rganic Chemistry I (Chem340), pring 2005 Final Exam This is a closed-book exam. No aid is to be given to or received from another person. Model set and calculator may be used, but cannot be shared. Please write your name and WU ID number at the top of each page. The exam time is 180 minutes. and in your exam papers to the proctor and sign your name at the end of the test. Good luck and have fun! For graders only: Q1 out of 15 Q2 out of 16 Q3 out of 20 Q4 out of 20 Q5 out of 14 Q6 out of 15 Bonus out of 6 Total out of 100 Checked by Page 1 of 12
Question 1 (15 points) a. List the pk a values for the following acids. (5 points) I -10 C 3 C 2 4.75 (~ 5) N 3 36 Me 2 + -2.5 3 + -1.7 Cl -7 Water 15.7 N 4 + 9.7 C C 25 C 3 C 2 16 b. Explain why aniline is a weaker base than cyclohexyl amine. (5 points) N 2 N 2 Aniline Cyclohyexyl Amine The pka values of protonated aniline and cyclohexyl amine are 4.6 and 11.2, respectively. The lone pair electrons on the nitrogen of aniline can be delocalized into the benzene ring. owever, once protonated, there is no lone pair of electrons available for delocalization in the protonated form of aniline (conjugated acid). ee page 284-285 for detailed discussion. Page 2 of 12
c. In the lecture, unny mentioned a solution with p value of 20. Eric said that it was not possible to have an aqueous solution with that high p value. aley said it was possible but unlikely. o what is the highest p value of an aqueous solution you can make? And what is the lowest p value of an aqueous solution you can make? All p values are measured at room temperature. Clearly state the reagents you will use and concisely discuss your calculation. (5 points) p = - log [ 3 + ], [ 3 + ] is the molar concentration of hydronium ion. o the higher the hydronium ion concentration, the lower the p value of a solution, and the lower the hydronium ion concentration, the higher the p values of a solution. There are, however, limits of the hydronium ion and hydroxide ion concentrations in aqueous solution. First, the concentration of water is 55.5 M. econd, for aqueous solutions, [ 3 + ]X[ - ]=10-14 M 2. ence, if all the water molecules are converted into hydronium ion by adding strong acids, such as I, 2 4 (sulfuric acid), the highest concentration of hydronium ion is 55 M or less (if we ignore density changes), so the lowest p is log (55) = -1.7. In practice, you can buy concentrated Cl at 12 M and sulfuric acid at 36 M, both with p below -1.0. imilarly, as one student wrote, a saturated Li (a strong base) should give a high p value. A typical concentrated solution of Na is 10 M. For this solution, the hydronium ion concentration is 10-14 M 2 /10 M= 10-15 M, and p is log (10-15 ) = 15. Use bases (such as amide anion, N - 2 ) stronger than hydroxide ion do not make the solution more basic, as the stronger base will react with water to generate hydroxide ion. o the p range is not limited to 0 to 14. In aqueous solution, p range of -1 to 15 can be easily accomplished. If one needs higher or lower p, different solvents (most time aprotic solvents) should be used. In essence, water becomes a buffer at p values equal to its two pka s, -1.7 and 15.7. Page 3 of 12
Question 2 (16 points) In 1951, the great Kurt Alder himself examined the reaction of a mixture of dienes 1 and 2 with maleic anhydride. e discovered that the diene 1 reacted at 35 ºC, but diene 2 reacted only at 150 ºC. (Problem 12.52 on p613, Maitland Jones, Jr.) Diene 1 Diene 2 Maleic anhydride a. Why is reaction of diene 1 so much easier than reaction with diene 2? (8 points) The question indicates the reaction is a Diels-Alder reaction. The generic form of the reaction is shown below, with the diene in s-cis-configuration (see p318-321). Diene 1 Diene 2 For both dienes 1 and 2, the compounds must rotate around the single bond highlighted in red. When the s-cis configuration is adopted, there is mild steric interference between the two hydrogen atoms highlighted in blue for diene 1; however, for diene 2, there is a stronger steric repulsion between the methyl group and hydrogen atom highlighted in blue. As a result, reaction with diene 2 required more energy to form the s-cis configuration than reaction with diene 1. Page 4 of 12
b. Give all the product(s) expected for the reaction with diene 2. how the absolute configuration (R or ) at each chiral center using perspective formulas with solid and hatched wedges. c. ow are each of these stereoisomers (in part b) related to the others? Are they enantiomers or diastereomers? (8 points) For stereochemistry, you can fix the orientation of one reactant, say diene 2, and alter the orientation of the other reactant, in this case, the dienophile, maleic anhydride. ince diene 2 is asymmetric but planar and the maleic anhydride is symmetric, the dienophile can approach the diene in four orientation directions (endo or exo; and top of bottom). Use one of your hands to mimic the diene and a book to mimic the dienophile. o total four products are possible. Two pairs are enantiomers to each other, and two different pairs are diastereomers. R R enantiomers R R R enantiomers R R R (3a,4R,7R,7aR)-3a,4,7,7a-tetrahydro- 4,7-dimethylisobenzofuran-1,3-dione Page 5 of 12
Question 3 (20 points) Give all the product(s) expected when 1-bromo-2-methyl-cyclohexane undergoes either E1 or E2 reactions. If more than one product can be formed, indicate the major product and the most likely minor products. Concisely show the mechanism leading to the formation of each product and provide rationale for the formation of the major product. a. E1 elimination (Et 3 N in Et) C 3 E1 C 3 Fast 1,2-hydride shift C 3 Major product more substituted Minor prodcut less substituted Major product more substituted Minor prodcut less substituted Triethylamine is to neutralize the generated, so the overall equilibrium will favor the products (see exam 1, question 6b). Page 6 of 12
b. E2 elimination (tbuk in tbu) First, one should recognize that there are four stereoisomers of the reactant. Two pairs are enantiomers, so their activities are the same. But the cis and trans isomers react differently, as illustrated below. For the trans-isomer, only the hydrogen shown in red can be eliminated, as E2 reaction requires anti or syn-configuration of the leaving group and proton being abstracted. C 3 C 3 nly product For the cis-isomer, two hydrogen atoms shown in red can be eliminated. Again, the more substituted olefin is the major product. C 3 C 3 C 3 Major product more substituted Minor prodcut less substituted Page 7 of 12
Question 4 (20 points) For the following two questions, all carbon atoms in the final products should be derived from the starting materials specified. ther reagents may be used when necessary. For each step of your proposed synthesis, the product that will be used as the starting material for the next step should be the major reaction product for that step (i.e., over 50% yield). a. (10 points) 2 Base NaN2 or tbuk NB or 2, light or Base NaN2 or tbuk Base NB Base NaN2 or tbuk or 2, light NaN2 or tbuk Page 8 of 12
b. (10 points) There are several ways to do this, see quiz 8 ether formation. N2 substitution acid-catalyzed addition Na, ammonia 2 Lindalar's catalyst, 2 NaN 2 Na, ammonia Lindalar's catalyst, 2 2, 2 4 (catalytic) RR 2 2 4 (catalytic) 1. Borane 2. Na, 2 2 4 (catalytic) Page 9 of 12
Question 5 (14 points) Give the major product expected when each of the following compounds reacts with one equivalent of. Reactant Major Product or olefin 1 intermediate 1 Major Minor olefin 2 Intermediate 2, most stable, more substituted olefin, allylic secondary carbocation Major b. Discuss which compound reacts faster with (intrinsic reactivity, i.e., higher rate constant). int: reaction coordinate diagrams may assist your discussion. The reaction rates depend on the energy difference (activation energy) between the reactant and the corresponding transition-state, which is similar to the carbocation intermediate in this case. While intermediate 2 is more stable than intermediate 1, olefin 2 is also more stable than olefin 1. o we cannot estimate the difference in activation energy, and thus cannot predication which olefin is more reactive based on these analyses. n the other hand, the conjugated olefin is electron richer, so it will be more reactive based on electrophilicity. Again, how do you present your rationales is most important. Page 10 of 12
Question 6 (15 points) Propose a mechanism for the following transformation. F 3 CC 2, 2 room temperature, 1 hour + In essence, this is an acid-catalyzed ether hydrolysis. The reason that the two ether bonds of the 1,2-diol are hydrolyzed, but not the ether bond in the upper position is due to the additional stability of intermediate 1, which is a tertiary carbon cation and also stabilized by the lone pair electrons on the oxygen. Remember ammond postulate: the more stable intermediate forms faster. The final step involves different forms of acetone (ketone), which is crucially important for Chem 342. + 2 + Intermediate 1 or Remember this for Chem 342 Last note is that all steps are reversible. The overall equilibrium can be shifted by controlling water concentration. The reaction is useful in protecting carbonyl groups, a topic will be covered in Chem 342. Page 11 of 12
Bonus questions (6 points) a. Are the following chemicals good or bad for you? And why? 1, aspartame; 2, phosphoric acid; 3, potassium benzoate; 4, citric acid; 5, caffeine. All from the Diet Coke bottle. All chemicals may be bad at high concentration or to different people. For example, you don t want to give a whole bottle of caffeine to a one-year old. b. Josh, a graduate student in unnyland, obtained a proton (1) NMR spectrum of the following compound in deuterated methanol (CD3D). Besides the two sets of signals originating from the solvent, he observed two more sets of peaks with relative integrals of 2:3. Propose mechanisms to account for his observation. Also propose additional experiments to test your proposed mechanism and how the experimental results will corroborate or disprove your mechanisms. The real question is why Josh did not see the two protons highlighted in blue (depicted in the scheme below). The reason is that ketone can tautomerize to enol, resulting in proton exchange (exchangeable proton, as deuterium (D) is invisible in 1 NMR). Normally, ketone form is much more stable, and the tautomerization is kinetically slow for isolated ketones. owever, for 1,3-diketone, the enol is stabilized by conjugation with the other carbonyl group. The result is the 1,3-diketone compounds become very acidic, with pk values around 13, just like normal acids (e.g. R). All the steps are reversible, but since methanol is the solvent, which is in large excess, the residual form with proton (highlighted in blue) is of low concentration and thus not detectable. The reduced pka and enhanced nucleophilicity of 1,3-dicarbonyl compounds will be used in Chem 342. CD 3 D D CD 3 D CD 3 D repeat several times in the NMR tube D D Page 12 of 12