Can we predict the trend in reactivity of fluoromethane, chloromethane, bromomethane and iodomethane based on any of the information below?
So what do we know about halogenoalkanes?
- the C-F bond is the shortest and therefore the strongest – it takes so much energy to break it fluoroalkanes are virtually inert
- as we move down the group, the C-X bond becomes weaker and so takes less energy / is easier to break
- bromine, chlorine and fluorine are all more electronegative than carbon – as we move up the group, the C-X bond becomes more polar
The reactivity of the halogenoalkanes depends entirely on the strength of the carbon-halogen bond which means that iodomethane will be the most reactive and chloromethane the least reactive.
The polarity of the carbon-halogen bond explains why halogenoalkanes react with nucleophiles in nucleophilic substitution reactions.
*** It is a common misconception amongst students that the polarity of the C-X bond explains the trend in reactivity and that chloromethane is more reactive than iodomethane (they tend to argue that if the bond is more polar, Cl is more likely to leave than I but this is not the case).
Even though the the C-I bond is essentially non-polar (iodine and carbon have a similar electronegativity), iodoalkanes still react with nucleophiles. This is because as the nucleophile approaches the C-I bond, the bond is polarised which causes the electrons in the C-I bond to move towards the iodine creating a 𝛅+ carbon that is open to attack.
The basic nucleophilic substitution mechanism shown above is called a SN2 reaction which is the mechanism of choice for primary halogenoalkanes.
SN2: S (substitution); N (nucleophilic); 2 (bimolecular)
the rate of the reaction depends the concentration of both the halogenoalkane and the nucleophile
Predicting the trend in reactivity of different halogenoalkanes 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.
SN1: S (substitution); N (nucleophilic); 1 (unimolecular)
the rate of the reaction depends only on the concentration of the halogenoalkane
Let’s look at the differences between the SN2 and SN1 mechanisms:
Primary halogenoalkanes are unable to react by an SN1 mechanism as they have no alkyl groups to stabilise the carbocation transition state.
Practice questions
- 1-iodo-3-methylbutane can be hydrolysed by refluxing gently with NaOH(aq).
CH2ICH2CH(CH3)CH3 + OH– ⇾ CH2OHCH2CH(CH3)CH3 + I–
(a) Name the mechanism.
(b) State and explain how the rate of hydrolysis would differ if this reaction were carried out using 1-chloro-3-methylbutane instead of the iodoalkane.
- A student devised an experiment to compare the relative reactivities of 2-chloropropane, 2-bromopropane and 2-iodopropane. The method is outlined below:
- 1 cm3 of ethanol and 4 cm3 of aqueous silver nitrate were measured into a test tube. The test tube was placed in a water bath at 45°C for 10 minutes
- 3 drops of 2-chloropropane were added to the test tube which was returned to the water bath and a stopwatch was started
- the time taken for a pale white precipitate was recorded
- steps i) – iii) were repeated with the bromoalkane and the iodoalkane
(a) State what you would expect to see in the test tube containing the bromoalkane and the iodoalkane.
(b) Why was ethanol added to each test tube?
The precipitates were the result of sparingly soluble silver halides forming in each test tube.
(c) State the order (in terms of time taken) in which the precipitates appear.
(d) Silver halides are ionic compounds. Explain how the reaction occurring in each test tube produces halide ions.
(e) Write an ionic equation with state symbols for the reaction between silver nitrate and bromide ions.
(f) Explain the order of reactivity of 2-chloropropane, 2-bromopropane and 2-iodopropane.
Answers
- (a) nucleophilic substitution
(b) the rate of hydrolysis is slower with the chloroalkane as the C-Cl bond is stronger (higher bond enthalpy) so it takes more energy to break the C-Cl bond than the C-I bond
2. (a) the test tube containing bromoethane would show a cream precipitate and the test tube containing iodoethane would show a yellow precipitate
(b) ethanol increase the miscibility of the reactants / increases the solubility of the halogenoalkane
(c) the precipitate shows fastest with 2-iodopropane and slowest with 2-chloropropane
(d) This is a hydrolysis reaction. Water is behaving as a nucleophile attacking the 𝛅+ carbon of the C-X bond. The C-X bond breaks heterolytically producing halide ions / Cl–, Br–, I– ions.
(e) Ag+(aq) + Br–(aq) ⇾ AgBr(s)
(f) 2-iodopropane is the most reactive producing a precipitate of AgI in the fastest time because the C-I bond is the weakest C-X bond and it takes less energy to break it.