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Questions on free-radical substitution
The reaction of methane with chlorine in ultraviolet light, radicals and homolytic fission, the initiation, propagation and termination steps written as equations and with fishhook arrows, why the reaction is a chain, the bond enthalpies that make it run, why fluorine explodes and iodine does not react, further substitution, the isomer mixture from propane and the selectivity of chlorine against bromine, atom economy, and the same chain in the ozone layer.
20 original questions · 65 marks · the free-radical substitution notes · Alkanes and halogenoalkanes · Flashcards for this unit
To print: the written questions (PDF, 78 KB) · the mark schemes (PDF, 84 KB)
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Outline the mechanism by which propane reacts with chlorine in ultraviolet light to form 1-chloropropane, and explain why 1-chloropropane is only one of many products.
Mark scheme
Indicative content
- initiation: ultraviolet light breaks the Cl–Cl bond homolytically, Cl₂ → 2Cl•, each atom taking one electron of the bonding pair
- first propagation step: Cl• + CH₃CH₂CH₃ → HCl + •CH₂CH₂CH₃, the chlorine radical taking a hydrogen from an end carbon
- second propagation step: •CH₂CH₂CH₃ + Cl₂ → CH₃CH₂CH₂Cl + Cl•, so the chlorine radical is regenerated and the chain continues
- termination when two radicals combine, for example •CH₂CH₂CH₃ + Cl• → CH₃CH₂CH₂Cl, or two propyl radicals giving hexane, C₆H₁₄
- a chlorine radical can also take a hydrogen from the middle carbon, giving 2-chloropropane, a position isomer with a similar boiling point that is hard to separate
- further substitution of the chloropropanes gives dichloropropanes and beyond, so the product is a mixture and the yield of 1-chloropropane is low
Levels
- Level 3 (5–6 marks) all three stages are named and each has a correct equation with radicals shown, the regeneration of the chlorine radical is pointed out, and both reasons for the mixture are given, substitution at the middle carbon and further substitution
- Level 2 (3–4 marks) initiation and both propagation steps are correct; termination is thin, a radical dot or a species is missing from one equation, or only one of the two reasons for the mixture of products is given
- Level 1 (1–2 marks) ultraviolet light is said to split chlorine, sometimes into ions rather than radicals, or one step is written; the chain is not followed and the other products are not accounted for
State the condition needed for methane to react with chlorine and name the type of mechanism, then state what is meant by a radical and describe how homolytic fission forms one.
Mark scheme
- B1 ultraviolet light, or sunlight; reject 'heat' or a catalyst
- B1 the mechanism is free-radical substitution
- B1 a radical is a species with an unpaired electron; reject 'a charged atom'
- B1 in homolytic fission a covalent bond breaks with one electron of the bonding pair going to each atom
Outline the mechanism for the reaction of methane with chlorine to form chloromethane, using equations.
Mark scheme
- B1 initiation: Cl₂ → 2Cl•
- B1 first propagation step: Cl• + CH₄ → HCl + •CH₃
- B1 second propagation step: •CH₃ + Cl₂ → CH₃Cl + Cl•
- B1 a termination step, such as •CH₃ + Cl• → CH₃Cl, or 2Cl• → Cl₂, or 2•CH₃ → C₂H₆
Calculate ΔH for each propagation step of the chain reaction between hydrogen and chlorine, Cl• + H₂ → HCl + H• and H• + Cl₂ → HCl + Cl•, and for the overall reaction H₂ + Cl₂ → 2HCl. The bond enthalpies are given in the table.
Bond Mean bond enthalpy / kJ mol⁻¹ H–H 436 Cl–Cl 243 H–Cl 432 Mark scheme
- M1 ΔH = bonds broken − bonds made: step 1 breaks H–H and makes H–Cl; step 2 breaks Cl–Cl and makes H–Cl
- A1 step 1: ΔH = 436 − 432 = +4 kJ mol⁻¹; the sign is required
- A1 step 2: ΔH = 243 − 432 = −189 kJ mol⁻¹
- A1 overall: ΔH = (436 + 243) − (2 × 432) = −185 kJ mol⁻¹, the sum of the two steps; error carried forward from either step
Give the two propagation steps by which chloromethane is converted into dichloromethane, and the termination step that forms a hydrocarbon with two carbon atoms, and identify what that hydrocarbon shows about the mechanism.
Mark scheme
- B1 Cl• + CH₃Cl → HCl + •CH₂Cl
- B1 •CH₂Cl + Cl₂ → CH₂Cl₂ + Cl•
- B1 •CH₃ + •CH₃ → C₂H₆, two methyl radicals combining; ignore the position of the dot in this equation
- B1 ethane shows that methyl radicals are present and that termination occurs by two radicals combining
Evaluate free-radical substitution as a method for making pure 1-chloropropane from propane.
Mark scheme
- B1 the reagents are cheap and the reaction needs only ultraviolet light at room temperature
- B1 the first substitution gives both 1-chloropropane and 2-chloropropane, isomers with similar boiling points that are hard to separate
- B1 further substitution and termination add more products, and hydrogen chloride is a by-product, so the atom economy and yield of the wanted product are low
- B1 it is a poor method for one pure product; a route from propan-1-ol would be better
Draw the initiation step and the first propagation step of the reaction of methane with chlorine, using fishhook arrows to show the movement of single electrons.
Mark scheme
- B1 two fishhook arrows start on the Cl–Cl bond, one ending on each chlorine atom; reject full-headed arrows
- B1 the products are two Cl• radicals, a dot on each Cl and no charge, with ultraviolet light over the reaction arrow
- B1 in propagation, one fishhook from the dot of Cl• and one from the C–H bond both end between Cl and H, forming H–Cl
- B1 a third fishhook from the C–H bond ends on the carbon, giving •CH₃ with the dot on carbon, and HCl
Draw the second propagation step of the reaction of methane with chlorine, and the termination step that forms ethane, using fishhook arrows.
Mark scheme
- B1 one fishhook from the dot of •CH₃ and one from the Cl–Cl bond both end between C and the nearer Cl, forming C–Cl
- B1 a third fishhook from the Cl–Cl bond ends on the far Cl, giving CH₃Cl and Cl• with the dot on Cl
- B1 for termination, one fishhook runs from the dot of each methyl radical to the space between the two carbon atoms
- B1 the product is ethane, CH₃–CH₃, with the new C–C bond and no dots or charges remaining
Explain why the reaction of methane with chlorine is described as a chain reaction.
Mark scheme
- B1 in each propagation step a radical reacts with a molecule and a new radical is formed
- B1 the chlorine radical used in the first propagation step is regenerated in the second, so the cycle repeats
- B1 one initiation step therefore leads to many molecules of product before termination removes the radicals
Calculate the percentage atom economy for making dichloromethane by the reaction CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl. Ar: H 1.0, C 12.0, Cl 35.5.
Mark scheme
- M1 Mr(CH₂Cl₂) = 12.0 + 2.0 + 71.0 = 85.0
- M1 total Mr of reactants = 16.0 + (2 × 71.0) = 158.0, and atom economy = 85.0 ÷ 158.0 × 100
- A1 atom economy = 53.8%, to three significant figures
Explain why the reaction of methane with chlorine gives a mixture of products.
Mark scheme
- B1 chloromethane still has C–H bonds, so a chlorine radical can take a hydrogen from it as well as from methane
- B1 further substitution gives dichloromethane, trichloromethane and tetrachloromethane
- B1 termination steps give other products, such as ethane from two methyl radicals, and its chlorinated derivatives
Suggest, using the bond enthalpies C–H 413 and C–Br 290 kJ mol⁻¹, why a bromine radical does not react with methane by the step Br• + CH₄ → CH₃Br + H•.
Mark scheme
- B1 this step breaks a C–H bond and makes a C–Br bond
- B1 ΔH = 413 − 290 = +123 kJ mol⁻¹, so the step is strongly endothermic
- B1 the step that does occur, Br• + CH₄ → HBr + •CH₃, is far less endothermic (+47 kJ mol⁻¹ with H–Br 366), so it happens in preference; allow 'lower activation energy'
Suggest equations for the initiation step and the two propagation steps in the reaction of propane with bromine in ultraviolet light to form 2-bromopropane.
Mark scheme
- B1 initiation by homolytic fission of bromine: Br₂ → 2Br•
- B1 Br• + CH₃CH₂CH₃ → HBr + CH₃ĊHCH₃, the hydrogen taken from the middle carbon; accept •CH(CH₃)₂
- B1 CH₃ĊHCH₃ + Br₂ → CH₃CHBrCH₃ + Br•, the bromine radical regenerated
Explain, using the bond enthalpies C–H 413 and H–I 299 kJ mol⁻¹, why iodine does not react usefully with methane in ultraviolet light.
Mark scheme
- B1 the first propagation step, I• + CH₄ → HI + •CH₃, breaks C–H and makes the weaker H–I bond
- B1 ΔH = 413 − 299 = +114 kJ mol⁻¹, so the step is strongly endothermic with a very high activation energy
- B1 iodine radicals almost never complete it and recombine to I₂ first, so no chain is propagated
Calculate how many times more readily the hydrogen on the central carbon of 2-methylpropane, (CH₃)₃CH, is replaced than each hydrogen on a methyl group, using the monochlorination results in the table.
Product Hydrogens it comes from Percentage of product 1-chloro-2-methylpropane 9 on the three CH₃ groups 64 2-chloro-2-methylpropane 1 on the central carbon 36 Mark scheme
- M1 each methyl hydrogen accounts for 64 ÷ 9 = 7.1% of the product
- M1 the central hydrogen accounts for 36 ÷ 1 = 36% of the product, and the ratio is 36 ÷ 7.1
- A1 the central hydrogen is replaced 5.1 times as readily, a ratio with no unit, to two significant figures; allow 5
Explain why ultraviolet light starts the reaction by breaking the Cl–Cl bond rather than a C–H bond.
Mark scheme
- B1 the Cl–Cl bond is the weakest bond present, 243 kJ mol⁻¹ against 413 kJ mol⁻¹ for C–H
- B1 ultraviolet light provides enough energy to break it homolytically, giving two chlorine radicals
Suggest how the proportion of chloromethane in the product could be increased, and explain why your change works.
Mark scheme
- B1 use a large excess of methane
- B1 a chlorine radical is then much more likely to collide with a methane molecule than with a chloromethane molecule, so less further substitution occurs
Calculate the percentage of the monochlorinated product of butane, CH₃CH₂CH₂CH₃, that would be 1-chlorobutane if every hydrogen atom were equally likely to be replaced.
Mark scheme
- M1 butane has six hydrogen atoms on its end carbons and four on its two middle carbons, ten in all
- A1 6 ÷ 10 × 100 = 60%, to two significant figures
Suggest why the measured percentage of 1-chlorobutane in the monochlorinated product of butane, about 28%, is lower than the figure predicted by counting hydrogen atoms.
Mark scheme
- B1 each hydrogen on a middle carbon is replaced more readily than each hydrogen on an end carbon
- B1 the C–H bonds on a carbon joined to two other carbons are weaker, so a chlorine radical removes those hydrogens faster
Explain why the reaction between methane and chlorine stops almost at once when the ultraviolet lamp is switched off.
Mark scheme
- B1 no more chlorine radicals are made by initiation once the light is off
- B1 the radicals already present are removed by termination, so no chains remain to carry the reaction