CHM 171Q - Final - 2004 Key

CHM 171Q - Final - 2004 Key - Chemistry 171Q, Final Exam,...

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Unformatted text preview: Chemistry 171Q, Final Exam, 2004 Answers 1. (40 Points) Illustrated below are similar reactions of similar pairs of compounds. In each case, circle the one that reacts more rapidly. Then provide a brief explanation for the difference in rate. (a) Br CH3OH OCH3 Br CH3OH CH3O The reaction conditions polar, protic solvent with no strong nucleophile favor an SN1 substitution mechanism. In the case of the tertiary bromide, heterolytic cleavage of the C-Br bond will lead to the more stable tertiary carbocation, which is stabilized to a greater degree of hyperconjugation. According to the Bell-Evans-Polanyi (BEP) principle, the activation energy required to form the more stable carbocation will be lower for the bottom reaction and, therefore, the rate of this reaction will be greater. (b) -40 Et2O -40 Et2O H Cl Cl H + HCl + HCl Protonation at the terminus of the diene gives the more stable resonance-stabilized allylic carbocation (vs. a secondary, alkyl carbocation for the alternate protonation regiochemistry). As in part (a), the BEP principle predicts that the activation energy will be lower for the reaction that forms the more stable carbocation intermediate. -1- (c) CH3 Cl Me + NaSCH3 CH3 O S Me SCH3 + Cl NaSCH3 Me O S SCH3 Me Both reactions are expected to proceed by an SN2 substitution mechanism (dipolar aprotic solvent with a good nucleophile). The chloride in the top reaction contains three b-branching substituents and is, therefore, expected to experience significant nonbonded steric interactions between these substituents and the nucleophile in the substitution transition state. These steric interactions will raise the energy of the transition state (an lower the reaction rate) relative to the primary chloride. (d) (CH3)3C (CH3)3C H H + + I Cl (CH3)3C (CH3)3C + + H-I H-Cl The difference in thermodynamic driving forces for the two reactions is equal to the differences in the H-Cl vs. H-I bond dissociation energies. The H-Cl bond is significantly stronger than the H-I bond (103 vs. 71 kcal/mol), which explains why free radical iodination of hydrocarbons with I2 is synthetically useless. The top reaction is endothermic and the bottom reaction exothermic. According to the BEP principle, the bottom reaction should have the lower activation energy and, therefore, the greater reaction rate. (2) (20 points) Alcohols typically show broad characteristic infrared (IR) absorptions at 3400 cm-1 due to the O-H stretching vibration. However, if the OH group is intramolecularly hydrogen bonded to a Lewis basic group within the molecule, the IR absorption is typically shifted to slightly higher energy. The IR spectra of the two diastereoisomers of butane-2,3-diol both show two IR absorptions in the OH stretching region. Interestingly, the intensity of IR resonance assigned to the intramolecular, hydrogen-bonded OH group is significantly larger in one of the stereoisomers, indicating a greater fraction of the isomer exists in the hydrogen-bonded state. (a) Using Newman projections, draw the most stable conformers of the two diastereoisomers of butane-2,3-diol that would be expected to show an intramolecular, hydrogen bonded -2- OH group. It may be helpful to show the hydrogen bond in each structure with a dashed line (i.e. O-----H). Hydrogen bonding can only occur between the hydrogen of one OH group and the oxygen of the other OH group if the two groups are proximal. It should be clear that this can only occur in staggered conformations where the OH groups are gauche. Of the various staggered conformations in which the OH groups are gauche, the most stable ones will be those that minimize torsional strain. These are shown below. CH3 H H CH3 O H O H CH3 H 3C H H O H O H 1 (b) 2 Which of the two isomeric conformers drawn above (1 or 2) would you expect to be more stable? (Hint: A CC/CC gauche interaction is more destabilizing than a CC/CO gauche interaction.) Explain your reasoning. 1 contains two CC/CO gauche interactions. 2 contains one CC/CO gauche interaction and one CC/CC gauche interaction. Since the CC/CC gauche interaction is more destabilizing than a CC/CO gauche interaction, 1 contains less torsional strain and is, therefore, more stable. (c) Which diastereoisomer of butane-2,3-diol (1 or 2) would you expect to show the greater fraction of the IR resonance due to the intramolecular, hydrogen-bonded OH group? Briefly explain your reasoning. Since 1 and 2 are the most stable hydrogen bonded conformers for each stereoisomer, their relative population will be related to their relative stability. Since 1 is more stable than 2, the equilibrium population of 1 will be greater than that of 2, consequently the IR resonance due to the hydrogen-bonded OH group will contribute a greater fraction to the total OH stretching absorptions in the IR spectrum of 1. 3. (45 points) Complete each of the following: (a) (10 points) Free radical bromination of 1,1-diphenylpropene (3) or 3,3-diphenyl-1-propene (4) with N-bromosuccinimide (NBS) in CCl4 give identical ratios of monobromination products. -3- Ph Ph 3 Ph Ph 4 Clearly explain why 3 and 4 give identical product ratios with NBS. Reaction of 3 and 4 will both proceed via the same allylic radical intermediate (shown below) and, therefore, must give the same product(s). Ph Ph CH2 Ph Ph The enthalpies of formation of 3 and 4 are 52 and 62 kcal/mol, respectively. If both isomers were present in solution at equal concentrations, which would you expect to react more rapidly with NBS? Explain your reasoning. The relative rates of reaction of 3 and 4 with Br will be determined by the relative activation energies which, in turn, will be related to the enthalpy changes for both reactions. Since the products of both reactions are the same (the allylic radical), the relative energy changes will be governed by the relative energies of the starting materials. The enthalpies of formation show that 3 is more stable than 4, therefore, the energy change for hydrogen atom abstraction will be greater for 3 than for 4. Consequently, 4 is expected to react more rapidly than 3. Ph Ph Ph 3 10 Ph 4 Ph Ph CH2 Ph Ph -4- (b) (5 points) The reaction of n-chlorobutane with sodium hydroxide to give n-butanol in aqueous THF is found to be significantly accelerated by the addition of a small amount of sodium iodide. Careful analysis shows that sodium iodide is not consumed in the reaction, therefore, it acts as a reaction catalyst. n-C4H9Cl + NaOH THF H2O n-C4H9OH + NaCl Provide an explanation for how sodium iodide catalyzes the conversion of n-chlorobutane to n-butanol. It is apparent, after the fact, that iodide must compete with hydroxide as a nucleophile for reaction with n-chlorobutane. The n-iodobutane so formed will be more reactive in an SN2 reaction with hydroxide than n-chlorobutane due to the greater leaving group ability of iodide vs. chloride anion (which is related to the relative proton affinities of both; the smaller the proton affinity, the better the leaving group ability). Reaction of n-idobutane with hydroxide produces n-butanol and regenerates iodide anion, which can react again with n-chlorobutane, thus serving as a catalyst. (c) (30 points) For each of the three following interconversions, circle the compounds that would predominate at equilibrium. Then provide a brief explanation for your choice. K CH3OH + HNEt3 CH3OH2 + NEt3 The base with the greater proton affinity will predominate at equilibrium. Triethyl amine has a greater proton affinity (pK a 11 for triethyl ammonium ion) than methanol (pKa 2 for protonated methanol), therefore, the equilibrium will lie to the left. K CH3Br + OH CH3OH + Br The energy change for the reaction (and therefore the equilibrium constant) will be determined by the relative heterolytic bond dissociation energies, C-Br vs. C-OH. These bond dissociation energies are expected to follow the relative proton affinities of bromide ion vs. hydroxide ion. Hydroxide ion has the greater proton affinity (greater pK a of its conjugate acid) and should therefore have the stronger bond, which will make the products on the right hand side of the equilibrium more stable, leading to K >1. -5- CH3O + O K CH3OCH2CH2O The equilibrium constant will be determined by the relative energies of the C-O bonds that are made and broken in the reaction. Due to bond angle strain, the C-O bond in the epoxide will be weaker than the C-O bond in the acyclic ether. Therefore, upon reaction of methoxide with the epoxide, the C-O bond being made will be stronger than the CO bond being broken. Thus, the equilibrium will lie to the right. 4. (30 points) Supply the missing starting material, reagent(s), or product(s) for each of the following reactions. Be careful to show the appropriate stereochemistry where relevant. (a) CH3CH2 H H CH3 CH3CH2 C C CH3 Na NH3 (b) HgSO4, H2O H2SO4 O (c) CH3 CH2OH HCl Cl CH2CH3 -6- (d) O (S)-pent-4-en-2-ol Cl S Cl N H Cl (e) 1) Hg(OAc)2 H2O, THF CH3 CH3 2) NaBH4, NaOH OH CH3 CH3 mixture of cis- and trans-isomers (f) Br CH3CH2 H3C D H Br2 CH3OH CH3O D CH3 CH2CH3 H 5. (20 points) As discussed in class, one procedure for the conversion of alkyl halides (R-X) to hydrocarbons (R-H) is to transform the halide into a Grignard reagent and then treat it with water. This procedure is not useful, however, if the starting halide contains substituents that react with Grignard reagents. A useful alternative procedure is to react the halide with tri-n-butyl tin hydride (HSnBu3 ), e.g. Br + H-SnBu3 ButOOBut H + Br-SnBu3 -7- Bond Dissociation Energies (kcal/mol) Br 71 Bu3Sn H 74 ButOO ButO H 100 Bu3Sn Br 85 ButO ButO But OBut H Br 66 38 105 55 (a) Estimate H for the reaction of bromocyclohexane with HSnBu3 . Show how you arrived at your estimate. H = BDE(C 6 H11Br) + BDE((Bu3 SnH) [BDE(C6 H11H) + BDE((Bu3 SnBr)] H = 71 + 74 [100 + 85] = 40 kcal/mol The direct reaction of cyclohexane with HSnBu3 proceeds very slowly. The reaction is greatly accelerated, however, by the addition of small amounts of di-tert-butyl peroxide (ButO-OBut). The di-tert-butyl peroxide is ultimately converted to tert-butanol during the reaction. Mechanistic studies have demonstrated that the reaction proceeds via a free radical chain mechanism. (b) Write a plausible mechanism (with single-barbed arrows to show electron movement) for the reaction of bromocyclohexane with HSnBu3 . Include all relevant initiation and propagation steps. Label the initiation steps as I1, I2, I3, ... and the propagation steps as P1, P2, P3, ... (You need not show termination steps.) I1 I2 (CH3)3CO (CH3)3CO OC(CH3)3 H SnBu3 2 (CH3)3CO (CH3)3CO H + SnBu3 P1 Br SnBu3 H + Br SnBu3 P2 H H H SnBu3 + H SnBu3 -8- 6. (15 points) The reaction of cyclohexene with HCl in acetic acid as solvent leads to the formation of two products, chlorocyclohexane (5) and cyclohexyl acetate (6). HCl CH3CO2H 5 (25%) Cl + 6 (75%) OAc (a) It is found that the product percentages change when tetramethylammonium chloride (TMAC) is added to the solution (note: TMAC dissolves in CH3 CO2 H). For example, when [TMAC] = 0.1 M, the reaction produces 50% 5 and 50% 6. When [TMAC] = 0.9 M, the 5:6 product ratio is 80:20. Control experiments show that 5 and 6 are stable under the reaction conditions. Explain how adding TMAC changes the product ratio. If, as shown below, the reaction proceeds via a carbocation intermediate, then the partitioning of this intermediate to form 5 and 6 will depend on the concentrations of chloride ion and acetic acid, respectively (note that acetic acid is the solvent so its concentration will be essentially constant). Thus, the mechanism predicts that as [Cl] increases so will the 5/6 ratio. Cl Cl H H H Cl H AcOH H OAc (b) What dependency, if any, would you expect on the product ratio if [HCl] was changed? (note: Ka for HCl in CH3 CO2 H is 3 x 10-9 M) Explain your answer. The low value for K a(HCl) in acetic acid means that very little of the HCl will be dissociated in this solvent (as compared with water, where HCl is nearly completely dissociated). Therefore, as [HCl] increases, [Cl] will not increase significantly. Thus, while the overall reaction rate will increase with increasing [HCl], the product ratio should be unaffected. -9- (c) The reaction with cyclohexene has also been run with DCl in CH3 CO2 D. When the reaction was stopped at partial conversion and the cyclohexene reactant recovered, it was found the cyclohexene contained no deuterium. What is the mechanistic implication of this observation? Based on the mechanism shown above, the fact that recovered cyclohexene had no deuterium incorporation requires that the protonation of the alkene be irreversible. 7. (65 points) Specify the reagents that would be needed to carry out each of the following synthetic transformations. You may use any organic building block with 3 carbons for your syntheses. However, the reagents that you use that are not incorporated into the products may contain more than 3 carbon atoms. The structure of the reactant and the product should be shown for each chemical transformation. Retrosynthetic analysis will be considered for partial credit. (a) H OH CH3C 1) NaNH2, NH3 2) 3) H OH CH3C C H2 Lindlar's catalyst CH3 OH H H O OsO4, H2O2 CH H3C HO H OH (b) CH3 O Br2 hn CH3 H2CrO4 CH2Br CH3 OH 1) Mg, Et2O 2) CH3CHO 3) H - 10 - (c) CH3 OH HC 1) NaNH2 O 2) 3) H CH3 HC C OH CH3 1) BH3, THF 2) H2O2, NaOH O H CH3 OH CH3 1) NaBH4 2) H CH HO CH3 (d) OH OH Cl2 hn Cl Cl KOBut HOBut CO3H HO or H3O O - 11 - 8. (a) (45 points) Compound A and B have formula C5 H10. A and B react with H2 /Pt to give a single compound, C5 H12. A reacts with BH3 THF followed by treatment with a basic solution of hydrogen peroxide to give C (C5 H12O). B reacts with BH3 THF followed by treatment with a basic solution of hydrogen peroxide to give D (C5 H12O). When C or D is treated with chromic acid the solutions turn blue-green. Preparative reaction of C with chromic acid gives E (C5 H10O2 ). E has a broad IR absorption at 3000 cm-1 and a strong, sharp absorption at 1710 cm-1 , and the following 1 H NMR spectrum: d 11.9 (1H, singlet), 2.23 (2H, doublet), 2.12 (1H, multiplet), and 0.99 (6H, doublet). Based on these observations, propose structures for A-D. Your observation/deduction reasoning will be considered for partial credit. Observation A & B, C5 H10 A & B + H2 same C 5 H12 A + BH3 /H2 O2 C, C 5 H12O B + BH3 /H2 O2 D, C 5 H12O C or D positive chromic acid tests C + H2 CrO4 E, C5 H10O2 E, IR broad peak at 3000 cm-1 E, IR strong peak at 1710 cm-1 E, 1 H NMR 11.9, 1H, singlet 2.23, 2H, doublet 2.12, 1H, multiplet 0.99, 6H, doublet A fl B fl C fl Deduction 1 degree of unsaturation A and B are alkenes A & B have same atom connectivity C, 0 degrees of unsaturation C is an alcohol D, 0 degrees of unsaturation D is an alcohol C & D are 1 or 2 alcohols E, 1 degree of unsaturation E is likely a carboxylic acid O-H stretch C=O E is a carboxylic acid H-O -CH2 -CH-CH- with multiple couplings (CH3 )2 -CH D fl E fl or CH2OH or OH CO2H HOCH2 - 12 - (b) Compound F has formula C6 H12 and an IR absorption at 1640 cm-1. Reaction of F with either Hg(OAc)2 /aqueous THF followed by NaBH4 , or with BH3 THF followed by treatment with a basic solution of hydrogen peroxide, gives the same compound G, C6 H14O. Reaction of F with H2 O2 in the presence of a catalytic amount of OsO4 gives H, C6 H14O2 , which can not be resolved into a pair of enantiomers. Reaction of F with O3 followed by reduction with (CH3 )2 S gives only one product I, C3 H6 O. Based on these observations, propose structures for F-I. Your observation/deduction reasoning will be considered for partial credit. Observation F, C6 H12 F, IR peak at 1640 cm-1 F + BH3 /H2 O2 G F + Hg(OAc)2 G G, C6 H14O F + OsO4 /H2 O2 H, C6 H14O H can not be resolved F + O3 /Me2 S I, C3 H6 O Deduction 1 degree of unsaturation C=C F is a symmetrical alkene F can not have inversion symmetry based on IR C=C stretch 0 degrees of unsaturation G is an alcohol H, 0 degrees of unsaturation H is a 1,2-diol H is achiral F is a symmetrical alkene F fl G fl H fl I fl O OH HO OH H - 13 - ...
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This test prep was uploaded on 04/17/2008 for the course CHM 171Q taught by Professor Goodman during the Fall '04 term at Rochester.

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