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CHEMISTRY 313-03 MIDTERM # 1 – answer key October 05, 2010 Statistics: • Average: 73 pts (73%); • Highest: 99 pts (99%); Lowest: 31 pts (31%) • Number of students performing at or above average: 61 (52%) • Number of students performing below 55%: 8 (7%) • Number of students performing at or above 90%: 17 (15%) 1. (12 pts) Mark as true (T) or false (F) the following statements. Do not explain! • (T) Single bonds are always σ-bonds; • (F) Resonance structures are always in a state of rapid equilibrium; • (F) The gauche conformation of butane is a transition state; • (T) The chair conformation of cyclohexane is a global minimum; • (F) Cyclopropane does not have any torsional strain; • (F) Molecules with polar bonds always polar; • (T) All Arrhenius acids are also Brønsted acids; • (F) Lewis acids are proton donors; • (T) Increasing oxidation number indicates an oxidation process; • (T) Intermediates are not formed in concerted reactions; • (T) The Hammond postulate relates the energies and structures of two adjacent species on the potential energy profile; • (T) Carbocations are electron-deficient species; 2. (5 pts) Provide the structural formula for each of the following molecules. 3. (2 pts) Provide a structure for each of the following compounds: a. Bicyclo[1,1,0]butane. b. Spiro[2,2]pentane. 4. (2 pts) Draw curved arrows to rationalize the following conversion. 5. (2 pts) Draw the structure of the carbocation produced from 2-methyl-2-butanol. 6. (6 pts) For each of the following molecules, complete the Lewis structure and provide two additional valid resonance forms. Use the curved arrow formalism to show the flow of electrons. Rank the resultant resonance structures. a. Acetonitrile oxide; b. 1 2 O C O C 3 Methyl acetate; H3C O CH3 H3C 1 7. H3C C N O H3C C N O H3C C N O O CH3 H3C 3 O C O CH3 2 (4 pts) Each of the following species has a resonance structure that is higher ranking than the one shown. Show that structure and very briefly explain why it is higher ranking. N CH3 N CH3 more bonds O O more bonds, less charge separation 8. (6 pts) For some of the substances listed below the acid-base equilibrium for the reaction with KOH is shifted to the left, while for others it is shifted to the right. Using Table 1.8, decide the equilibrium shift for each structure. 9. (4 pts) Predict the shift of equilibrium (to the left or right) for the following acid – base reactions. HC CH + NH2 + HCN HC C + + NH3 to the right CN to the left N H N 10. (6 pts) Label the reactants in the following acid – base reactions as Lewis acids or Lewis bases. Cl Lewis base + AlCl3 Lewis acid Cl AlCl3 H H H + O O Lewis acid Lewis base CH3NH2 + N Cl H H + Cl Lewis acid Lewis base 11. (6 pts) For each of the following reactions, determine if it is a reduction, oxidation or not redox with respect to the organic compound. H3C + O hν Cl2 H3C oxidation O CH3OH H3O+ OH O C + HCl Cl MCPBA CH3 H3C O C OCH3 not redox CH3 oxidation O 12. (4 pts) Draw an example of each of the following classes of compounds: a. An amine with 4 C-atoms; b. An acid chloride with a 5-membered ring; c. A carboxylic acid with 5 C-atoms; 13. (4 pts) For each of the following compounds, draw an isomer that has the same functional group. More than one possibility exists in each case! SH O O SH O O 14. (7 pts) Circle and name all functional groups in the following structures. amine amide sulfide S sulfide NH2 H N benzene ring carboxyl O C OH NH2 S O alkene N O amide amine methionine, an amino acid O C OH carboxyl cephalexin, an antibiotic 15. (6 pts) Give the relationship between the following pairs of structures. There are four (4) possible relationships: same compound, constitutional isomers, cis-trans isomers, not isomers (i.e. different molecular formula). and and a. Br and b. not isomers d. same compound Br Br Br same compound e. same compound and O O and HO and c. HO constitutional isomers f. cis-trans isomers 16. (4 pts) Draw the structure of the principle organic product of each of the following reactions. OH PBr3 SOCl2 Br OH heat Cl 17. (6 pts) Using Newman projections, draw the conformations arising upon a full 360o turn (in 60o steps) around the C1 – C2 bond in isobutyl bromide. Represent the energy changes on a qualitative potential energy/dihedral angle diagram. Assign the proper term (i.e. minimum, transition state, etc.) to each conformation. See next page!! Br θ = 0o CH3 Br H H CH3 H eclipsed, one Br --- CH3 eclipsing interaction E Br H CH3 H staggered, two Br --- CH3 gauche interactions H H CH3 Br eclipsed, one Br -- CH3 eclipsing interaction transition state H H CH3 Br staggered, one Br -- CH3 gauche interaction BrH 60o CH3 H H CH3 H staggered, one Br --- CH3 gauche interaction transition state 180o H H CH3 H eclipsed, one Br --- CH3 eclipsing interaction transition state global minimum 120o CH3 Br H Br eclipsed, zero Br -- CH3 eclipsing interactions transition state θ = 360o = 0o CH3 CH3 H H local minimum 0o θ = 300o θ = 240o CH3 CH3 H CH3 H θ = 180o θ = 120o θ = 60o global minimum 240o 300o 360o θ 18. Consider menthol: CH3 HO a. (3 pts) Draw the most stable chair conformation of menthol. CH3 OH H3C a,a,a OH e,e,e most stable conformation b. (4 pts) How many stereoisomers of menthol are possible? Draw their structures using a bold-and-dashed wedge representation. 19. Consider 1-methylcyclohexanol and cyclohexylmethanol. a. (2 pts) Does the energy below better describe the reaction of 1-methylcyclohexanol or cyclohexylmethanol with HBr? Briefly explain! The reaction with HBr would result in substitution and formation of the corresponding alkyl bromide. The process could occur following either an SN1 or an SN2 mechanism. Based on the structure of each of the alcohols, one can make a prediction on the mechanism of its reaction with HBr, as shown above to the right. The profile, on the other hand, definitely reflects a stepwise reaction, i.e. one that follows an SN1 mechanism. The conclusion is that the profile must describe the reaction of 1-methylcyclohexanol with HBr. b. (5 pts) Assign the correct chemical structure(s), corresponding to each minimum in the diagram. A= + H Br + Br E= + H2O + Br OH C= + H2O G= Br OH2 20. (2 pts) BONUS PROBLEM (In order to receive credit for this problem, it has to be solved entirely!!). Ethylene glycol (see below!) is unusual in that the gauche conformation is more stable than the anti conformation. Offer an explanation. H H H HO C C OH H H ethylene glycol intramolecular hydrogen bond O OH H H H OH H H H H OH anti conformation H gauche conformation The gauche conformation of ethylene glycol is strongly stabilized by hydrogen bonding. Ethylene glycol is a good example of a species whose conformational preference is affected by intramolecular hydrogen bonding.