Chemistry Organic Chemistry II: Reactivity and Synthesis FINAL EXAM Winter 2003

Transcription

1 Name: Student No: Page 1 of 13 Chemistry rganic Chemistry II: Reactivity and Synthesis FINAL EXAM Winter 2003 Paper Number 546 Thursday April 24, :00 9:00 pm Frank Kennedy own Gym Students are permitted to bring into the exam room NE SEET 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 PART I are to be entered in the indicated spaces on the exam paper itself. An answer to NE of the Challenge Questions in PART II may be written in the exam booklet for EXTRA CREDIT. PART I: Question 1 (20 Marks) Question 2 Question 3 Question 4 Question 5 (10 Marks) (15 Marks) (15 Marks) (10 Marks) SUB-TTAL: PART II (EXTRA CREDIT) (70 Marks) (7 Marks) TTAL:

2 Page 2 of 13 PART I: D ALL QUESTINS There is choice in question 1 only. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions to correctly complete TEN of the following reactions. All reactions do in fact lead to products. Clearly indicate which TEN responses you want marked! (a) 1) 2, Al 3 C 2 Cl 2 2) Sn, Cl (b) LDA, TF, -78 o C then (c) (d) g(ac) 2 (cat.) 2 S 4 (aq.) (e)

3 Page 3 of 13 2, Al 3 C 2 Cl 2 (f) N LiAl 4, TF ( 3 + workup) (g) NaMe Me Me Me (h) Me reagent, solvent, workup (i), C 2 Cl 2 Me (j)

4 Page 4 of 13 (k) C 6 5 N 2, Ether N (l) (m) C (n) N 2 1) NaN2, Cl (aq.) 2) 2 S 4 (aq.) heat (o) LDA, TF -78 o C then heat ( 3 + workup)

5 Page 5 of (10 MARKS) Propose a synthetic route to prepare 5-methylheptane-4-one (A) starting from ethyl acetoacetate (B). You may use any additional organic starting materials having three or fewer carbons, as well as any reagents or solvents you require. This transformation can be accomplished in fewer than 5 steps. Remember that although a retrosynthetic analysis may be useful to you in solving this problem, the answer I want is the forward synthesis complete with reagents and products. Make starting from Et A B

6 Page 6 of (7 MARKS) Provide a detailed stepwise mechanism for the following transformation. NaC 3 C 3 +

7 Page 7 of 13 (b) (8 MARKS) The Wieland-Miescher Ketone is an important starting material in steroid synthesis. It can be prepared by the Robinson Annulation strategy shown here. Provide a detailed stepwise mechanism for this reaction. C 3 C 3 Na + 2, 95 o C

8 Page 8 of (a) (5 MARKS) Compound A below is suggested to be a trace impurity formed when 4,4'- di-(tert-butyl)biphenyl is prepared from biphenyl and t-butylchloride in C 2 Cl 2, using FeCl 3 as a catalyst. Explain how A could be formed under these conditions. + Cl FeCl 3 major product C 2 Cl 2 + C A - trace impurity (b) (5 MARKS) The aldehydes A and B below both undergo base-catalyzed condensation reactions with acetone (C). The reaction of B under these conditions is much faster than that of A. iefly explain why this is so, taking into account electronic effects. 3 C 2 N 3 C C 3 A B C

9 Page 9 of 13 (c) (5 MARKS) Draw and label a diagram showing a reflux apparatus set-up. Give 2 advantages of using reflux conditions in a chemical reaction.

10 Page 10 of (10 MARKS) The IR, 13 C NMR and 1 NMR spectra of an organic compound having the molecular formula C 7 15 N are shown below. NB: In the 1 NMR spectrum, m=multiplet, d=doublet, tr=triplet, q=quartet m m d q tr Draw the structure of the compound in the box below.

11 Page 11 of 13 PART II: CALLENGE PRBLEM (EXTRA CREDIT). D NE PRBLEM NLY. Part II is worth up to XX additional marks. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Insert the booklet into Part I when you hand it in. A. SYNTESIS There is something seriously wrong with the following proposed synthesis of the drug terfenadine. Identify the problem, and propose specific steps needed to make this route workable. N Mg Ether N terfenadine B. MECANISM When carvone (A) is boiled in acidic water, it is converted to carvacrol (B). Give a mechanism to explain this transformation o C A B C. SPECTRSCPY The unusual compound diketene has the molecular formula C Its 1 and 13 C NMR spectra are shown on the next page. eating diketene in the presence of cyclopentadiene leads to the formation of bicyclo[2.2.1]hept-5-ene-2-one. What is the structure of diketene? Diketene + heat

12 PPM PPM PPM Page 12 of 13 Diketene 13 C NMR Four signals, δ , , and ppm. PPM Diketene 1 NMR Three signals, δ 4.877, and ppm. Double doublet Double triplet Double triplet PPM

13 Spectroscopy Crib Sheet for Introductory rganic Chemistry II 1 NMR 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 Cl C C Aryl I C RC 2 Aromatic, heteroaromatic X C X =, N, S, halide R 3 C Aliphatic, alicyclic Y =, NR, S Y Y Y =, NR, S Low Field δ igh Field 13 C NMR Typical Chemical Shift Ranges Ketone, Aldehyde Alkene Carbox. Acid Amide Aryl Ester C x -Y Y =, N 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 2 N Medium, broad R C C R Medium (R R ) C N 1230, 1030 Medium Aryl Medium C N Medium Aryl C=C 1600, 1500 Strong RN Strong δ

14 ANSWER KEY Page 1 of 15 Chemistry rganic Chemistry II: Reactivity and Synthesis FINAL EXAM Winter 2003 Paper Number 546 Thursday April 24, :00 9:00 pm Frank Kennedy own Gym Students are permitted to bring into the exam room NE SEET 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 PART I are to be entered in the indicated spaces on the exam paper itself. An answer to NE of the Challenge Questions in PART II may be written in the exam booklet for EXTRA CREDIT. PART I: Question 1 (20 Marks) Question 2 Question 3 Question 4 Question 5 (10 Marks) (15 Marks) (15 Marks) (10 Marks) SUB-TTAL: PART II (EXTRA CREDIT) (70 Marks) (7 Marks) TTAL:

15 Page 2 of 15 PART I: D ALL QUESTINS There is choice in question 1 only. 1. (20 MARKS) Provide the missing product, starting compound or reagent/solvent/conditions to correctly complete TEN of the following reactions. All reactions do in fact lead to products. Clearly indicate which TEN responses you want marked! (a) 1) 2, Al 3 C 2 Cl 2 2) Sn, Cl C 2 I 2, Zn, C 2 Cl 2 (I-C 2 -Zn-I) (b) LDA, TF, -78 o C then (c) mcpba, C 2 Cl 2 (d) or any other specific peroxyacid reagent g(ac) 2 (cat.) 2 S 4 (aq.) (e)

16 Page 3 of 15 2, Al 3 C 2 Cl 2 (f) ortho isomer is also acceptable, although this would be the minor product N LiAl 4, TF ( 3 + workup) N (g) Me Me NaMe Me Me (h) 2 PhMg, Ether w/u with aq. N 4 Cl Me reagent, solvent, workup (i), C 2 Cl 2 Me Me (j)

17 Page 4 of 15 (C 3 C 2 ) 2 CuLi TF, -78 o C (acid w/u) (k) You could also use C 3 C 2 Mg/CuI (cat.)/ether/acid workup Cl or C 6 5 N 2, Ether N (l) Na 2 (m) C N 2 1) NaN 2, Cl (aq.) 2) 2 S 4 (aq.) heat (n) LDA, TF -78 o C then heat ( 3 + workup) (o)

18 Page 5 of (10 MARKS) Propose a synthetic route to prepare 5-methylheptane-4-one (A) starting from ethyl acetoacetate (B). You may use any additional organic starting materials having three or fewer carbons, as well as any reagents or solvents you require. This transformation can be accomplished in fewer than 5 steps. Remember that although a retrosynthetic analysis may be useful to you in solving this problem, the answer I want is the forward synthesis complete with reagents and products. Make starting from Et A B NaEt NaEt Et Et C 3 C 2 Et Et C 3 Et NaEt Et C 3 C 2 or LDA, TF -78 o C C 3 C 2 1) Na (aq) 2) 3 + or 3 +, heat Et ( and C 2 and Et) ( The Acetoacetic Ester synthesis of ketones)

19 Page 6 of (7 MARKS) Provide a detailed stepwise mechanism for the following transformation. NaC 3 C Me - Me Me Me Me Me Me - - Me

20 Page 7 of 15 (b) (8 MARKS) The Wieland-Miescher Ketone is an important starting material in steroid synthesis. It can be prepared by the Robinson Annulation strategy shown here. Provide a detailed stepwise mechanism for this reaction. C 3 C 3 Na + 2, 95 o C 3 C - - C C 3 C 3 C

21 Page 8 of (a) (5 MARKS) Compound A below is suggested to be a trace impurity formed when 4,4'- di-(tert-butyl)biphenyl is prepared from biphenyl and t-butylchloride in C 2 Cl 2, using FeCl 3 as a catalyst. Explain how A could be formed under these conditions. + Cl FeCl 3 major product C 2 Cl 2 + C A - trace impurity The C 2 grouping comes from the solvent, C 2 Cl 2. Dichloromethane is not very prone to reaction of this kind, but it may undergo reaction with Lewis acids to a slight extent, especially if the reaction is left for an extended period of time. This is probably what happened here. C Cl Cl FeCl 3 δ Cl FeCl δ 3 Cl C C 2 FeCl 4 Cl Cl C 2 Cl 2 C and then do the Friedel-Crafts alkylation over again with the other phenyl group to form the product 2 C Cl (b) (5 MARKS) The aldehydes A and B below both undergo base-catalyzed condensation reactions with acetone (C). The reaction of B under these conditions is much faster than that of A. iefly explain why this is so, taking into account electronic effects. 3 C 2 N 3 C C 3 A B C These condensation reactions are Aldol processes, in which the enolate of acetone attacks the aldehyde. The rates of the two processes will depend on the degree of electrophilicity of the two aldehydes. Since B is substituted with a strong electron-withdrawing group, whereas A carries a strong electron-donating group, the carbonyl group of B will have a greater degree of partial positive charge. Thus, reaction of B will be faster. 2 N 2 N 2 N These resonance forms show that the nitro group can directly withdraw electron density from the carbonyl. 2 N 2 N

22 Page 9 of 15 (c) (5 MARKS) Draw and label a diagram showing a reflux apparatus set-up. Give 2 advantages of using reflux conditions in a chemical reaction. Top of condenser is PEN! Water hose UT Condenser Water hose IN Clamp attached to stand Flask Liquid Boiling chips eat source Advantages: 1) Reflux apparatus keeps solvent (and possibly also reagents and products) from boiling away by recondensing it (them) and returning the condensate to the flask. 2) Refluxing maintains a steady reaction temperature, which is the boiling temperature of the solvent used.

23 Page 10 of (10 MARKS) The IR, 13 C NMR and 1 NMR spectra of an organic compound having the molecular formula C 7 15 N are shown below. NB: In the 1 NMR spectrum, m=multiplet, d=doublet, tr=triplet, q=quartet m m d q tr Draw the structure of the compound in the box below. 3 C 3 C N C 2

24 Page 11 of 15 PART II: CALLENGE PRBLEM (EXTRA CREDIT). D NE PRBLEM NLY. Part II is worth up to 7 additional marks. Write your answer in the exam booklet provided. Be sure your name and student number are on the booklet. Insert the booklet into Part I when you hand it in. A. SYNTESIS There is something seriously wrong with the following proposed synthesis of the drug terfenadine. Identify the problem, and propose specific steps needed to make this route workable. N Mg Ether terfenadine Answer: You cannot form a Grignard reagent from a halide that also contains an alcohol group. The solution to the problem is to protect the alcohol with a group that is unaffected by Grignard formation or reaction. This group must be removed from the final product. Protection/deprotection is not necessarily straightforward, however. A benzyl ether could be easily installed in the starting material, which would permit Grignard formation. This answer (with appropriate reagents specified) would be acceptable for 6 out of 7 marks. (NB: making an ester would NT be suitable, because you can t make a Grignard that contains a carbonyl group either). N N N Na TF group incompatible with Grignard formation Benzyl ether protects, will not interfere with Grignard formation owever, in fact the removal of the benzyl group from the final product would likely damage the product. ydrogenation will cleave benzylic C- bonds, but there are actually two such groups in this molecule. We do not want to remove the existing hydroxyl group! N 2 (g) Pd cat. Me N You can also cleave benzylic ethers using very strong acids, particularly. owever, this would certainly lead to elimination reactions. N N Either or both of the hydroxyl groups would eliminate The 7 th bonus mark would be given for commenting on the problem of deprotection. I don t require you to have a solution to this problem. You could get around the difficulty by completely re-designing the synthesis. That is actually not what the problem asked for, but I will evaluate any proposed alternates on their merits.

25 Page 12 of 15 B. MECANISM When carvone (A) is boiled in acidic water, it is converted to carvacrol (B). Give a mechanism to explain this transformation o C A B There are actually a couple of ways you could have written this process. It requires only a series of protonation/deprotonation steps to move the double bonds around. ( + + ) ( - + ) (1,2-shift) ( + + ) + ( - + ) The 1,2-shift step could also be written as an elimination/re-protonation sequence. + +

26 Page 13 of 15 C. SPECTRSCPY The unusual compound diketene has the molecular formula C Its 1 and 13 C NMR spectra are shown on the next page. eating diketene in the presence of cyclopentadiene leads to the formation of bicyclo[2.2.1]hept-5-ene-2-one. What is the structure of diketene? Diketene + heat Diketene has the structure: The two vinylic hydrogens are double triplets. This is because each is distinct, splitting the other into a doublet, which is then split by the two other hydrogens (which are both equivalent). The C 2 group is a double doublet because the signals are split by the two different vinylic hydrogens. The 13 C NMR is harder to interpret. The signal at 165 ppm is the carbonyl carbon, and the one at 42 ppm is the C 2 group. The alkene portion of the structure is a bit strange. The carbon in the ring is at 147 ppm, but the terminal carbon is upfield at 87 ppm. This spectrum really just told you that all 4 carbons were different, and that one was probably a carbonyl As its name implies, diketene is a dimer of the molecule ketene, and they are interconverted by an electrocyclic reaction. It is the monomer that undergoes a Diels-Alder reaction with cyclopentadiene. You could see that if you worked backwards from the product in the reaction shown. 2 C

27 PPM PPM PPM Page 14 of 15 Diketene 13 C NMR Four signals, δ , , and ppm. PPM Diketene 1 NMR Three signals, δ 4.877, and ppm. Double doublet Double triplet Double triplet PPM

28 Spectroscopy Crib Sheet for Introductory rganic Chemistry II 1 NMR 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 Cl C C Aryl I C RC 2 Aromatic, heteroaromatic X C X =, N, S, halide R 3 C Aliphatic, alicyclic Y =, NR, S Y Y Y =, NR, S Low Field δ igh Field 13 C NMR Typical Chemical Shift Ranges Ketone, Aldehyde Alkene Carbox. Acid Amide Aryl Ester C x -Y Y =, N 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 2 N Medium, broad R C C R Medium (R R ) C N 1230, 1030 Medium Aryl Medium C N Medium Aryl C=C 1600, 1500 Strong RN Strong δ