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

1 Answer
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 #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.