Please provide clear and concise answers to all of the following questions. Use equations and/or drawings to support your answers where appropriate.

Transcription

1 Organic hemistry I ( ) Examination II October 27, 2004 Key Name (PINT LEGIBLY): Please provide clear and concise answers to all of the following questions. Use equations and/or drawings to support your answers where appropriate. Please answer questions in the area provided, the back of an exam page or on the clearly labeled spare page. Problem core 1. (a-e) /25 2. (a-e) /25 3. (a-c) /18 4. (a-b) /10 5. (a-b) /10 6. (a-c) /12 Total /100 PLEA E observe the following. 1) Write LAGE and LEGIBLY. If I can t read what you write, I can t give you full or partial credit. 2) ead each question carefully before answering. If you don t answer the question I posed, you will not receive credit. 1

2 1. (25 pts) (a, 4 pts) Assign the configurations of each stereo center in the compound below: 3 O 2 3 (b, 4 pts) Draw a configurational enantiomer of the compound above O (c, 4 pts) Identify each of the stereocenters in the structure below with an asterisk 3 * * O O * * 2

3 (d, 6 pts) ircle the chiral compounds in the following list: 3 O O O O 3 3 O 3 O (e, 7 pts) Indicate whether each of the following is true or false. 1. If a compound has an enantiomer it must be chiral - true 2. If a compound has a diasteromer it must be chiral - false 3. Any molecule containing only one stereocenter must be chiral - true 4. ome chiral molecules are optically inactive - false 5. ome diastereomers have a mirror-image relationship - false 6. If a structure has no plane of symmetry it is chiral - false 7. Mirror-image molecules are in all cases enantiomers - false 3

4 2. (25 pts, 5 pts) Describe the relationship of each pair of structures as specifically as possible and in no more than two words (a) flip 180 onfigurational enantiomer (b) N 2 N 2 onfigurational diasteromer 4

5 (c) onfigurational enantiomers (d) N 2 N 2 O O 2 O 2 O onfigurational diastereomers (e) rotate 180 Identical 5

6 3. (18 points) (a, 6 pts) Draw trans-1,4-dimethylcyclohexane in its lowest energy conformation (b, 6 pts) Draw cis-1,4-dimethylcyclohexane in its lowest energy conformation (c, 6 pts) Based on the structures you drew in parts (a) and (b), which is lower in energy, cis- or trans-1,4-dimethylcyclohexane? iefly explain why. trans-1,4-dimethylcyclohexane because it contains no strain due to 1,3-diaxial interactions since its groups are equatorial. onversely, cis-1,4-dimethylcyclohexane has additional strain due to two / 1,3-diaxial interactions. 6

7 4. (10 points, 5 points each) (a) While reorganizing his laboratory, Joe Greenhorn found an enantioenriched sample of 2-butanol containing a 3:1 mixture of enantiomers. What is the ee of Joe s sample of 2-butanol? ee = [(er 1) / (er + 1)] x 100, where er is the ratio of the two enantiomers. ee = [(3-1)/3+1] x 100 = 50% (b) Joe measured the specific rotation, [α], of his sample of 2- butanol to be -7. Enantiopure ()-2-butanol is known to have [α] D 25 = +14. What is the configuration at the stereogenic atom in the major enantiomer of 2-butanol in Joe s sample? The major major enantiomer in Joe s sample rotates plane-polarized light in the opposite direction for ()-2-butanol. Therefore, its configuration is. 5. (10 pts) Living systems put glucose into other forms for specific purposes. The molecules are A: glucopyranose α-anomer, B: glucopyranose β-anomer and : glucofuranose (below). Using what you know about ring strain and the conformation of substituted cycloalkanes, briefly explain (no more than two sentences): O O O A O O O O O O B O O O O O O O O 7

8 (a) Which molecule is most stable and why? B is most stable because all of the substituents of the sixmembered ring are equatorial hence it contains no 1,3-diaxial interactions. (b) Which molecule is least stable and why? is least stable since its five-membered ring results in more strain (relative to a six-membered ring) 6. (12 points, 4 points each) (a) ircle the two chair structures of cis- 1-bromo-3-methylcyclohexane. A B D (b) One structure is more stable than the other by 3.7 kcal/mol. Which is it? iefly explain why. onformer D is more stable than A by 3.7 kcal/mol because its substituents are equatorial and there are no 1,3 diaxial interactions. Whereas A is more strained because it contains three (/-, /-, and /) 1,3-diaxial interactions. 8

9 (c) Using the table below, compute the approximate energy cost of a 1,3-diaxial interaction between a bromide and a methyl group (show your work). Group ost of single 1,3-diaxial ost of a gauche interaction with interaction with kcal/mol 0.25 kcal/mol kcal/mol 0.85 kcal/mol O 0.50 kcal/mol 0.50 kcal/mol Approximate energy cost of a 1,3-/ diaxial interaction = 2.6 kcal/mol The strain energy of conformer A (3.7 kcal/mol) is due to the three 1,3-diaxial interactions mentioned above. Thus, the energy costs of a / 1,3-diaxial interaction = [Total strain energy (/ 1,3-diaxial interaction + / 1,3- diaxial interaction)] 3.7 ( ) = 2.6 kcal/mol 9