Alkanes are saturated hydrocarbons

Alkanes are a class of hydrocarbons that have the general formula C_nH_2n+2_, where 'n' represents the number of carbon atoms.

Key features of alkanes include:

• They consist solely of carbon and hydrogen atoms (thus called hydrocarbons).
• Each carbon atom forms four single bonds.
• They are fully "saturated" with hydrogen, meaning they contain no double or triple bonds.

Examples of alkanes:

Cycloalkanes are a type of alkane in which the carbon atoms form a ring structure. Their general formula is C_nH_2n_, where 'n' is the number of carbon atoms in the ring. Despite the ring structure, cycloalkanes remain saturated.

Photochemical halogenation of alkanes

Halogens react with alkanes in special light-induced reactions called photochemical reactions.

For the reaction to occur:

• Ultraviolet (UV) light must be present.
• This UV light provides the activation energy to start the reaction.

The overall reaction is a substitution, where a hydrogen atom in the alkane molecule is replaced by a halogen atom like chlorine or bromine.

Free radical substitution mechanism

Photochemical halogenation of alkanes follows a three-step free radical substitution mechanism:

1. Initiation - UV light produces reactive radicals.
2. Propagation - Radicals react in a chain reaction.
3. Termination - Radicals join to form stable molecules.

Methane reacts vigorously with chlorine gas in the presence of UV light, as represented by the equation:

CH₄ + Cl₂ ➔ CH₃Cl + HCl

The three-stage mechanism for this reaction is detailed below.

Stage 1 - Initiation

• UV light breaks the Cl-Cl bond in chlorine via homolytic fission to give two chlorine radicals (Cl•): Cl₂ ➔ 2Cl•

• The unpaired electron makes Cl• highly reactive.

Stage 2 - Propagation

• The chlorine radical attacks a methane molecule in a substitution reaction: Cl• + CH₄ ➔ CH₃• + HCl

• The methyl radical (CH₃•) attacks another chlorine molecule: CH₃• + Cl₂ ➔ CH₃Cl + Cl•

• This propagation cycle continues until reagents are used up.

Stage 3 - Termination

• Two radicals join to form a stable covalent bond: Cl• + CH₃• ➔ CH₃Cl

• Other combinations like 2CH₃• ➔ C₂H₆ or 2Cl• ➔ Cl₂ are also possible.

The end products of the termination step depend on which reagent is in excess:

• Excess chlorine - Further substitution on products like chloromethane occurs, producing a mixture of products such as CH₃Cl, CH₂Cl₂, CHCl₃, and CCl₄.
• Excess methane - Predominantly single substitution occurs to form chloromethane.

Problems with free radical substitution

Using free radical halogenation to make specific haloalkane products has two key problems.

1. Production of product mixtures

• With an excess of halogen, additional substitution reactions can occur.
• For example, chloromethane (CH₃Cl) can undergo further substitution: Cl• + CH₃Cl ➔ CH₂Cl• + HCl

CH₂Cl• + Cl₂ ➔ CH₂Cl₂ + Cl•

• This results in a mixture of halogenoalkanes (e.g., CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄) that must be separated.

2. Formation of multiple isomers

• The propagating radical can substitute at any position along a carbon chain.
• This leads to the production of various positional isomers.
• For example, halogenation of propane yields a mixture of 1-chloropropane and 2-chloropropane.

To maximise the yield of a desired halogenoalkane, it's advisable to use a significant excess of the alkane relative to the halogen, minimising further substitution reactions.