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Practice Test 5

Practice Test 5 - addition circle the correct...

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Practice Exam 5 (9/30/09) (10 pts) I. Predict the major product(s) for the following reactions and show the stereochemistry if any. Br CH 3 COO room temperature (10 pts) MeOH 25 °C Me Cl OCOCH 3 Me OMe Me OMe (10 pts) NaOH 25 °C ( R )-2-chlorobutane OH (15 pts) II. In elimination reactions, t -butyl chloride reacts faster than isobutyl chloride. The difference in reactivities is accounted for by the stabilities of the incipient carbocations. Explain why the carbocation formed from t -butyl chloride is more stable than the one from isobutyl chloride. Cl Cl tertiary: more stable primary; less stable A tertiary carbocation is more stable than the primary one because a tertiary carbocation has more hyperconjugation than the primary. The alkyl substituents offer the delocalization of electrons from C-H bonds to the empty p-orbital in the carbocation center. C C H orbital overlap KO t Bu KO t Bu
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III. Draw the major product from each reaction, with appropriate stereochemistry. In
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Unformatted text preview: addition, circle the correct stereochemistry, R or S, of each substrate and product. (10 pts each) Br CN R o r S R o r S PCl 5 R o r S R o r S Me H H H Cl O CN Me H H H Cl Cl Cl IV. Circle which reaction would take place faster. Then explain why the reaction of your choice would be faster. Use drawings and narratives (concise), as appropriate. (45 pts) Cl NaOEt IVa. Circle one reaction. Cl NaOEt Explain here. Cl H H Me Cl Me H anti periplanar Me Me H syn periplanar MeOH IVb. Circle one reaction. Br MeOH Explain here. Br OMe OMe IVc. Circle one reaction. Explain here. Br I I Cl I I S N 1 rxn: tertiary carbocation is more stable than the secondary. Bromide is a better leaving group than chloride. Because bromide is more stable due to larger polarizability....
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