Haloalkanes - how to explain order of reactivity?

:marchmell I know that haloalkanes of iodide ions react faster than bromide and chloride (in that order) with a nucleophile. Im not quite certain of how to explain this.

Could i say:
The larger iodide cation is more polarisable due to its lower charge density and it also has a lower value of electronegativity than bromide and chloride. This means that the C-I bond is more polar and is therefore more susceptible to attack from a nucleophile.

Should i mention that the C-I bond has a lower bond enthalpy?
Why does having greater polarity make a bond easier to break - or have i got it all the wrong way round

Reply 1

oxymoron

No ... first of all, a haloalkane does not conatin an iodide ion (and iodide is an anion not a cation anyway). It is an iodine atom COVALENTLY bonded to the carbon.

Also, because the iodine is not as electronegative as Br or Cl ... the C-I bond is LESS POLAR than the C-Cl bond.

So - to answer the question, there are two factors involved here. Firstly the polarity of the C-X bond. As C-Cl bond is more polar than C-I bonds the C will have a greater delta-postive charge so will be more succeptible to attack by nucleophile. Looking at this factor alone threfore you would expect chloroalkanes to react FASTER than iodoalkanes.

However, the other factor involved is the C-X bond enthalpy. C-Cl bond has a more negative bond enthalpy indicating that it is a stronger bond than C-I. This would suggest that the C-I bond is broken more easily, so iodoalkanes will react FASTER than chloroalkanes.

You can see that these two factors are acting against each other. Experiments show that the bond enthalpy argument "wins" over polarity (ie the reactivity depends on the bond enthalpy rather than polarity of the bond) so iodoalkanes do react the fastest.

As to why C-I bonds are weaker: Iodine atoms are much larger than Carbon, so the overlap is not as good when forming bonds. Chlorine atoms are a much more similar size to carbon so the overlap is better and a stronger bond is formed.
(Actually due to the difference in orbital energies but that is beyond AS/A2)