AS Chemistry 9701: Free-Radical Substitution Mechanism of Ethane and Bromine

Ethane, $\text{CH}_3\text{CH}_3$ , reacts slowly with bromine vapour when the mixture is exposed to ultraviolet light, forming bromoethane as the main organic product alongside hydrogen bromide gas. A small amount of butane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ , is also detected among the products.

(a) State the type of mechanism occurring, and the essential condition needed to initiate it. [2]

(b) Write an equation for the initiation step of this mechanism, and state the type of bond fission that occurs. [2]

(c) Write equations for the two propagation steps that together convert ethane and bromine into bromoethane and hydrogen bromide. [2]

(d) Write an equation for one termination step that would account for the butane detected among the products, and explain why several different termination products (not just bromoethane) are obtained overall. [2]

Part (a): Identifying the mechanism and its essential condition

The reaction of an alkane with a halogen, forming a halogenoalkane and a hydrogen halide, is a free-radical substitution. It only proceeds at a useful rate in the presence of ultraviolet light, which supplies the energy needed to split the weak $\text{Br}-\text{Br}$ bond into reactive radicals. (Heating alone, without UV light, is not sufficient to start the reaction.)

Part (b): The initiation step

$\text{Br}_2 \xrightarrow{h\nu} 2\text{Br}^\bullet$

Each bromine atom in $\text{Br}_2$ takes one electron from the shared covalent pair, so this is homolytic fission, the bond breaks symmetrically to give two neutral bromine free radicals (each with a single unpaired electron), rather than a pair of oppositely charged ions.

Part (c): The two propagation steps

Step 1. A bromine radical abstracts a hydrogen atom from ethane, generating hydrogen bromide and an ethyl radical:

$\text{Br}^\bullet + \text{CH}_3\text{CH}_3 \rightarrow \text{HBr} + \text{CH}_3\text{CH}_2^\bullet$

Step 2. The ethyl radical then reacts with a molecule of bromine, forming the organic product and regenerating a bromine radical to continue the chain:

$\text{CH}_3\text{CH}_2^\bullet + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Br} + \text{Br}^\bullet$

Check: adding steps 1 and 2 together and cancelling the radical intermediates gives the overall equation $\text{CH}_3\text{CH}_3 + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Br} + \text{HBr}$ , which matches the reaction described in the stem, and the bromine radical regenerated in step 2 allows the cycle to repeat. This is why the two steps are called “propagation”.

Part (d): A termination step and why several products form

Termination occurs whenever two radicals collide and combine, removing both from the chain. One possibility that accounts for the butane detected is two ethyl radicals combining directly:

$\text{CH}_3\text{CH}_2^\bullet + \text{CH}_3\text{CH}_2^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$

Other termination steps are equally possible, for example $\text{Br}^\bullet + \text{Br}^\bullet \rightarrow \text{Br}_2$ or $\text{CH}_3\text{CH}_2^\bullet + \text{Br}^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{Br}$ . Because the reaction mixture contains a mixture of ethyl radicals and bromine radicals at any moment, and termination is simply any radical meeting any other radical, several different combinations, and therefore several different termination products, occur alongside the intended propagation cycle. This is also why the reaction gives a mixture of products in practice, rather than a single, perfectly clean substitution.

Final answers

(a) Free-radical substitution; requires UV light.
(b) $\text{Br}_2 \xrightarrow{h\nu} 2\text{Br}^\bullet$ , homolytic fission.
(c) $\text{Br}^\bullet + \text{CH}_3\text{CH}_3 \rightarrow \text{HBr} + \text{CH}_3\text{CH}_2^\bullet$ then $\text{CH}_3\text{CH}_2^\bullet + \text{Br}_2 \rightarrow \text{CH}_3\text{CH}_2\text{Br} + \text{Br}^\bullet$ .
(d) $\text{CH}_3\text{CH}_2^\bullet + \text{CH}_3\text{CH}_2^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ ; random radical-radical combinations give several different termination products.