How do you tell #"S"_"N"1# and #"S"_"N"2# reactions apart?
1 Answer
See below.
Explanation:
There are several key differences between
#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# 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:
