Exam 3 Key - Bocknack CH 310M/318M Spring 2008 MIDTERM...

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Part I Answers & Scoring How to read this table : In the first three columns, you’ll find the question number, the maximum points possible, and the correct answer choice. The columns labeled “A” through “J” indicate the points awarded for the problem if this letter was the choice indicated on the answer sheet. Please check the Blackboard gradebook to see how you answered the Part I questions, and to check your raw score (out of 150 points possible) for Part I of the exam. Part I answer sheets will not be returned, and no regrades are possible on Part I of the exam!!! Question Max. Points Correct A B C D E F G H I J 1 5 G 0 0 0 0 0 0 5 0 0 0 2 5 A 5 0 0 0 0 0 0 0 0 0 3 5 F 0 0 0 0 0 5 0 0 0 0 4 5 B 0 5 0 0 0 0 0 0 0 0 5 5 D 0 0 0 5 0 0 0 0 0 0 6 5 B 0 5 0 0 0 0 0 0 0 0 7 5 C 0 0 5 0 0 2 0 0 0 0 8 5 B +4 for A if Q7 answer = E or F or G or H +4 for B if Q7 answer = A or B or C or D 9 5 E If Q8 answer = A : 0 0 0 2 0 5 0 0 0 0 If Q8 answer = B : 0 0 2 0 5 0 0 0 0 0 10 5 D 0 2 0 5 0 0 0 0 0 0 11 5 F 0 4 4 0 0 5 2 0 0 0 12 5 C 1 3 5 0 0 0 0 0 0 0 13 5 B 3 5 0 0 0 0 0 0 0 0 14 5 D 0 0 0 5 0 0 0 0 0 0 15 5 A 5 0 0 3 0 0 0 0 0 0 16 5 E 0 2 0 0 5 0 0 0 0 0 17 5 B 0 5 0 0 0 0 0 0 0 0 18 5 C 0 0 5 0 0 0 0 0 0 0 19 5 A 5 0 0 0 0 0 0 0 0 0 20 5 B 0 5 0 0 0 0 0 0 0 0 21 5 G 0 3 0 3 0 0 5 0 0 0 22 5 B 0 5 0 0 0 0 0 0 0 0 23 5 A 5 0 0 0 0 0 0 0 0 0 24 5 C 0 0 5 0 0 0 0 0 0 0 25 5 C 0 0 5 0 0 0 0 0 0 0 26 5 D 0 2 0 5 0 0 3 0 0 0 27 5 D 0 0 2 5 0 0 0 0 0 0 28 5 E 0 0 0 0 5 0 0 0 0 0 29 5 F 3 0 3 0 0 5 0 0 0 0 30 5 C 0 0 5 0 0 0 0 0 0 0 Bocknack CH 310M/318M – Spring 2008 MIDTERM EXAM #3 ANSWER KEY
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Part II – Write your answers to all questions on this page in the space provided. If you use a pen, only blue or black ink is acceptable!!! PLEASE READ EACH QUESTION VERY CAREFULLY !!! 31. (20 points) Compounds A and B each have molecular formula C 6 H 12 . When A or B is exposed to H 2 over a Pd catalyst, the only product isolated is 3-methylpentane. When A or B is treated with HBr, the same optically inactive compound, C (C 6 H 13 Br), is obtained as the major product. When A or B is treated with O 3 followed by (CH 3 ) 2 S, the same mixture of products is obtained. This mixture is comprised of compound D , which has molecular formula C 4 H 8 O, and compound E , which has molecular formula C 2 H 4 O. When A is treated with BH 3 , followed by H 2 O 2 , NaOH, H 2 O, the chiral alcohol indicated in the reaction scheme below is obtained as a racemic mixture. When B is treated with BH 3 , followed by H 2 O 2 , NaOH, H 2 O, a different chiral alcohol is obtained, also as a racemic mixture. The products obtained from hydroboration-oxidation of A are diastereomers of the products obtained from hydroboration-oxidation of B . Complete the reaction scheme given below by proposing structures for the unknown compounds A , B , C , D , and E . For full credit, stereochemistry needs to be shown clearly and unambiguously where indicated!
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