Chemistry 35 Exam 2 Answers - April 9, 2007 Problem 1. (14 points) Provide the products for the reactions shown below. If stereochemistry is important, be sure to indicate stereochemistry clearly. If no reaction would occur, write N REACTIN. A) (2 points) No partial credit P 3, DEAD Intramolecular Mitsunobu reaction. B) (2 points) No partial credit MUST PRVIDE EXPLANATIN!!! Ms NaN 3 DMF N REACTIN This is similar to a neopentyl system. Beta branching does not allow an SN2 reaction to occur. C) (2 points) No partial credit N Li S The base is very strong, with a pka around 35. owever, it is a poor nucleophile because it is very sterically hindered due to the two tbu groups. This base deprotonates the thiol to generate a thiolate, which is a strong nucleophile. The thiolate initiates an intramolecular SN2 reaction with the benzyl bromide to give the cyclic thioether. S 1
D) (2 points) 2 points for either answer, but 2 points total. No partial credit. Ts This is an SN1 reaction. The tosylate leaves to generate a carbocation. Resonance distributes the positive charge over two different carbon atoms, either of which can be attacked by acetic acid to generate the indicated products. E) (2 points) No partial credit. 1 equivalent of Na 1 equivalent of I Na deprotonates the most acidic position, which is the phenolic rather than the benzylic. The resulting phenoxide ion does an SN2 reaction with I (Williamson ether synthesis) to generate the methyl ether product. F) (2 points) No partial credit. K to show the Na salt of the product. S NaN 3 C 3 CN N 3 S The starting material is an internal sulfonic ester, and behaves just like a tosylate or mesylate. Azide is a strong nucleophile, and it attacks the sulfonic ester (SN2 reaction) to give the ring opened product. G) (2 points) No partial credit R P 3, DEAD N REACTIN N 2 The Mitsunobu reaction only works with nucleophiles that have a conjugate acid with a pka between 3 and 12. The primary amine in this reaction has a pka of approximately 35, which is outside of this range. 2
Problem 2 (14 points) A) (12 points) Provide a detailed, step-by-step mechanism for the following reaction. In this question, we provide the reactant and reaction conditions. YUR JB IS T: 1) Show the individual chemical steps and all intermediates formed in the process of converting the reactant to the product. 2) Use curved arrows to show all changes in bonding and lone pair electrons. 3) Show all formal charges and all contributing resonance structures. Remember, the sequence of steps you propose must convert the reactant into the product using only the starting material provided. You may not use additional acids or bases, since they are not present in the reaction mixture. The solvent for the reaction is TF, the structure of which is provided below. Ts TF solvent + Ts Ts - Ts + Ts Ts Arrows (4 points): 1 point for each of the four arrows shown in the 1 st, 4 th, and 5 th structures. The arrows must clearly start and end in the proper place (bond or atom). Formal charge (4 points): 1 point for each of the + formal charges shown in the 2 nd, 3 rd, 4 th and 5 th structures. Resonance structures (3 points): 1 point for each resonance structure that is completely correct. They do not have to show the arrows in the 2 nd and 3 rd structures that show movement of electrons to interconvert resonance structures. Deprotonation (1 point): 1 point if the deprotonation step is drawn completely right: deprotonation occurs AFTER the alcohol reacts with the carbocation. No point if the student shows deprotonation of the alcohol BEFRE reaction with the carbocation. They do not have to show the lone pair on oxygen. It is K if they use TF rather than tosylate as the base during the deprotonation. 3
B) (2 points) Based on your mechanism, predict one other ether that might be formed as a product during this reaction. R 2 points for either answer, but 2 points total. No partial credit. Problem 3. (14 points) For each set of reactions shown below, circle the reaction that is fastest, and cross out the reaction that is slowest. Grading: 1 point for each correct circled reaction, and 1 point for each correct crossed out reaction. A) (2 points) NaS solvent solvent Na solvent S B) (2 points) Et, Et N Et, N Et Et, Et 4
C) (2 points) Ts NaCN, DMF CN Ts NaCN, DMF CN Ts NaCN, DMF CN D) (2 points) N Cl NaS, DMS 3 N + 2 S Cl NaS, DMS 2 + 2 S S Cl NaS, DMS 2 S + 2 S E) (2 points) Ms NaN 3 N 3 N 2 (DMF) is the solvent Ms NaN 3 N 3 N is the solvent Ms NaN 3 N 3 2 solvent 5
F) (2 points) S N 2 Na, DMS S CF 3 Na, DMS S Na, DMS G) (2 points) The products for these three reactions are intentionally not provided. N 2 P 2, DMS, DMS C N, DMS 6
Problem 4. (a) 8 points S Starting Material Final Product Requires - substitution (S - ) with inversion at less substituted epoxide carbon, producing alkoxide at more substituted epoxide carbon (4 points properly completed) - ether formation (-X electrophile) at 3 o alkoxide (or alcohol) (4 points properly completed) S - Na + TF S -X (X=Cl,, I, Ms, Ts) S or 1.S - Na + 2. dil 3 + S 1. Na (TF or other polar aprotic solvent) 2. -X (X=Cl,, I, Ms, Ts) S 7
(b) 12 points S Starting Material Final Product Requires - substitution with inversion (S - ) at less substituted epoxide carbon - substitution with inversion ( -??) at more substituted epoxide carbon SUBSTITUTIN WIT INVERSIN AT 3 o epoxide carbon only possible using 2 S 4 / 2 S 4 Need to convert to a good leaving group w/o inversion, followed by Sn2 with inversion using S -. MsCl / pyridine or TsCl / pyridine Ms or Ts S S - Na + P 3 DEAD S 4 points proper epoxide ring opening, AND 4 pts proper Ms or Ts, 4 points proper S - R 8 points proper Mitsunobu silly errors 2 S 4 / - - reaction works but - is protonated by 2 S 4 MsCl / TsCl without pyridine - pyridine prevents formation of Cl which will open epoxide 0 points if use SCl 2 or P 3 to convert alcohol into good leaving group - inversion, not retention 8
Problem 5. (20 points) For the following reactions, (1) Give the absolute configuration, (R) or (S), of all stereocenters in the reactants and products PRVIDED BY US. (2) If the reaction is 100% correct as written, write CRRECT in the box. If the reaction product(s) provided are not entirely correct, draw ALL TE CRRECT PRDUCTS in the box and SPECIFY TE SUBSTITUTIN MECANISM(S). If no reaction occurs, write N REACTIN in the box. (a) Na - (R) 1 pt TF 0 C (R) 1 pt Sn2 will not occur at a tertiary carbon. The solution temperature (0 o C) greatly slows Sn1 carbocation formation. If one waits long enough, Sn1 racemic product forms. N REACTIN - (must explain that Sn1 is slow because TF is not very polar) or RACEMIC PRDUCT 2pts ( 1 pt each enantiomer or squiggly line) 4 pts total for part (A) (b) 50 o C (S) (R) 1 pt each center (S) (S) 1 pt 1 pt Sn1 reaction: obtain both absolute configurations at reacting C (NT RACEMIC left stereocenter (S)) (squiggly line okay for right stereocenter) 6 pts total for part (B) Na CN TF 50 o C 9 CN
(c) (R) 1 pt (S) 1 pt Sn1 reaction good leaving group, resulting carbocation has a resonance structure with complete octets on every atom. RACEMIC PRDUCT 2pts (1 pt each enantiomer or for squiggly line) NC CN CN 4 pts total for part C (d) S Na TF 20 C S (S) (R) (S) 1 pt for labeled 4 ( ) ( ) (R) Slow Sn2 inversion of configuration, Sn2 at cyclohexane requires axial bromide (WY? when the is equatorial, the axial hydrogens on the C- backside block backside access. When the is axial, the molecule acts as a normal 2 o halide. ) CRRECT 2pts r N REACTIN if student states that this reaction required extensive heating in lab (No points if student does not give explanation) 6 pts total for part D 10
Problem 6. (18 points) Provide the missing starting materials (left box), reagents (box around arrow), product of first reaction (middle box) or product of second reaction (right box) that complete the following reaction sequences. (a) 6 pts total or int: C 7 13 (b) 6 pts total Na heat 2pts 2 S 4 1 pt - acetic acid 1 pt - sulfuric acid or Ts (catalyst) (Cl,, I not acceptable acids as catalyst since the X- reacts D TsCl / pyridine @ 0 o C or MsCl / pyridine @ 0 o C either Sulfonate 1 pt if not at 0 C D Ms or Ts - must be retention Na D (c) 6 pts total Best - Ms-Cl pyridine @ 0 C followed by - of I - (no Sn1) ------ K- 2pts SCl 2 heat or P 3 heat (may give some Sn1) ----- Cl / ZnCl2, / 2 S4, or I - greatest chance for mixed Sn2 / Sn1 Cl, - substitution of group must occur with inversion NaCN / DMS NC 0 pts for 2 S 4 + NaCN 11