Halogenoalkanes (7) Elimination reactions

Reactions of the Halogenoalkanes

How does competition between substitution and elimination reactions in haloalkanes work out?

What we find is this:

A) In ethanol solvent where OH^— acts as a base rather than a nucleophile then the corresponding alkene forms by elimination of HBr like this:

CH₃CH₂CH₂CH₂Br     +    OH^—       =      CH₃CH₂CH=CH₂    +   H₂O   +   Br^—

1-bromobutane                    base               but-1-ene

B) But if the solvent is changes for water then the substitution reaction dominates as the hydroxide ion acts as a nucleophile in the more polar solvent:

CH₃CH₂CH₂CH₂Br   +          OH^—   =          CH₃CH₂CH₂CH₂OH   +   Br^—

1-bromobutane                                            butan-1-ol

The important point to remember is that both substitution and elimination mechanisms occur together.

The use of a solvent favours one over the other but does not mean that one works exclusively instead of the other. 

So ethanolic potassium hydroxide allows the elimination mechanism to predominate

Aqueous potassium hydroxide favours the substitution mechanism in the haloalkane.

I have discussed the substitution mechanism here

The elimination mechanism is a concerted mechanism, one step following on the next…

First, the base abstracts a proton attached to the carbon atom attached to the C—Br bond.

Second, the two electrons remaining from the C—H bond form a double carbon-carbon bond

Third, these two electrons repel the electrons in the C—Br bond (rather than the C—CH₃ bond) because the bromine atom forms a good leaving group as a bromide ion (Br^—). 

You ought also to realise that breaking a C—H or a C—CH3 bond costs much more energy than breaking a C—Br bond.

The result is an alkene see below:

In this example, 2-bromo-2-methyl propane is converted into methylpropene.

You should be able to see why this is called an elimination reaction.

HBr is removed (eliminated) from the haloalkane molecule  (It is product and should be in the equation).

The experiment can be carried out in the laboratory by use of a simple test tube apparatus collecting the gaseous alkene over water.

You should be able to draw the apparatus used as it is essentially the same kit as that used below in cracking liquid paraffin.

You should of course be able accurately to add labels to your diagram

You should be able to give the equation for the reaction of 2-bromobutane and ethanolic KOH in which two different alkenes are formed. 

You should be able to name and draw the displayed formulae of these two alkene products.