Dr. Steven Pedersen July 28, Chemistry 3A. Midterm 2

Transcription

1 Dr. Steven Pedersen July 28, 2015 hemistry A Midterm 2 Student name: ANSWER KEY Student signature: Problem 1 Problem 2 Problem Problem 4 Problem 5 Problem 6 Problem 7 Total Points (16 pts) (4 pts) (2 pts) (24 pts) (15 pts) (20 pts) (9 pts) (150 pts) No alculators Allowed No Molecular Models Allowed Be Sure Your Exam has 10 Pages Do Not Write Any Answers on the Back of a Page 1 Li 11 Na 19 K 4 Be 12 Mg 20 a 5 B 1 Al 6 14 Si 7 N 15 P 8 16 S 9 F 17 l e 10 Ne 18 Ar 6 Kr 54 Xe 1

2 1. There will be N partial credit for this problem. Avoid careless errors by checking over your answers. (16 pts) A. Provide a systematic name for the following compounds (include R/S when appropriate). Use common nomenclature for any branched substituents. (S)-1,-dibromopentane R (S)-,5-dibromopentane (S)-2-chloro-1,1-dimethylcyclohexane l R (S)-1-chloro-2,2-dimethylcyclohexane 2,6-dimethylheptane (1R,2R)-1,2-diethylcyclobutane B. Draw a structure for each of the following names. For cycloalkanes use flat rings.! (1R,S)-1,-dipropylcycloheptane! (1S,S)-1-tert-butyl--isobutylcyclohexane! (R)-1,1,4-triiodohexane! meso-1,-difluorocyclopentane F F 2

3 2. Predict all of the possible organic product(s) from the following reactions. Where relevant, show all stereoisomers. Pay particular attention to any information given in the product boxes. Each redundant or wrong answer within a box cancels one correct in the same box. (4 pts) S S S NLY NE RGAN PRDUT Me, heat l 2 hν l l l ontinued on next page nly monochlorinated products.

4 2 S 4 heat 2 2 hν 1. nly products resulting from reaction at centers. 2. nly monobrominated products.. ALL products must be shown with at least one chair. 4

5 . Write logical arrow-pushing mechanisms for the following reactions. Be sure that your mechanism accounts for all products shown. (2 pts) P heat P P P hν 2 2 ontinued on next page 5

6 heat your mechanism(s) must account for all the products shown. 2 S 4, heat 2 S 2 S 6

7 4. (24 pts) A. Write a logical arrow-pushing mechanism for the following reaction. S heat S S S B. Based on the mechanism you wrote for Part A, draw a reaction coordinate diagram for this reaction. ASSUME the overall reaction is ENDTERM. Label all relevant energy levels on your diagram. Energy S S S Reaction oordinate. For one transition state you have drawn on your diagram in Part B, draw a transition state picture using a dotted line(s) to represent bond breaking and/or bond formation. Based on your diagram in Part B, label your transition state picture as EARLY or LATE (based on ammond s postulate). First Step Second Step S δ S LATE EARLY δ D. For whatever transition state you drew in Part, draw the opposite transition state picture below. That is, if you drew an EARLY transition state picture, now draw what the same transition state would look like if it were LATE. First Step Second Step S δ S EARLY LATE δ 7

8 5. For all the questions below, you must attach each methyl group using a wedge or a dash to indicate relative orientation of the group(s). (15 pts) A. The compound shown below is achiral. Add NE methyl group to one of the carbons labeled 1-5 that results in a new ARAL molecule. R B. The compound shown below is achiral. Add two methyl groups to one of the carbons labeled 1-5 that results in a new ARAL molecule. WEVER, YU ANNT ADD ETER METYL GRUP T TE SAME ARBN YU ADDED A METYL GRUP T N PART A. R R methyl's can also be on 4 or 5 (if Part A was not answered correctly). The compound shown below is achiral. Add two methyl groups to any of the carbons labeled 1-5 that results in a RAL molecule. You can place the methyl groups on the same or different carbons. n the last four examples, the methyls can all be "dashed" or they can be a mixture of both. D. The compound shown below is achiral. Add two methyl groups to any of the carbons labeled 1-4 that results in a new ARAL molecule. You can place the methyl groups on the same or different carbons E. The compound shown below is achiral. Add NE methyl group to one of the carbons labeled 1-5 that results in a new ARAL molecule. 8

9 6. (20 pts) A. ompound A is chiral and compound B is achiral. Explain how this is possible using terminology you have learned in lecture. Your explanation must be specific. For example, just using the term atropeisomer without any explanation will receive no credit. l f compounds A and B could exist in the plane they are drawn in, they would both be achiral. owever, the two chlorines in compound A are too large and prevent the compound from being able to achieve this planar conformation (see picture below). Therefore, the preferred conformation is one where the two rings are perpendicular and in compound A, the previously mentioned steric hinderance is significant enough that there is no free rotation around the bond (atropeisomerism). n compound B, the only steric hinderance is between the four hydrogens shown, which is not significant enough to stop free rotation, which means the planar geometry can be achieved, leading to a plane of symmetry in an accessible conformer. l l A l B l l l l l l no free rotation about this bond l l free rotation about this bond B. Underneath each molecule shown below is three terms. ircle each term that applies to that molecule. Wrong answers cancel right answers so don t guess. chiral meso achiral chiral meso achiral chiral meso achiral chiral meso achiral. For each pair of molecules use one term that best describes their relationship to one another. Use the abbreviations for the terms shown below. The terms are: dentical (), Diastereomer (D), Enantiomer (E), None of These (N) l l l l l l l l A B D A and B: D A and D: E A and : B and D: D 9

10 7. (9 pts) A. The reported optical rotation of a compound is shown below. There is an important piece of information missing. What is it? [α] 25 = 18.1 (1.4 g/ml) D The name of the solvent that was used is missing. The following questions are True or False. ircle the correct term. B. The optical rotation of compound A is (). The optical rotation of compound B must be (-). True False A B. f the enantiomeric excess of a solution is 96%, the composition of enantiomers is 96:4. True False D. ompounds and A and B are enantiomers. True False l l A B 10