double bond. And this is where I know I said to write this down, I'm going to keep saying this is where it gets important. Because what happens is we break the carbon-oxygen double bond first that produces this negatively charged oxygen. Once that's happened, we had a choice because we have a leaving group. We know that these processes are reversible. If we reform the carbon oxygen double bond rather than the nucleophile acting as the leaving group, we have a perfectly settled leaving group. So that can then leave. And what we've done is we've substituted the leaving group for the nucleophile. what we haven't done is have the nucleophile attack. And the leaving group leaves straightaway. We're not going to break that single bond first. if The carbon is going to accept those electrons the first thing it will do is break that c=o. Once it's done, then we can think about breaking this single bond. But in reality it's not actually a nucleophilic substitution. What it actually is, first is an addition. we add to our carbonyl group And then once we've added to our carbonyl group then we can eliminate our nucleophile

Answer 42

We go from an acetal chloride to a carboxylic acid

Answer 43

We're going to break carbonyl and double bond first. So this is the correct sequence. We take our nucleophile, it reacts withThe electron deficient carbon. Electron deficient carbon to accept those electrons. Not just going to break this bond to the leaving group, it's going to break that double bond. That's the weakest bond out of the lot. so it breakes, the carbon-oxygen double bond. That point we get our negative charge on oxygen. We add our nucleophile, we have our Lg and tetrahedral intermediate. Then we reform the carbonyl bond. And at that point we break the single bond to the LG. So that is the correct addition elimination. We end up with substitution and our LG. Its not a nucleophillic substitution So attacking the carbonyl carbon and breaking the leaving group bond, that's not going to happen. Because that doesn't show us that the carbonyl is important. It's really important that we can break that double bond down. First, That's a weaker bond. Once that's broken, then there's a driving force for the leaving group to leaves. Equally,it's not an SN1. We're not going to have this leaving group leave to produce this positively charged carbon. We already know that carbon is electron deficient, not going to allow the electrons tto get away easily, iIt needs something else for that to happen. Don't want to see any of this kind of leaving group producing this positive charge. What you do sometimes see it is acceptable. Heres our nuclophile, reacts with the electron deficient carbon and We break the carbon-oxygen double bond, but we know from up here, the next thing we do is immediately reform it. So you can use this kind of double headed arrow to say this break then it reforms and then the leaving group leaves. So this one at the bottom is okay. This one at the top is okay. These two in the middle are not.

Answer 44

In order for this to happen, the nature of the leaving group is crucial So the first thing is why alderhydes and ketones not do this? it's basically the groups that attack, the hydrogen or carbon, they're really poor LGs. If you have hydrogen as gas it doesnt want to spontaneously chnage to H + and H- it wants to stay as H2, so H- isn't a very stable model. So what we need to know is H and R are not good leaving groups and I'll see you in the next table that laid out for you. So in that case, for these simple carbonyls you don't do that addition elimination. You do get an addition by adding to the carbonyl and deprotonating the oxygen. To get Substitution we need a good LG. So good leaving groups that thinngs like Cl- That's what this is telling us, is that if you have HCl, the equilibrium favors if they are H+ and Cl minus, Cl minus exist on its own is a nice stable anion. And so therefore, you can make CL leave, fairly electronegative lots of electron more than happy to have a few extra electrons. Carboxylic acid is the same thing with the OH, we will protonate that, So if we have a little bit of acid in our system, we can protonate that. water makes a good leaving group. Water is a very stable molecule The most important thing is to say to recognize these ones here are good leaving groups where what you've got halogens, thios, anhydrides, good leaving groups, water, alcohols amines make moderate leaving groups. But Hydrogen and R very poor.

Answer 45

So we've got our carboxylic acid and the first thing we can do is we can protonate. And remember if you go right back, we talked about fundamentally, we can have a really strong charged nucleophile attack an uncharged or unactivated carbonyl. Or we can have an uncharged nucleophile requiring some kind of charged carbon. This is an example where a fairly weak uncharged nucleophile. So we need to make this more electron deficient, make it more reactive. We can do that byprotonating the oxygen. Same as previously done, it pulls the electrons away from the carbon to mak it more electron deficient. Exactly the same thing happens. Our nucleophile attacks to accept those electrons would break the carbon-oxygen double bond. And we get our tetrahedral intermediate here. So that is exactly the same . It's just instead of having the carbon or hydrogen here, we've got this OH group, but at this point, it's exactly the same. Once we form this tetrahedral intermediate bit like when we have nitrogen attacking, we now have some choices to make. The choices to make are. We can use our terminology of proton transfer. It's just kind of all encompassing term that what it really means is we recognize that, this oxygen is currently protonated. (Bottom left ) But it's all in equilibrium. We can deprotonate that. And that proton will be kicked off by a number of these hydrogen or the oxygens. Once we protonate one ofthe OHs and we thought water, That allows us to move forward into a different reaction. Otherwise, it's just the reverse. So eventually we get to a point where You protonate this OH group you form ia Very good leaving group now. And so your molecule has a choice. It doesn't have to just pick up a proton here, it can use its lone pair of electrons, re-form that C=H double bond. I can kick out water. What we've done is we've substituted the OH, group here for OEthyl / alcohol. (we form an esther)

Answer 46

when you see acid that means I can protonate things differently. They think, well, we've got methanol, it's not the best nucleophile and it's not charged strong nucleophile so anything we can do to make the carbon are more electrophilic means that our nucleophile will hopefully attack. we can do that by protonating the oxygen that produces a much more activated carbonyl. Then we can use our alcohol. So again, a lone pair of electrons on the oxygen is our nucleophile. Again. The carbon is the electron deficient areas and that's going to receive those electrons. In order for those electrons to go into the carbon, it has to release some electrons, it's got to break carbon oxygen double bond. That also neutralizes this positive charge. We then end up with this nucleophile is positively charged. We know we can do these proton transfers and we dont need to show where that proton goes. Somebody comes in and takes it. You can just show an equilibrium arrow is showing the proton moving around. So I just said, you know, proton transfer in equilibrium means that if we can go from here to being protonated up here as this is what we need for the reaction. We then set up our leaving group. We then re-form our carbonyl bond. And we kick out for that bond to re-form and we need to lose some electrons. We lose this leaving group which is ready to take away the oxygen + charge. it would have drawn the electrons towards itself anyway. So it's no much of a push to send it out the door. We end end up adding alcohol, then we deprotonate it here and we end up with our Ester

Answer 47

So we can undergo hydrolysis. 2 possible ways. We've looked at with the ketones and aldehydes, you've got acid catalyzed or base catalyzed. The acid-catalyzed version, it's the same mechanism which we protonate the oxygen to make that esther bit more attractive to our water molecule because our water isn't the strongest nucleophile. So we it need something to just help it along to make it an electron deficient carbon now and is really looking for some electrons. The water's got the electrons. It's now willing to give those up to that carbon. So we get that nucleophilic attack. To accept those electrons, the carbon has to do something. Now what it's going to do, these guys break that carbon-oxygen double bonds. So we end up in this situation where we have the OH of the waters added to neutralize this charged. And again, we do our our proton transfer. some point, we're all going to end up with our OCH3 protonated And when that happens, that's when we can move on to lose the Esther and form a carboxylic acid. So we can reform our carbonyl group and we can lose the alcohol as a leaving group. So we've gone from an ester to a carboxylic acid,

Answer 48

Youcould take something like sodium hydroxide. you've got an OH group that's a nice charged strong nucleophile It doesn't mean any activation of this now, they can just go straight in and attack that electron deficient carbon, just like we saw with the aldehydes and ketones. We then break the Carbon-oxygen double bond, then we form negative charges here. Now, we don't have protons. Yeah, doing that proton transfer is not possible now. We don't have protons. So how do we do it? Well, we have to reform the carbon-oxygen double bond. We'd have to kick out the O Methyle (OCH3) We end up with a carboxylic acid. Now what would happen is that on its own, isn't going to be a very good leaving group. But we can stabilize this because we now do have a acid in the form of a Carboxylic acid. So we can maybe take that OH, group here to form our alcohol and our carboxylic anion with a question mark based capitalist, because if they get from here is not regenerate the catalyst. The end product is stable and irreversable as we have used up the base and the nucleophile is not wtrong enough to do the reverse action . The negative charge can here actually help stabalise the carbon

PM1B - Autumn Functional Group Chemistry Flashcards

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