PM1B - Reaction Mechanisms Flashcards

Question

How do Solvent effects help determine SN1 or SN2

Answer 1

With SN1 we are looking to stabalise that negative charge and the way to do this is to add other things which have positive and negative charges in there, as an example something like a polar solvent. This is a solvent where the molecule that makes them the solvent has some polarisation within its bonds. That means we will have areas of electron deficiency, electron richness and if we put in a negative charge that will attract the areas of electron deficiency ( delta positive), which is going to help us stabilise that negative charge. Polar solvents are things like water, the OH groups, alcohols, ethanol, methanol, these have an OH group and a CO bond which is polarised. Polar solvents include protic means there is a hydrogen that has been deprotonated?? Aprotic substances means that they cant, there is no proton, there is no hydrogen that can be lost. The ones that are polar are for example things like Acetone, acetonitrile, DMF and DMSO. What we need to know from these is, these are the kind of solvents we might put in to a mechanism, we might say this has been done in water, this is being done in DMF and so for example we need to understand DMF is an aprotic solvent. For SN2, the solvents are important but in a different way. For example the nucleophile, the nucleophilic is based on how available the electrons are. Anything which reduces the availability of the electrons, its going to stop our reactions from working. So if we have polar protic solvents, they can help us stabilise that nucleophile by reducing the nucleophilicity of it. What we might want to do is use something like an aprotic solvent because they can dissolve our nucleophiles but they don’t reduce the nucleophillicity. It is possible we can switch between an SN1 and a SN2 purely by choosing the solvent. For one example if we did it in water that would favour an SN1 reaction. But for example if we did it in something like acetonitrile, the acetonitrile will help the SN2 because it would help dissolve our neucleophile but not change its nucleophilicty.

Answer 2

Not all nucleophilic displacement reactions give us 100% yields of the desired product. What can happen is there can be a side reaction. For this we need the right substrate to start with, but if we have a substrate where there is for example a hydrogen that’s attached to a carbon, that’s attached to our nucleophilic substitution centre, we can get a elimination reaction. So instead of giving us the nucleophilic substitution that we want, we will actually end up with an elimination. Which means the proton also gets removed as well as the leaving group and this will form an unsaturated double bond.

Answer 3

*The H-C and C-Y bonds can be broken separately in two-step processes. *If the C-Y bond is broken first a carbocation intermediate is involved. *This occurs in the E1 mechanism (Elimination, unimolecular) and it is reminiscent of the SN1 reaction. *The rate determining steps for both the E1 and SN1 reactions are the same. NB : With an E1 mechanism we see this H-C bond which is essentially a hydrogen attached to the Carbon, and the carbon leaving group bond can be broken in too 2 separate processes. What this means is if for example we do C-Y, in this case Y being the leaving group, If we do that bind breaking first we end up with a carbocation. We recognise in an SN1 reaction that we also start by making that carbocation. At that point we can do the breaking of the hydrogen carbon bond ( which means removing the proton). What happens then is we use a base, a base is a nucleophile that’s happy to accept a proton. Weve got a loan pair of electron or a negative charge, so we can think it can offer those electrons to the hydrogen, and we can consider that the hydrogen doesn’t need to take the electrons that its currently bonded to the carbon with, it can leave those electrons for the carbon atom. The carbon has a positive charge next to it, in which it needs help to neutralize it, so it can use those electrons and make a double bond ( see diagram), which mean it can get rid of that positive charge and we end up with the base with a hydrogen attached to it and we end up with a leaving group. A key thing to consider is the rate determining step for both E1 and SN1 reactions, the generation of the carbocation which is the rate determining step for our SN1 reaction, is exactly the same for the E1 mechanism ( rate is determined by the concentration of the substrate), once the leaving group has gone and once we have formed the carbocation, that base will come in very quickly, take away the protons and form the double bond.

Answer 4

Note the rate equation for E1 is the same as that for SN1 Here we look at these 2 side by side. Sn1 at the top E1 at the bottom. We can straight away see that movement of electrons (those curved arrows) are the same in both reactions. In both of these reaction mechanisms, although they are different, they both start with the same thing which is the loss of a leaving group. So the electrons move from this bond, between the carbon and the leaving group, and they go with the leaving group, this leaves us with a carbocation. Once we have our carbocation, the same thing happens the nucleophile comes in and attacks the carbocation, in this instance it’s the electrons between the carbon and the hydrogen that will do that movement of electrons, NB its not the nucleophile but it’s the same movement of electrons. It’s a way of neutralising by providing electrons to where that positive charge is. In the other reaction we just have to add a little bit more, The fundamentals are te same but we reach the concept of elimination has that extra bits that go on top. We have to recognise, for that hydrogen to give up these electrons, there has got to be a base somewhere to take them away or offer the electrons to that hydrogen and take that away. Once we have done that, in an SN1 the nuclephile can attack above or below the plane which gives our racemic mixture and we end up with a double bond. Both of them end up with a leaving group, and the elimination reaction ends up with the base.

Answer 5

They slightly differ in their stereochemical outcomes When we talk about an SN1 reaction we mention the racemic mixture, the fact that we form this carbocation, and we can get attacked from above or below the plane of that carbocation and that gave us our racemic mixture. In the case of an E1 we don’t have that, because we don’t have the nucleophile attack the carbocation. In an E1 the electrons will move to this positive charge from wherever the hydrogen is, but what we have to recognise is, in this particular instance, there is rotation about that bond, its not in a fixed place, and if we have another R group, so we have something attached to our carbocation, we have this other R group ( which could be anything or any kind of group), theres an equilibrium that exists where that rotates around this bond, so R1 could be sticking up or sticking down ( similar to Cis and trans from Dr allman lecture). In theory, one molecule of the starting material could give a mixture of the E or Z isomer products. So on the right (top) here we can get the trans and here we can get the cis

Answer 6

It is generally true that E1 reactions afford predominantly E-alkenes as the transition state leading to it is lower in energy. However, some of the Z-alkene will also result, and the reaction is stereoselective. There is still some stereochemistry to be decided, so when we talked the SN1 we talked about sletreoselectivity where we might be able to help favour 1 reaction over the other depending on if we blocked it with some steric bulk. In this example there is actually fundamentally a more stereoselective process happening. In this example we have our di-stereo isomer, our 2 groups here but one isn’t a chiral centre as it has 2 hydrogens on it, but we have got a methyl group on one and a phenyl group on the other (phenyl is an 6 membered aromatic ring). We do this in water with abit of acid and we end up with this carbocation. The way we d this is, the water gts protonated and it becomes a very good leaving group, The OH leaves as water and we get this carbo cation, then this is what we are saying is an equilibrium, meaning it can flip between this trans and cis isomer, so in theory its 50/50 and we end up with this 50/50 mixture of these 2 double bonds. In reality here we are more likely to get the E isomer and the trans isomer, this could be down to the electronic structure stability that when these 2 groups are furthest apart from one another in that molecule, that is a more stable structure. If they are together there can be some steric hinderance, so we would like to get a molecule in which we get far apart from one another so we get this lower energy ( bottom diagram). So in reality this equilibrium lies more to whats effectively the Trans E system then the Z or sis system over there. So this is what we mean by being stereoselective as well, theres a chance to form both, but the reaction itself, part of the chemistry, the structure behind it says we are more likely to get the E, the trans isomer, its not necessarily specific as we could get a bit of both, but its selective because of the way it works and that’s the one that’s more likely to form. If we imagine something where there was not a phenyl group and only 2 methyl groups, it would not matter as much as we might get 50/50. If there are both great big aromatic groups in there, we are more Likely to get the trans isomer group ( E isomer) because of that steric bulk.

Answer 7

The synthetic route of this is to start with the chiral centre here and knock off this OH group. ( we add some acid, we protonate this, and we end up with a carbocation). This particular carbocation is really stable because what actually happens here is, the aromatic rings here have a lot of electron density, and they can actually push this electron density, not jus through the hyperconjucation, but through resonance structures, i.e we can take some of these PI electrons and share them into this carbocation which makes it really stable. So once we have this carbocation, then we can abstract this proton and form a double bond on our way to making the tamoxifen.

Answer 8

*Alternatively the H-C and C-Y bond breaking could be concerted, suggesting a one step process which passes through a single transition state. This is referred to as the E2 mechanism (Elimination bimolecular) and is reminiscent of the SN2 reaction We will find the E2 mechanism is very similar to the SN2 mechanism. We can also do this reaction all in one go in a one step process, so the base will attack the proton, that frees up the electrons from the carbon-hydrogen bond to form a double bond, once that double bond is formed, it allows the leaving group to take those electrons away. We could potentially look at this by the leaving group wanting to take the electrons way first, which leads to this forming a double bond which leads to the base first, which ever way we look at this. We then end up with the transition state, and essentially what we form at the end of it is the same products what we had before. ( wev got our double bond, our base and our leaving group which should be negatively charged but that’s gone missing.

Answer 9

In E2 eliminations, both the base and substrate are involved in this single step process and the reaction displays second order kinetics. *i.e. Rate = k [base] [substrate] *The rates of E2 reactions are affected by the strength of the base employed, which is a similarity with SN2 reactions where the rate of reaction is affected by the nature and concentration of the nucleophile. NB: Just like the SN2, we can look at the reaction rate and say, both the base and the substrate are involved in that way, as if we looked at the SN2, it was the nucleophile and the substrate, In an E2 it’s the base and the substrate that were involved, so therefore the rate has both the base and the substrate, and therefore that’s why we have the E2, because there are 2 elements that make up the rate determining step. It worth noting because it relies on the base, the reaction rate is actually effected by the strength of the base that we use. So if we use a stronger base, the reaction will happen a lot quicker, similarly with SN2, if we use a strong neucleophile, that is going to make the reaction happen a lot quicker. ( the difference is the product between the 2 reactions, but the similarities are how it gets to that product).

Answer 10

Another example of an E2 mechanism. We can take this cyclic structure, and what we are looking at here, we have an hydroxyl group and turn it in to a tosyle group and finally turn it in to a double bond. We start of with this OH group, we then make this tosyle group. This tosyle group is a very good leaving group, this is due to resonance structures. ( look at diagram under tosyle group on the middle / right) . Resonance is not necessarily changing the structure of what we are looking at, but recognizing where we have charges and where we have things like double bonds and Pi electrons, we can move the electrons around the structure. This is a tosyle group, and important to remember here the oxygens are different. When it reacts or we lose a tosyl group and replace it with an OH, the oxygens are actually different atoms, so the oxygen is actually part of the tosyl group. When this leaves it takes the electrons with it, and that negative charge sits on the oxygen. Resonance tells us that this is going to be in a P orbital, We have got orbitals in the Sulphur and P orbitals in the oxygen. The electrons in the P orbital can form a new double bond, a new Pi bond between the Sulphur and the oxygen. Sulphur cannot have too many bonds to it will have to break one at that point, If we have a choice to break a single bond, or a double bond, we always in chemistry break the double bond first as it’s the weaker bond. So if the suphur has too many bonds, its not going to break the single bond connected to the aromatic ring,, its going to break the double bond. Now what happens is, this comes in and the electrons that are in the sulphur oxygen double bond, actually relocate to the oxygen ( Furthest right diagram) and its that diagram what we end up with, structurally it’s the same only the negative charge has moved to the top. The double headed arrow means it can go back and forth, we can also move the negative charge to the bottom of the group. ( last diagram). What this means is, the negative charge is shared over the/4 different atoms, A negative charge on one atom is going to be unstable as it’s a very localized negative charge, a very electron rich area that is likely to react, because that’s what electron rich areas like to do. But in this instance we can take that electron rich area and we can spread it, and delocalize it across multiple atoms, but as a result the electrons are no longer as available as they were because there not just sat on one ayom they are spread out to the different atoms.

Answer 11

Going back to an SN2 reaction, it also did all of the bond breaking and making at the same time, its also concerted mechanism. We can say if we have some stereochemical information in our starting material, we bring that through in to our product in SN2, which is this inversion of stereochemistry. E2 mechanism is a little more complicated. As we have discussed before electrons don’t like going around corners so if we can line up our orbitals, which the electrons are in and moving in between, if we can line these up this had this energetic advantage. SO if we can get the hydrogen, these carbons and the leaving group and the base, all in the same plane as one another, all of the P orbitals the electrons are moving between, if there all lined up and overlapped then we can maximise the overlap of the P orbitals and we can energetically make this more favorable to us. This is the idea drawn on the paper, all of it is in the plane of the board, but we also know these atoms are not just flat planar atoms, they are SP3 hyberdized carbons, so they are tetrahedral in nature. So the above is showing what it would like to look like but in reality, other stuff can go on, its not just planar

Answer 12

For acylic molecules, ( in a chain mainly in this situation and not joined in a ring), there are mainly 2 confirmations of what we are thinking off. If we have to line up our atoms and orbitals there are 2 ways we can do that. We can do this via the anti-periplanar orientation (first diagram). What this means is, we have our hydrogen which our base is going to abstract, ( the base isn’t there at the moment, as the base is in the solution), and that at some point will eventually line up. We are not worried about getting the base to line up in this instance. We’re knowing that its out there in the solution, when the reaction is ready to go and everything is lined up, the base will appear probably through random probability and at some point eventually get in to the right position and when it does its going to add to the reaction. What we are looking at there is the hydrogen, and Y, Y here is our leaving group. We can also see here that the hydrogen and the Y are 180 degrees to one another, so they are in the same plane. So In our first red highlighted diagram, this is coming up, what we mean by that is we can see its all lined up in a straight line and its also in a staggered conformation as well, the Hydrogen is pointing up, and the leaving group is pointing down. This is also whats called a new objective, in other words another way of showing what we are looking at here, that is, lets take this molecule and we got our hydrogen, R group and a R group prime on it. If we were to look down that bond to the carbon behind it, what would we see? When we look at that first carbon , the circle in the middle represents the orbital behind it ( that’s the single bond). And if we look the hydrogen is pointing upwords, the R group in this case is going in to the plane ( that’s on our left), and the R prime group will be coming out of the plane towards us that’s on our right. So all we have done is translated that H, R and R prime in to this circle and we can see the relationship of them towards one another. We can then say, what do we see of that carbon behind. The answer is we cant see th carbon as its behind the front one, but we can see the substituents sticking out. And we can see the Y group ( leaving group pointing down, we can see our double prime is here going in to the plane of the board, ( in this example its on our left). Our triple prime will be coming out towards us, which is on the right. All it is, taking the molecule, rotating it and looking down the bond to establish where are the substituents and the groups attached to it. Its also a staggered conformation as all the groups are far apart as they can be too one another. (antiperiplanar left diagram ) Syn-Periplaner (right diagram ). H and Y they are in the same plane as each other, this now called an eclipsed conformation, what we can see is the groups are not staggered anymore, they are not as far apart from one another, infact they over lap each other. This is denoting, imagine looking down that bond (looking where we got the H, R and the R prime again, Hydrogen is up, R is left and R prime is on the right. ( same as the first one). The only difference to the first one is we have rotated that back carbon (180*) so the leaving group is now pointing up, the hydrogen and the leaving group are in plane, our 3 prime going away from us, (left) and our 2 prime on the right. Effectively our 2 prime and 3 prime have swapped sides. But they haven’t swapped, we have just rotated, now what as on the right is now on the left, and only that back carbon has changed. From the 2, the anti-periplanar is the one we want ( bottom right writing). Anti-periplanar is favoured for a couple of reasons. 1. this staggered conformation is energetically better, as everything is far apart from one another we don’t get the interactions or interruptions or electron density close to one another repelling each other, so it’s a lower energy system and a better system then the one on the right. ( 2nd reason next slide ).

Answer 13

The other reason is, the base and the leaving group are far apart to one another as well. That makes sense because there the bits where there is lots of electron density. With this proton we are going to have a base coming in with all of its electrons, to take that proton away, The leaving group is a good leaving group as its where we end up with a concentration of electron density, so having those 2 things as far as apart from each other, is also a real positive. If we look at the previous slide to compare, the eclipsed example, the H and Y are close together, so that base coming in to abstract the proton, the hydrogen is going to come in close contact with Y, Y has got a lot of electron density, so they will both repell one another. But, in the staggered anti-periplanar form, they are far aoart. What this means for us is the base will come in and abstract the proton, and the leaving group is going to leave and this is the preferred conformation. The reaction will work via this reaction, it wont work via the syn-proplanar, as the molecule will re-orientate it self to make it anti-periplanar. What we can see then is, if we abstract the hydrogen, form the double bond and lose the leaving group, If we take the line ( the plane) through that H and Y, one one side of it we have R R double prime, and one one side R prime R3 Prime. Whats going to happen is, When this occurs we are going to form our double bond, and those R groups are going to end up on a specific side of the double bond. Herw we can have thing like E/Z Cis etc (Dr Allmans content). And so the actual reaction will give us one product, because everything has to line up, that forces these groups to be on a specific side of that double bond.

Answer 14

That means the configuration of the substrate will control the geometry of the alkene that we produce, this is why its stereo specific. A stereo specific reaction is where, a single product ( a single structural isomer) is produced, and its produced because of the direct mechanism that’s happening. The stereochemistry of the starting material follows a specific mechanism that leads to only one possible isomer. ( A specific isomer). If we know what we start with, we can predict what we end with, so it’s a specific product. Stereo selective reactions are where, a reaction has a choice. That choice can determine what product we end up with. There is a choice that can lead too one product or another. Something about the reaction, whether it be the conditions, the substrate, we use directs it towards one rather then the other. The examples we have had are in the E1 mechanism, we form a carbo cation. In that carbo cation, if we have other substituents, they are going to arrange themselves predominantly trans to one another because it’s a more energetically favourable one. But, there is a choice here, does it produce a cis product or a trans product. The reaction mechanism, the substrate will selectively choose to be trans if it can. Another example is an SN1 reaction, where we form a carbocation, at that point, the neucleophile can attack from either face of that cation, and if there is something in our molecule, in our substrate to start with, that can block one face, it can direct the neucleophile so it can give us selectivity. When we talk about E2 and SN”, because all of the bonds are broken at the same time , there is no chance for the reaction pathway to make a choice, it follows one pathway, so what ever we start off with is what we end up with.

Answer 15

Due to the nature of the carbocation, its likely to be E1, if we don’t form a stable carbocation its most likely not E1. ( same difference between E1 and E2 is similar between SN! And SN2). The nature of the solvent, it supports the carbocation. If we have a carbocation and with it something ionising that’s highly polar that’s going to help stabilise the carbocation, that would favour an E1 mechanism and not have much effect for an E2 mechanism. The base is very important also in E1 and E2 mechanisms ( in Sn1 and Sn2 it’s the nucleophile). A nice strong base ( as an example sodium hydroxide) or by putting lots of base in the reaction ( a high concentration ) that will favour an E2 mechanism. Steric effects that have an effect for SN1 and SN2 play less of a role here, the steric/ substation pans more around carbocation stability and less around the sterics of blocking because the elimination works on 2 different carbons, the base abstracts that proton from one carbon and the leaving group goes from the other carbon. Sterics don’t play much of a role. So we need to understands as pharmacists by changing the solvent we can favour one over the other.

Answer 16

How do we choose whether it will be elimination or substitution. A base and a neucleophile are very similar, and theres a lot of chemistry that goes on behind whether it’s a neucleophile or a base. A basic nucleophile would effect elimination more. SO for example strong bases are going to take the proton rather then be a neucleophile attacking a carbon. An example would be ethanol. Ethanol is not a strong base, it already has a proton, so to protonate ethanol again will be really challenging but it can act as a nucleophile because it has a lone pair of electrons, so if we had something like ethanol, that is a nucleophile but a very weak base, so its likely to go for a substitution reaction, so we would end up with this oethoxide rather then the leaving group ( the chlorine ). Now if we were able to deprotonate ethanol, which we can do to form this ethoxide anion ( so a negatively charged O Ethyl group ). The negative charge sits on the oxygen. It wants to go back to being ethanol, so it doesn’t want to react as a nucleophile, if it sees that proton, it wants to grab that proton, its would be a very quick process, the proton is nice and small and it can get to it very easily, so something like this is likely to undergo an elimination reaction, where it would abstract a proton from one of these carbons, form the double bond and kick the chlorine out ( the leaving group ). So something that’s a strong base is likely to do an elimination reaction. Something that is a nucleophile, a weak base is more likely to do an substitution reaction.

Answer 17

Size of the nucleophile is the other factor. It goes back to that steric hindrance. We often talked about steric hindrance of the substrate when we are looking at SN1 and SN2, but of course there is this steric bulk associated with the nucleophile itself. So if we have a small nucleophile, ( for example a potassium hydroxide in the diagram), that will give us an OH minus. It’s a small nucleophile so it can very easily get to that carbon, attack the carbon in an SN2 reaction and kick out the bromine. Something like the potassium t-butoxide we have this tertiary butyl group, a tertiary butyl group is a big carbon skeleton for carbon atoms with all those hydrogens ( its bulky). So this big bulky nucleophile getting in to attack the carbon it has to get past the hydrogens. But it will think its easier to take the hydrogens so it doesn’t have to get close to the carbon, so something like this is going to abstract a proton so abstract the hydrogen from this carbon here, form the double bond and kick out the bromine as a leaving group. NB this is different infact the opposite to the previous steric bulk example for SN1 and SN2.

Answer 18

The reaction temperature is the other thing we can use to change whether wev got an elimination or a substitution reaction. It does involve counting of some molecules. In a nucleophilic substitution, we start with 2 molecules and we end up with 2 molecules, but in an elimination reaction we start with 2 molecules and we end up with 3 molecules. Entropy is a key word here, entropy is a measure of disorder, so te more disordered the system the greater the entropy. To go from 2 molecules to 2 molecules doesn’t change the disorder of the system, but to go from having 2 molecules to 3 molecules we have increased the disorder meaning theres less order now in the system. Entropy is related to whats called Gibbs free energy, this is linked to the key term energetically favourable. What we are talking about is, is there free energy in that system. If theres free energy that comes out that system its favourable. Delta G is the change of energy, if its negative it’s a good thing, which means the reaction actually releases free energy and that’s a positive thing, nd if we get a negative delta G that gives us a product. Now the entropy is related to that, it’s a negative, so you take the enthalpy and subtract the entropy, but the entropy is changed by the temperature. So if we have a positive change in entropy, ( which is what we hae here) it will be then multiplied by the temperature. So therefore if we want to increase our entropy, if this number goes up ( which it does in elimination) then ultimately because its subtracted from delta G, it will increase the free energy and that delta G will become more negative and that’s more favourable. SO if we heat up a reaction, if its entropically favourable ( which this one is) its going to make the reaction more favourable, as the T delta S becomes bigger and we are now subtracting a much bigger number from our delta H, our delta H didn’t change, so therefore delta G is going to be determined by what ever the T delta S is. So all this means is, if its entropically favourable which the elimination is, we heat it up, we make that entropically favourable term even bigger and therefore makes the reaction more favourable. So if we increase the reaction temperature, we increase the chances of elimination rather then nucleophilic substitution. So if we were asked how would we favour an elimination reaction – We could say, well we need a strong base, I would use a bulky base and I would heat it up.

Answer 19

Conditions can be found to favour SN2 reactions over SN1 reactions, or to favour E2 reactions over E1 reactions, or to favour substitution reactions over elimination reaction. However, often more than one product will still result. SN1 reactions generally produce racemic mixtures of products whilst SN2 reactions produce single isomers (with inverted stereochemistry) from single isomers of starting material. SN2 reactions are therefore stereospecific. E1 Reactions are considered to be stereoselective, and normally allow the formation of E-alkenes. E2 Reactions are considered to be stereospecific, and the stereochemistry of the starting material will influence the stereochemistry of the resultant alkene. Therefore, careful choice of the diastereoisomer to be utilised in the reaction will allow specific entry to either the E- or the Z-alkene

PM1B - Reaction Mechanisms Flashcards

(43 cards)