Help me explain #"S"_N1# reactions? I was trying to explain it to my daughter.

1 Answer
Aug 22, 2016


#"S"_N1# reactions are first-order nucleophilic substitution reactions, which means that their first step is the slow step in which the bond with the leaving group is strong and the nucleophile is slow.

That means in general, the leaving group leaves before the nucleophile attacks, and a detectable carbocation intermediate forms.

This is in line with the definition of a first-order reaction since the slow (rate-determining) step is significantly unimolecular (i.e. the nucleophile is too slow to assist the leaving group and be a second participant in the rate-determining step).


The thermodynamic side of things is that a carbocation is in general less stable than the starting compound, for the predictable reason that the central carbon is electropositive due to missing two electrons.

A carbocation increases in stability with an increasing number of alkyl groups (which are electron-donating/releasing) surrounding it.

That is because the donated (negative) electron density stabilizes the electropositive central carbon (an effect called hyperconjugation).

Hence, methyl (#-"CH"_3#) carbocations are least stable, primary (#1^@#) carbocations are less stable than secondary (#2^@#) and those are less stable than tertiary (#3^@#) carbocations, as shown below:

So, tertiary carbocations form more easily than secondary carbocations, and so on, from a thermodynamic standpoint.


A carbocation is planar (its central carbon's #2p_z# orbital is empty).

Due to the planarity of the carbocation, nucleophilic attack can occur on either face of the carbocation, and therefore, a racemic mixture of products occurs.

i.e. If you are to form a chiral product (for carbon compounds, four different functional groups - even if one of them is #"H"#)...

...You form both enantiomers in equal quantities because either face is as likely to be attacked.

So if you have one #(R)-# enantiomer, you also have the #(S)-# enantiomer (non-superimposable mirror images).


In addition, the kinetic side of things is that the greater the number of alkyl groups around the central carbon, the bulkier and more sterically-hindered the carbocation is.

That means the more alkyl groups there are surrounding the central carbon, the more crowded the leaving group is and the more quickly it wants to leave to form the carbocation.


Hence, #"S"_N1# reactions occur more slowly and least easily for methyl carbocations, and more quickly and most easily with tertiary carbocations.

Similarly, #"S"_N2# occurs when #"S"_N1# does not, if both reactions are under the same conditions otherwise (temperature, solvent, etc) - though technically, both can occur to the same substrate.