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Name: Student No: of 14

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

FINAL EXAM – Winter Session 2016R

Wednesday April 13, 2016 9:00 am – 12:00 Frank Kennedy Gold Gym

Students are permitted to bring into the exam room ONE SHEET of LETTER-SIZE (8½ x 11 inches) 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 (32 Marks)

Question 2 – Synthesis (10 Marks)

Question 3 – Mechanism (12 Marks)

Question 4 – Mechanism pot-pourri (30 Marks)

Question 5 – Laboratory (10 Marks)

Question 6 – Spectroscopy (6 Marks)

TOTAL: (100 Marks)

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NAME: CHEM 2220 Final 2016R of 14

1. (32 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction

conditions to correctly complete the following reactions. Show stereochemistry when necessary. If a

compound is racemic, indicate this by writing “racemic” or “+/-“. Simple aqueous acid or base workups can be

assumed but any “special” workup conditions should be specified.

(a) (2 Marks)

(b) (2 Marks)

(c) (2 Marks)

(d) (2 Marks)

(e) (4 Marks)


(f) (2 Marks)

(g) (2 Marks)

(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (2 Marks)


(l) (4 Marks)

(m) (4 Marks)


2. (10 MARKS) Propose a practical synthetic route to prepare the ester shown. All carbon atoms in your

product must come from either ethyl alcohol or ethyl acetate. You may use any reagents or solvents you wish.

The entire sequence can be accomplished in 5 or 6 steps.

Although a retrosynthetic analysis will be useful, your answer must show the specific reactions needed, with

reagents and conditions, in the forward synthetic direction.


3. (12 MARKS TOTAL) Carmegliptin is a drug used to treat Type 2 diabetes. The reaction shown below is part of

an industrial-scale synthesis of carmegliptin. Everything happens in one reaction pot but the second set of

reagents is only added once all the starting material is converted to the intermediate.

(a) (8 Marks) Write a stepwise mechanism for the first part of the process leading to the intermediate (for

6 Marks). Your mechanism should allow you to determine the structure for the intermediate (for 2 marks).

NB: the intermediate is a neutral organic compound!

(b) (4 Marks) Write a stepwise mechanism for the second part of the process, in which the intermediate

reacts with ammonia in the presence of acetic acid to form the final product.


5. (30 MARKS TOTAL) Mechanism pot-pourri! Assorted shorter mechanisms from across the course.

(a) (8 Marks) When the diketone shown is treated with base in ethanol solution, it undergoes a series of

reactions to form an interesting polycyclic product. Provide a stepwise mechanism to explain this process.


(b) (6 Marks) Heating anthranilic acid in acetic anhydride forms the heterocyclic compound shown.

Provide a mechanism for this process, which is the first step in a synthesis of the drug imiquimod, which is

an immune response modifier used to treat various skin afflictions.

(c) (6 Marks) It can be difficult to make very large ring lactones (cyclic esters) by conventional

esterification methods. Here is a very simple alternative route involving expansion of a smaller ring.

Provide a mechanism for the process, which only requires a catalytic amount of NaH base!


(d) (4 Marks) Adding 1 molar equivalent of Br2 to alkynes in the presence of aqueous acid leads to the

formation of α-bromoketones. Provide a plausible mechanism for this transformation.

(e) (6 Marks) Treating the compound shown with aqueous acid causes a decarboxylation reaction.

Provide a stepwise mechanism for this reaction.


6. Lab Questions (10 MARKS Total)

Many esters are used in food flavourings and in perfumes because of their fruity aromas and tastes. Because

of your expertise in performing Fischer-type esterification reactions, you have been asked to make up a batch of

ethyl benzoate for a local perfume shop. Ethyl benzoate has an aroma that is described as “heavy fruit, musty

and minty”.

Your client wants about 15 grams of product in as pure a state as possible. Assume that you can obtain a yield

of 50%. You have available all the equipment found in the CHEM 2220 laboratory, as well as the following

chemicals.

Formula MW

Density

(g/mL)

Mp

(°C)

Bp

(°C) Solubility

Benzoic acid C7H6O2 122.12 N/A 122 249 EtOH 56 g/100 mL; water 0.34 g/100 mL; very soluble in

Et2O, EtOAc, DCM.

Ethyl alcohol

(EtOH) C2H6O 46.07 0.789 -114 78 Miscible with water, acetone, Et2O, EtOAc, DCM.

Ethyl ether

(Et2O) C4H10O 74.12 0.713 -116 35

Miscible with EtOH, acetone, EtOAc, DCM. Minimal

water solubility.

Ethyl acetate

(EtOAc) C4H8O2 88.11 0.897 -84 77

Miscible with EtOH, acetone, Et2O, DCM. Minimal water

solubility.

Dichloromethane

(DCM) CH2Cl2 84.93 1.33 -97 40

Miscible with EtOAc, EtOH, acetone, Et2O. Minimal

water solubility.

Acetone C3H6O 58.08 0.791 -95 56 Miscible with water, Et2O, EtOH, DCM, EtOAc.

Sulfuric acid

(conc). H2SO4 98.08 1.84 10 337 Soluble in water, EtOH

Na2SO4

(anhydrous)

Insoluble in organics, minimal solubility in water

CaCl2

(anhydrous)

You also have brine (saturated aqueous NaCl); 1 M aqueous NaOH; saturated aqueous NaHCO3; and 1 M

aqueous HCl.

(a) (8 Marks) On the following page, write out the procedure you would follow to obtain about 15 g of

ethyl benzoate. Include specific amounts of the reagents and reactants you would use to obtain this

amount assuming 50% isolated yield.

Be sure to mention any special glassware you would require for various operations in your procedure.

NOTE: just as in the CHEM 2220 lab, you only have the equipment to distill compounds that boil below

about 100 °C. You have to use other methods to separate the product from any unreacted starting

compound(s), solvent(s) or reagent(s).

(b) (2 Marks) You can use the IR spectrometers in the CHEM 2220 lab to confirm the structure of your

product as well as to detect impurities. Briefly explain what IR signals you would see if your compound was

contaminated with unreacted benzoic acid.


WRITE YOUR ANSWERS TO QUESTION 6 HERE


7. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C12H16O2.

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 Marks) An obvious polar functional group can be seen in the IR spectrum. Additional signals in the

other spectra permit us to identify it. What is this functional group and what signals in the spectra tell us

this?

(c) (1.5 Marks) A carbon-based skeletal fragment stands out in the 13C NMR spectrum and also in the 1H

NMR spectrum. What is this fragment and what evidence supports your conclusion? Be as specific about

the structure of the fragment as you can!

(d) (1.5 Marks) In the 1H NMR spectrum, there is a 5-proton multiplet near 2.3 ppm. In the expansion the

shape of this signal can be seen more clearly and it should be evident that there are two different types of

protons here. What is the correct interpretation of this multiplet?

(e) (1 Mark) Draw the structure of this compound in the box below.

Structure for C12H16O2

CHEM 2220 Final 2016R THIS PAGE MAY BE REMOVED of 14

Spectra for Question 6

IR

C12H16O2

13C NMR C12H16O2 NB: solvent peaks have been edited out.

1H NMR

C12H16O2

2H

s

5H

m

2H

m

3H

tr

2H

d

2H

d

Expansion of 0.8 – 2.4 ppm region

THIS PAGE MAY BE REMOVED of 14

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 14

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

FINAL EXAM – Winter Session 2016R

Wednesday April 13, 2016 9:00 am – 12:00 Frank Kennedy Gold Gym

Students are permitted to bring into the exam room ONE SHEET of LETTER-SIZE (8½ x 11 inches) 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 (32 Marks)

Question 2 – Synthesis (10 Marks)

Question 3 – Mechanism (12 Marks)

Question 4 – Mechanism pot-pourri (30 Marks)

Question 5 – Laboratory (10 Marks)

Question 6 – Spectroscopy (6 Marks)

TOTAL: (100 Marks)

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ANSWER KEY CHEM 2220 Final 2016R Page 2 of 14

1. (32 MARKS) Reactions and Products. Supply the missing molecular structure or reagent/solvent/reaction

conditions to correctly complete the following reactions. Show stereochemistry when necessary. If a

compound is racemic, indicate this by writing “racemic” or “+/-“. Simple aqueous acid or base workups can be

assumed but any “special” workup conditions should be specified.

(a) (2 Marks)

(b) (2 Marks)

(c) (2 Marks)

(d) (2 Marks)

(e) (4 Marks)


(f) (2 Marks)

(g) (2 Marks)

(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (2 Marks)


(l) (4 Marks)

(m) (4 Marks)


2. (10 MARKS) Propose a practical synthetic route to prepare the ester shown. All carbon atoms in your

product must come from either ethyl alcohol or ethyl acetate. You may use any reagents or solvents you wish.

The entire sequence can be accomplished in 5 or 6 steps.

Although a retrosynthetic analysis will be useful, your answer must show the specific reactions needed, with

reagents and conditions, in the forward synthetic direction.

Here is the best answer, 5 steps all directly from chemistry covered in class.

This is the retrosynthetic analysis for another route, not as efficient but still fairly straightforward.


3. (12 MARKS TOTAL) Carmegliptin is a drug used to treat Type 2 diabetes. The reaction shown below is part of

an industrial-scale synthesis of carmegliptin. Everything happens in one reaction pot but the second set of

reagents is only added once all the starting material is converted to the intermediate.

(a) (8 Marks) Write a stepwise mechanism for the first part of the process leading to the intermediate (for

6 Marks). Your mechanism should allow you to determine the structure for the intermediate (for 2 marks).

NB: the intermediate is a neutral organic compound!

(b) (4 Marks) Write a stepwise mechanism for the second part of the process, in which the intermediate

reacts with ammonia in the presence of acetic acid to form the final product.


5. (30 MARKS TOTAL) Mechanism pot-pourri! Assorted shorter mechanisms from across the course.

(a) (8 Marks) When the diketone shown is treated with base in ethanol solution, it undergoes a series of

reactions to form an interesting polycyclic product. Provide a stepwise mechanism to explain this process.


(b) (6 Marks) Heating anthranilic acid in acetic anhydride forms the heterocyclic compound shown.

Provide a mechanism for this process, which is the first step in a synthesis of the drug imiquimod, which is

an immune response modifier used to treat various skin afflictions.

(c) (6 Marks) It can be difficult to make very large ring lactones (cyclic esters) by conventional

esterification methods. Here is a very simple alternative route involving expansion of a smaller ring.

Provide a mechanism for the process, which only requires a catalytic amount of NaH base!


(d) (4 Marks) Adding 1 molar equivalent of Br2 to alkynes in the presence of aqueous acid leads to the

formation of α-bromoketones. Provide a plausible mechanism for this transformation.

(e) (6 Marks) Treating the compound shown with aqueous acid causes a decarboxylation reaction.

Provide a stepwise mechanism for this reaction.


6. Lab Questions (10 MARKS Total)

Many esters are used in food flavourings and in perfumes because of their fruity aromas and tastes. Because

of your expertise in performing Fischer-type esterification reactions, you have been asked to make up a batch of

ethyl benzoate for a local perfume shop. Ethyl benzoate has an aroma that is described as “heavy fruit, musty

and minty”.

Your client wants about 15 grams of product in as pure a state as possible. Assume that you can obtain a yield

of 50%. You have available all the equipment found in the CHEM 2220 laboratory, as well as the following

chemicals.

Formula MW

Density

(g/mL)

Mp

(°C)

Bp

(°C) Solubility

Benzoic acid C7H6O2 122.12 N/A 122 249 EtOH 56 g/100 mL; water 0.34 g/100 mL; very soluble in

Et2O, EtOAc, DCM.

Ethyl alcohol

(EtOH) C2H6O 46.07 0.789 -114 78 Miscible with water, acetone, Et2O, EtOAc, DCM.

Ethyl ether

(Et2O) C4H10O 74.12 0.713 -116 35

Miscible with EtOH, acetone, EtOAc, DCM. Minimal

water solubility.

Ethyl acetate

(EtOAc) C4H8O2 88.11 0.897 -84 77

Miscible with EtOH, acetone, Et2O, DCM. Minimal water

solubility.

Dichloromethane

(DCM) CH2Cl2 84.93 1.33 -97 40

Miscible with EtOAc, EtOH, acetone, Et2O. Minimal

water solubility.

Acetone C3H6O 58.08 0.791 -95 56 Miscible with water, Et2O, EtOH, DCM, EtOAc.

Sulfuric acid

(conc). H2SO4 98.08 1.84 10 337 Soluble in water, EtOH

Na2SO4

(anhydrous)

Insoluble in organics, minimal solubility in water

CaCl2

(anhydrous)

You also have brine (saturated aqueous NaCl); 1 M aqueous NaOH; saturated aqueous NaHCO3; and 1 M

aqueous HCl.

(a) (8 Marks) On the following page, write out the procedure you would follow to obtain about 15 g of

ethyl benzoate. Include specific amounts of the reagents and reactants you would use to obtain this

amount assuming 50% isolated yield.

Be sure to mention any special glassware you would require for various operations in your procedure.

NOTE: just as in the CHEM 2220 lab, you only have the equipment to distill compounds that boil below

about 100 °C. You have to use other methods to separate the product from any unreacted starting

compound(s), solvent(s) or reagent(s).

(b) (2 Marks) You can use the IR spectrometers in the CHEM 2220 lab to confirm the structure of your

product as well as to detect impurities. Briefly explain what IR signals you would see if your compound was

contaminated with unreacted benzoic acid.


WRITE YOUR ANSWERS TO QUESTION 6 HERE

Approximately 15 grams of ethyl benzoate (MW 150.17) is 0.1 mole or 100 mmol. Since the yield is assumed to

be 50%, we need to scale our process for 0.2 moles or 200 mmol.

It makes sense to use EtOH in excess (to drive the equilibrium) since it is cheap and quite volatile. It can be the

solvent for the reaction. If we wanted 0.2 moles (1:1) of EtOH we would need 9.2 grams. From the density

(0.789) 10 mL of EtOH would be 7.89 grams. Thus, about 12 mL of EtOH would be roughly 9.2 grams (9.47 for

calculator-obsessed people). Anything between a 5-10 fold excess would probably be good, so 60 to 120 mL of

solvent can be used – even as much as 500 mL might be practical if the glassware was big enough.

The acid is a catalyst for this reaction, but we don’t want to have too much of it because it makes purification

later on a problem. Anything between 0.01 – 1.0 equivalents (i.e. 2 – 200 mmoles) of H2SO4 would work, but a

smaller amount would be better. Note that 1 mL of conc. H2SO4 = 1.84 g = 18.8 mmoles.

PART (a) Procedure:

Dissolve 24.4 grams (0.2 moles) of benzoic acid in 120 mL of ethanol in a large round-bottom flask. Add 0.5 mL

of conc. H2SO4 (9.4 mmoles), fit a reflux condenser and heating unit, and boil the mixture under reflux for

several hours. We could use TLC to monitor consumption of benzoic acid, but the question did not tell us we

had any hexanes to make an EtOAc/hexanes solvent for TLC so this might not be possible.

Replace the reflux condenser with a distillation head and condenser, and distill away the bulk of the ethanol. Be

careful not to overheat the mixture!

Cool the remaining material and add water (ca 100-200 mL would be fine). Pour the mixture into a separatory

funnel.

You now have some choices. If the mixture forms two layers (the product is not miscible with water) you could

just separate the organic layer from the water, and then wash the organic layer with several portions of water to

remove remaining H2SO4 and ethanol.

If you do not get two layers, you can extract the ethyl benzoate from the water with a solvent (3 x 100 mL

extractions would be good). DCM would be the best solvent, but Ether or Ethyl Acetate would also work. Note

which layer is which! Then, wash the combined organic layers with water (3 x 50-100 mL) to ensure that all the

H2SO4 has been removed.

The question did not say you had any NaHCO3 solution but if you did, you could use this instead of water to

remove acid. This would also remove any remaining benzoic acid. Likewise it did not specify that you had

brine, which could be used to remove bulk water at the end of the extraction.

Dry the organic material with anhydrous Na2SO4. Filter the solids. If you used solvent extraction, use distillation

to remove all the solvent. Note that the product is a liquid that boils at 212 °C so you can easily remove any of

the extraction solvents leaving the product in the still-pot.

Transfer the product to a pre-weighed storage bottle and determine the mass of the bottle plus product to find

your actual product yield.

PART (b) IR Analysis:

If there were any benzoic acid remaining in the product, the IR spectrum would show the OH stretch of the acid

group as a broad peak around 3400 cm-1. The C=O band would probably not be apparent because the product

also has a carbonyl and they would overlap.


7. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C12H16O2.

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 Marks) An obvious polar functional group can be seen in the IR spectrum. Additional signals in the

other spectra permit us to identify it. What is this functional group and what signals in the spectra tell us

this?

(c) (1.5 Marks) A carbon-based skeletal fragment stands out in the 13C NMR spectrum and also in the 1H

NMR spectrum. What is this fragment and what evidence supports your conclusion? Be as specific about

the structure of the fragment as you can!

(d) (1.5 Marks) In the 1H NMR spectrum, there is a 5-proton multiplet near 2.3 ppm. In the expansion the

shape of this signal can be seen more clearly and it should be evident that there are two different types of

protons here. What is the correct interpretation of this multiplet?

(e) (1 Mark) Draw the structure of this compound in the box below.

Structure for C12H16O2

Unsaturation = 5

There is a carbonyl band at ~1730 cm-1 in the IR. The 13C NMR signal for this

group is at ~170 ppm, so it must be a carboxyl carbonyl. There is no OH band in

the IR, so it isn’t RCOOH. It has to be an ESTER!

There are 4 13C NMR signals between ~125 and ~140 ppm, and two 1H NMR doublets each integrating

for 2 protons at around 7.2 – 7.3 ppm. With all the unsaturation in the compound, this is clearly a para

disubstituted phenyl ring.

This multiplet consists of a 3-proton singlet and a 2-proton triplet. The

chemical shift of the singlet suggests CH3 next to C=C or C=O, and

likewise the triplet looks like -CH2CH2 next to C=C or C=O.

CHEM 2220 Final 2016R THIS PAGE MAY BE REMOVED Page 13 of 14

Spectra for Question 6

IR

C12H16O2

13C NMR C12H16O2 NB: solvent peaks have been edited out.

1H NMR

C12H16O2

2H

s

5H

m

2H

m

3H

tr

2H

d

2H

d

Expansion of 0.8 – 2.4 ppm region

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