Chemistry Organic Chemistry II: Reactivity and Synthesis ANSWERS FOR FINAL EXAM Winter 2004

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1 ASWER KEY Page 1 of 18 Chemistry rganic Chemistry II: Reactivity and Synthesis ASWERS FR FIAL EXAM Winter 2004 Paper umber 631 Thursday April 22, :30 4:30 pm University Centre Rm Students were permitted to bring into the exam room E SEET of 8½ x 11 paper, with any ADWRITTE notes they wished (both sides). Molecular model kits were also permitted but no other aids could be used. TE EXAM WAS I TW PARTS. PART I was mandatory, with choice only in Question 1. E of the Challenge Questions in PART II could be answered for EXTRA CREDIT. PART I: Question 1 (20 Marks) Question 2 Question 3 Question 4 Question 5 (10 Marks) (20 Marks) (10 Marks) (10 Marks) SUB-TTAL: PART II (EXTRA CREDIT) (70 Marks) (7 Marks) TTAL:

2 PART I: D ALL QUESTIS There is choice in question 1 only. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions to correctly complete TE of the following reactions. All reactions do in fact lead to products. Clearly indicate which TE responses you want marked! 1. C 3 C 3 2. ptos (cat) in C 3 C 3 or C 3 (a) Epoxidation followed by acid-catalyzed opening of the epoxide by attack of methanol. Either regioisomer of the product will be accepted. You might have reasoned that attack would occur at the tertiary centre, if you thought that the opening was more S1-like, or you might have supposed attack at the less-hindered secondary centre if you were thinking of S2. 1. ac 2 C 3 C 3 C 2 Ph workup Ph or Ph Ph (b) This is a variation on Review Problem 13.7d from Fox and Whitesell. We will accept either product shown, since it does depend on whether you interpreted step 2 as just a quench, or actually hydrolysis of the ester under more vigorous conditions. 1. LiAl 4 /TF (C 3 ) workup (C 3 ) 2 (c) C 2 I 2, Zn C 2 (d) / 2 2. ab 4 /C 3 3. PBr 3 Br (e)

3 1. B 2 6, TF , - (f) 1. 3, C 2 Cl 2 2. Zn/Ac 3. "Cr 6+ ", pyridine-water Cl (g) 4. SCl 2 2 Br 2, a (aq.) 2 (h) MgBr (i) CuI (cat.) Ether ( 3 + workup) LDA, TF, -78 o C Br (j)

4 heat (k) 2 1. a 2, Cl (aq). 2. (l) Cl AlCl 3 C 2 Cl 2 3, 2 S 4 2 (m) (n) 2 1. Cl, Et, C eat 2. a (aq.) C 3 2 S 4 (aq.) C 3 (o) C 3

5 2. (10 MARKS) Provide a rational and effective synthesis of the following compound starting from cyclohexanone and any other required inorganic or one-carbon containing organic reagents. Remember that although a retrosynthetic analysis may be useful to you in solving this problem, the answer we want is the forward synthesis complete with reagents and products. starting from The key is the enamine functional group - the only functional grouping present in the target molecule. Recall that enamines are formed from ketones plus secondary amines (this is the only way we have discussed to make this type of group), so the bulk of the synthesis must be to form the desired secondary amine. 3 C + C 3 C 3 Br The corresponding "forward" synthesis is: 1. ab 4, Me 2. PBr 3 Br Mg, Ether then C 2 (s) ( 3 + w/u) + C 2 C 2 SCl 2 Cl C 3 2 C 3 LiAl 4 2 S 4 (cat.) TF C 3 ere is another way of making the secondary amine you need. Technically, triphenylphosphine is an organic reagent, so you might have excluded this possibility, but we will accept it nonetheless. C 2 + C C 3 Br + P(Ph) 3 C 3 Br 1) P(Ph) 3 2) MeLi or a TF, then cyclohexanone C 2 1) B 2 6, TF then 2 2, a 2) Cr +6, pyridine C 3 2 ab 3 C p 5 C 3

6 3. (a) (7 MARKS) Provide a detailed stepwise mechanism for the following transformation. 3 + / 2 C 3 C 3 1,2-shift - + C 3 C 3 It is not necessary to explicitly draw the resonance forms. This is Supplementary Problem (page 740) from Fox and Whitesell. It was one of the recommended questions. It is also very similar to Groutas problem #17.

7 (b) (8 MARKS) Provide a detailed stepwise mechanism for the following transformation. a + - C 3 C 3 (Sorry about the error in the starting structure on the actual test. The correct structure is shown here. This is just an Aldol Condensation + C 3 C 3 C 3 + C 3 + C 3 +

8 (c) (5 Marks) Consider the following two base-induced aldol condensation reactions: + a (aq.) % Yield + a (aq.) < 20% Yield Why does the first reaction proceed in very good yield, but the second reaction only gives the indicated product in less than 20% yield? + 2 These reactions are crossed aldol condensations. Recall that for a successful crossed aldol you need to have one aldehyde that is enolizable and one aldehyde that is not enolizable. This is true in the first instance, but not in the second case. The low yield in the second reaction reflects the formation of several other aldol products. a a + 2 (this process is common to both reactions) This compound cannot be enolized, as it has no aliphatic hydrogens a This enolate can also act as a nucleophile, forming alternative products. This is Review Problem 13.8 (page 698) from the course textbook, Fox and Whitesell. Problem 13.8 was one of the practice problems on the list of recommended questions.

9 4. (a) (4 MARKS) Salicaldehyde (1) gives two different products under slightly different reaction conditions. Briefly explain how each product is formed. Detailed mechanisms are not necessary, but be specific in your explanation. K 2 C 3 C 3 C 2 C 2 C 2 Br acetone, heat 2 1 K C 3 C 2 C 2 C 2 Br acetone, heat The difference between the two processes is the nature of the base that was used. The top process used potassium carbonate, while the bottom process used potassium hydroxide. K is a much stronger base than K 2 C 3, and it is able to deprotonate acetone, the solvent used in these reactions, to a certain extent. The enolate of acetone will cause an aldol condensation with the aldehyde group, leading to product 3. Potassium carbonate is too weakly basic to deprotonate a ketone to any significant extent. In both processes, the phenol group is deprotonated and reacts with butyl bromide. ote that phenols are quite acidic and even a weak base like carbonate ion can deprotonate them. 3 (b) (3 MARKS) A thin layer chromatography separation is done on a mixture of ethylbenzene, acetophenone and 1-phenylethanol. Silica gel plates are used, and the developing solvent is pentane. List the compounds from lowest to highest R f values, and provide a short explanation of your reasoning. ethylbenzene acetophenone 1-phenylethanol Lowest R f : ext: ighest R f : 1-phenylethanol acetophenone ethylbenzene In TLC, the R f is an approximate measure of the polarity and hydrogenbonding ability of an organic molecule. Alcohols are more polar than ketones, which are more polar than alkanes. The more polar the molecule, the slower it migrates on a TLC plate.

10 (c) (3 MARKS) You have carried out the following reaction between acid chloride 1 and an excess amount of amine 2, to form amide 3. The reaction was quenched by adding water. You now need to work it up, to separate unreacted excess amine from the product amide. Cl + 2 C 2 Cl Describe the specific liquid-liquid extraction you would use to achieve this: i) What aqueous reagent would you use? Dilute aqueous acid, such as Cl (aq.) or 2 S 4 (aq.) ii) Would the organic layer be on top or on the bottom in your separatory funnel? The organic solvent is dichloromethane which is more dense than water. It will thus be on the bottom. iii) Which compound(s) would be in each layer? The amide will be in the organic phase, while the excess amine will be in the aqueous layer

11 5. (10 MARKS) The spectra of an organic compound with the formula C 9 13 are shown below. What is the structure of this compound? ppm region expanded septet doublet Draw the structure of the compound in the box below. Formula: 4 degrees of unsaturation. IR: stretches ( ) C= (the band at 1600 is a C=C stretch, carbonyls are not found below about 1685 or so) 13 C MR: 8 signals. E signal must represent TW identical carbons. ote 6 signals between ppm. 2 C MR: 7 signals, 1:1:1:1:1:2:6. ote 4 aryl signals, one of which is a SIGLET. Broad peak at 3.6 ppm suggests 2 (along with IR). Septet-Doublet, 1:6 ratio must be an isopropyl group (C(C 3 ) 2 ). Downfield chemical shift of septet says group must be attached to oxygen.

12 PART II: CALLEGE PRBLEM (EXTRA CREDIT). D E PRBLEM LY. Part II is worth up to 7 additional marks. Write your answer on the pages headed Answer to Challenge Problem that are provided. A. SYTESIS The drug ocfentanil is a narcotic painkiller with effects similar to morphine but without the addictive properties. Devise a reasonable synthesis of ocfentanil starting from any organic material(s) having 6 or fewer carbon atoms. You may use any reagents you wish. C 3 F cfentanil C F 2 2 B. MECAISM The following reaction has several features in common with reactions we have discussed in detail during this course. Write a detailed stepwise mechanism that shows how each of the three organic products is formed. Et 3 Cl S + + S C. SPECTRSCPY The spectra of the drug Flexazone, an analgesic and anti-inflammatory that is no longer used, are shown on the following pages. In addition to these spectra, laboratory tests have confirmed that there is nitrogen present, but no halogens. For full bonus marks, what is the structure of Flexazone and what features of the spectra support your assignment? Part bonus marks can be earned by identifying in detail key structural fragments. The more detail you can provide, the more part marks you can earn.

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15 ASWER FR SYTESIS CALLEGE PRBLEM Retrosynthetic Analysis: "starting" compounds are highlighted C 3 + Cl C 3 F F + Br + 2 F + Br The "forward" synthesis might look like this: Br Mg, Ether PBr 3 then Br F 2 F ab 3 C p 5 K 2 C 3 Cl pyridine C 3 F C 3

16 ASWER FR MECAISM CALLEGE PRBLEM Part 1: Activate - bond to be a leaving group Et 3 Cl S "Tosyl" group a.k.a "Ts" S (- Cl ) Cl S Part 2: form benzaldehyde by a process related to the offmann and Benzilic Acid rearrangements Ts Et 3 + Et 3 C + + S C C Et 3 This is Groutas problem #200. The chemistry of sulfonyl chlorides like tosyl chloride is found in Fox and Whitesell Chapter 12, section 12.6 pages

17 ASWER FR SPECTRSCPY CALLEGE PRBLEM The correct structure for flexazone is shown at right: This was by far the hardest of the three challenge problems, because of the unusual - linkage, the high symmetry of the compound, and the fact that one carbon signal was not visible. We expect that most people who attempted this problem would be able to identify fragments, but might not be able to put them together correctly. Thus, most of the marks here were for the logic of the solution you proposed, and could be earned whether you had the right structure or not. What spectroscopic information do we have? You know that the molecular ion is m/z 308. Even mass for a nitrogen-containing molecule implies an even number of nitrogens. In the IR, you can see carbonyl signals. There are no - or - stretches. The IR gives us little else to go on. In the 13 C MR you can see 9 signals, 4 clearly aliphatic, 4 in the aromatic region, plus a carbonyl signal at about 170 ppm. The carbonyl is obviously a member of the carboxylic acid family, but it can t be an acid because there is no signal in the IR. Possibly an ester or an amide, but if it is an amide, the amide must be tertiary since there are no - IR signals either. An ester is unlikely since there is no 13 C signal at the right chemical shift for a C- grouping. If there were only 9 carbons, it would be difficult to come up with a molecular mass of 308. bviously there must be considerable symmetry in the molecule, so that some of the carbon signals represent more than one carbon atom. In the 1 MR, you can see 7 signals. There are two signals between 7.0 and 7.5 ppm that are clearly consistent with the aromatic 13 C signals we saw. There is a triplet at about 3.3 ppm, a quartet at about 2 ppm, two multiplets around ppm, and a triplet at 0.9 ppm. The integration ratio is 8:2:1:2:2:2:3. We see no 1 signal consistent with ester structure. It looks like we must have a tertiary amide in our molecule. ow, what can we deduce? There must be more than one phenyl ring in the molecule, because of the 1 integration (10 total equals two monosubstituted phenyl rings). The two phenyls must also be equivalent because we see only 4 13 C signals for aromatic carbons. The aliphatic 1 signals and the 4 aliphatic 13 C signals suggest a C 2 C 2 C 2 C 3 group. The quartet-triplet pattern could be an ethyl group, but then what would the two 2- multiplets be? The triplet signal integrating for 3- is obviously the methyl. We thus have 2 phenyls and a butyl, for a total of 16 carbons and 19 hydrogens. We also have at least 1 carbonyl, making 17 carbons, 19 hydrogens and 1 oxygen minimum. We also know that there is an even number of nitrogens, probably 2. If we look at a possible formula equalling a molecular mass of 308, we find: C: 17 x : 19 x 1 19 : 2 x 14 (?) 28 : 1 x 16 (?) 16 Total: 267 This leaves us with a mass deficit of = 41. If we have two nitrogens, and one of them is a tertiary amide, it is possible that both of them are, given the symmetry implied by our spectra. What would we get for a formula if we added another carbonyl to the mix? C: 18 x : 19 x 1 19 : 2 x 14 (?) 28 : 2 x 16 (?) 32 Total: butyl-1,2-diphenyl-3,5-pyrazolidinedione C Mol. Wt.:

18 k, now our mass deficit is = 13, which is consistent with a missing C- group. The 1- signal at 3.3 ppm is still unassigned, and here it is. So, our formula is probably C , which has 11 degrees of unsaturation. The two phenyls account for 8 of these, the two carbonyls for another 2, so there is one unsaturation left to be either a double bond or a ring. The lack of any 13 C signals for an alkene strongly suggests a ring. ow can we have two identical tertiary amides? There is only one butyl group, but there are two phenyls, so we can deduce that there is most likely one phenyl and one other group on each. These are the fragments we have so far, and they account for all the atoms in our formula, so we have to find a way to put them together. 2 x C Ph 1 x C 2 C 2 C 2 C 3 1 x C The correct structure is obviously one way, but there is another isomer that would work as well: Ph Ph 2-Butyl-1,3-diphenyl-imidazolidine-4,5-dione