Chem210-1-2008-Exam1-Answers

Chem210-1-2008-Exam1-Answers - Table 1 Bond Dissociation...

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5 5. (20 pts total) The methyl group of propene (CH 2 =CH-CH 3 ) can be brominated in the presence of bromine and light. A) (10 pts) Write out the mechanism for the bromination of propene including the initiation step, the propagation steps, and at least one termination step. Answer: Initiation: Br 2 2Br Propagation: CH 2 =CH-CH 3 + Br CH 2 =CH-CH 2 + HBr Propagation: CH 2 =CH-CH 2 + Br 2 CH 2 =CH-CH 2 -Br + Br Termination: CH 2 =CH-CH 2 + Br CH 2 =CH-CH 2 -Br Termination: 2Br Br 2 Termination: 2 CH 2 =CH-CH 2 CH 2 =CH-CH 2 -CH 2 -CH=CH 2 Termination: reactions of radicals with the walls of the reaction container B) (10 pts) Using the data in Table 1, calculate the enthalpy change for the bromination of propene considering only the propagation steps.
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Unformatted text preview: Table 1 Bond Dissociation Energies: CH 2 =CH-CH 3 CH 2 =CH-CH 2 + H 364 kJ/mol Br 2 2Br 192 kJ/mol HBr H + Br 368 kJ/mol CH 2 =CH-CH 2-Br CH 2 =CH-CH 2 + Br 196 kJ/mol Answer: Step 1: CH 2 =CH-CH 3 CH 2 =CH-CH 2 + H 364 kJ/mol H + Br H-Br-368 kJ/mol CH 2 =CH-CH 3 + Br CH 2 =CH-CH 2 + HBr -4 kJ/mol Step 2: CH 2 =CH-CH 2 + Br CH 2 =CH-CH 2-Br -196 kJ/mol Br 2 2Br 192 kJ/mol CH 2 =CH-CH 2 + Br 2 CH 2 =CH-CH 2-Br + Br -4 kJ/mol Total energy change ( H ) for the two steps = -4kJ/mol + (-4kJ/mol) = -8 kJ/mol...
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This note was uploaded on 11/23/2008 for the course CHEM 210-1 taught by Professor Trzupek during the Fall '08 term at Northwestern.

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Chem210-1-2008-Exam1-Answers - Table 1 Bond Dissociation...

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