Now, let's talk about eliminations. These are essentially the reverse of substitutions. We're going to lose two substituents on a chain and their vacancies leave behind them a double bond. So if we lose this one and this one, they're going to collapse into a double bond plus H2O or some other compound. Read full transcript
In this case, it's water. There are two subsets of eliminations. E1 reactions and E2 reactions. Let's start with E1. Just like in the Sn1 and Sn2 you essentially read this backwards. So this is a unimolecular elimination.
Now, an E1 reaction occurs when this X group, for some reason, dissociates from the carbon. This creates a carbocadion. Now in our second step, a base comes along and abstracts this hydrogen and then we collapse down into the alkene. Now in this reaction, since we had two steps, the first step is the Rate Determining Step.
Once this X group is left, the base pulls off the hydrogen fairly quickly. So it just depends on how quickly this thing can decellciate. Now the rate only depends on one of these compounds just the amount of this substance right here. So it's a 1st Order reaction with the rate = k times the amount of it's substance. And the stereochemistry on this is going to be mixed because we have a single bond, here.
This can rotate around and we can get to see H3 group on this side. So we can also have the transproduct. Our stereochemistry will be mixed because we have this free rotation. And since we are going through a carbocadion, we need to remember that we're going to have some competition from SN1. So what could we do to favor the E1 mechanism over the SN1 mechanism?
Well, in general, the same conditions that promote SN1 reactions also promote E1 reactions. So the most reliable thing we can do is just to raise the temperature. Now let's look at E2 reactions. Again, we read this this way, where this is a bimolecular elimination. Now on an E2 reaction, we need to have our leaving group here and a hydrogen oriented into an anti position.
Once they're oriented this way, its strong base will come in and grab that hydrogen which then causes those electrons to collapse into the double bond and simultaneously kick off the X functionality. All this happens extremely rapidly and we're going to be left then with our alkene. Now the reaction for this one has only one step, so that one step becomes the Rate Determining Step.
Essentially once this base attacks, everything happens simultaneously. Now, the Reaction Order is 2nd Order because it's gonna depend on the concentration of our substance here and on the concentration of the base. So our rate is going to equal k times the concentration of this substance here times the concentration of the base. Now the stereochemistry for an E2 reaction is going to be fixed by the spatial orientation of the alkene substituents.
When we looped X and H into the anti position, our methyl groups were fixed in these positions. If the methyl group happened to be on this side, we would've had the cis product. As-is, we have the transproduct. So the way these things are oriented in space will determine what we get in the elimination product.
Now, how can we favor E2 reactions over SN2 reactions? Well, these are easier to control than E1 versus SN1. One thing we can do is, like the E1, is just raise the temperature. These eliminations run at higher temperatures. The second thing we can do is to have highly substituted and bulky chains. Remember that in an SN2 reaction we essentially had five bonds on carbon for a moment, as the one bond broke and the other formed.
So if there are lots of bulky groups on this carbon, SN2 would be less likely because it would be hard for the nuclear file to get in there. So highly substituted carbons would favor an elimination. The last thing we could use is a strong base. Because strong bases will also just attack hydrogen as the nucleus instead of the carbon.
In general, we want a nuclear file that will go after carbon for substitution as the strong bases will be perfectly happy to take a hydrogen. Now, how can we favor E1 over E2? Well, in a unimolecular elimination we are going to go through a carbo cad ion, so anything that would stabilize the carbo cad ion will help us to promote E1. Highly substituted carbo cad ions are most stable.
Also, if we use polar protic solvents, those polar solvents will also stabilize the carbo cad ion and encourage an E1 mechanism. If we have a strong base in low concentration, that would also favor E1. Remember that the rate of an E1 elimination is equal to just k times the substance that's undergoing the elimination and E2 is equal to k times the substance undergoing the elimination times concentration of the base.
So if there's lots of base, that would start to favor E2. So we want to keep this number, base concentration, as low as possible to make E2 a less likely occasion. And the last thing we need are good leaving groups because remember, there are two steps in E1 and the Slow Step is the leaving of the leaving group. So whatever we can do to get that leaving group off quickly will make it so that an E1 can happen faster.
So in summary, E1 has two steps, E2 has a single step. The rate for an E1 is just dependent on the concentration of the substance that's eliminating. In E2 it's dependent on both the substance and the concentration of the base. We would use a polar solvent in E1 and nonpolar solvent in E2. Tertiary carbon in E1, primary carbon in E2.
And the stereochemistry for an E1 reaction will be mixed because there is rotation of a single bond. And in E2 it will be fixed, based on where things are located when they are in the anti position. So we have a big difference. Carbo cad ion for the E1 not for E2, And that concludes this portion of organic chemistry one.