chemistry 2220 final exam

Name: Student No: 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)

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CHEM 2220 Final Exam 2013R 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) (2 Marks)

(d) (2 Marks)

(e) (2 Marks)

(f) (4 Marks)

(g) (2 Marks)


(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (4 Marks)

(l) (4 Marks)


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.


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.


4. (32 MARKS TOTAL) Mechanism grab-bag!

(a) (5 MARKS) 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.


(c) (5 MARKS) Amides can be converted to nitriles by heating in thionyl chloride (SOCl2). 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.


(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.


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 Diphenylacetone Benzil TPCD

MW (g/mol) 210 210 384

MP (oC) 30-34 95 220

BP (oC) 330 346 N/A

Solubilities (g/mL) Ethanol – 1 g/mL DCM – 0.5 g/L Hexanes – 1 g/mL

Ethanol – 1 g/mL DCM – 2 g/mL Hexanes – 0.2 g/L

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.


(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.


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.

(c) (1 MARK) What can you conclude from the number of 13C NMR signals?

(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


Spectra for Question 6

IR

1H NMR

13C NMR

NB: all signals are single lines.

m m s s s

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

1H NMR – Typical Chemical Shift Ranges

Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)

C CH3 0.7 – 1.3 C C H 2.5 – 3.1

C CH2 C 1.2 – 1.4

O

H

9.5 – 10.0

C H

C

C

C

1.4 – 1.7

O

OH

10.0 – 12.0

(solvent dependent)

C H

1.5 – 2.5 C OH

1.0 – 6.0

(solvent dependent)

O

H

2.1 – 2.6 CO H

3.3 – 4.0

Aryl C H 2.2 – 2.7 CCl H

3.0 – 4.0

H

4.5 – 6.5 CBr H

2.5 – 4.0

Aryl H 6.0 – 9.0 CI H

2.0 – 4.0

13C NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands

Group Frequency

(cm-1) Intensity Group

Frequency (cm-1)

Intensity

C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad

C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong

C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong

C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad

R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium

Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium

Aryl C=C 1600, 1500 Strong RNO2 1540 Strong

12 11 10 9 8 7 6 5 4 3 2 1 0

R3C–H

Aliphatic, alicyclic X–C–H

X = O, N, S, halide

Y

H H

Aromatic,

heteroaromatic

Y

H

RCO2H

Y = O, NR, S Y = O, NR, S

“Low Field” “High Field”

220 200 180 160 140 120 100 80 60 40 20 0

CH3-CR3

CHx-C=O

CR3-CH2-CR3

CHx-Y

Y = O, N Alkene

Aryl

Amide

Ester

Ketone, Aldehyde

Acid RCN

RCCR

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)

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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) (2 Marks)

(d) (2 Marks)

(e) (2 Marks)

(f) (4 Marks)


(g) (2 Marks)

(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (4 Marks)

(l) (4 Marks)


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.


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.


4. (32 MARKS TOTAL) Mechanism grab-bag!

(a) (5 MARKS) 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.


(c) (5 MARKS) Amides can be converted to nitriles by heating in thionyl chloride (SOCl2). 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.


(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.


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 Diphenylacetone Benzil TPCD

MW (g/mol) 210 210 384

MP (oC) 30-34 95 220

BP (oC) 330 346 N/A

Solubilities (g/mL) Ethanol – 1 g/mL DCM – 0.5 g/L Hexanes – 1 g/mL

Ethanol – 1 g/mL DCM – 2 g/mL Hexanes – 0.2 g/L

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!


(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.

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.

Atom economy, catalysis, prevention of waste


6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C8H14O2. The

IR, 13

C 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.

(c) (1 MARK) What can you conclude from the number of 13

C NMR signals?

(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

2

There is a carbonyl (IR ~1730 cm-1

) which is probably an ester (13

C NMR ~173 ppm and NO OH in

IR).

There is also an alkene (13

C NMR ~110 and ~148 ppm).

There are 7 13

C NMR signals but 8 carbons in the formula, therefore there is one pair of symmetry-

related carbons.

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.

Other relevant observations:

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

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

Multiplet 1H 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 CH3 groups with no nearest neighbors.

Note that this is consistent with 13

C NMR observation about a pair of symmetric groups.


Spectra for Question 6

IR

1H NMR

13C NMR

NB: all signals are single lines.

m m s s s

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

1H NMR – Typical Chemical Shift Ranges

Type of Proton Chemical Shift (δ) Type of Proton Chemical Shift (δ)

C CH3 0.7 – 1.3 C C H 2.5 – 3.1

C CH2 C 1.2 – 1.4

O

H

9.5 – 10.0

C H

C

C

C

1.4 – 1.7

O

OH

10.0 – 12.0

(solvent dependent)

C H

1.5 – 2.5 C OH

1.0 – 6.0

(solvent dependent)

O

H

2.1 – 2.6 CO H

3.3 – 4.0

Aryl C H 2.2 – 2.7 CCl H

3.0 – 4.0

H

4.5 – 6.5 CBr H

2.5 – 4.0

Aryl H 6.0 – 9.0 CI H

2.0 – 4.0

13C NMR – Typical Chemical Shift Ranges

IR – Typical Functional Group Absorption Bands

Group Frequency

(cm-1

) Intensity Group

Frequency (cm

-1)

Intensity

C–H 2960 – 2850 Medium RO–H 3650 – 3400 Strong, broad

C=C–H 3100 – 3020 Medium C–O 1150 – 1050 Strong

C=C 1680 – 1620 Medium C=O 1780 – 1640 Strong

C≡C–H 3350 – 3300 Strong R2N–H 3500 – 3300 Medium, broad

R–C≡C–R′ 2260 – 2100 Medium (R ≠ R′) C–N 1230, 1030 Medium

Aryl–H 3030 – 3000 Medium C≡N 2260 – 2210 Medium

Aryl C=C 1600, 1500 Strong RNO2 1540 Strong

12 11 10 9 8 7 6 5 4 3 2 1 0

R3C–H

Aliphatic, alicyclic X–C–H

X = O, N, S, halide

Y

H H

Aromatic,

heteroaromatic

Y

H

RCO2H

Y = O, NR, S Y = O, NR, S

“Low Field” “High Field”

220 200 180 160 140 120 100 80 60 40 20 0

CH3-CR3

CHx-C=O

CR3-CH2-CR3

CHx-Y

Y = O, N Alkene

Aryl

Amide

Ester

Ketone, Aldehyde

Acid RCN

RCCR

chemistry 2220 final exam

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