Question

How do you tell `"S"_"N"1` and `"S"_"N"2` reactions apart?

Answer 1

See below.

Explanation:

There are several key differences between `S_N1` and `S_N2` reactions. I've outlined a comparison below, with the assumption that the reader has some basic knowledge of both reaction types.

`S_N2`

`S_N2` stands for "substitution, nucleophilic, bimolecular," or bimolecular nucleophilic substitution. This implies that there are two molecules (bimolecular) involved in the transition state or rate-determining (slow) step. This also tells us that the rate of the reaction depends upon both reactants (nucleophile and electrophile); if you double the concentration of one of the reactants, you double the rate of the reaction.

`S_N2` reactions require a strong nucleophile. Strong nucleophiles are strong bases, so it may be easier to identify them this way at first. For example, strong nucleophiles bear a negative charge. `NaOCH_3` is a strong nucleophile, as it breaks apart into `Na^+` and `OCH_3^-` in solution.

This strong nucleophile forces what is called a backside attack. This is fairly literal. The nucleophile attacks the carbon opposite the leaving group as the two repel each other.

`S_N2` products show inversion of stereochemistry, a result of the backside attack. For example, if the leaving group was once represented as a wedge in the perspective drawing of the molecule, the nucleophile which replaces it will now be shown as a dash. The stereochemistry of any other substituents are left alone.

`S_N2` reactions are concerted, which means that the nucleophile attacks the electrophilic carbon at the same time that the leaving group leaves. There is no intermediate.

`S_N2` reactions prefer polar aprotic solvents, where polar protic solvents hinder `S_N2` reactions. Examples include `DMSO` and acetone.

`S_N2` reactions favor electrophilic carbon atoms which are least highly substituted, so `1^o>2^o>3^o`. You won't see a tertiary carbon undergo an `S_N2` reaction. This is the big barrier for `S_N2` reactions. It is due to the fact that the reaction is concerted, and a backside attack must take place. The steric hinderance on a tertiary carbon is too great to allow this.

This is a reaction diagram for a general `S_N2` reaction, with the reaction coordinate on the `x`-axis and energy on the `y`-axis. The reactants are represented by `color(blue)(Nu^-)+Rcolor(green)(LG)`. This symbolizes the nucleophile `(color(blue)(Nu^-))` plus the leaving group `(color(green)(LG))` attached to some `R`. The transition state or rate-determining step is represented by `[color(blue)(Nuc)---R---color(green)(LG)]^-`, which symbolizes the concerted reaction, the backside attack of the nucleophile while the leaving group leaves. Note that the transition state cannot be isolated! This is just a visualization of what is happening at this point. We see two molecules involved here: the nucleophile and the compound it attacks (electrophile). Finally, we have the products all the way to the right of the diagram.

`S_N1`

`S_N1` stands for "substitution, nucleophilic, unimolecular," or unimolecular nucleophilic substitution. This implies that there is only one molecule (unimolecular) involved in the transition state or rate-determining (slow) step. This also tells us that the rate of the reaction depends upon only one reactant at a time.

`S_N1` requires a weak nucleophile. This is because the `S_N1` reaction is step-wise, or occurs in two steps. First, the slow step: the leaving group leaves and the formation of the carbocation. If a strong nucleophile is present, this slow step does not occur because the nucleophile quickly attacks the electrophile. Additionally, because the carbocation formed is such a reactive electrophile, a weak nucleophile is all that is required. We also see solvolysis in `S_N1` reactions, meaning that the nucleophile and the solvent are the same. A common example is `CH_3OH`. You might also see heat is used, given by `Delta`.

`S_N1` products show both inversion and retention of stereochemistry. You will usually get a mixture of stereoisomers in your products.

Because a carbocation is formed in `S_N1` reactions, rearrangement is possible. Rearrangement will only occur if a more substituted product is possible through a hydride or alkyl shift (must be adjacent). Because no carbocation is formed in `S_N2` reactions, no rearrangement is possible.

`S_N1` reactions prefer polar protic solvents. Examples include `CH_3OH` and acetic acid.

`S_N1` reactions favor electrophilic carbon atoms which are most highly substituted, so `3^o>2^o>1^o`. You will not see a primary carbon undergo an `S_N1` reaction.

In this diagram for an `S_N1` reaction, we see that there are two separate transition states. One represents the formation of the carbocation, which is the slow, rate-determining step (notice it requires more energy), while the second represents the nucleophilic attack on the newly-formed carbocation. This is the fast step. Note that `SM` stands for starting material and `P` for products.

Here is a comparison chart: