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Chem 634 February 10, 2004Pre-Problem Set Prof. FoxKEY1. Assign all stereochemistry (R or S; E or Z)HCH3CHOCH2BraOHBrOOClClcd(2R)-3-bromo-2-methylpropanalb(1R,6R)-6-bromo-2-cyclohexenol(2E, 5S)-5-methyl-2-heptene(1R)-3-chloro-1-methylpropyl (3R)-3-chlorobutanoatet-But-But-But-But-But-BuOHBrBrOH2. For each pair, which is more stable?abct-B uMeHHHMet-B uMeHHMeHt-B uHaxial methylcosts 1.7 kcalMeHt-B uHHMeaxial methylt-B uHHOHBrt-B uHBrHOHaxial Brcosts 0.55 kcalaxial OHcosts more:0.94 kcalmore stablemore stablemore stable3. Indicate the relationship for each pair below: enantiomers, diastereomers, or mesoHOCH3HHHOCH3HCH3OHOHHCH3abHOOHHO OHmesoSS S RdiastereomersBrBrBrBrt-But-Bu4) For each pair, indicate which is more stable. Use a clear picture to explain whyBreqBreqBraxBraxBraxBreqBraxBreqHHHHHHHHmost stable form has noaxial Br'st-BuHCH3CH3tBuHHHsp2 centerThere is noaxial group here,and therefore there is only one1,3-diaxial interactionmake a model to convince yourself!two 1,3-diaxial interactionsBoth cyclohexanes are 'locked' with equatorial tBu's and axial methyl groupsso the answer lies in the double bond5. Provide a detailed arrow pushing mechanism for the following reactionBrO OHHHBu3SnHAIBNNC N N CNCN H SnBu3Bu3SnN::N •+H-atom extractionAIBNBrOOHHH•SnBu3•O•OH•OHHH•Br-atom extractionmultiplebond additionmultiplebond additionHBu3SnH-atom extractionInitiationSnBu3•+6. Draw Newman projections for the staggered and eclipsed forms of n-butane. Give therelative energies of all of the eclipsed and staggered conformers.7. Gauche butane is ~ 0.8 kcal/mol higher in energy than anti butane. The A-strain of amethyl group on cyclohexane is ~1.7 kcal/mol. Use Newman projections to explain therelationship in detail.HHH CH3CH3HH0.8 kcal/mola gauche butaneinteraction costs0.8 kcal/molCCH3HHHaxial methylcyclohexane has two gauche butane type interactions• the first can be seen from this projectionCH3HCH2HH2C Hgauche• the second can be seen from this projectionCCH3HCH3CH2HHHH2Cgauche(your eye)HUse Models if you are having trouble seeing this!since each gauche interaction costs ~0.8 kcal.mol, 2 x 0.8 =1.6 kcal/mol. The actualvalue of 1.7 kcal/mol is a close fit.8.9.HOHBrBrABOHBrHBrEtEtHOHOHBrABHBrBrEtEtBr10) Which reaction is faster (A or B)? Why?11) Which reaction is faster (A or B)? Why?HBrOH H+SLOW+Remember: carbocation formation is rate determining!The ideal geometry for a carbocation is planar with 120° angles.Thus, the carbocation formation is slow because it is severly distorted by the 60 ° angle of the cyclopropane ring.Br–Severe distortions are not required in order to put a cyclopentane in a planar arrangement awith 120 ° angles.This is the faster reactionEtEt+EtEt+EtEt+EtEt+The carbocation intermediate is resonance stabilized (with the benzene ring— see below). This is the faster reactionThe arrow pushing is the same as for question 2. The rate differences again relate to carbocation stability.Such resonance structures do not apply for the cation belowside view+the π orbital of the carbocation cannot overlap with theπ system of the benzene ring. No resonance stabilization+Br–BrBrX12) Provide a detailed explanationbutNaOtBuNaOtBuBrMeMeHHO–E2 elimination can occur via an anti-periplanar conformationi.e. both the hydrogenand the bromine are axialHMeMeBr(eq)H(ax)Br(ax)MeMeHH(eq)the molecule cannot adopt an conformation where Br and H are anti-periplanar. At least one group must be equatorial.13) Provide a detailed arrow pushing mechanism.HOTscatalytic H+H2OΔHHCH3CH3HOTsCH3HHCH3H+HCH3H+Halkyl


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