06_ - P1: PBU/OVY P2: PBU/OVY JWDD052-06...

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-06 JWDD052-Solomons-v1 March 20, 2007 19:11 6 IONIC REACTIONS — NUCLEOPHILIC SUBSTITUTION AND ELIMINATION REACTIONS OF ALKYL HALIDES SOLUTIONS TO PROBLEMS 6.1 Substrate Nucleophile Leaving group I I O O CH 3 CH 3 CH 2 CH 3 CH 3 CH 2 ++ (a) Substrate Nucleophile Leaving group I Br Br CH 3 CH 2 CH 3 CH 2 I (b) Substrate Nucleophile Leaving group Cl Cl (CH 3 ) 3 C O (CH 3 ) 3 C CH 3 OH CH 3 + (c) Substrate Nucleophile Leaving group CN Br C N (d) (e) Substrate Nucleophile Leaving group H NH 3 + + H + Br + + NH 2 Br Br 6.2 I (CH 3 ) 3 C Br I + (CH 3 ) 3 C Br 80
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-06 JWDD052-Solomons-v1 March 20, 2007 19:11 IONIC REACTIONS 81 6.3 (a) We know that when a secondary alkyl halide reacts with hydroxide ion by substitution, the reaction occurs with inversion of conFguration because the reaction is S N 2. If we know that the conFguration of ( )-2-butanol (from Section 5.7C) is that shown here, then we can conclude that ( + )-2-chlorobutane has the opposite conFguration. HO H OH S N 2 Cl H ( R )-( )-2-Butanol = − 13.52 ° [ α ] 25 ° D ( S )-( + )-2-Chlorobutane = + 36.00 ° [ α ] 25 ° D (b) Again the reaction is S N 2. Because we now know the conFguration of ( + )-2-chlorobut- ane to be ( S ) [cf., part (a)], we can conclude that the conFguration of ( )-2-iodobutane is ( R ). Cl H I S N 2 H I ( R )-( )-2-Iodobutane ( S )-( + )-2-Chlorobutane ( + )-2-Iodobutane has the ( S ) configuration 6.4 (a) (b) H 2 O S N 1 I + (CH 3 ) 3 C (a) (b) (CH 3 ) 3 C CH 3 OH 2 + A HA CH 3 By path (a) By path (b) + (CH 3 ) 3 C OH CH 3 (CH 3 ) 3 C CH 3 OH 6.5 and (CH 3 ) 3 C OCH 3 CH 3 (CH 3 ) 3 C CH 3 OCH 3 6.6 (a) Being primary halides, the reactions are most likely to be S N 2, with the nucleophile in each instance being a molecule of the solvent (i.e., a molecule of ethanol).
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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU Printer: Hamilton JWDD052-06 JWDD052-Solomons-v1 March 20, 2007 19:11 82 IONIC REACTIONS (b) Steric hindrance is provided by the substituent or substituents on the carbon β to the carbon bearing the leaving group. With each addition of a methyl group at the β carbon (below), the number of pathways open to the attacking nucleophile becomes fewer. C H H H H H Br Nu H 3 C C H H H H Br Nu H 3 C H 3 C C H H H Br Nu H 3 C H 3 C H 3 C C H H Br Nu 6.7 Protic solvents are those that have an H bonded to an oxygen or nitrogen (or to another strongly electronegative atom). Therefore, the protic solvents are formic acid, HCOH O ; formamide, HCNH 2 O ; ammonia, NH 3 ; and ethylene glycol, HOCH 2 CH 2 OH. Aprotic solvents lack an H bonded to a strongly electronegative element. Aprotic sol- vents in this list are acetone, CH 3 CCH 3 O ; acetonitrile, CH 3 C N; sulfur dioxide, SO 2 ; and trimethylamine, N(CH 3 ) 3 . 6.8 The reaction is an S N 2 reaction. In the polar aprotic solvent (DMF), the nucleophile (CN ) will be relatively unencumbered by solvent molecules, and, therefore, it will be more reactive than in ethanol. As a result, the reaction will occur faster in N , N -dimethylformamide.
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This note was uploaded on 10/20/2011 for the course CHEMISTRY ES110 taught by Professor Te.li during the Spring '11 term at Central Texas College.

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06_ - P1: PBU/OVY P2: PBU/OVY JWDD052-06...

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