sn1

An Explanation Of The Mechanism:

In the first step the leaving group leaves. Notice, how the carbon has only three bonds at this point which is called a carbocation. Notice how the carbon is a tertiary carbon which make the molecule a tertiary carbocation(remember that a tertiary carbocation is very stable). SInce the carbocation intermediate is so stable it will last for a short time longer than a transition state. Finally, the nucleophile (-OH) attacks the carbocation and forms a bond with the carbon. Since a carbocation is a flat (like a pancake) molecule the nucleophile can attack from the top or the bottom of the molecule. Therefore, both the "R" and "S" configuration of the molecule is produced.

*Make sure that you can draw this mechanism from memory.

The Rate Law:

You will see by looking at the mechanism, that the rate of the reaction (how quickly the reaction occurs) depends on one thing; how quickly the leaving group can leave. No matter how hard the nucleophile tries it cannot get in until the leaving group leaves because the center carbon is surrounded by three big carbon groups.This is the reason why the reaction is called SN1. Additionally since the rate depends on only one thing the reaction is said to show "first order kinetics".You should be able to draw the rate law on an exam for a SN1 reaction. It looks like the following:

Rate = k [R-X]

*R-X represents the original molecule with the leaving group attached to it (molecule (A).

* k is the rate constant (this is all you need to know about this)

When Do Molecules Undergo An SN1 Reaction:

Molecules undergo an SN1 reaction when the carbon attached to the leaving group is a tertiary carbon(meaning it is attached to three other carbons). Please note that secondary carbons can undergo a SN1 reaction but the reaction will not occur as quickly. A primary carbon can undergo a SN1 reaction but it is very unlikely. Let's make sure that you understand this concept by attempting a practice problem.

Example: Label the following organic compounds with respect to how quickly they will react in an SN1 reaction

(#1 being the best).

Notice how I have placed a gold star next to the carbons where the leaving group is attached. These are the carbons you should be focusing on when trying to answer this question. Notice that molecule (A) is the worst because the carbon with the star attached to it has only one other carbon directly attached to it. Therefore, it is primary carbon and should react the slowest in an SN2 reaction. In molecule (B), the carbon with the star next to it has two other carbons directly attached to it. Therefore, it is a secondary carbon and is not as fast as molecule (C) in an SN1 reaction. Finally, notice that the carbon in molecule (C) has three other carbons directly attached to it. Therefore it is a tertiary carbon and will react very quickly in an SN1 reaction. Did you get the answer correct? I hope so!

You will notice that the starting material has an "S" configuration and the products have both an "S" and "R" configuration. In all SN1 reactions this phenomenon occurs because when the nucleophile comes in to attack the carbocation intermediate(remember the carbocation is flat like a pancake) it is equally as likely that the nucleophile can come in and attack from the top as it is from the bottom. If the nucleophile comes in from one direction it will produce the "R" configuration and if it comes in from the direction, it will produce the "S" configuration. Since, both directions of attack are very likely the two products are produced in equal amounts. Now you may notice that the relationship between the two products are that they are enantiomers. If you can also remember an equal 50/50 mixture of enantiomers is called a racemic mixture. In almost all SN1 reactions, a racemic mixture is produced.

Therefore, when answering an SN1 reaction question make sure that you do two things:

1. Replace the leaving group with the nucleophile.

2. Make sure that no matter what the configuration is in the starting material that you draw both the "R" and "S" configurations of the product in equal amounts.

Solution: Notice how the carbon with the leaving group (I) attached to

it has three other carbons directly attached to it. Therefore, it is a tertiary carbon. This should be your first hint that the reaction should proceed via a SN1 mechanism. SInce the reaction is SN1, a racemic mixture must be formed. Therefore, there should be two products formed in equal amounts; the "S" and "R" configuration. Also make sure to replace the leaving group (I) with the nucleophile (-CN).

You should first notice that at the beginning of the reaction you just have the starting material (S.M.). SInce the carbon in the starting material has four bonds, it is happy. When a molecule is happy, it is stable and thus low in energy(remember that a stable molecule is low in energy and that an unstable molecule is high in energy). When the leaving group is beginning to leave the

molecule is very confused because it doesn't know if it has four bonds on the carbon or not. This is very bad,and the molecule becomes very unhappy and in turn very unstable. Therefore the energy of the molecule rises and this is called the transition state one (T.S. #1). Finally the leaving group leaves and the carbon has three bonds again. At this point the carbon has three bonds and is thus a carbocation. However, since it is a tertiary carbocation it is a little more stable that just the average carbocation. Therefore,the molecule becomes a little happier and the energy lowers a little bit. This is called the carbocation intermediate (C.I.) Then the nucleophile begins to form a bond with the carbocation intermediate and the molecule becomes confused again because it doesn't know if it has three of four bonds. Therefore, the energy rises and this is called transition state two (T.S. #2). Finally, the nucleophile is completely bonded to the carbon and the carbon becomes happy again (stable) and the energy goes down.

-The End

(Make sure that you can explain the diagram in a way that is similar to this!)

You will see that the nucleophile has a negative charge. Therefore the protic solvent will surround the nucleophile and form a cage around it just like in SN2. However remember that in SN1 reactions that the rate of the reaction depends on only one thing: how quickly the leaving group can leave. Even if the nucleophile is surrounded by a cage formed by the protic solvent it doesn't matter. So what else has a negative or positive charge in the SN1 mechanism? The carbocation intermediate! Therefore the protic

solvent will surround the carbocation intermediate. When it does this, it tricks the carbocation into thinking that it has four bonds again. This makes the carbocation very happy and thus more stable. When this happens the energy level of the carbocation goes down and the energy diagram looks more like the one below:

When you run a reaction you can think of the energy diagram as a hill that you have to climb over in order for the reaction to occur. If the hill is very big, the reaction will take a long time. If the hill is made into a smaller one (as it is when an SN1 reaction is run in a protic solvent) then the reaction is run much faster because it takes a shorter amount of time to cross a smaller hill. Which is great because this means that you don't have to sit in the lab for such a long time waiting for the reaction to run.

Therefore, always run an SN1 reaction in a protic solvent!