What are `S_N1, S_N2, E1 and E2` reactions? How to identify which reaction is occurring in any given reaction.

You should consider several factors:

• Solvent (may or may not deactivate nucleophile)
• Sterics (of substrate or nucleophile; more sterics = less second-order reactions, and vice versa)
• Reagent pKas (high pKas allow elimination as a possible side mechanism)
• Temperature (high temperatures favor elimination)
• Leaving group propensity (great leaving groups favor first-order reactions, such as `E1` or `S_N1`)

This table summarizes what we'll look at.

![http://www.aceorganicchem.com/]()

The following examples assume a great leaving group.

SN2 REACTIONS

As a random example, consider a general primary alkyl halide reacting with `"NaOH"` in `"DMSO"` (dimethylsulfoxide):

`"OH"^(-)` is small, so it is fast, and therefore, it is a good nucleophile. It also is quite a strong base.

In DMSO, which is a polar aprotic solvent (has no protons, and can dissolve both `"NaOH"` and the alkyl halide), `"OH"^(-)` cannot be deactivated because it cannot accidentally act as a base and grab a proton from anything.

That means in this context, `"OH"^(-)` is a good nucleophile.

Furthermore, the alkyl halide is not sterically-hindered. That is, it is easy to attack it from behind, because there is no bulkiness around the `"C"-"Br"` carbon, the electrophilic center.

Therefore, this situation would be mostly `S_N2`, or bimolecular nucleophilic substitution, because

• the nucleophile is strong
• the primary alkyl halide is sterically unhindered

allowing backside-attack that is typical to `S_N2` reactions.

This creates a stereochemical inversion.

SN1 REACTIONS

Now modify our example, and use a protic solvent instead, like ethanol. We can still use `"NaOH"` I suppose, but since ethanol is protic, we instead have competition:

Since `"OH"^(-)` is a strong base, it steals a proton from the protic solvent, and now we instead have `"H"_2"O"`, `"EtOH"`, and `"EtO"^(-)` in solution. So, we'd consequently have a mixture of products.

• The major product forms from `S_N1` instead (unimolecular nucleophilic substitution), since `"OH"^(-)` has become `"H"_2"O"`, which is in general a weak nucleophile.
• Some minor side products form when remaining `"EtOH"` coordinates with any leftover alkyl halide, or when the formed `"EtO"^(-)` reacts with the leftover alkyl halide via `S_N2`.

It's not entirely clear what the product mixture will give us in terms of percentages, but it will not be just `S_N2`, that is for sure. We will certainly get some `S_N1` products.

Furthermore, if the alkyl halide happens to be secondary or tertiary, you will get some carbocation intermediate. That would also facilitate `S_N1`---a bulky, tertiary alkyl halide usually does.

`S_N1` forms a racemic mixture.

E2 REACTIONS

Now if we modify the `S_N2` example by raising the temperature and changing our nucleophile to tert-butoxide, we have both bulkiness of the nucleophile and temperature-dependence of the mechanism to consider.

At higher temperatures, elimination is favored over substitution.

Furthermore, the greater sterics on the nucleophile make it a poor nucleophile, but since the alkyl groups are electron-donating, tert-butoxide is still a great base:

This bimolecular elimination reaction removes one proton and one leaving group (`"Br"^(-)` in this case), generating a new `pi` bond.

Most `E2` of this sort has no detectable intermediate, and requires that the proton and leaving group be antiperiplanar to each other.

E1 REACTIONS

`E1` is similar to `E2`, but has a carbocation intermediate, similar to `S_N1`. `E1` more easily occurs at higher temperatures relative to `S_N1`, similar to `E2` relative to `S_N2`.

This more often occurs when the alkyl halide is also sterically-hindered, but it also somewhat occurs even when `E2` occurs, just like how some `S_N1` can occur, even when `S_N2` primarily occurs.

Like `S_N1`, a carbocation intermediate would form, and like `E2`, the base acts to remove one proton and one leaving group (`"Br"^(-)` in this case), generating a new `pi` bond on the original alkyl halide.