10.2 Alkanes

Which of these is an alkane? A. C 6 H 14 B. C 4 H 8 C. C 12 H 24 D. C 102 H 204

Assessment Objectives Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity Describe using equations the complete and incomplete conbustion of alkanes Describe using equations the reactions of methane and ethane with chlorine and bromine Explain the reactions of methane and ethane with chlorine and bromine in terms of a free radical mechanism.

References Textbook Textbook on moodle Powerpoint on moodle Assessment objectives on moodle

Assessment Objective Explain the low reactivity of alkanes in terms of bond enthalpies and bond polarity.

A bit of revision………. BOND ENTHALPY The standard molar enthalpy change of bond dissociation is the energy change when 1 mole of bonds is broken. At a gas state at 298 K and a pressure of 100 kPa. Use a data table for values

Bond Enthalpy You have to be able to draw the structural chemical formula to see all the bonds and add them up properly. Bond strength is associated with how much energy it takes to break the bond

A Bit of Revision Breaking bonds requires energy (Endothermic) Making bonds releases energy (Exothermic) Reactants – products this time!

∆H = [ (5 x O=O) + ( 8 x C-H) + (2 x C-C)] – [(3 x 2 x C=O) + (4 x 2 x H-O)] ∆H = [(5 x 498) + (8 x 413) + (2 x 347)] – [ (3 x 2 x 805) + (4 x 2 x 464)] ∆H = kJ

If the enthalpy of reaction is negative it is an exothermic reaction. The bonds broken in the starting material are weaker than the bonds formed in the product.

11 Calculate  H for each of the following reactions, knowing  H of O 2 and O-H = 119 kcal/mol,  H of C-H = 104 kcal/ml and  H of one C=O = 128 kcal/mol. a) Bonds Broken C-H = 4 x 104 kcal/mol = 416 kcal/mol O-O = 2 x 119 kcal/mol = 238 kcal/mol  H = = +654 kcal/mol Bonds Formed C-O = 2 x -128 kcal/mol = -256 kcal/mol O-H = 4 x -119 kcal/mol = -476 kcal/mol  H = = -732 kcal/mol  H = kcal/mol = -78 kcal/mol

Why are Alkanes Unreactive? 1. C-C and C-H bonds are very strong. 1. The bonds have no (C-C) or a very low polarity (C-H) C-H bond has only a difference in electronegtivity of 0.4 which means that an electrophile or nucleophile are not greatly attracted to the molecule.

Assessment Objective Describe using equations the complete and incomplete combustion of alkanes.

Combustion of Alkanes Alkanes are not especially reactive but they do have one very important reaction: combustion. With an adequate supply of air they react to form carbon dioxide and water. Methane + oxygen  water + carbon dioxide CH 4 + 2O 2  2H 2 O + CO 2

Incomplete Combustion of Alkanes In the absence of an adequate supply of air, alkanes may react to form carbon monoxide and water. Carbon monoxide is highly poisonous and this is one reason why gas boilers must be serviced regularly. Methane + oxygen  water + carbon monoxide 2CH 4 + 3O 2  4H 2 O + 2CO A carbon monoxide detector

Complete the equations below assuming an adequate supply of oxygen for complete combustion. (These are quite tricky!) 1. 2C 2 H 6 + 7O 2  2. C 3 H 8 + 5O 2  3. 2C 4 H O 2  1. 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O 2. C 3 H 8 + 5O 2  3CO 2 + 4H 2 O 3.2C 4 H O 2  8CO H 2 O Activity

Assessment Objective Describe using equations the reactions of methane and ethane with chlorine and bromine.

Methane and ethane react with chlorine and bromine ONLY with UV light. The product is a chlormethane, chloroethane and hydrogen chloride (HCl) All the hydrogens can be replaced  SUBSTITUTION REACTION

Free Radical Substitutions Many organic molecules undergo substitution reactions. In a substitution reaction one atom or group of atoms is removed from a molecule and replaced with a different atom or group. Example: Cl 2 + CH 4  CH 3 Cl + HCl 19

Exercise Draw the following reactions under UV light. 2moles of chlorine and 1 mole of methane 1 mole of bromine with 1 mole of ethane

Assessment Objective Explain the reactions of methane and ethane with chlorine and bromine in terms of a free radical mechanism.

Reactions of Alkanes: with Halogens Energy absorbed from light allows homolytic fission: each resulting atom receives one unpaired electron, known as free radicals Cl-Cl → Cl + Cl

Three Basic Steps in a Free Radical Mechanism Chain initiation The chain is initiated (started) by UV light breaking a chlorine molecule into free radicals. Cl 2  2Cl. Chain propagation reactions These are the reactions which keep the chain going. CH 4 + Cl.  CH 3. + HCl CH 3. + Cl 2  CH 3 Cl + Cl Chain termination reactions These are reactions which remove free radicals from the system without replacing them by new ones. 2 Cl.  Cl 2 CH 3. + Cl.  CH 3 Cl CH 3. + CH 3.  CH 3 CH 3 23

Free Radical Mechanism-The Initiation Step The ultraviolet light is a source of energy that causes the chlorine molecule to break apart into 2 chlorine atoms, each of which has an unpaired electron The energies in UV are exactly right to break the bonds in chlorine molecules to produce chlorine atoms. 24

Free Radical Propagation The productive collision happens if a chlorine radical hits a methane molecule. The chlorine radical removes a hydrogen atom from the methane. That hydrogen atom only needs to bring one electron with it to form a new bond to the chlorine, and so one electron is left behind on the carbon atom. A new free radical is formed - this time a methyl radical, CH 3. 25

Free Radical Propagation II If a methyl radical collides with a chlorine molecule the following occurs: CH 3. + Cl 2  CH 3 Cl + Cl. The methyl radical takes one of the chlorine atoms to form chloromethane In the process generates another chlorine free radical. This new chlorine radical can now go through the whole sequence again, It will produce yet another chlorine radical - and so on and so on. 26

Termination Steps The free radical propagation does not go on for ever. If two free radicals collide the reaction is terminated. 2Cl.  Cl 2 CH 3. + Cl.  CH 3 Cl CH 3. + CH 3.  CH 3 CH 3 27

Exercise Write the steps in the free radical mechanism for the reaction of chlorine with methyl benzene. The overall reaction is shown below. The methyl group is the part of methyl benzene that undergoes attack. 28

Solution Initiation Cl 2  2Cl. Propagation Termination 2Cl.  Cl 2 29

Reactions of Alkanes: with Halogens Substitution of an alkane with a halogen has 3 steps: 1. Initiation 2. Propagation 3. Termination Rate of reaction: Cl 2 > Br 2 > I 2 … Why???

Activity Write the reaction mechanisms for the reaction between methane and bromine. Write the reaction mechanisms for the reaction between ethane and chlorine.