The Sn1 Reaction – Unimolecular Substitution

How does it work?

The Sn1 reaction occurs in two steps.  They are shown below.

Step One:

                                                                             Carbocation           Leaving Group

The reactant is sp³ hybridized.  It is a chiral molecule with tetrahedral geometry.  In this first step, the carbocation is formed.  The geometry of the carbocation is trigonal planar. 

Step Two:   

The negatively charged nucleophile (R) attacks the positively charged carbon.  The product is going to be sp³ hybridized with tetrahedral geometry. 

What type of Nucleophile does this reaction prefer?

A gentle or less effective nucleophile is the most effective because it will allow the reaction to occur more readily.  Recall, a nucleophile has a lone pair of electrons.  A weak Lewis base is a good nucleophile for this reaction.

Why is it called unimolecular?

It is called unimolecular because the rate of the reaction depends on only one species.  Generally the rate-determining step is the initial step, the dissociation of the leaving group, to form a stable, positively charged ion.  This ion is called a carbocation intermediate.

What type of carbocation does this reaction prefer?

In order to answer this question, it is important to remember that the more stable the carbocation, the quicker the reaction.  Therefore, the Sn1 reactions prefer tertiary substrates.  The following is the order of reactivity – 3^o >>> 2^o > 1^o > methyl groups

Why are tertiary carbocations the most stable?

They are most stable because the three alkyl groups share electrons with the positively charged carbon and therefore disperse the charge.  The more delocalized charge stabilizes the intermediate, allowing for lower activation energy.  The lower the activation energy is, the faster the reaction will proceed. 

Do Sn1 prefer protic solvents?

Yes, because protic solvents are polar and contain a hydrogen that can participate in hydrogen bonding.  Molecules of protic solvents can form hydrogen bonds to nucleophiles.

What is going to make a good leaving group for this reaction?

Good leaving groups are going to be halide ions, found in column seven on the periodic table.  The following are the halides (in order from strongest to weakest leaving group) – I > Br > Cl > F

Can this reaction produce a racemic mixture of products?