Halogenoalkanes (6) Nucleophilic substitution in tertiary haloalkanes

#3 Nucleophilic substitution in halogenoalkanes

In a previous post I discussed the basics of this reaction between haloalkanes and aqueous alkali. 

Here is the general reaction scheme:

We made some criticisms of this image from the web.

The curly arrows are pretty random and sloppily drawn.

We’d like to see them start on a lone pair of the nucleophile and end on a specific positive centre.

We’d also like to see them start on the R—X bond and end as they do on the halide lone pair. 

These generic mechanisms do not indicate anything about the rate at which these reactions proceed. 

Let’s look at another representation of this hydrolysis reaction mechanism from the web:

Now this is the hydrolysis of 2-bromo-2-methyl propane: a tertiary haloalkane

Note how the arrows are drawn, where they start and where they end but we don’t like to see the arrow from the hydroxide ion coming from the negative charge rather it should come from a lone electron pair drawn on the OH. 

Let’s consider what is happening here.

We have a two step reaction in which step one involves the formation of a tertiary carbocation.

The formation of this carbocation is possible because the three electron pushing methyl groups stabilise (delocalise) the positive charge that remains on the ion after the bromide ion has left. 

The use of a polar solvent such as silver nitrate solution (e.g. 0.02M AgNO3) facilitates the formation of the carbocation. 

Not only is it a polar solvent the Ag+ ion is able to form strong bonds with halide ions according to ROC Norman (p130)

There is no intermediate transition state as in the hydrolysis of a primary haloalkane.

The second step, which is faster than the first, involves the attack of the nucleophile (OH^–) on the positive carbocation.

This reaction is always going to be much quicker than the first step because the second step involves the combination of oppositely charged ions. 

The slowest or rate determining step of the reaction involves just the haloalkane and not the hydroxide ion i.e. it is monomolecular. 

We can summarise the reaction mechanism as follows: substitution (S), nucleophilic(N) and monomolecular(1)

That’s why it is called an Sn1 reaction!!

The stereo chemistry of this reaction is important too but in a different way to the hydrolysis of a primary haloalkane.

The hydroxide ion can attack from either side of the carbocation.

Let’s look at this more closely:

See how the nucleophile can attack the carbocation from either side.

The attack of the nucleophile leads to two distinctively different molecules if R¹, R² and R³ are different.

You should try re-writing this equation using actual structures for the R groups to see the effect here.

The point is this: the two different products are stereoisomers of each other.

Their carbon atom is a chiral centre.

They possess one of the structural features that gives rise to optical isomerism in organic chemistry that is they have four different functional groups attached to the chiral carbon atom. 

Solutions of each product will rotate the plane of plane polarised light. 

However, because there is a 50:50 chance of the formation of either isomer the resulting product mixture is a non-optically active substance: it is a racemic mixture. 

One molecule of the D- form cancels out the rotation effect of a molecule of the L- form. 

In my next post, I’ll discuss competing elimination reaction in tertiary halogenoalkane hydrolysis.