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IB Chemistry R3.3: Electron sharing reactions worked solutions

Worked solutions to every practice question in R3.3: Electron sharing reactions of Photon Chemistry: Study Guide for the IB Diploma (SL & HL). Try each question, then compare your working with the full solution.

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Basic practice

Question A1

[1 mark]
Which species is a radical? (A) \ce{Cl-} (B) \ce{CH3.} (C) \ce{CH4} (D) \ce{CH3+}
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\ce{CH3.} has seven valence electrons in total (4 + 3), an odd number, so one is unpaired. The other three all have paired electrons.

Final answer

B

Question B1

[1 mark]
Bromine is broken apart by ultraviolet light. Which are the particles formed? (A) \ce{Br+} and \ce{Br-} (B) two \ce{Br.} (C) \ce{Br-} only (D) \ce{Br2^2+}
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UV splits the bond evenly, one electron to each atom (homolytic fission), giving two bromine radicals.

Final answer

B

Question C1

[1 mark]
Ethane reacts with chlorine in UV light. Which equation represents a propagation step? (A) \ce{Cl2 -> 2Cl.} (B) \ce{C2H5. + Cl. -> C2H5Cl} (C) \ce{Cl. + C2H6 -> HCl + C2H5.} (D) \ce{Cl. + C2H6 -> C2H5Cl + H.}
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A propagation step has one radical on each side. \ce{Cl2 -> 2Cl.} has none on the left (initiation) and the second option has two on the left and none on the right (termination). The last option has one radical on each side but makes a lone \ce{H.}, which a chlorine radical never releases; it removes H as \ce{HCl}, as in the third option.

Final answer

C

Question C2

[1 mark]
Propane reacts with bromine in UV light. The first propagation step makes \ce{HBr} and which species? (A) \ce{C3H7.} (B) \ce{C3H7Br} (C) \ce{Br-} (D) \ce{H.}
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A bromine radical takes a hydrogen atom from propane and leaves a propyl radical, \ce{Br. + C3H8 -> HBr + C3H7.}. The alkyl radical is what carries the chain into the second propagation step.

Final answer

A

Question D1

[1 mark]
Ethane reacts with chlorine in UV light. Which product forms only in a termination step? (A) \ce{HCl} (B) \ce{C4H10} (C) \ce{Cl.} (D) \ce{C2H5.}
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Two ethyl radicals joining give butane and end the chain. \ce{HCl} forms in propagation, and the other two are radicals.

Final answer

B

Question E1

[1 mark]
Hexane and hex-1-ene are each shaken with bromine water in a darkened room. What is seen? (A) both stay orange-brown (B) only hex-1-ene turns colourless (C) only hexane turns colourless (D) both turn colourless
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The alkene adds bromine across C=C without light. An alkane needs UV to form bromine radicals, so in the dark it does not react.

Final answer

B

Question F1

[1 mark]
Which reaction goes by a free-radical mechanism? (A) \ce{CH3Br + NaOH(aq)} warmed (B) \ce{C2H4 + Br2} in the dark (C) \ce{CH4 + Br2} in UV light (D) \ce{CH3CH2Cl + NH3} in ethanol
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Free-radical substitution needs a saturated C–H bond and a halogen with UV light. The first and last options use a nucleophile with a halogenoalkane, and the second is electrophilic addition to a double bond.

Final answer

C

Question G1

[1 mark]
How many different monochloro products can form from butane, \ce{C4H10}, in UV light? Ignore optical isomers. (A) 1 (B) 2 (C) 3 (D) 4
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Butane has two kinds of hydrogen: those on the end carbons and those on the middle carbons. Removing one of each gives 1-chlorobutane and 2-chlorobutane, so there are two products.

Final answer

B

Tutorial practice

Question H2

[4 marks]
Ethane reacts with chlorine in UV light to form chloroethane. (a) Write the equation for the initiation step. [1] (b) Write the two propagation steps. [2] (c) Write an equation for a termination step that forms butane, \ce{C4H10}. [1]
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(a) \ce{Cl2 -> 2Cl.} (1). (b) \ce{Cl. + C2H6 -> HCl + C2H5.} (1) and \ce{C2H5. + Cl2 -> C2H5Cl + Cl.} (1). (c) \ce{C2H5. + C2H5. -> C4H10} (1).

Final answer

(a) Cl2 -> 2Cl. (b) Cl. + C2H6 -> HCl + C2H5. ; C2H5. + Cl2 -> C2H5Cl + Cl. (c) C2H5. + C2H5. -> C4H10

Question H3

[3 marks]
Propane reacts with bromine in UV light and 2-bromopropane is the main product. Use \ce{C3H7.} for propyl. (a) Write the equation for the initiation step. [1] (b) Write the two propagation steps. [2]
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Initiation: \ce{Br2 -> 2Br.} (1). Propagation: \ce{Br. + C3H8 -> HBr + C3H7.} (1) and \ce{C3H7. + Br2 -> C3H7Br + Br.} (1). The bromine radical is regenerated, so the chain continues.

Final answer

Br2 -> 2Br. ; Br. + C3H8 -> HBr + C3H7. ; C3H7. + Br2 -> C3H7Br + Br.

Question H4

[4 marks]
Ethane reacts with bromine in UV light to form bromoethane. Initiation, \ce{Br2 -> 2Br.}, is given. (a) Deduce the two propagation steps. [2] (b) Deduce a termination step that forms bromoethane. [1] (c) Explain why UV light is needed. [1]
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(a) \ce{Br. + C2H6 -> HBr + C2H5.} (1) and \ce{C2H5. + Br2 -> C2H5Br + Br.} (1). (b) \ce{C2H5. + Br. -> C2H5Br} (1): two radicals join and none are left. (c) Ultraviolet photons carry enough energy to break the Br–Br bond homolytically, so radicals form and the chain can start (1).

Final answer

(a) Br. + C2H6 -> HBr + C2H5. ; C2H5. + Br2 -> C2H5Br + Br. (b) C2H5. + Br. -> C2H5Br (c) UV supplies the energy to split Br2 into radicals, which start the chain.

Question H5

[6 marks]
Methane reacts with bromine to form bromomethane. Explain the free-radical mechanism. (a) State the condition needed to start the reaction. [1] (b) Write equations for the initiation step and the two propagation steps. [3] (c) Write a termination step that forms ethane, and explain why it ends the chain. [2]
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(a) Ultraviolet light (1). (b) \ce{Br2 -> 2Br.} (1), \ce{Br. + CH4 -> HBr + CH3.} (1), \ce{CH3. + Br2 -> CH3Br + Br.} (1). (c) \ce{CH3. + CH3. -> C2H6} (1). It ends the chain because two radicals are removed and no radical is formed (1).

Final answer

(a) UV light. (b) Br2 -> 2Br. ; Br. + CH4 -> HBr + CH3. ; CH3. + Br2 -> CH3Br + Br. (c) CH3. + CH3. -> C2H6 ; two radicals are used and none made, so no carrier is left.

Question I2

[3 marks]
Methane and chlorine react in UV light. Chloromethane is the target product. (a) Write a propagation step that turns \ce{CH3Cl} into \ce{CH2Cl2}, and explain why it is possible. [2] (b) State how the mixture is arranged to favour chloromethane. [1]
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(a) \ce{Cl. + CH3Cl -> HCl + CH2Cl.} followed by \ce{CH2Cl. + Cl2 -> CH2Cl2 + Cl.} (1). Chloromethane still has C–H bonds that a chlorine radical can attack (1). (b) A large excess of methane makes a chlorine radical far more likely to meet methane than chloromethane (1).

Final answer

(a) Cl. + CH3Cl -> HCl + CH2Cl. ; then CH2Cl. + Cl2 -> CH2Cl2 + Cl. ; CH3Cl still has C-H bonds. (b) use a large excess of methane.

Question I3

[3 marks]
Propane reacts with excess chlorine in UV light and several products form. (a) State how many dichloropropane isomers can form and name two. [2] (b) A student claims only one monochloropropane forms, as every C–H bond is the same. Give the flaw. [1]
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(a) Four isomers (1): 1,1-dichloropropane, 1,2-dichloropropane, 1,3-dichloropropane and 2,2-dichloropropane; two correct names (1). (b) The end carbons carry six equivalent hydrogens and the middle carbon carries two others, so a chlorine radical can form either 1-chloropropane or 2-chloropropane (1).

Final answer

(a) 4; any two of 1,1-, 1,2-, 1,3- and 2,2-dichloropropane. (b) The CH3 and CH2 hydrogens are in different environments, so Cl. can remove either and two isomers form.

Question J2

[5 marks]
Cyclohexane and cyclohexene are each mixed with bromine water. (a) State what is seen for each liquid in the dark. [2] (b) Name the type of reaction of cyclohexene and give the organic product. [2] (c) State why cyclohexane reacts only after exposure to UV light. [1]
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(a) Cyclohexane: no change in colour (1); cyclohexene: decolourises (1). (b) Addition (1); 1,2-dibromocyclohexane (1). (c) UV light is needed to split \ce{Br2} into radicals, which start the substitution (1).

Final answer

(a) cyclohexane: stays orange-brown; cyclohexene: turns colourless. (b) addition; 1,2-dibromocyclohexane. (c) UV is needed to form Br radicals.

Question J3

[3 marks]
Bromine is exposed to UV light; separately, bromomethane is attacked by hydroxide ions. (a) State the type of bond fission in the bromine molecule and name the particles it forms. [1] (b) Contrast the electron movement in \ce{Br-Br} with that in \ce{C-Br} of bromomethane. [2]
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(a) Homolytic fission, giving two bromine radicals \ce{Br.} (1). (b) In \ce{Br-Br} the shared pair is split evenly, one electron to each atom, shown by two half-headed arrows (1). In \ce{C-Br} the bromine takes both bonding electrons, shown by one full-headed arrow, and forms \ce{Br-} with a positive carbon centre left behind (1).

Final answer

(a) homolytic fission; two bromine radicals. (b) Br-Br: one electron goes to each atom (half-headed arrows) and radicals form. C-Br: both electrons go to Br (full-headed arrow) and Br- forms.

Advanced practice

Question K3

[6 marks]
Methane reacts with bromine in UV light. Use bond enthalpies from the data booklet: C–H 414, H–Br 366, Br–Br 193 and C–Br 285, all in \mathrm{kJ\,mol^{-1}}. (a) Calculate \Delta H for \ce{Br. + CH4 -> HBr + CH3.}, treating the step as breaking one C–H bond and making one H–Br bond. [2] (b) Calculate \Delta H for \ce{CH3. + Br2 -> CH3Br + Br.}. [2] (c) Show that the two values add to the enthalpy change of \ce{CH4 + Br2 -> CH3Br + HBr}. [2]
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(a) 414-366=+48\,\mathrm{kJ\,mol^{-1}} (1 for method, 1 for value with sign). (b) Break Br–Br, make C–Br: 193-285=-92\,\mathrm{kJ\,mol^{-1}} (1, 1). (c) +48+(-92)=-44 (1). Direct: break C–H and Br–Br, make C–Br and H–Br: 414+193-285-366=-44, equal (1).

Final answer

(a) +48 kJ/mol. (b) -92 kJ/mol. (c) +48 + (-92) = -44; check 414 + 193 - 285 - 366 = -44.

Question L4

[7 marks]
A photon splits one \ce{Cl-Cl} bond, which has bond enthalpy 242\,\mathrm{kJ\,mol^{-1}}. Use h=6.63\times10^{-34}\,\mathrm{J\,s}, c=3.00\times10^{8}\,\mathrm{m\,s^{-1}} and L=6.02\times10^{23}\,\mathrm{mol^{-1}} from the data booklet. (a) Calculate the longest wavelength, in nm, that can split a \ce{Cl-Cl} bond. [3] (b) A flask absorbs 2.00\times10^{-6} mol of photons and 0.0300 mol of chloromethane forms. Assuming each photon splits one \ce{Cl2}, calculate the number of \ce{CH3Cl} molecules formed per photon. [2] (c) Explain why this ratio is far above 1, and why it is not infinite. [2]
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(a) Energy per bond E=\dfrac{242\times10^{3}}{6.02\times10^{23}}=4.02\times10^{-19}\,\mathrm J (1). \lambda=\dfrac{hc}{E} (1) =\dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{4.02\times10^{-19}}=4.95\times10^{-7}\,\mathrm m=495\,\mathrm{nm} (1). (b) Moles of product per mole of photons =0.0300\div(2.00\times10^{-6})=1.5\times10^{4} (1 for method, 1 for value). (c) Propagation regenerates the chain carrier, so each photon-made pair of radicals drives many cycles (1). Termination joins two radicals and removes the carrier, so every chain ends after a finite number of cycles (1).

Final answer

(a) 495 nm. (b) 1.5 x 10^4. (c) Each propagation cycle regenerates the chlorine radical, so one photon starts thousands of cycles; termination steps remove radicals and end each chain eventually.

Question M3

[6 marks]
A student proposes this chain for ethane and chlorine: \ce{Cl2 -> 2Cl.}; \ce{Cl. + C2H6 -> C2H5Cl + H.}; \ce{H. + Cl2 -> HCl + Cl.}. Bond enthalpies in \mathrm{kJ\,mol^{-1}}: C–H 414, C–Cl 324, H–Cl 431. (a) Show that the two propagation steps add to the correct overall equation. [1] (b) Estimate \Delta H for the student's first propagation step and for \ce{Cl. + C2H6 -> HCl + C2H5.}, and use the values to explain which step is preferred. [3] (c) Write the correct pair of propagation steps. [2]
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(a) Adding the steps and cancelling \ce{H.} and \ce{Cl.} gives \ce{C2H6 + Cl2 -> C2H5Cl + HCl} (1). (b) Student's step: break C–H, make C–Cl: 414-324=+90\,\mathrm{kJ\,mol^{-1}} (1). Accepted step: break C–H, make H–Cl: 414-431=-17\,\mathrm{kJ\,mol^{-1}} (1). The accepted step is exothermic and the student's is endothermic, so the chlorine radical removes H as \ce{HCl} (1). (c) \ce{Cl. + C2H6 -> HCl + C2H5.} (1) and \ce{C2H5. + Cl2 -> C2H5Cl + Cl.} (1).

Final answer

(a) sum: C2H6 + Cl2 -> C2H5Cl + HCl. (b) student's step: 414 - 324 = +90; accepted step: 414 - 431 = -17. The accepted step is exothermic, the student's is strongly endothermic, so Cl. takes H to form HCl. (c) Cl. + C2H6 -> HCl + C2H5. ; C2H5. + Cl2 -> C2H5Cl + Cl.

Question N3

[3 marks]
Propane reacts with bromine in UV light and the product contains a trace of \ce{C6H14}. Write an equation for its formation, name the stage, and state why more than one monobromo isomer also forms.
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\ce{C3H7. + C3H7. -> C6H14} (1); termination, because two radicals join and none remain (1). A bromine radical can remove a hydrogen from an end carbon or from the middle carbon, giving two different propyl radicals and so two isomers (1).

Final answer

C3H7. + C3H7. -> C6H14; termination; Br. can remove H from the CH3 carbon or the CH2 carbon, giving two different propyl radicals.

Question P4

[7 marks]
Methane and chlorine are kept at 298\,\mathrm K. Nothing happens in the dark, but the mixture reacts rapidly in strong light. The activation energy is modelled as 242\,\mathrm{kJ\,mol^{-1}} for splitting \ce{Cl2} and 16\,\mathrm{kJ\,mol^{-1}} for \ce{Cl. + CH4 -> HCl + CH3.}. Use R=8.31\,\mathrm{J\,K^{-1}\,mol^{-1}}. (a) Calculate the fraction e^{-E_a/RT} at 298\,\mathrm K for each activation energy. [3] (b) Use your values to explain the dark result. [2] (c) Explain why doubling the light intensity increases the rate of formation of chloromethane. [2]
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(a) RT=8.31\times298=2476\,\mathrm{J\,mol^{-1}} (1). Splitting \ce{Cl2}: e^{-242000/2476}=3.6\times10^{-43} (1). Propagation: e^{-16000/2476}=1.6\times10^{-3} (1). (b) The fraction for splitting \ce{Cl2} is vanishingly small, so no radicals form in the dark (1). Once a radical exists the propagation barrier is crossed easily, but nothing starts the chain (1). (c) Twice the light makes twice as many photons absorbed per second, so more radicals form and more chains run at the same time (1), and the rate of product formation rises (1).

Final answer

(a) 3.6 x 10^-43 and 1.6 x 10^-3. (b) Almost no Cl2 molecules have enough energy to split, so no radicals form; the propagation step would be feasible if radicals existed. (c) More photons make more radicals, so more chains run at once.

Connected questions

Question X1

Connected questions[8 marks]
Methane reacts with chlorine in UV light: \ce{CH4 + Cl2 -> CH3Cl + HCl}. Use bond enthalpies in \mathrm{kJ\,mol^{-1}}: C–H 414, Cl–Cl 242, C–Cl 324, H–Cl 431. (a) Calculate \Delta H for the overall reaction. [2] (b) Calculate \Delta H for each propagation step, \ce{Cl. + CH4 -> HCl + CH3.} and \ce{CH3. + Cl2 -> CH3Cl + Cl.}. [2] (c) Calculate the energy released when 3.21\,\mathrm g of methane (M=16.05\,\mathrm{g\,mol^{-1}}) reacts completely. [2] (d) State why the answer to (a) is only an estimate, and why UV light is still needed although the reaction is exothermic. [2]
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(a) \Delta H=(414+242)-(324+431)=656-755=-99\,\mathrm{kJ\,mol^{-1}} (1 method, 1 value). (b) Step 1 breaks C–H and makes H–Cl: 414-431=-17 (1). Step 2 breaks Cl–Cl and makes C–Cl: 242-324=-82 (1); the two add to -99. (c) n=3.21\div16.05=0.200\,\mathrm{mol} (1); energy =0.200\times99=19.8\,\mathrm{kJ} (1). (d) Bond enthalpies are averages over many compounds and apply to gases (1). A negative \Delta H does not provide radicals; the first step needs the 242\,\mathrm{kJ\,mol^{-1}} of the Cl–Cl bond, which the UV photon supplies (1).

Final answer

(a) -99 kJ/mol. (b) -17 and -82 kJ/mol. (c) 19.8 kJ. (d) Mean bond enthalpies for gaseous molecules are averages. The reaction needs radicals first; UV breaks Cl-Cl (+242) to start the chain.

Question X2

Connected questions[8 marks]
A student mixes 3.21\,\mathrm g of methane with 7.09\,\mathrm g of chlorine in UV light and collects 3.50\,\mathrm g of chloromethane, \ce{CH3Cl}. Use M(\ce{CH4})=16.05, M(\ce{Cl2})=70.90 and M(\ce{CH3Cl})=50.49\,\mathrm{g\,mol^{-1}}. (a) Determine the limiting reagent and the maximum mass of \ce{CH3Cl} that could form. [3] (b) Calculate the percentage yield. [2] (c) Calculate the atom economy for \ce{CH3Cl} in \ce{CH4 + Cl2 -> CH3Cl + HCl}. [2] (d) Give one reason, from the radical mechanism, why the real yield is below the maximum. [1]
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(a) n(\ce{CH4})=3.21\div16.05=0.200 mol and n(\ce{Cl2})=7.09\div70.90=0.100 mol (1). The ratio is 1:1, so chlorine is limiting (1). Maximum \ce{CH3Cl}=0.100\times50.49=5.05\,\mathrm g (1). (b) \dfrac{3.50}{5.049}\times100=69.3\% (1 method, 1 value). (c) \dfrac{50.49}{16.05+70.90}\times100=58.1\% (1 method, 1 value). (d) Chloromethane still has C–H bonds, so \ce{Cl.} can make \ce{CH2Cl2} and further products; termination also removes radicals into traces of \ce{C2H6} (1).

Final answer

(a) chlorine; 5.05 g. (b) 69.3%. (c) 58.1%. (d) Further substitution forms CH2Cl2 and other products, or radicals combine in termination steps.

Question X3

Connected questions[8 marks]
Propane is converted to propan-2-ol in two steps. Step 1: \ce{C3H8 + Br2 -> C3H7Br + HBr} in UV light, with 2-bromopropane as the main product. Step 2: 2-bromopropane is warmed with aqueous sodium hydroxide. (a) Name the type of reaction in each step. [2] (b) Explain why propane does not react directly with aqueous sodium hydroxide. [1] (c) Identify the nucleophile in step 2 and state what the full-headed curly arrow from it represents. [2] (d) Contrast the way the Br–Br bond breaks in step 1 with the way the C–Br bond breaks in step 2, including the arrow heads used. [3]
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(a) Step 1: free-radical substitution (1); step 2: nucleophilic substitution (1). (b) Propane has strong, almost non-polar C–H and C–C bonds and no \delta+ site for a nucleophile (1). (c) The nucleophile is \ce{OH-} (1); the full-headed arrow shows a lone pair of electrons moving to form a bond with carbon (1). (d) In \ce{Br-Br} the bond pair is shared out evenly, one electron to each atom (1), shown by two half-headed arrows, giving two radicals (1). In \ce{C-Br} bromine takes both bonding electrons (1), shown by a full-headed arrow, giving \ce{Br-} and not a radical.

Final answer

(a) 1: free-radical substitution; 2: nucleophilic substitution. (b) C-H and C-C are strong and almost non-polar, so a nucleophile has no delta+ atom to attack. (c) OH-; the movement of a lone pair to carbon to form a bond. (d) Br-Br: homolytic, half-headed arrows, one electron each, two radicals. C-Br: heterolytic, full-headed arrow, both electrons to Br, giving Br-.

From the book

Photon Chemistry: Study Guide for the IB Diploma (SL & HL)

Electron sharing reactions from the core ideas to full exam questions, with question types, worked examples and graded practice.

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