Chemistry 2220 Final Exam [PDF]

Apr 13, 2013 - tetraphenylcyclopentadienone (TPCD) from materials available in the organic chemistry lab to use as his f

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Name:

Student No:

Page 1 of 13

CHEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. Hultin, Dr. H. Luong

FINAL EXAM – Winter Session 2013R Saturday April 13, 2013 1:30 pm – 4:30 pm Frank Kennedy Gold Gym

Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper with any HANDWRITTEN notes they wish (both sides). Molecular model kits and calculators (no text or graphics memory!) are permitted but no other aids may be used.

Question 1 – Reactions and Products

(30 Marks)

Question 2 – Synthesis Road-Map

(10 Marks)

Question 3 – Mechanism

(12 Marks)

Question 4 – Mechanism grab-bag

(32 Marks)

Question 5 – Laboratory

(10 Marks)

Question 6 – Spectroscopy

(6 Marks)

TOTAL:

(100 Marks)

CHEM 2220 Final Exam 2013R

Page 2 of 13

1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction conditions to correctly complete the following reactions. Show stereochemistry when necessary.

(a)

(2 Marks)

(b)

(2 Marks)

(c)

(d)

(e)

(2 Marks)

(2 Marks)

(2 Marks)

(f)

(g)

(4 Marks)

(2 Marks)

CHEM 2220 Final Exam 2013R

(h)

(i)

(j)

(k)

(l)

Page 3 of 13

(2 Marks)

(2 Marks)

(2 Marks)

(4 Marks)

(4 Marks)

CHEM 2220 Final Exam 2013R

Page 4 of 13

2. (10 MARKS) The polyether antibiotic monensin is produced by a particular strain of Streptomyces. Its structure was determined in 1967, but the molecule was not successfully synthesized until the work of Y. Kishi and his co-workers in 1979. The following sequence depicts the first steps in this successful effort. Fill in the missing information in the boxes provided. Note: for this problem, R/S stereochemistry has been removed but in fact the stereochemistry of monensin was a vital consideration in the actual synthesis.

CHEM 2220 Final Exam 2013R

Page 5 of 13

3. (12 MARKS) Mechanism. Nitrogen-containing rings are common structures in pharmaceuticals. One convenient way to prepare 4-piperidinones is shown below. Fill in the structures of the missing intermediates, and provide a detailed stepwise mechanism for each step.

CHEM 2220 Final Exam 2013R 4. (32 MARKS TOTAL) (a) (5 MARKS)

Page 6 of 13

Mechanism grab-bag!

Give a mechanism for the formation of the cyclobutanone in the following reaction.

(b) (5 MARKS) Acrolein (propenal) is made commercially from glycerol by heating in the presence of sulfuric acid. Propose a stepwise mechanism to explain this reaction.

CHEM 2220 Final Exam 2013R

Page 7 of 13

(c) (5 MARKS) Amides can be converted to nitriles by heating in thionyl chloride (SOCl 2). Propose a stepwise mechanism for the formation of acetonitrile from acetamide.

(d) (5 MARKS) The Vilsmeier Reagent offers an alternative method for forming acid chlorides from carboxylic acids. Suggest a stepwise mechanism for this transformation.

CHEM 2220 Final Exam 2013R

Page 8 of 13

(e) (6 MARKS) The amino group in α-amino acids must be protected at certain stages of the synthesis of peptides. The protecting group must be removed at other points, however, and the conditions for removal must avoid extremes of acid or base. The so-called Alloc derivatives can be removed by warming with diiodine in a mixture of acetonitrile and water as shown below. Provide a stepwise mechanism to explain how this deprotection occurs.

(f) (6 MARKS) Here is another way to make piperidine rings, starting from a β-lactam such as 1. The intermediate 2 from the first step is isolated before being submitted to the second step. Draw the structure of 2 and provide detailed stepwise mechanisms for both steps in this process.

CHEM 2220 Final Exam 2013R

Page 9 of 13

5. Lab Questions (10 MARKS Total) Local physicist Dr. Sheldon Copper has decided to dress up as one of the Twilight characters for Comic Con 2013. Of course the outfit is not complete unless there’s some fake blood. The fake blood at the party store was pretty expensive, so to save money Sheldon decided he could make tetraphenylcyclopentadienone (TPCD) from materials available in the organic chemistry lab to use as his fake blood. This compound is a deep red colour because of its extended -conjugation. TCPD can be synthesized by aldol condensation using diphenylacetone, benzil and sodium hydroxide, as shown below. Sheldon has written out the following lab procedure intended to make 4 g of TPCD. Your instincts developed from working in the CHEM 2220 laboratory tell you that although the reaction is possible, Sheldon’s procedure will not work and modifications are necessary. Note: The story, data and procedures in this question are fictitious. None of this should be used for experimental purposes.

Data MW (g/mol) MP (oC) BP (oC) Solubilities (g/mL)

Diphenylacetone 210 30-34 330 Ethanol – 1 g/mL DCM – 0.5 g/L Hexanes – 1 g/mL

Benzil 210 95 346 Ethanol – 1 g/mL DCM – 2 g/mL Hexanes – 0.2 g/L

TPCD 384 220 N/A Ethanol – 1 g/mL DCM – 0.1 g/L Hexanes – 2 g/mL

Dissolve 50 g of diphenylacetone in 50 mL hexanes and sodium hydroxide (1 kg). Add 40 g of benzil. Reflux the reaction for the length of time equivalent to one TV episode of Big Bang Theory (about 30 minutes) and the product might have precipitated out. Filter off the product and recrystallize the crude material with ethanol and dichloromethane. (a) (7 Marks) Critique Sheldon’s procedure and indicate how you would modify the procedure to make it work better so that approximately 4 grams of product can be obtained. You will need a slight excess of one of the reactants to push the equilibrium to favor the product. The goal is to obtain TPCD in the most pure form possible. You are free to use any reagents and chemicals you wish. Assume that your proposed reaction will produce a 50% yield and that there is residual starting material (the one used in excess). State all assumptions. Of course there is some math involved but we have chosen numbers that can be easily approximated.

CHEM 2220 Final Exam 2013R

Page 10 of 13

(b) (1.5 Marks) Sheldon performed a TLC of his crude product (shown at right). Unfortunately he did not label his origins for starting material and product. Can you, once again, help Sheldon figure out which is likely the product and which is the starting material (benzil). Please justify your reasoning.

(c) (1.5 Marks) Let’s face it: Sheldon isn’t the ideal candidate to work in the synthetic chemistry lab. As the dear friend that you are, you suggested that he consider being an advocate for green chemistry rather than working in the lab. Using your proposed reaction and comparing it to Sheldon’s, name at least three principles of green chemistry that are observed.

CHEM 2220 Final Exam 2013R

Page 11 of 13

6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C8H14O2. The IR, 13C NMR and 1H NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (0.5 MARK)

What is the unsaturation number for this compound?

(b) (1.5 MARK) What functional group(s) does this compound contain? Indicate the specific evidence for your conclusion.

13C

(c) (1 MARK)

What can you conclude from the number of

(d) (1 MARK)

What can you conclude from the 1H NMR signal at 2.3 ppm?

(e) (2 MARKS)

Draw the structure of this compound in the box below. Structure for C8H14O2

NMR signals?

CHEM 2220 Final Exam 2013R

Page 12 of 13

Spectra for Question 6

IR

13C

NMR NB: all signals are single lines.

1H

NMR

m

m

s

s

s

Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II 1H

NMR – Typical Chemical Shift Ranges Chemical Shift (δ)

Type of Proton C

0.7 – 1.3

CH3

Chemical Shift (δ)

Type of Proton C

C

2.5 – 3.1

H

O

C

CH2

1.2 – 1.4

C

9.5 – 10.0 H

C C

O

C

10.0 – 12.0 (solvent dependent)

1.4 – 1.7

H

OH

C C

H

1.5 – 2.5

C

OH

1.0 – 6.0 (solvent dependent)

O H

Aryl

C

H

2.1 – 2.6

O

C

H

3.3 – 4.0

2.2 – 2.7

Cl

C

H

3.0 – 4.0

4.5 – 6.5

Br

C

H

2.5 – 4.0

6.0 – 9.0

I

H

Aryl

H

Aromatic, heteroaromatic

RCO2H

Y = O, NR, S

11

2.0 – 4.0

H

X–C–H X = O, N, S, halide

Y

10

H

9

8

7

6

5

4

3

2



160

CR3-CH2-CR3 CHx-C=O RCCR CH3-CR3 RCN 140

120

100

80

60

40

 13C

NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands Group C–H C=C–H C=C C≡C–H R–C≡C–R′ Aryl–H Aryl C=C

Frequency (cm-1) 2960 – 2850 3100 – 3020 1680 – 1620 3350 – 3300 2260 – 2100 3030 – 3000 1600, 1500

0

CHx-Y Y = O, N

Aryl Ketone, Aldehyde Ester Amide Acid 180

1

“High Field”

Alkene

200

Y = O, NR, S

H

“Low Field”

220

R3C–H Aliphatic, alicyclic

Y H

12

C

Intensity Medium Medium Medium Strong Medium (R ≠ R′) Medium Strong

Group RO–H C–O C=O R2N–H C–N C≡N RNO2

Frequency (cm-1) 3650 – 3400 1150 – 1050 1780 – 1640 3500 – 3300 1230, 1030 2260 – 2210 1540

Intensity Strong, broad Strong Strong Medium, broad Medium Medium Strong

20

0

ANSWER KEY

Page 1 of 13

CHEM 2220 Organic Chemistry II: Reactivity and Synthesis Prof. P.G. Hultin, Dr. H. Luong

FINAL EXAM – Winter Session 2013R Saturday April 13, 2013 1:30 pm – 4:30 pm Frank Kennedy Gold Gym

Students are permitted to bring into the exam room ONE SHEET of 8½ x 11 paper with any HANDWRITTEN notes they wish (both sides). Molecular model kits and calculators (no text or graphics memory!) are permitted but no other aids may be used.

Question 1 – Reactions and Products

(30 Marks)

Question 2 – Synthesis Road-Map

(10 Marks)

Question 3 – Mechanism

(12 Marks)

Question 4 – Mechanism grab-bag

(32 Marks)

Question 5 – Laboratory

(10 Marks)

Question 6 – Spectroscopy

(6 Marks)

TOTAL:

(100 Marks)

CHEM 2220 Final Exam Answers 2013R

Page 2 of 13

1. (30 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction conditions to correctly complete the following reactions. Show stereochemistry when necessary.

(a)

(2 Marks)

(b)

(2 Marks)

(c)

(d)

(e)

(f)

(2 Marks)

(2 Marks)

(2 Marks)

(4 Marks)

CHEM 2220 Final Exam Answers 2013R

Page 3 of 13

(g)

(2 Marks)

(h)

(i)

(j)

(k)

(l)

(2 Marks)

(2 Marks)

(2 Marks)

(4 Marks)

(4 Marks)

CHEM 2220 Final Exam Answers 2013R

Page 4 of 13

2. (10 MARKS) The polyether antibiotic monensin is produced by a particular strain of Streptomyces. Its structure was determined in 1967, but the molecule was not successfully synthesized until the work of Y. Kishi and his co-workers in 1979. The following sequence depicts the first steps in this successful effort. Fill in the missing information in the boxes provided. Note: for this problem, R/S stereochemistry has been removed but in fact the stereochemistry of monensin was a vital consideration in the actual synthesis.

CHEM 2220 Final Exam Answers 2013R

Page 5 of 13

3. (12 MARKS) Mechanism. Nitrogen-containing rings are common structures in pharmaceuticals. One convenient way to prepare 4-piperidinones is shown below. Fill in the structures of the missing intermediates, and provide a detailed stepwise mechanism for each step.

CHEM 2220 Final Exam Answers 2013R 4. (32 MARKS TOTAL) (a) (5 MARKS)

Page 6 of 13

Mechanism grab-bag!

Give a mechanism for the formation of the cyclobutanone in the following reaction.

(b) (5 MARKS) Acrolein (propenal) is made commercially from glycerol by heating in the presence of sulfuric acid. Propose a stepwise mechanism to explain this reaction.

CHEM 2220 Final Exam Answers 2013R

Page 7 of 13

(c) (5 MARKS) Amides can be converted to nitriles by heating in thionyl chloride (SOCl 2). Propose a stepwise mechanism for the formation of acetonitrile from acetamide.

(d) (5 MARKS) The Vilsmeier Reagent offers an alternative method for forming acid chlorides from carboxylic acids. Suggest a stepwise mechanism for this transformation.

CHEM 2220 Final Exam Answers 2013R

Page 8 of 13

(e) (6 MARKS) The amino group in α-amino acids must be protected at certain stages of the synthesis of peptides. The protecting group must be removed at other points, however, and the conditions for removal must avoid extremes of acid or base. The so-called Alloc derivatives can be removed by warming with diiodine in a mixture of acetonitrile and water as shown below. Provide a stepwise mechanism to explain how this deprotection occurs.

(f) (6 MARKS) Here is another way to make piperidine rings, starting from a β-lactam such as 1. The intermediate 2 from the first step is isolated before being submitted to the second step. Draw the structure of 2 and provide detailed stepwise mechanisms for both steps in this process.

CHEM 2220 Final Exam Answers 2013R

Page 9 of 13

5. Lab Questions (10 MARKS Total) Local physicist Dr. Sheldon Copper has decided to dress up as one of the Twilight characters for Comic Con 2013. Of course the outfit is not complete unless there’s some fake blood. The fake blood at the party store was pretty expensive, so to save money Sheldon decided he could make tetraphenylcyclopentadienone (TPCD) from materials available in the organic chemistry lab to use as his fake blood. This compound is a deep red colour because of its extended -conjugation. TCPD can be synthesized by aldol condensation using diphenylacetone, benzil and sodium hydroxide, as shown below. Sheldon has written out the following lab procedure intended to make 4 g of TPCD. Your instincts developed from working in the CHEM 2220 laboratory tell you that although the reaction is possible, Sheldon’s procedure will not work and modifications are necessary. Note: The story, data and procedures in this question are fictitious. None of this should be used for experimental purposes.

Data MW (g/mol) o MP ( C) o BP ( C) Solubilities (g/mL)

Diphenylacetone 210 30-34 330 Ethanol – 1 g/mL DCM – 0.5 g/L Hexanes – 1 g/mL

Benzil 210 95 346 Ethanol – 1 g/mL DCM – 2 g/mL Hexanes – 0.2 g/L

TPCD 384 220 N/A Ethanol – 1 g/mL DCM – 0.1 g/L Hexanes – 2 g/mL

Dissolve 50 g of diphenylacetone in 50 mL hexanes and sodium hydroxide (1 kg). Add 40 g of benzil. Reflux the reaction for the length of time equivalent to one TV episode of Big Bang Theory (about 30 minutes) and the product might have precipitated out. Filter off the product and recrystallize the crude material with ethanol and dichloromethane. (a) (7 Marks) Critique Sheldon’s procedure and indicate how you would modify the procedure to make it work better so that approximately 4 grams of product can be obtained. You will need a slight excess of one of the reactants to push the equilibrium to favor the product. The goal is to obtain TPCD in the most pure form possible. You are free to use any reagents and chemicals you wish. Assume that your proposed reaction will produce a 50% yield and that there is residual starting material (the one used in excess). State all assumptions. Of course there is some math involved but we have chosen numbers that can be easily approximated. Critique: Despite Dr. Copper’s intelligence, there are a number of issues with his proposed synthesis: 1. Amounts – To synthesize 4 g (about 0.01 mol) of TPCD, Sheldon would require a minimum of 4 g of each of benzil and diphenylacetone (keeping in mind the 50% yield). Choose benzil to be in excess since it should be easier to remove at the end. Sodium hydroxide is a catalyst and therefore should be used in much smaller amounts. 2. Solvent – The choice of solvent is critical to help the reaction proceed AND to help with the workup. Sheldon chose hexanes as the solvent and this is inappropriate for two reasons: 1) sodium hydroxide has limited solubility in hexanes and 2) benzil is insoluble in hexanes. Out of the three solvents listed, the most appropriate is ethanol (at a minimum volume of 5 mL, adding more as necessary) since it can dissolve all start materials and products. Dichloromethane unfortunately cannot dissolve diphenylacetone and will likely not be able to dissolve sodium hydroxide either. 3. Reaction time – It is unclear from the way the procedure is written whether there is certainty in the yielding of product after refluxing for 30 minutes. The best thing to do is to monitor the reaction by TLC (co-spot with starting materials and Sheldon should watch for the disappearance of diphenylacetone). 4. Work up – Notice that the solvent chosen for the synthesis dissolves all starting material and product. As well, in the monitoring of the reaction, all the diphenylacetone must be consumed to allow for an ease of workup. The next step is to remove the solvent (and water byproduct) by distillation (to concentrate the reaction down). Once the solvent has been removed, DCM is added to the reaction to precipitate out the crude product. 5. Purification – no issues here!

CHEM 2220 Final Exam Answers 2013R

Page 10 of 13

(b) (1.5 Marks) Sheldon performed a TLC of his crude product (shown at right). Unfortunately he did not label his origins for starting material and product. Can you, once again, help Sheldon figure out which is likely the product and which is the starting material (benzil). Please justify your reasoning.

The spot on the left is more polar than the spot on the right. Therefore, it’s likely that the spot on the left is benzil while the spot on the right is TCPD.

(c) (1.5 Marks) Let’s face it: Sheldon isn’t the ideal candidate to work in the synthetic chemistry lab. As the dear friend that you are, you suggested that he consider being an advocate for green chemistry rather than working in the lab. Using your proposed reaction and comparing it to Sheldon’s, name at least three principles of green chemistry that are observed.

Atom economy, catalysis, prevention of waste

CHEM 2220 Final Exam Answers 2013R

Page 11 of 13

6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C8H14O2. The 13 1 IR, C NMR and H NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (0.5 MARK)

What is the unsaturation number for this compound? 2

(b) (1.5 MARK) What functional group(s) does this compound contain? Indicate the specific evidence for your conclusion. -1

13

There is a carbonyl (IR ~1730 cm ) which is probably an ester ( C NMR ~173 ppm and NO OH in IR). 13

There is also an alkene ( C NMR ~110 and ~148 ppm).

(c) (1 MARK)

What can you conclude from the number of

13

C NMR signals?

13

There are 7 C NMR signals but 8 carbons in the formula, therefore there is one pair of symmetryrelated carbons.

(d) (1 MARK)

1

What can you conclude from the H NMR signal at 2.3 ppm?

This is a 2-proton singlet. The chemical shift suggests it might be next to a double bond. It could be a CH2C=O structure with no neighboring protons.

(e) (2 MARKS)

Draw the structure of this compound in the box below. Structure for C8H14O2

Other relevant observations: 1

H NMR Integral ratio: 1:2:3:2:6. 1

3-Proton H Singlet at ~3.7 ppm is OCH3, from the ester. 1

Multiplet H NMR signals between 4.9 and 5.9 ppm are from alkene. There are three protons in this region, so the alkene is monosubstituted R-CH=CH2. 6-Proton singlet at ~1 ppm is probably 2 identical CH 3 groups with no nearest neighbors. 13 Note that this is consistent with C NMR observation about a pair of symmetric groups.

CHEM 2220 Final Exam Answers 2013R

Page 12 of 13

Spectra for Question 6

IR

13

C NMR NB: all signals are single lines.

1

H NMR

m

m

s

s

s

Spectroscopy “Crib Sheet” for CHEM 2220 – Introductory Organic Chemistry II 1

H NMR – Typical Chemical Shift Ranges Chemical Shift (δ)

Type of Proton C

0.7 – 1.3

CH3

Chemical Shift (δ)

Type of Proton C

C

2.5 – 3.1

H

O

C

CH2

1.2 – 1.4

C

9.5 – 10.0 H

C C

O

C

10.0 – 12.0 (solvent dependent)

1.4 – 1.7

H

OH

C C

H

1.5 – 2.5

C

OH

1.0 – 6.0 (solvent dependent)

O H

Aryl

C

H

2.1 – 2.6

O

C

H

3.3 – 4.0

2.2 – 2.7

Cl

C

H

3.0 – 4.0

4.5 – 6.5

Br

C

H

2.5 – 4.0

6.0 – 9.0

I

H

Aryl

H

Aromatic, heteroaromatic

RCO2H

Y = O, NR, S

11

2.0 – 4.0

H

X–C–H X = O, N, S, halide

Y

10

H

9

8

7

6

5

4

3

2



160

CR3-CH2-CR3 CHx-C=O RCCR CH3-CR3 RCN 140

120

100

80

60

40

 13

C NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands Group C–H C=C–H C=C C≡C–H R–C≡C–R′ Aryl–H Aryl C=C

Frequency -1 (cm ) 2960 – 2850 3100 – 3020 1680 – 1620 3350 – 3300 2260 – 2100 3030 – 3000 1600, 1500

0

CHx-Y Y = O, N

Aryl Ketone, Aldehyde Ester Amide Acid 180

1

“High Field”

Alkene

200

Y = O, NR, S

H

“Low Field”

220

R3C–H Aliphatic, alicyclic

Y H

12

C

Intensity Medium Medium Medium Strong Medium (R ≠ R′) Medium Strong

Group RO–H C–O C=O R2N–H C–N C≡N RNO2

Frequency -1 (cm ) 3650 – 3400 1150 – 1050 1780 – 1640 3500 – 3300 1230, 1030 2260 – 2210 1540

Intensity Strong, broad Strong Strong Medium, broad Medium Medium Strong

20

0

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