Can we predict the trend in reactivity of fluoromethane, chloromethane, bromomethane and iodomethane based on the information above?
- the C-F bond is the shortest and strongest – so much so that it takes a large amount of energy to break it and fluoroalkanes are virtually inert
- as we move down the group, the C-X bond becomes weaker, easier to break and so the haloalkane is more reactive
- iodomethane will react fastest
- bromine, chlorine and fluorine are all more electronegative than carbon – as we move up the group, the C-X bond becomes more polar explaining why halogenoalkanes react with nucleophiles in nucleophilic substitution reactions
- the C-I bond is highly polarizable and an incoming nucleophile will cause electrons in the C-I bond to move towards the iodine creating the 𝛅+ carbon that is open to attack
Reactivity also depends on whether the halogenoalkane is primary, secondary or tertiary …
The basic nucleophilic substitution mechanism shown above is a SN2 reaction.
S (substitution); N (nucleophilic); 2 (bimolecular – the rate depends the concentration of both the halogenoalkane and the nucleophile).
However, predicting the reactivity of different halogenoalknaes is not that straightforward.
All of these factors mean that the rate of reaction is slower in secondary halogenoalkanes than it is in primary halogenoalkanes. But what about tertiary?
The steric hindrance and positive inductive effect of having three alkyl groups bonded to the carbon of the C-X bond is so great that tertiary halogenoalkanes react by an SN1 mechanism.
S (substitution); N (nucleophilic); 1 (unimolecular – the rate of the reaction depends only on the concentration of the halogenoalkane).
Primary halogenoalkanes are unable to react by an SN1 mechanism as they have no alkyl groups to stabilise the carbocation transition state.