ANSWER KEY Page 1 of 11

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1 ANWER KEY Page 1 of 11 Chemistry rganic Chemistry II: Reactivity and ynthesis FINAL EXAM Winter ession 06R Paper 325 Frank Kennedy own Gym aturday April 15, :30 pm 4:30 pm tudents are permitted to bring into the exam room NE EET of 8½ x 11 paper with any ANDWRITTEN notes they wish (both sides). Molecular model kits are also permitted but no other aids may be used. The exam is in two parts. Answers to all questions in PART I are to be entered in the indicated spaces on the exam paper itself. An answer to the Challenge Question in PART II may be written for EXTRA CREDIT. PART I: Question 1 (28 Marks) Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 (15 Marks) (10 Marks) (7 Marks) (10 Marks) (15 Marks) (5 Marks) UB-TTAL: PART II (EXTRA CREDIT) (90 Marks) (10 Marks) TTAL:

2 ANWER KEY Page 2 of 11 PART I: D ALL QUETIN 1. (26 MARK) Provide the missing product, starting compound or reagent/solvent/conditions to correctly complete the following reactions. All reactions do in fact lead to products. g(ac) 2 2 4, 2 (a) Mercury-assisted hydrolysis of terminal alkyne to methyl ketone - text pp KMn 4, 3 + (heat optional) (b) xidative cleavage of alkene. Note formation of carboxylic acid plus ketone indicates that we need a stronger oxidizing system - acidic permanganate is discussed in text p Cl 2 2. N , Na, 2 N 2 Formation of acid chloride using thionyl chloride - text pp Preparation of amide from ammonia and acid chloride - text p offmann rearrangement of amide - text pp (c) This example appears in the text at the bottom of p CC 3 Δ CC 3 3 CC CC 3 (d) Diels-Alder reaction - text pp Note that product has an "extra" C=C bond - this indicates that dienophile must be an alkyne rather than an alkene. This is review problem 14.1f. Δ (e) Claisen rearrangement - text p This example is shown at the top of this page in the text.

3 ANWER KEY Page 3 of 11 (f) C 3 C 2 Mg CuI (cat.) Ether ( 3 + workup) Addition of cuprous iodide to make a "soft" carbon nucleophile, resulting in 1,4-alkylation rather than 1,2-addition - text pp Note that the isolated alkene is not altered by nucleophiles. (g) 3, C 2 Cl 2 then Zn, Ac xidative cleavage of an alkene to make a pair of carbonyls - text pp This example is almost identical to exercise 10.22d. (h) Et C 3 Et NaEt Et, heat 3 C Et Dieckmann condensation - text pp This example is shown at the top of page 660 to illustrate the requirement for enolization of the product to avoid reversibility. (i) C 3 1. Nitration para to methyl - text N C 3 3, NaN 2, C 3 Cl (aq.) 2. n, Cl (aq.) N 2 2. K, Cu 2 ( 2 ) pp and Reduction of nitro to amine - text p Diazotization and andmeyer reaction to introduce nitrile - text pp NTE: ortho nitration would also be acceptable, but only one N 2 will be added because nitro groups are strongly deactivating. base heat Cl pyridine any non-nucleophilic base is ok Formation of an ester from alcohol plus acyl chloride - text p Ester enolate Claisen rearrangement - text p.698. This process is similar to exercise 14.17c. (j) NTE: the preferred conditions for rearrangement would be 1) LDA, TF, -78 o C; 2) heat. NaC 3 3 C C 3 Ac or C 3, + 3 C (k) Baeyer-Villiger oxidation/rearrangement - text p This example is very similar to the one shown in the middle of page 697. The opening of the lactone to make an alcohol plus an ester is seen in review problem 14.11b. Transesterification is discussed on p. 604.

4 ANWER KEY Page 4 of (10 MARK) Propose a synthetic route to the molecule shown below. The compounds shown in the box are the only ones you can use to provide carbon atoms to be incorporated in your target. Note that you will only require some of these compounds choose carefully. You may employ any solvents and reagents you wish. how each reaction in your proposed sequence with the appropriate reagent(s), starting material and product. Mechanisms are NT required. Target: 3 C Et C 3 C 2 C 3 I C 3 These compounds are your only sources of carbon atoms. Choose the one(s) you want to use as starting materials for your synthesis. The structure of the target should have suggested a Robinson annulation approach. The Robinson annulation is in the text on pp As a note, examination of the starting compounds you have available should also have indicated that you should be thinking about enolate chemistry as the key method, since just about everything in the box of structures looks like something from chapter C 3 C 3 C 3 C C 3 C 2 C 3 C 3 3 C 3 C Note that you don't have suitable starting materials for most other strategies to make 6-membered rings (Diels-Alder for example). Thus, the forward synthetic route would be C 3 C 2 NaEt Et C 3 3 C NaEt Et 3 C 3 C 3 C C 3 2 (g) Pd/C Me 3 C C 3

5 ANWER KEY Page 5 of (10 MARK) Write a detailed stepwise mechanism to explain the following reaction. Recall that the 2 Ph group will be electron-withdrawing. + Ph NaC 3 C 3 Ph This is essentially identical to Groutas #143, which was one of the recommended study problems. Remember to start any problem by figuring out WAT needs to be accomplished. In mechanism problems, it is a good idea to find the starting compounds in the product. 3 2 Ph 1 a b c d e Ph ab c d e The 2 Ph and CC 3 groups in the product should be a dead giveaway here. Notice now what your mechanism must do: 1) Form a C-C bond between carbons "c" and "3" with loss of. 2) Form a C- bond between oxygen "b" and carbon "2" with loss of. 3) Make sure the C=C bonds are in the right places in the new ring. The fact that we have a base present and a 1,3-dicarbonyl compound starting material tells us this reaction is all based on enolate chemistry from Chapter 13, and the alkene conjugated to an EWG also tells us to look for a "conjugate addition" somewhere. C 3 Ph Ph C 3 Ph Ph Ph C 3 Ph C 3 Ph

6 ANWER KEY Page 6 of (7 MARK) Provide a detailed stepwise mechanism for the following transformation, showing all steps (arrow pushing). iefly, explain what drives the reaction thermodynamically. This is Groutas problem 113. Again, finding the starting compound in the product, and identifying just what has to change is the key to solving the problem. - The driving force behind this reaction is the formation of the aromatic ring in the product. Note that protonation of the ester group in the starting material is also possible but we get into trouble very soon because in order to get nucleophilic attack at the right location we would have to make a dication or a zwitterion. Either alternative is likely to be a high-energy pathway and we know to avoid such choices.

7 ANWER KEY Page 7 of (10 MARK) The local anaesthetic proparacaine is made by this sequence of reactions. Deduce the structure for each product. 2 N NEt 2 Proparacaine N 3, 2 4 A 2 N K 2 C 3 Cl 2 N Cl 2 C N C 3 7 Cl Et 2 N 4 NEt 2 B 5 C 2 N C NEt 2 2 /Pd/C Proparacaine ince you know what the final product is, and you know all the reagents used to make it, one key strategy for solving this problem was to identify the pieces of the puzzle, before trying to see how they might go together. Notice that the product of the first three steps (compound 4) enters into the sequence again in the second part of the lower row of reactions. Compound 5 reacts with 4 to make 6.

8 ANWER KEY Page 8 of (15 MARK total) a) (5 MARK) The reaction of 4-phenylcyclohexanecarboxylic acid ethyl ester 1 with lithium aluminium hydride gives (4-phenylcyclohexyl)methanol 2. The progress of the reaction was monitored throughout by thin layer chromatography (TLC, silica gel) using ethyl acetate/hexane = 1/4 as eluent (solvent). What would the TLC look like after half of the ester is consumed (50%), and once the reaction is complete (100%)? Draw both TLC plates and label the spots corresponding to starting material (1) and product (2) in each. C 2 5 LiAl 4, TF pure 1 pure 1 50% 100% Note that the product (2) should run lower than the starting compound (1) because an alcohol is more polar than an ester. rxn rxn b) (10 MARK) (2,2-Dimethylcyclopentylidene)acetic acid ethyl ester 3 was synthesised by the orner- Emmons reaction (also known as orner-wadsworth-emmons or orner-wittig reaction). You performed a similar experiment in the laboratory. In point form, describe step-by-step an experimental procedure to synthesise 3 based on the method you used in lab. (Things to consider: What starting materials would you use? What solvent(s)? ow would you prepare the stabilized ylide necessary for E-alkene formation? What reagents are needed? What workup would be appropriate?) 3 The reaction you did in the lab was: C Ph Ph + P Ph Et C 2 Et C 2 Cl 2 0 o C Et + P Ph Ph Ph To prepare the desired compound using a similar method, you should use the same ylide but simply change the carbonyl compound to: Method: 1) Dissolve ketone in dichloromethane, cool on ice. 2) Add the ylide slowly. Warm to room temperature. 3) Evaporate the dichloromethane solvent using simple distillation on a steam bath. 4) Add hexanes to the residue to precipitate the triphenylphosphine oxide. 5) Remove the triphenylphosphine oxide by filtration - may take several repeated filtrations. 6) Evaporate the hexanes from the filtrate, leaving the product as residue. 7) Collect the product and weigh. Note: the ylide was commercially available so you did not need to do anything special to prepare it. owever, if you did need to make the ylide it could be done in a separate process as follows: Et Ph 3 P Et Ph Ph P Ph Et Na, TF Ph Ph P Ph Et

9 ANWER KEY Page 9 of (5 MARK) The spectra for a compound having the formula C 9 10 are shown below. What is the structure of this compound? IR Formula gives 5 degrees of unsaturation. IR shows alcohol present. NMR clearly indicates aryl ring (NB 4 deg. unsat.). Note 13 C NMR shows 1 aliphatic but downfield (C- ), can see 5 signals in aryl/alkene region. 1 integration 5:2:2:1. Monosubst. aryl, disubst. alkene, C C NMR 1 NMR Although this structure is not correct, it is a reasonable alternative based on what you should know from this course. It is also being given full marks. The key difference between this and the correct structure is the size of the coupling between the vinylic hydrogens. The geminal hydrogens of a terminal alkene show very little coupling to one another, whereas vicinal alkene hydrogens show couplings of around 10 z (cis) or z (trans).

10 ANWER KEY Page 10 of 11 PART II: CALLENGE PRBLEM (EXTRA CREDIT). Part II is worth up to 10 additional MARK. II. 1) The reaction of 4-methyl-pent-3-en-2-one with acrylonitrile under basic conditions gives the product A. Deduce the structure of A by giving the detailed mechanism of the reaction as well as using the provided 1 and 13 C NMR spectra. Please note that we call this a challenge problem for a reason. It is not meant to be easy, but that is why it can earn bonus points. As always, partial credit is given for reasonable analysis 4-Methyl-pent-3-en-2-one base N N A, C N 2 1 NMR of A C NMR of A

11 ANWER KEY Page 11 of 11 (ptional Challenge Question) Write the mechanism for the formation of unknown A here First, it is a good idea to see what can be learned from the spectra. You did not need to record any of this for marks, but you likely had to go through this analysis to have something to build a mechanism from. The formula for A tells you it has 6 degrees of unsaturation. The NMR clearly shows that there is no aromatic ring. You can see 9 13C NMR signals, including one that is either ketone or aldehyde at 208 ppm. The 1 NMR tells you that it is NT an aldehyde, so it must be a ketone. ince the formula has 12 carbons, and we only see 9 signals, some of these must be related by symmetry. The 1 integration is 1:1:4:3:4:3. Notice that there is a singlet integrating for 3 at 2.2 ppm - this suggests C 3 C- and you can see that this fragment is present in the starting material. The other 3 signal, at about 1.68 ppm is a complicated multiplet. This is a C 3 with more than one kind of neighbor proton. The pair of 4-proton signals are both triplets, which suggests a symmetrical set of C 2 C 2 fragments with no other neighboring protons. They are at 2.0 and 2.5 ppm, which also suggests that they are close to electron-withdrawing group(s). Also, you have signals at 4.4 and 4.7 ppm, in a 1:1 ratio, which are probably alkene hydrogens. This suggests the possibility of a disubstituted alkene. : what do we have? 3 C C 2 x C 2 C 2 X equivalent C 3 with nearby coupled s C C GIVEN the structures of the starting materials, the C 2 C 2 X fragments should immediately indicate that X =. Acrylonitrile is an electrophilic alkene, and it could be attacked by a nucleophile in a 1,4-fashion to give the observed R-C 2 C 2 - structure. Maybe now is a good time to consider a mechanism to see what the structure should look like. base or perhaps or C C ince we know the product has a C 3 C fragment, the first enolate is unlikely because if it were alkylated there would not be a C 3 next to the ketone! Let's consider the second anion, which is sort of an extended enolate: protonate the anion from solvent R R base base The molecular formula of each of these is C 9 13 N, which differs from the target formula by C 3 3 N. This equals one acrylonitrile unit! Each enolate can be alkylated at two sites. There are three possible products. A B C ption B does not have the symmetry implied by the spectra. A and C have the right symmetry, but A does not have two alkene C-s and also the C2C2 groups have another neighbor. Thus only structure C can be correct.