Now let's dive into SN1 Reactions. Now we essentially read this backwards like this. So SN1 is a uni molecular nucleophilic substitution. In an SN1 reaction, for whatever reason, this X pops off in the first step. After this is off, we get a carbocation. Next, a nucleophile will approach, and add in to the carbon structure in our second step and now we're left with our product which has the nucleophile as part of the carbon chain. Read full transcript
Now the kinetics for an SN1 reaction are such at the rate determining step is this first step. Once this X has fallen off, the nucleophile adds in almost instantaneously. So this is the slow step and it determines the rate. The reaction order is first order.
Unimolecular Nucleophilic Substitution indicates a first order reaction, and the rate is then rate equals k times substant, where the substant is just the amount of this initial compound. The amount of nucleophile does not matter for an SN1 reaction. So even though there are two steps it's an SN1 reaction. The 1 just indicates reaction order.
Now the Stereochemistry for an SN1 reaction is to result in a Racemate. Because if we have this carbocation, it's essentially trigonal planar. So the nucleophile can add from either the front or the back, which would give us our other enantiomer. Now in this case, is this R or S, and is our top one R or S? The way we look at this.
We have our central carbon. We've got a CH3, a CH3, a CH3, and a nucleophile. Well, in order to have RS, we need to have four different substituents. And we've got three that are identical, so this has no RS. So this particular compound wouldn't have an estimate at all because there's no optical activity.
The way we would get our Racemate is if we started something that was optically active. So supposedly change to C2H5 and this bottom one to a hydrogen. We would then end up over here with a hydrogen CH3, C2H5 in our nucleophile and up here, we would have C2H5, CH3, and our hydrogen. One of these would be R, one of these would be S, so there would be no net optical activity.
One of the easiest ways to tell us something goes to the SN1 mechanism is to take the original compound and measure it's optical activity. If it is optically active and then you undergo the reaction and the product is not optically active most likely, it was an SN1 reaction, if it's a substitution. Cuz that would indicate that you've destroyed the optical activity by creating a Racemate.
There's one last practice, let's assume that our X here is chlorine. What is the configuration of this reactive? Well, what we need to do is get our lightest compound back, so we can swap the H and the chlorine and rewrite this. As this. Now remember that by swapping these, we've just inverted the configuration, but if we start here, we'd go around this way because chlorine is number one.
The ethyl group's two and the methyl group's three. This is going to be S. But since we've switched them, the reactant here is going to be R. We could have done a second swap to verify this. We don't think a one swap is just going to be the opposite of what it actually is. Since this one is R.
At the end here, we'll have an R and and S mixed. So we have no net optical activity. Now how can we increase the rate of SN1 reactions? What are some factors that make SN1 reactions happen? First is to have highly substituted carbons. We need to go through a carbocation and in order to stabilize a carbocation, we want lots of metal groups that can donate electrons.
So large alkyl groups of lots of electrons will balance the charge in the carbocation. So a tertiary carbon is going to be the best. Primary carbons are not very good for SN1 reactions. The next thing we can do is to have a polar protic solvent. We said that light dissolves light and polar or protic solvents, will also stabilize the carbocation.
Another thing we can do is to have a weak base for the leaving group. Since SN1 is two steps, and the first step is the slow step, anything we can do to speed that first step, where our leading group leaves, will increase the rate of SN1. So the weaker the leaving group is or the more likely that group is to leave spontaneously, the better it is for the reaction rate as whole.
If we can speed up the slow step we can speed up the whole reaction. Now let's talk about SN2 reactions again, we read this backward so this is going to be a Bimolecular Nucleophilic Substitution. In a Bimolecular Nucleophilic Substitution or SN2, we can start with a similar compound, we've got some X attached to this carbon chain and the nucleophile will be brought in here.
Unlike SN1 where this x left first in Sn2, the step of the nucleophile adding in and the X leaving occur simultaneously. We get a five bond carbon complex or one of the bonds is forming and one of the bonds is breaking. And notice SN2 has no carbocation. And after the X bond is done, we're left with our new product.
The nucleophile added into our carbon chain. Let's look at our reaction kinetics. Our rate determining step is this first step, because it's really the only step where we are both breaking and forming a bond. Our reaction order however is second order, because we have to know both concentration of this nucleophile and our substance.
So our rate is going to be equal to k times the concentration of substance times the concentration of the nucleophile. A Bimolecular Nucleophilic Substitution indicates that it is a second order rate or two molecules are important to the rate. Now in an SN2 reaction, the stereochemistry is such that we will result in an inversion of the configuration generally.
This means that, since we have five bonds essentially on this carbon. The nucleophile has to attack from the opposite side of this leaving group. So the nucleophile is forced to attack from only one direction. So let's suppose our nucleophile is And our X is a fluorine. What is going to be the final configuration of our product? Now, we look at this.
We've Here, we've got a hydrogen back. This is going to be the heaviest, second and third. So, one, two, three. This is going to be counter clockwise or S. Hold on I just pulled the fasten on you again. Look at this right here.
One hydrogen, two hydrogen. Since we have two of the same substitutions on these, this is not chrial. So there is no S or R. So when we say there's an inversion of configuration generally, there has to be an R or S configuration to start with to actually invert configuration.
Let's suppose we started out with an R(+) compound. Now, if we did an SN2 reaction with our R(+) reactant, what would be the most likely product? An R(+) product, an R(-) product, an R(?) product, an S(+) product, an S(-) product, or an S(?) product? Well, we're saying the SN2 reactions generally invert configuration.
So we'd expect it to most likely be an S. And now the question is, is it S(+), S(-) or S(?)? Since this is a plus, we probably expect to do invert to the minus. However, we don't know. So instead we say, we have S(?), remember that to determine plus or minus to actually measure things in the lab.
So even though the reaction starts as a plus and we invert the RS configuration, we don't know which way it rotate light until we have actually measured it. So don't assume that inverting R and S will also invert plus minus. A new substituent could maintain the same positive rotation just with a different degree of rotation. Now suppose we have competing SN2 and SN1 reactions.
What can we do to favor the SN2 reaction over the SN1? The first thing we can do is attack primary carbons. Remember that SN1 go through a carbocation. And carbocation's are best for the tertiary carbon. Well, a primary carbon is a polar carbocation. The second is to use non-polar aprotic solvents.
Again, carbocations are stabilized by polar solvent. So nonpolar solvents would destabilize them. Now a third thing we could do is to use small nucleophiles. We need to form and break a bond simultaneously in SN2. Well the bigger the nucleophile, the harder it is to force that fifth bond on the carbon.
So small nucleophile makes it easier for SN2 to happen. A fourth thing we can do is to have lots of nucleophile. Remember that we have a rate for SN1=k. Our rate for SN2 is, k=. If we increase the amount of the nucleophile, we can make the rate of SN2 reaction go through up the roof and it will have no effect on SN1.
So the most ideal thing we could do, would be to add our substance into a large vat of nucleophiles. So there's essentially infinite nucleophile. This would almost guarantee an SN2 reaction.