Chemistry 30A Spring 2004 Exam 2 Solutions Page 1 Statistics High score 97 Average 65 0 Low score 00 Standard deviation Irrelevant does not influence grades in this course A note about exam keys The answers presented here are usually significantly longer than expected from a student taking the exam An exam key serves not only to reveal what was expected but to instruct you as well To see the grade curve consult the Chem 30A Home Page Cl I NaI 1 DMF CH3 CH3 I 2 I I Cl CH3 CH3 Cl I I NaI 3 CH3 CH3 DMF CH3 OCH3 OCH3 I 4 CH3OH H 5 Ph I rds Ph CH3OH HOCH3 OCH3 Ph OCH3 Ph H Ph 6 Ph CH3 CH3O Cl CH3OH CH3OH CH3OH Ph CH3 major product OCH3 OCH3 Ph CH2 Chemistry 30A 7 CH3O Spring 2004 CH3O H Exam 2 Solutions Page 2 H CH3 CH3 CH3 Cl Cl H OCH3 CH2 CH2 Cl 8 Rate k R Cl CH3O 9 CH3 OSO2 CF3 CH3 CH3OH CH2 major product 10 CH3 OSO2 CF3 CH3 rds H CH2 H CH3 HOCH3 CH2 HOCH3 11 Note that the base CH3OH is also the solvent for this reaction Changing to a less polar solvent tert butanol retards the carbon leaving group bond ionization but still allows the deprotonation to occur CH3 OSO2 CF3 CH3 3 COH CH3 CH2 12 The most probable mechanism for this reaction is SN2 13 When choosing between elimination and substitution mechanisms we examine E2 first except in the case of primary alkyl halides E2 requires among other things a strong base typically hydroxide or an alkoxide Methanethiolate CH3S is a weak base This disfavors E2 We next consider SN2 which requires a moderate nucleophile a moderate leaving group and the carbon bearing the leaving group cannot be tertiary Methanethiolate is a good nucleophile formal negative charge low electronegativity of sulfur chloride is a moderate leaving group and the carbon bearing the chlorine is secondary The SN2 conditions are reasonably met so we predict the main reaction pathway is SN2 Chemistry 30A Spring 2004 Exam 2 Solutions Page 3 14 Fate 1 Capture a nucleophile CH3OH CH3OH Fate 2 Lose proton form bond H CH3OH Fate 3 Rearrange H 15 a CF3SO3 b Methanol c CH2N CH3 2 16 In CH3OH fluoride is a poor nucleophile The reason for this is hydrogen bonding 17 R H and X Br 18 Br 19 Rates about equal 20 Reason 1 Fluoride is a very poor leaving group Reason 2 A methyl carbocation CH3 is very unstable Reason 3 The solvent CH3 CH2 10CH2OH is of very low polarity 21 There are many possible structures that meet the criteria of the question Any one is acceptable as long as it has a good leaving group such as iodide or triflate exactly five carbons but cannot engage in E2 elimination For example E2 can be prevented by replacing the hydrogens with methyl groups CH3 I H3C CH3
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