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University of British Columbia
CHEM 203

CHEM 203 Midterm #2-Nov. 14, 2006 surname given names last initial student number signature CIRCLE YOUR LECTURE SECTION: DAKE: MWF 9 AM CIUFOLINI MWF 2 PM MOLECULAR MODELS MAY BE USED 1 2 3 4 5 6 raw total total 18 10 14 6 10 12 70 10 1. (18 marks) Answer each of the following parts. a) Draw the structure of: e.g.:CH 3I, Br I , Br e.g.: Cl e.g.: Br Br etc. • an alkyl halide that is likely• an alkyl halide that is likely• an alkyl halide that is likely to undergo SN2 reaction when to undergo E 2reaction when to undergo nucleophilic treated with CH3ONa treated with CH3ONa substitution by the N1 mechanism Page 1 of 11 b) Circle any compound in the series below that has only two singlets in its H NMR spectrum. CH CH none of these OCH 3 OCH 3 3 3 H CO C OCH 3 3 CH 3 H CO OCH 3 O all of these 3 c) Provide the structures of: i) a hydrocarbon containing at least 5 carbon ii) a hydrocarbon containing at least 5 carbon atoms that is a good substrate for radical atoms that is a poor substrate for radical chlorination chlorination e.g.: e.g.: , , CH 3 , etc. CH C CH 3 3 CH 3 d) Treatment of 1,2-dibromoethane with magnesium metal in an ether solvent produces ethylene gas and magnesium bromide (below). Using arrows to represent electron movement, provide a reasonable mechanistic rationale for this known process. Mg Mg° Br CH 2CH 2 + MgBr 2 Br ether SET Br Br Br Br then, either: Mg Br Br Br Br Mg Br MgBr Mg or Br SET Br Br Br MgBr MgBr Mg followed by: Br MgBr ≈ Br CH =CH + Br + MgBr MgBr2 2 2 MgBr Page 2 of 11 e) The H NMR spectrum of unknown organic compound A is shown below. Circle the structural formula of A. relative ratio under 2 2 3 2 3 signals: O O O O O O CH3O O O O O O O O O O O CH 3 O Page 3 of 11 2. (10 marks) Compound A undergoes reaction CH3 CH3 Cl MeOH Cl with methanol as shown in the equation to the right: Cl OMe O O CH3 B CH3 A a) Provide a reasonable mechanism for this transformation: CH3 CH3 Cl Cl MeOH O Cl O OMe CH3 CH3 A B CH 3 CH 3 CH 3 Cl Cl Cl O CH3 Cl O CH3 O O CH 3 H Me CH 3OH H O Me b) Using considerations from your mechanism in part i (above), provide a short explanation as to why only compound B is formed. Compound B is formed selectively because dissociation of the Cl situated on the oxygen- bearing carbon can produces a resonance-stabilized carbocation. Such a carbocation is much better stabilized, and it will form considerably faster, than the alternative carbocation (see below), which is not resonance stabilized: CH3 CH3 CH 3 Cl Cl Cl resonance stabilized Cl O O CH3 O CH 3 CH3 CH3 CH 2 Cl no possibility of resonance stabilization Cl O O Cl CH3 CH3 Page 4 of 11 3. (14 marks) Provide the requested information for each of the following reactions. Clearly indicate stereochemistry when appropriate. a) OH O Br Br 2 IR: no diagnostic absorptions organic product K O O diagnostic selected H NMR data: d 6.1 (1H), 5.8 (1H), 3.9 (2H) organic product b) 1. NaBH4 OCH CH H OCH 2H 3 2 3 O
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