4 Free radical substitution

Synthesis of chloroalkanes Reaction of alkanes with bromine / chlorine in UV light In the presence of UV light alkanes react with chlorine to form a mixture of products with the halogens substituting hydrogen atoms. In general, alkanes do not react with many reagents. This is because the C-C bond and the C-H bond are relatively strong Overall Reaction CH4 + Cl2  CH3Cl + HCl methane chloromethane This is the overall reaction, but a more complex mixture of products is actually formed To understand this reaction fully we must look in detail at how it proceeds step by step. This is called its mechanism The MECHANISM for this reaction is called a FREE RADICAL SUBSTITUTION It proceeds via a series of steps: STEP ONE: Initiation STEP TWO: Propagation STEP THREE: Termination STEP ONE Initiation Cl2  2Cl. Essential condition: UV light The UV light supplies the energy to break the Cl-Cl bond. It is broken in preference to the others as it is the weakest. The bond has broken in a process called homolytic fission. each atom gets one electron from the covalent bond When a bond breaks by homolytic fission it forms Free Radicals. Free Radicals do not have a charge and are represented by a DEFINITION A Free Radical is a reactive species which possess an unpaired electron CH4 + Cl.  HCl + .CH3 STEP TWO Propagation .CH3 + Cl2  CH3Cl + Cl. The chlorine free radicals are very reactive and remove an H from the methane leaving a methyl free radical The methyl free radical reacts with a Cl2 molecule to produce the main product and another Cl free radical All propagation steps have a free radical in the reactants and in the products. As the Cl free radical is regenerated, it can react with several more alkane molecules in a CHAIN REACTION STEP THREE Termination .CH3 + Cl .  CH3Cl .CH3 + .CH3  CH3CH3 Collision of two free radicals does not generate further free radicals: the chain is TERMINATED. Minor step leading to impurities of ethane in product. Write this step using structural formulae and don’t use molecular formulae N Goalby chemrevise.org 4 Applying the mechanism to other alkanes The same mechanism is used: Learn the patterns in the mechanism STEP ONE Initiation Br2  2Br . Essential condition: UV light Example: Write mechanism of Br2 and Propane Br2 splits in the same way as Cl2 CH3CH2CH3 + Br.  HBr + CH3CH2CH2 . STEP TWO Propagation CH3CH2CH2 . + Br2  CH3CH2CH2Br + Br. Remove one H from the alkane to produce a radical To the radical produced in the previous step add a Br STEP THREE Termination CH3CH2CH2 . + Br.  CH3CH2CH2Br CH3CH2CH2 . + CH3CH2CH2 .  CH3CH2CH2CH2CH2CH3 5 N Goalby chemrevise.org Further substitution Excess Cl2 present will promote further substitution and could produce CH2Cl2 , CHCl3 and CCl4 These reactions could occur CH3Cl + Cl2  CH2Cl2 + HCl CH2Cl2 + Cl2  CHCl3 + HCl CHCl3 + Cl2  CCl4 + HCl CH3Cl + Cl.  HCl + .CH2Cl . CH2Cl + Cl2  CH2Cl2 + Cl . Example propagation steps that would lead to further substitution You should be able to write overall reaction equations for various reactions Example 1. Write the overall reaction equation for the formation of CCl4 from CH4 + Cl2 CH4 + 4 Cl2  CCl4 + 4 HCl Example 2. Write the overall reaction equation for the formation of CFCl3 from CH3F + Cl2 CH3F + 3 Cl2  CFCl3 + 3 HCl Note HCl is always the side product – never H2