Chemistry 3719, Fall 2002 Exam 1 Name: This exam is worth 100 points out of a total of 600 points for Chemistry 3719/3719L. You have 50 minutes to complete the exam and you may use molecular models as needed. Good Luck. 1. (6 pts) Label the hybridization of all of the C, N and atoms in the following compounds: sp 2 sp2 all ring C sp 2 Ibuprofen sp3 sp sp sp 2 N 2 sp 2 Pent-3-ynoic acid amide 1
2. (15 pts) Give each of the following compounds acceptable names. You may use either the IUPAC system or the common names for substituents. a. cis-1,2-diethylcyclohexane b. 4-ethyl-2-methylnonane c. trans-3-isopropylcyclohexanol d. Br 2-bromo-5-methylhexane e. bicyclo[3.2.0]heptane 2
3. (12 pts) Molecular models may help with this question. a) For cis-1,4-di-isopropylcyclohexane draw two chair forms that are related by a ring flip and indicate which (if either) of the two chair forms will be the more stable. Explain your choice. The two ring-flipped conformations are of equal energy since in both one of the large isopropyl groups is axial and one is equatorial. The axial substituents in both will cause 1,3-diaxial interactions as shown above. b) For trans-1,4-di-isopropylcyclohexane draw two chair forms that are related by a ring flip and indicate which (if either) of the two chair forms will be more stable. Explain your choice. The two ring-flipped conformations are now vastly different in energy since the large isopropyl groups are both axial in one and both equatorial in the other. The axial substituents will cause 1,3-diaxial interactions as shown above and so the conformation above on the left is greatly favoured. c) Compare your pictures for the cis- and trans- isomers in parts a) and b) above and decide which of the two isomers should be thermodynamically more stable. Explain your choice. The trans isomer will definitely be more stable since >99% of the molecules will have both large substituents equatorial and thus avoiding any destabilizing 1,3-diaxial interactions, which are unavoidable in the cis isomer. 3
4. (12 pts) For each of the following acid-base reactions, complete the equation and, using your knowledge of pka values, decide which side of the reaction is favoured. a. Na right favoured + Na + 2 (Ka ~ 10 6 ) pka ~ 10 pka ~ 16 b. right favoured + NaN 2 + N Na 3 (Ka ~ 10 20 ) pka ~ 16 pka ~ 36 c. N 2 NLi right favoured + C3Li + C 4 (Ka ~ 10 12 ) pka ~ 36 pka ~ 48 5. (6 pts) Draw all of the possible isomers of C 6 14. Points will be subtracted for repeat answers. 6. (4 pts) Use the δ+/δ- notation to show any dipoles within the following molecules. δ+ δ- δ- δ- F δ+ δ+ δ+ 4
7. (12 pts) For the following molecule, construct Newman projections that are centered on the C-2 C-3 bond and that depict the conformations listed below: Br Br Lowest energy staggered C 2 Br C 2 Br ighest energy eclipsed Br 2 C C 2 Br A gauche conformation C 2 Br C 2 Br A second eclipsed conformation C 2 Br C 2 Br 5
8. (6 pts) Add any formal charges that are missing from the following structures, then draw a second resonance form for each: C 2 C 2 9. (5 pts) Explain the dramatic difference between the boiling point of ethane (-89 o C) and ethanol (78 o C) in terms of the forces holding the molecules together in each. Ethane molecules are only attracted to each other via relatively weak van der Waal s forces and so are easy to break apart. Ethanol has the functional group, which is capable of intermolecular hydrogen bonding, which causes a much stronger attraction between molecules than van der Waal s alone. This translates into the ethanol molecules being harder to break apart from each other (i.e. going from solid to liquid and liquid to gas) and thus more energy being needed to do this. Since the energy is usually supplied as heat, the ethanol requires a higher temperature to break these forces and therefore boils at a higher temperature. 6
10. (6 pts) Draw structures for each of the following compounds. a) cis-3-methylcyclohexanol b) 4-bromo-3-chloro-2-methyloctane Br Cl c) bicyclo[3.2.0]heptane 11. (9 pts) For each of the following pairs, circle the one that is the least stable of the two. Give a brief explanation for your choice in each case. 3 C C 3 vs 3 C C 3 one of the C 3 groups has to be axial in the cis isomer; they can both be equatorial in the trans isomer thus avoiding 1,3-diaxial interactions, which are unfavourable C 3 C 3 vs C 3 C 3 having both C 3 groups on the same face of the cyclopropane ring will cause unfavourable steric interactions 3 C C 3 vs C 3 C 3 the gauche conformer has the large groups interacting whereas in the anti conformer (right) they are as far apart as possible 7
12. (7 pts) For the following substitution reaction, give structures for the appropriate species in each box. (C 3 ) 3 C + Cl (C 3 ) 3 C- 2 Cl - 2 (C 3 ) 3 C-Cl + Cl (C 3 ) 3 C 8