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5 2-Bromobutane, CH3CH₂CHBrCH3, can be prepared by three different methods. The relative molecular mass, M,, of 2-bromobutane is 136.9. (a) 2-Bromobutane can be prepared by reacting butane with bromine (Reaction 5.1). CH₂CH₂CH₂CH₂ + Br₂ → CH₂CH₂CHBrCH3 + HBr Reaction 5.1 The reaction is initiated by the formation of bromine radicals from bromine. (i) State the conditions for the formation of bromine radicals from bromine. (ii) Write two equations for the propagation steps in the mechanism for Reaction 5.1. Use structural formulae for organic species and dots (•) for unpaired electrons on radicals. CH3CH₂CH₂CH3 (iii) The yield of CH3CH₂CHBRCH3 is only 30%. Suggest two reasons why the yield of CH3CH₂CHBrCH3 is so low. 1 2

5 2-Bromobutane, CH3CH₂CHBrCH3, can be prepared by three different methods. The relative molecular mass, M,, of 2-bromobutane is 136.9. (a) 2-Bromobutane can be prepared by reacting butane with bromine (Reaction 5.1). CH₂CH₂CH₂CH₂ + Br₂ → CH₂CH₂CHBrCH3 + HBr Reaction 5.1 The reaction is initiated by the formation of bromine radicals from bromine. (i) State the conditions for the formation of bromine radicals from bromine. (ii) Write two equations for the propagation steps in the mechanism for Reaction 5.1. Use structural formulae for organic species and dots (•) for unpaired electrons on radicals. CH3CH₂CH₂CH3 (iii) The yield of CH3CH₂CHBRCH3 is only 30%. Suggest two reasons why the yield of CH3CH₂CHBrCH3 is so low. 1 2

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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5 2-Bromobutane, CH₂CH₂CHBRCH3, can be prepared by three different methods. The relative molecular mass, M₁, of 2-bromobutane is 136.9. (a) 2-Bromobutane can be prepared by reacting butane with bromine (Reaction 5.1). CH₂CH₂CH₂CH₂ + Br₂ → CH₂CH₂CHBrCH3 + HBr Reaction 5.1 The reaction is initiated by the formation of bromine radicals from bromine. (i) State the conditions for the formation of bromine radicals from bromine. (ii) Write two equations for the propagation steps in the mechanism for Reaction 5.1. Use structural formulae for organic species and dots (•) for unpaired electrons on radicals. CH3CH₂CH₂CH3 (iii) The yield of CH3CH₂CHBCH3 is only 30%. Suggest two reasons why the yield of CH3CH₂CHBCH3 is so low. 1 2 (b) 2-Bromobutane can also be prepared by reacting but-2-ene, CH₂CH=CHCH3, with hydrogen bromide, HBr (Reaction 5.2). CH₂CH=CHCH3 + HBr → CH3CH₂CHBrCH3 Explain, in terms of atom economy, why Reaction 5.2 is more sustainable than Reaction 5.1. Include calculations to justify your answer. Reaction 5.2
Transcribed Image Text:5 2-Bromobutane, CH₂CH₂CHBRCH3, can be prepared by three different methods. The relative molecular mass, M₁, of 2-bromobutane is 136.9. (a) 2-Bromobutane can be prepared by reacting butane with bromine (Reaction 5.1). CH₂CH₂CH₂CH₂ + Br₂ → CH₂CH₂CHBrCH3 + HBr Reaction 5.1 The reaction is initiated by the formation of bromine radicals from bromine. (i) State the conditions for the formation of bromine radicals from bromine. (ii) Write two equations for the propagation steps in the mechanism for Reaction 5.1. Use structural formulae for organic species and dots (•) for unpaired electrons on radicals. CH3CH₂CH₂CH3 (iii) The yield of CH3CH₂CHBCH3 is only 30%. Suggest two reasons why the yield of CH3CH₂CHBCH3 is so low. 1 2 (b) 2-Bromobutane can also be prepared by reacting but-2-ene, CH₂CH=CHCH3, with hydrogen bromide, HBr (Reaction 5.2). CH₂CH=CHCH3 + HBr → CH3CH₂CHBrCH3 Explain, in terms of atom economy, why Reaction 5.2 is more sustainable than Reaction 5.1. Include calculations to justify your answer. Reaction 5.2
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