Chemistry Organic Chemistry II: Reactivity and Synthesis FINAL EXAM Winter 2002

Transcription

1 ame: Student o: Page 1 of 12 Chemistry rganic Chemistry II: Reactivity and Synthesis FIAL EXAM Winter 2002 Paper umber 332 Friday April 19, :00 9:00 pm Frank Kennedy Gold Gym Students are permitted to bring into the exam room E SEET of 8½ x 11 paper, with any ADWRITTE 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 PART I are to be entered in the indicated spaces on the exam paper itself. An answer to E of the Challenge Questions in PART II is to be written in the exam booklet. PART I: Question 1 (20 Marks) Question 2 Question 3 Question 4 Question 5 Question 6 (4 Marks) (10 Marks) (12 Marks) (8 Marks) (8 Marks) PART II: TTAL: (8 Marks) (70 Marks)

2 ame: Student o: Page 2 of 12 PART I: D ALL QUESTIS TERE IS CICE I QUESTI 1 LY. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions to correctly complete TE (10) of the following. All reactions do in fact lead to a new product. Be sure to indicate clearly which ones you want marked. Br (C 6 5 ) 3 P Ethanol n-buli TF then add (a) heat (b) Br 2, C 2 2 (c) 1) aet/et (d) C Et Et 2) aet/et C 3 I C C 3 1) C 3, 2 S 4 (cat.) 2) 2 equiv. C 3 MgBr Ether ( 2 workup) (e) 2 (g), Pd/C Acetic acid (solvent) (f)

3 ame: Student o: Page 3 of 12 2 Al 3, C 2 2 (g) aet, Et heat (h) C CF 3 C 2 (i) C 2, S 4 (cat.) (j) toluene, heat (k) (l) Br + 1) 2 ac 3 C C 3 3 2) 2 S 4, 2 Br heat (m) (n) (Two Steps)

4 ame: Student o: Page 4 of 12 LiAl 4 TF followed by 2 workup (o) C 2 g(ac) 2 2, TF followed by ab 4 (p) 2 + heat (q) 2 S S (r) 2. (4 MARKS) Four steps of a synthesis of the plant terpene juvabione are shown below, with some items missing. All chemistry in this sequence has been covered in this course. Fill in the blanks. Keep your eyes on what is changing, don t be distracted by what stays the same! 3 C C 3 a, TF 3 C C 3 C C S 3 C S Et 3, C C C 3 3 C 3 C 3 C C 3

5 ame: Student o: Page 5 of (10 MARKS) Provide detailed stepwise mechanisms for both reactions in the following two-step synthetic transformation. C 3 C 2 S S C 2 2 Et 3, C 2 2 ( + C 2 + C + )

6 ame: Student o: Page 6 of (a) (1 MARK) The presence of nitrogen in an organic compound can be demonstrated by carrying out a sodium fusion. After dissolving the fusion product in water, the presence of nitrogen in the original compound is confirmed by adding FeS 4 and observing the formation of Prussian Blue, Fe 4 [Fe(C) 6 ]. n the basis of this information, what molecular or ionic nitrogen-containing species is formed when a nitrogen-containing compound is fused with sodium? (b) (2 MARKS) Silver halides (AgX; X =, Br or I), formed when testing a sodium fusion residue for halogens, are distinguished from one another in two ways: Physical appearance Solubility in aqueous ammonium hydroxide. Briefly describe how these two tests would distinguish Ag from AgI. (c) (2 MARKS) A student measured the boiling temperature of a liquid by distilling a sample. A value of 157 o C was obtained, but the literature value was found to be 138 o C. Indicate two experimental errors that might have led to the incorrect boiling temperature being measured. (d) (3 MARKS) An unknown compound (X) was dissolved in cold hydrochloric acid and the solution was treated with a cold aqueous solution of sodium nitrite. The mixture was then carefully added to a cold alkaline solution of β-naphthol. A red solid (Y) was formed. i) What type of compound was X? ii) Draw a reasonable structure for Y. β-naphthol

7 ame: Student o: Page 7 of 12 (e) (2 MARKS) A solution of sodium hydroxide in aqueous ethanol was treated with a 1 : 1 mixture of phenylethanal (C 6 5 C 2 C) and propanone. The mixture was shaken for about 20 minutes. A yellowish solid precipitated and was filtered off. The solid was shaken with water and this mixture was then treated with dilute hydrochloric acid until it was weakly acidic (p 4-7). The product of the reaction was worked up. Draw the structure of this product. (f) (2 MARKS) A student is given a sample of aniline and told to use it to prepare a sample of p-nitroaniline. Indicate the steps that would be required to ensure that the synthesis would work and that little or no m-nitroaniline by-product would be formed.

8 ame: Student o: Page 8 of (8 MARKS) The aisen Rearrangement is a common example of a sigmatropic process. Two very similar molecules shown below were subjected to identical conditions expected to lead to aisen Rearrangement. The outcomes of the two reactions were, however, rather different. Using your knowledge of organic structures and reaction mechanisms, EXPLAI why the product of reaction B does not appear to follow the standard aisen pattern. A toluene, heat B 3 C toluene, heat 3 C

9 ame: Student o: Page 9 of (8 MARKS) The IR, 13 C MR and 1 MR spectra of an organic compound having the molecular formula C 7 13 are shown below br s (a) Draw the structure of the compound in the box at right (4 Marks). (b) early indicate which atom(s) in your structure is(are) responsible for each of the signals in the 1 MR spectrum (4 Marks).

10 ame: Student o: Page 10 of 12 PART II: CALLEGE PRBLEM. D E PRBLEM LY. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Part II is worth 8 Marks. A. SYTESIS The simple alkaloid (R)-δ-coniceine has been synthesized from the intermediate A shown. The synthesis used reactions that have been discussed in this course. Construct a route to coniceine from intermediate A in 4 or 5 steps. The first reaction should form the bicyclic ring system found in the product. (R)-δ-coniceine 4 or 5 steps CEt A B. MECAISM The Benzoin Condensation, shown below, may appear surprising at first but its mechanism is actually quite simple. The process is catalyzed by cyanide ion stoichiometric KC is not needed. owever, the catalyst must be cyanide (or one of a very few other substances). ydroxide or similar basic reagents will T work. KC Et/ 2 heat Provide a mechanism for the Benzoin condensation and explain why cyanide is essential. Can you suggest what other kinds of substances besides cyanide might catalyze the Benzoin condensation? C. SPECTRSCPY You have been presented with an unknown organic compound to analyze. It is a clear liquid with a spicy odour similar to nutmeg. You perform several chemical tests, with the following results: Insoluble in water or dilute acid, soluble in aqueous a. Also soluble in C 3, ether, alcohol. Decolourises bromine water fairly rapidly. Forms a blue solution when treated with Fe 3. A mass spectrum shows the molecular ion at m/z 164, and a significant fragment at m/z 149, along with numerous other fragments. You then obtain IR, 1 and 13 C MR spectra, which are shown on the next page. 1. What is the structure of the unknown material? 2. Explain W the chemical and spectroscopic evidence supports your proposed structure.

11 ame: Student o: Page 11 of 12 There are 2 peaks in this signal Doublet of quartets Doublet of quartets Double doublet multiplet singlet singlet

12 Spectroscopy Crib Sheet for Introductory rganic Chemistry II 1 MR Typical Chemical Shift Ranges Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ) C C C C C C 2 C C C C C C C (solvent dependent) (solvent dependent) C Aryl C C Br C Aryl I C RC 2 Aromatic, heteroaromatic X C X =,, S, halide R 3 C Aliphatic, alicyclic Y =, R, S Y Y Y =, R, S Low Field δ igh Field 13 C MR Typical Chemical Shift Ranges Ketone, Aldehyde Alkene Carbox. Acid Amide Aryl Ester C x -Y Y =, CR 3 -C 2 -CR 3 C x -C= C 3 -CR IR Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (cm -1 ) Intensity C Medium R Strong, broad C=C Medium C Strong C=C Medium C= Strong C C Strong R Medium, broad R C C R Medium (R R ) C 1230, 1030 Medium Aryl Medium C Medium Aryl C=C 1600, 1500 Strong R Strong δ

13 ASWER KEY Page 1 of 12 Chemistry rganic Chemistry II: Reactivity and Synthesis FIAL EXAM Winter 2002 Paper umber 332 Friday April 19, :00 9:00 pm Frank Kennedy Gold Gym Students are permitted to bring into the exam room E SEET of 8½ x 11 paper, with any ADWRITTE 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 PART I are to be entered in the indicated spaces on the exam paper itself. An answer to E of the Challenge Questions in PART II is to be written in the exam booklet. PART I: Question 1 (20 Marks) Question 2 Question 3 Question 4 Question 5 Question 6 (4 Marks) (10 Marks) (12 Marks) (8 Marks) (8 Marks) PART II: TTAL: (8 Marks) (70 Marks)

14 ASWER KEY Page 2 of 12 PART I: D ALL QUESTIS TERE IS CICE I QUESTI 1 LY. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions to correctly complete TE (10) of the following. All reactions do in fact lead to a new product. Be sure to indicate clearly which ones you want marked. Br (C 6 5 ) 3 P Ethanol n-buli TF then add (a) heat (b) Br Br 2, C 2 2 Br (c) 1) aet/et Et C Et Et 2) aet/et C 3 I C C 3 (d) C C C 3 1) C 3, 2 S 4 (cat.) C 3 2) 2 equiv. C 3 MgBr Ether ( 2 workup) C 3 C 3 (e) 2 (g), Pd/C Acetic acid (solvent) (f)

15 ASWER KEY Page 3 of 12 2 Al 3, C (g) aet, Et heat (h) C CF 3 C 2 2 (i) C 2, S 4 (cat.) (j) toluene, heat (k) (C 3 ) 2 CuLi or C 3 MgBr, CuI (cat.) TF or Ether (l) Br + 1) 2 ac 3 C C 3 3 2) 2 S 4, 2 Br heat (m) (n) 3, C 2 2 then Zn/Acetic Acid (Two Steps)

16 ASWER KEY Page 4 of 12 LiAl 4 TF followed by 2 workup (o) C 2 g(ac) 2 2, TF followed by ab 4 (p) 2 + heat (q) 2 S S Al 3, C 2 2 (r) 2. (4 MARKS) Four steps of a synthesis of the plant terpene juvabione are shown below, with some items missing. All chemistry in this sequence has been covered in this course. Fill in the blanks. Keep your eyes on what is changing, don t be distracted by what stays the same! 3 C C 3 a, TF 3 C 3 C C 3 3 C C 3 C C S C 3 C 3 3 C C S 3 C S Et 3, C C ab 4, C 3 or 2 C 3 ac 3, C 3 3 C 3 C 3 C C 3

17 ASWER KEY Page 5 of (10 MARKS) Provide detailed stepwise mechanisms for both reactions in the following two-step synthetic transformation. C 3 C 2 S S C 2 2 Et 3, C 2 2 ( + C 2 + C + ) C 2 + C + - C 3 C 2 S S Et 3 - Et 3 + S - - S C 2 C 3 3 C 2 C 3 C 2 C

18 ASWER KEY Page 6 of (a) (1 MARK) The presence of nitrogen in an organic compound can be demonstrated by carrying out a sodium fusion. After dissolving the fusion product in water, the presence of nitrogen in the original compound is confirmed by adding FeS 4 and observing the formation of Prussian Blue, Fe 4 [Fe(C) 6 ]. n the basis of this information, what molecular or ionic nitrogen-containing species is formed when a nitrogen-containing compound is fused with sodium? The only nitrogen in Prussian Blue is in the ion [Fe(C) 6 ] 4- where it occurs as C. Therefore cyanide ion (C ) is formed during sodium fusion. (b) (2 MARKS) Silver halides (AgX; X =, Br or I), formed when testing a sodium fusion residue for halogens, are distinguished from one another in two ways: Physical appearance Solubility in aqueous ammonium hydroxide. Briefly describe how these two tests would distinguish Ag from AgI. Silver chloride (Ag) is white and soluble in aqueous 4. Silver iodide (AgI) is yellow and insoluble in aqueous 4. (c) (2 MARKS) A student measured the boiling temperature of a liquid by distilling a sample. A value of 157 o C was obtained, but the literature value was found to be 138 o C. Indicate two experimental errors that might have led to the incorrect boiling temperature being measured. 1) The thermometer bulb may have been immersed in the liquid rather than just the vapour. 2) Rapid boiling of the liquid may have caused heating of the vapour above the true boiling point. ther reasonable suggestions will be considered on their merits. (d) (3 MARKS) An unknown compound (X) was dissolved in cold hydrochloric acid and the solution was treated with a cold aqueous solution of sodium nitrite. The mixture was then carefully added to a cold alkaline solution of β-naphthol. A red solid (Y) was formed. i) What type of compound was X? ii) Draw a reasonable structure for Y. β-naphthol Compound X was a primary aromatic amine, Ar- 2. Y = this is a minimal structure. Any modified aryl ring is acceptable

19 ASWER KEY Page 7 of 12 (e) (2 MARKS) A solution of sodium hydroxide in aqueous ethanol was treated with a 1 : 1 mixture of phenylethanal (C 6 5 C 2 C) and propanone. The mixture was shaken for about 20 minutes. A yellowish solid precipitated and was filtered off. The solid was shaken with water and this mixture was then treated with dilute hydrochloric acid until it was weakly acidic (p 4-7). The product of the reaction was worked up. Draw the structure of this product. In lab, you did the aisen-schmidt reaction between acetone and benzaldehyde. ote that this is a crossed aldol, which is only feasible because benzaldehyde is T enolizable. In this question, the aldehyde IS enolizable, and thus the major product will not be derived from acetone it will be a self-condensation of the phenylethanal. (f) (2 MARKS) A student is given a sample of aniline and told to use it to prepare a sample of p-nitroaniline. Indicate the steps that would be required to ensure that the synthesis would work and that little or no m-nitroaniline by-product would be formed. 1) React aniline withacetic anhydride or acetyl chloride to form acetanilide. 2 (C3C)2 or C 3 C 2) React acetanilide with cold 3 / 2 S 4. 3 / 2 S 4 ice bath temperature 2 3) ydrolyse with aqueous a. 2 a/ 2 + C 3 C 2 2

20 ASWER KEY Page 8 of (8 MARKS) The aisen Rearrangement is a common example of a sigmatropic process. Two very similar molecules shown below were subjected to identical conditions expected to lead to aisen Rearrangement. The outcomes of the two reactions were, however, rather different. Using your knowledge of organic structures and reaction mechanisms, EXPLAI why the product of reaction B does not appear to follow the standard aisen pattern. toluene, heat A B 3 C toluene, heat 3 C The key difference between these two reactions is that the substrate in reaction B has an alkyl group at both ortho positions, which means that it is impossible for it to re-aromatize by loss of a hydrogen if the rearrangement stops at this position. C 3 cannot reform aromatic ring because C 3 group cannot be lost in a tautomeric equilibrium process WEVER, this intermediate product is also a 1,5-diene and thus can undergo a second Cope-type rearrangement! 3 C 3 C

21 ASWER KEY Page 9 of (8 MARKS) The IR, 13 C MR and 1 MR spectra of an organic compound having the molecular formula C 7 13 are shown below Vinylic (C=C-) hydrogens Amide - br s t-butyl group C 3 hydrogens (a) Draw the structure of the compound in the box at right (4 Marks). Full marks will be given for either of these structures, although the spectra are actually of the one on the left t-butyl Acrylamide. Part marks will be given for reasonable fragments. C 3 C 3 C 3 or C 3 C 3 C 3 (b) early indicate which atom(s) in your structure is(are) responsible for each of the signals in the 1 MR spectrum (4 Marks).

22 ASWER KEY Page 10 of 12 PART II: CALLEGE PRBLEM. D E PRBLEM LY. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Part II is worth 8 Marks. A. SYTESIS The simple alkaloid (R)-δ-coniceine has been synthesized from the intermediate A shown. The synthesis used reactions that have been discussed in this course. Construct a route to coniceine from intermediate A in 4 or 5 steps. The first reaction should form the bicyclic ring system found in the product. (R)-δ-coniceine 4 or 5 steps CEt A B. MECAISM The Benzoin Condensation, shown below, may appear surprising at first but its mechanism is actually quite simple. The process is catalyzed by cyanide ion stoichiometric KC is not needed. owever, the catalyst must be cyanide (or one of a very few other substances). ydroxide or similar basic reagents will T work. KC Et/ 2 heat Provide a mechanism for the Benzoin condensation and explain why cyanide is essential. Can you suggest what other kinds of substances besides cyanide might catalyze the Benzoin condensation? C. SPECTRSCPY You have been presented with an unknown organic compound to analyze. It is a clear liquid with a spicy odour similar to nutmeg. You perform several chemical tests, with the following results: Insoluble in water or dilute acid, soluble in aqueous a. Also soluble in C 3, ether, alcohol. Decolourises bromine water fairly rapidly. Forms a blue solution when treated with Fe 3. A mass spectrum shows the molecular ion at m/z 164, and a significant fragment at m/z 149, along with numerous other fragments. You then obtain IR, 1 and 13 C MR spectra, which are shown on the next page. 1. What is the structure of the unknown material? 2. Explain W the chemical and spectroscopic evidence supports your proposed structure.

23 ASWER KEY Page 11 of 12 There are 2 peaks in this signal Doublet of quartets Doublet of quartets Double doublet multiplet singlet singlet

24 Answer to Synthesis Challenge Problem You know that the conversion of A to coniceine can be achieved in 4 or 5 steps, and that you should start by forming the bicyclic ring system. This means making a C- bond, and the obvious way to do this is to make the nitrogen attack the electrophilic ester carbonyl. This is easily achieved using a base. otice that because of conjugation with the enone, this amine will be quite acidic. (R)-δ-coniceine 4 or 5 steps CEt A Base CEt CEt ow all that has to be done is to remove excess functional groups an alkene, a ketone carbonyl and an amide carbonyl. The order in which you do this matters to some extent however. The only way we 2 (g) Pd catalyst aet, Et Ethanol know to remove an amide carbonyl is or a, TF by reduction with LiAl 4 in TF or ether, but this will also reduce a ketone to a secondary alcohol. If the ketone CEt were reduced, we would have to reoxidize it back to the ketone because the only obvious way we know to completely remove an oxygen from an aliphatic molecule is the Wolff-Kishner LiAl reduction, which doesn t work with 4 TF alcohols. The best route is thus to hydrogenate the alkene, then use Wolff- Kishner to remove the ketone, and finally LiAl 4 to reduce the amide to an amine. ow, you could have treated the keto-amide with LiAl 4 and proceeded from the alcohol, if you had remembered from Chapter 8 that alcohols, when converted to sulfonate esters, can be attacked by nucleophiles. Thus, a possible 5-step synthesis would have involved the following steps: CEt CEt K, 2 2 S C 3 LiAl 4 TF 3 C S LiAl 4 TF Et 3 or pyridine There is considerable leeway in the details of these routes, and there are other alternatives that you may have come up with as well. Credit will be given for any reasonable approach or portion thereof.

25 Answer to Mechanism Challenge Problem KC Et/ 2 heat Looking at the reaction to find the starting material in the product, it should be evident that the product is a dimer of the starting aldehyde. You also can see that the cyanide ion is not incorporated into the product, but that should not be a surprise since you are told that it functions as a catalyst. The product has a C-C bond between the carbons that were the aldehyde carbonyls in the starting material. otice that one of the carbonyls has been converted to an alcohol in a pattern that suggests a possible nucleophilic addition. The other carbonyl is still present but has lost its terminal hydrogen, suggesting that somehow it has been deprotonated and is acting as a nucleophile. The problem is, aldehyde C- groups are T acidic. So, what is the cyanide doing to change this fact? Recall that cyano groups are strongly electron withdrawing. Moreover, we have seen (in Chapter 15) that this makes the C- bond adjacent to a C group quite acidic. We also know that cyanide reacts with aldehydes to make cyanohydrins. These facts provide the solution to the problem. C C Et C Et C C Et C C Et C You can see that the cyanide is important because it stabilizes the nucleophilic anion by resonance. This is why hydroxide etc. will not work in this reaction. Any other nucleophile that can add to a carbonyl AD can stabilize an adjacent anion by resonance would also work.

26 Answer to Challenge Spectroscopy Problem The mystery compound is trans-isoeugenol, shown at right. owever, from the spectra given it is actually very difficult to ascertain the positions of the and C 3 groups absolutely. This is because the 3 aromatic hydrogen signals all happen to overlap, so the pattern of a singlet and a pair of doublets that the structure would suggest is not visible. Thus, full marks will be given for isomers of trans-isoeugenol that differ in the positions of these groups. This structure is indicated by the following facts: 1) Solubility suggests a phenol (sol. in aqueous a but not in other aqueous solvents). 2) Decolourises bromine water suggests electron-rich aromatic ring (such as a phenol) and/or an alkene. 3) Blue colour with Fe 3 is a clear indication of a phenol. 4) Mass spectrum fragment at 149 indicates loss of a C 3 group from the parent ion at ) IR spectrum shows but C=. If you looked further you might have pointed out the presence of aromatic C=C stretches at 1600 and 1500 cm -1. 6) The 13 C MR shows 10 separate signals. nly two of these signals are above 60 ppm, and the signal at 55 ppm strongly suggests an -C 3 grouping. 7) The 1 MR shows 6 groups of signals in a ratio of 3:1:1:1:3:3. A 3 singlet at about 3.7 ppm definitely says C 3, while the 3 double doublet at about 1.9 ppm suggests another C 3 group, probably adjacent to an alkene, and coupled to two different hydrogens. There are only 3 aromatic protons, evidence for a trisubstituted ring. The 1-proton doublets of quartets are at the right chemical shift for vinylic hydrogens, and the splitting pattern suggests that they are coupled to each other (doublet) and to a C 3 group (quartet). The remaining 1 signal must be the phenolic. At this point it is possible to check for a formula. We have evidence for 10 carbons, at least 2 oxygens, and 12 hydrogens. This does in fact add to 164, so this is probably the formula. This formula has 5 degrees of unsaturation, consistent with having both an aromatic ring and an alkene. These deductions give us the following fragments: isoeugenol C Mol. Wt.: C 3 C3 with 2 other substituents Since the aromatic hydrogen signals don t offer much help, we really have to guess about the correct placement of the C 3 and alkenic groups.

27 Spectroscopy Crib Sheet for Introductory rganic Chemistry II 1 MR Typical Chemical Shift Ranges Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ) C C C C C C 2 C C C C C C C (solvent dependent) (solvent dependent) C Aryl C C Br C Aryl I C RC 2 Aromatic, heteroaromatic X C X =,, S, halide R 3 C Aliphatic, alicyclic Y =, R, S Y Y Y =, R, S Low Field δ igh Field 13 C MR Typical Chemical Shift Ranges Ketone, Aldehyde Alkene Carbox. Acid Amide Aryl Ester C x -Y Y =, CR 3 -C 2 -CR 3 C x -C= C 3 -CR IR Typical Functional Group Absorption Bands Group Frequency Frequency (cm -1 Intensity Group ) (cm -1 ) Intensity C Medium R Strong, broad C=C Medium C Strong C=C Medium C= Strong C C Strong R Medium, broad R C C R Medium (R R ) C 1230, 1030 Medium Aryl Medium C Medium Aryl C=C 1600, 1500 Strong R Strong δ