PracticeQuestionsChap8-10-key

PracticeQuestionsChap8-10-key - ANSWER KEY Practice...

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ANSWER KEY Practice Questions for Chapters 8-10 1 1. Indicate (circle) the substrate in each of the following groups which is most reactive under the reaction conditions given. (Chapter 8) MeOH NaSH NaI H 2 O acetone or or heat or or Cl Cl Cl acetone or or Cl Cl Cl or or Cl Cl heat Cl Cl a) b) c) d) Cl Cl 2. Indicate (in the box) for each of the following trends, whether the trend is observed for polar protic solvents, polar aprotic solvents, both types of solvent, or neither type of solvent. (Chapter 8) ( PP ) Polar Protic ( PA ) Polar Aprotic ( BO ) Both ( NE ) Neither I Br Cl F increasing nucleophilicity > > > (least nucleophilic) (most nucleophilic) a) PP F Cl I Br increasing nucleophilicity > > > (least nucleophilic) (most nucleophilic) b) PA H 2 N F HO H 3 C increasing nucleophilicity > > > (least nucleophilic) (most nucleophilic) c) NE F Cl Br I leaving group ability > > > (worst leaving group) (best leaving group) d) NE I Br Cl F leaving group ability > > > (worst leaving group) (best leaving group) e) BO
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ANSWER KEY Practice Questions for Chapters 8-10 2 3. Identify the product(s) in each of the following reactions. (Chapter 8) Na 2 S C 4 H 8 S Cl Cl + a) S S –2 S Cl The formula (C 4 H 8 S) tells you that both Cl leaving groups are replaced by a single S, and there is only one way for this to happen. The reason this happens instead of another S 2– reacting to replace the second Cl S Cl S –2 intramolecular (much faster) intermolecular (much slower) is that the intramolecular substitution is much faster (much lower entropy cost). H 2 O Cl C 4 H 8 O, 4 isomers b ) heat pair of enantiomers OH OH OH OH 4. Reaction of enantiomerically pure ( R )-2-chloropentane ( 1) with a solution of sodium cyanide in acetone gives enantiomerically pure ( S )-2-methylvaleronitrile ( 2 ), but the corresponding reaction with sodiumiodide gives a racemic mixture of ( S )- and ( R )-2-iodoropentanes. Explain these observations. (Chapter 8) NaCN I NaI I Cl acetone acetone CN I I + ( R )- 1 ( S )- 2 ( S )- 3 ( R )- 3 racemic enantiomerically pure I I and so on, leading to racemization of the iodide product I Cl ( R ) I ( S ) ( R ) I inversion inversion ( S ) since the nucleophile I is also a good leaving group , the reaction doesn't stop here the nucleophile I replaces the leaving group I giving back 2-iodopentane, but with inverted stereochemistry! The nucleophile NC is not also a good leaving group, so no funny business here. 5. Reaction of 1-bromopropane and NaOH in ethanol follows an S N 2 mechanism. What happens to the rate of the reaction if the: (Chapter 8) (a) Concentration of NaOH is doubled? Rate is doubled. (b) Concentrations of both NaOH and 1-bromopropane are doubled? Fourfold increase (4x rate) (c) Volume of the solution in which the reaction is run is doubled? Fourfold decrease (¼ rate).
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ANSWER KEY Practice Questions for Chapters 8-10 3 6. Rank the following alkyl chlorides in order of increasing reactivity (least reactive most reactive) in an S N 2 reaction where a chloride is converted to the corresponding iodide: (Chapter 8) NaI
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This note was uploaded on 04/21/2010 for the course CHEM 2261 taught by Professor Crowe during the Fall '08 term at LSU.

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PracticeQuestionsChap8-10-key - ANSWER KEY Practice...

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