Download Organic Chemistry I Examination II - October 2004 and more Exams Chemistry in PDF only on Docsity! 1 Organic Chemistry I (230-001) Examination II October 27, 2004 Key Name (PRINT 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 Score 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 PLEASE observe the following. 1) Write LARGE and LEGIBLY. If I can’t read what you write, I can’t give you full or partial credit. 2) Read each question carefully before answering. If you don’t answer the question I posed, you will not receive credit. 2 1. (25 pts) (a, 4 pts) Assign the configurations of each stereo center in the compound below: CH3 H HO2C CH3 S S (b, 4 pts) Draw a configurational enantiomer of the compound above CO2H H H3C CH3 R R (c, 4 pts) Identify each of the stereocenters in the structure below with an asterisk CH3 OH HO * * * * 5 (c) C C C H Cl Cl H CCC H ClCl H Configurational enantiomers (d) CO2H NH2 HO OH NH2 HO2C R S R R Configurational diastereomers (e) Me Me Cl Cl Me Me Cl Cl R R S S rotate 180˚ Identical 6 3. (18 points) (a, 6 pts) Draw trans-1,4-dimethylcyclohexane in its lowest energy conformation Me Me (b, 6 pts) Draw cis-1,4-dimethylcyclohexane in its lowest energy conformation Me Me (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? Briefly explain why. trans-1,4-Dimethylcyclohexane because it contains no strain due to 1,3-diaxial interactions since its Me groups are equatorial. Conversely, cis-1,4-dimethylcyclohexane has additional strain due to two Me/H 1,3-diaxial interactions. 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 (S)-2-butanol is known to have [α] 25 D = +14˚. What is the configuration at the stereogenic C 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 (S)-2-butanol. Therefore, its configuration is R. 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 C: glucofuranose (below). Using what you know about ring strain and the conformation of substituted cycloalkanes, briefly explain (no more than two sentences): O HO OH HO HO HO A O HO HO HO HO B OH O HO HO OH H HO C
