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

Nov 12, 2016

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 $E 1$ or ${S}_{N} 1$)

This table summarizes what we'll look at.

The following examples assume a great leaving group.

SN2 REACTIONS

As a random example, consider a general primary alkyl halide reacting with $\text{NaOH}$ in $\text{DMSO}$ (dimethylsulfoxide):

${\text{OH}}^{-}$ 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 $\text{NaOH}$ and the alkyl halide), ${\text{OH}}^{-}$ cannot be deactivated because it cannot accidentally act as a base and grab a proton from anything.

That means in this context, ${\text{OH}}^{-}$ 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 $\text{C"-"Br}$ carbon, the electrophilic center.

Therefore, this situation would be mostly ${S}_{N} 2$, or bimolecular nucleophilic substitution, because

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

allowing backside-attack that is typical to ${S}_{N} 2$ reactions.

This creates a stereochemical inversion.

SN1 REACTIONS

Now modify our example, and use a protic solvent instead, like ethanol. We can still use $\text{NaOH}$ I suppose, but since ethanol is protic, we instead have competition:

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

• The major product forms from ${S}_{N} 1$ instead (unimolecular nucleophilic substitution), since ${\text{OH}}^{-}$ has become $\text{H"_2"O}$, which is in general a weak nucleophile.
• Some minor side products form when remaining $\text{EtOH}$ coordinates with any leftover alkyl halide, or when the formed ${\text{EtO}}^{-}$ reacts with the leftover alkyl halide via ${S}_{N} 2$.

It's not entirely clear what the product mixture will give us in terms of percentages, but it will not be just ${S}_{N} 2$, that is for sure. We will certainly get some ${S}_{N} 1$ products.

Furthermore, if the alkyl halide happens to be secondary or tertiary, you will get some carbocation intermediate. That would also facilitate ${S}_{N} 1$---a bulky, tertiary alkyl halide usually does.

${S}_{N} 1$ forms a racemic mixture.

E2 REACTIONS

Now if we modify the ${S}_{N} 2$ 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 (${\text{Br}}^{-}$ in this case), generating a new $\pi$ bond.

Most $E 2$ of this sort has no detectable intermediate, and requires that the proton and leaving group be antiperiplanar to each other.

E1 REACTIONS

$E 1$ is similar to $E 2$, but has a carbocation intermediate, similar to ${S}_{N} 1$. $E 1$ more easily occurs at higher temperatures relative to ${S}_{N} 1$, similar to $E 2$ relative to ${S}_{N} 2$.

This more often occurs when the alkyl halide is also sterically-hindered, but it also somewhat occurs even when $E 2$ occurs, just like how some ${S}_{N} 1$ can occur, even when ${S}_{N} 2$ primarily occurs.

Like ${S}_{N} 1$, a carbocation intermediate would form, and like $E 2$, the base acts to remove one proton and one leaving group (${\text{Br}}^{-}$ in this case), generating a new $\pi$ bond on the original alkyl halide.