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

FINAL EXAM – Winter Session 2015R

Friday April 17, 2015 1:30 pm – 4:30 pm University Centre Room 210-224 (Multi-purpose room)

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 (32 Marks)

Question 2 – Synthesis (10 Marks)

Question 3 – Mechanism (12 Marks)

Question 4 – Mechanism grab-bag (30 Marks)

Question 5 – Laboratory (10 Marks)

Question 6 – Spectroscopy (6 Marks)

TOTAL: (100 Marks)

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CHEM 2220 Final Exam 2015R of 15

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. 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) (3 Marks)


(e) (3 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 synthetic route to prepare the diketone A from 1,3-cyclopentanedione plus any

other organic compounds of fewer than 5 carbon atoms. You can accomplish this in between 3 and 5 reaction

steps. You may find that retrosynthetic analysis is a useful way to plan, but your answer should be a sequence

of reactions complete with reagents and conditions, showing the intermediate products along the route.


3. (12 MARKS TOTAL) The Wolff-Kishner Reduction of ketones involves two stages: first the reaction of a ketone

with hydrazine to form a hydrazone (which we discussed in class) and then the reaction of this hydrazone with

hydroxide ion at high temperature to form a –CH2– group by the expulsion of N2 gas.

(a) (4 Marks) Write a stepwise mechanism for STAGE 1, assuming that there is a small amount of acid

present as a catalyst.


(b) (8 Marks) Propose a stepwise mechanism for STAGE 2. Show all relevant resonance structures for

the intermediates. Note that nothing else is present besides what is shown in the scheme, and that the

nature of the “R” groups is irrelevant. HINT: surprisingly, carbanion intermediates are involved!


4. (30 MARKS TOTAL) Mechanism grab-bag!

(a) (4 Marks) Briefly explain why Friedel-Crafts alkylation of aniline (Process A) fails, but the 3-step

sequence shown as Process B is successful. You do not have to show a detailed stepwise mechanism

but some structure diagrams may be helpful.

(b) (6 Marks) Potassium t-butoxide (KOtBu) is a poor nucleophile but a decent base. Suggest a detailed

stepwise mechanism for the following reaction.


(c) (6 Marks) In the course of synthesizing some potential anti-cancer compounds, Nakanishi and Suzuki

performed the following reaction. We are familiar with the mechanism although the conditions are a bit

different from those discussed in class. Write out a stepwise mechanism for this process. Note that 2

molar equivalents of the base LDA were essential – if only 1 equivalent was used, unchanged starting

material was obtained after workup. You may truncate the structures in your answer if you wish.

(d) (4 Marks) Provide a stepwise mechanism for the following reaction, performed in concentrated H2SO4

without any other solvent present.


(e) (6 Marks) The triene shown reacts with an excess of maleic anhydride when heated to form a product

whose molecular formula is C14H12O6. Based on mechanistic reasoning, draw the structure of this product,

including the relative stereochemistry at each of the 5 new stereocentres it contains.

(f) (4 Marks) Intramolecular Claisen condensations are referred to as “Dieckmann Cyclizations”. Briefly

explain why the following Dieckmann cyclization forms only the product shown, using mechanistic

arguments. You do not have to write a detailed stepwise mechanism.


5. Lab Questions (10 MARKS Total)

(a) (7 Marks) Until recently, bisphenol A was used in the manufacturing of plastics used in consumer products and food containers. It was found to display hormone-like properties and therefore in October 2010 Health Canada added bisphenol A to its list of known toxic substances.

Provide a procedure (of at least two steps) for the synthesis of bisphenol A using chemicals listed in the table below (you may not need all of them) and anything else commonly found in our organic chemistry labs. Your procedure should form the purest product possible.

Assume:

Each step in the reaction will yield 100% with the exception of the last step, which only yields 50%.

You must obtain about 110 grams of pure product.

You will use one reagent in excess and there will be some of this reagent remaining at the end of the reaction. You must explain how this excess will be separated from the product.

Please note:

Be very careful choosing which starting materials you use!

There is some arithmetic involved but numbers have been chosen which can be approximated.

Please state any reasonable assumptions made.

Some data in the table are fictitious and should not be used for experimental purposes!

Mol. Mass b.p. (oC) m.p. (oC) Solubility (g/mL)

4-(2-chloroisopropyl)phenyl acetate

214 280 120 EtOH: 5 CH2Cl2: 1.0 CHCl3: 3 Water: 0.1

phenylacetate


FeCl3 161 454 123 EtOH: Incompatible

CH2Cl2: 0.01 CHCl3: 6 Water: Incompatible

C(CH3)2Cl2

2,2-dichloropropane 85 69 -35 Miscible with EtOH, CH2Cl2, CHCl3 and water

CH3CO2H acetic acid

60 118 16 Miscible with EtOH, CH2Cl2, CHCl3 and water


(b) (1 Mark) After you isolated the product from your synthesis, you obtained the IR spectrum shown

below. What conclusions about your product can be drawn from this spectrum?

(c) (2 Marks) In preparing to do your reaction, you checked whether TLC could separate the compounds

involved. The plate on the left shows a mixture of 4-(2-chloroisopropyl)phenyl acetate and bisphenol A,

eluted with hexane. Complete the blank plate to show how the same mixture of compounds would appear if

it had been eluted with ethyl acetate. Be sure to label the spots. No explanation is required.


6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C8H16O2. 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) (1.0 Mark) What functional group(s) does this compound contain? Indicate the specific evidence for

your conclusion.

(c) (1.5 Marks) Between 0.88 and 1.25 ppm in the 1H NMR spectrum there are three very prominent

signals. What structures do each of these indicate?

(d) (1 Mark) There is a 1H NMR multiplet integrating for 1 hydrogen at 5.01 ppm. What type of

environment is this hydrogen in? HINT: This is a bit further downfield than we typically expect for this type

of hydrogen.

(e) (1 Mark) The two 1-hydrogen multiplets at 2.3 and 1.67 ppm are both attached to the same carbon

atom. Why do they have different chemical shifts?

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

Structure for C8H16O2


Spectra for Question 6

IR

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

1H NMR

1H

m

1H

m

1H

m

6H

d

3H

tr

3H

d

1H

m

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 15

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

FINAL EXAM – Winter Session 2015R

Friday April 17, 2015 1:30 pm – 4:30 pm University Centre Room 210-224 (Multi-purpose room)

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 (32 Marks)

Question 2 – Synthesis (10 Marks)

Question 3 – Mechanism (12 Marks)

Question 4 – Mechanism grab-bag (30 Marks)

Question 5 – Laboratory (10 Marks)

Question 6 – Spectroscopy (6 Marks)

TOTAL: (100 Marks)

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CHEM 2220 FINAL EXAM 2015R ANSWER KEY Page 2 of 15

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. 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) (3 Marks)


(e) (3 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 synthetic route to prepare the diketone A from 1,3-cyclopentanedione plus any

other organic compounds of fewer than 5 carbon atoms. You can accomplish this in between 3 and 5 reaction

steps. You may find that retrosynthetic analysis is a useful way to plan, but your answer should be a sequence

of reactions complete with reagents and conditions, showing the intermediate products along the route.


3. (12 MARKS TOTAL) The Wolff-Kishner Reduction of ketones involves two stages: first the reaction of a ketone

with hydrazine to form a hydrazone (which we discussed in class) and then the reaction of this hydrazone with

hydroxide ion at high temperature to form a –CH2– group by the expulsion of N2 gas.

(a) (4 Marks) Write a stepwise mechanism for STAGE 1, assuming that there is a small amount of acid

present as a catalyst.


(b) (8 Marks) Propose a stepwise mechanism for STAGE 2. Show all relevant resonance structures for

the intermediates. Note that nothing else is present besides what is shown in the scheme, and that the

nature of the “R” groups is irrelevant. HINT: surprisingly, carbanion intermediates are involved!


4. (30 MARKS TOTAL) Mechanism grab-bag!

(a) (4 Marks) Briefly explain why Friedel-Crafts alkylation of aniline (Process A) fails, but the 3-step

sequence shown as Process B is successful. You do not have to show a detailed stepwise mechanism

but some structure diagrams may be helpful.

(b) (6 Marks) Potassium t-butoxide (KOtBu) is a poor nucleophile but a decent base. Suggest a detailed

stepwise mechanism for the following reaction.

Amines are Lewis as well as Brønsted bases, so in Process A aniline will

form a complex with AlCl3. The nitrogen in this complex will be positively

charged. Therefore, it will become a strongly deactivating group, and

Friedel-Crafts reactions will not work on deactivated (electron-poor)

aromatic rings.

In Process B the amine is converted to an amide. Amides are much less

basic than amines and do not interact strongly with the Lewis acid. The

lone pair of the nitrogen is still somewhat available to activate the aromatic

ring for Friedel-Crafts alkylation.


(c) (6 Marks) In the course of synthesizing some potential anti-cancer compounds, Nakanishi and Suzuki

performed the following reaction. We are familiar with the mechanism although the conditions are a bit

different from those discussed in class. Write out a stepwise mechanism for this process. Note that 2

molar equivalents of the base LDA were essential – if only 1 equivalent was used, unchanged starting

material was obtained after workup. You may truncate the structures in your answer if you wish.


(d) (4 Marks) Provide a stepwise mechanism for the following reaction, performed in concentrated H2SO4

without any other solvent present.


(e) (6 Marks) The triene shown reacts with an excess of maleic anhydride when heated to form a product

whose molecular formula is C14H12O6. Based on mechanistic reasoning, draw the structure of this product,

including the relative stereochemistry at each of the 5 new stereocentres it contains.

(f) (4 Marks) Intramolecular Claisen condensations are referred to as “Dieckmann Cyclizations”. Briefly

explain why the following Dieckmann cyclization forms only the product shown, using mechanistic

arguments. You do not have to write a detailed stepwise mechanism.

The condensation process in a Dieckmann is thermodynamically unfavorable under basic

conditions, but the deprotonation shown is thermodynamically favorable. Thus, deprotonation

of the first-formed product drives the entire condensation forward.

The alternative product cannot be deprotonated in this way, so if it is formed, under the

basic conditions of the reaction it reverses back to starting material.


5. Lab Questions (10 MARKS Total)

(a) (7 Marks) Until recently, bisphenol A was used in the manufacturing of plastics used in consumer products and food containers. It was found to display hormone-like properties and therefore in October 2010 Health Canada added bisphenol A to its list of known toxic substances.

Provide a procedure (of at least two steps) for the synthesis of bisphenol A using chemicals listed in the table below (you may not need all of them) and anything else commonly found in our organic chemistry labs. Your procedure should form the purest product possible.

Assume:

Each step in the reaction will yield 100% with the exception of the last step, which only yields 50%.

You must obtain about 110 grams of pure product.

You will use one reagent in excess and there will be some of this reagent remaining at the end of the reaction. You must explain how this excess will be separated from the product.

Please note:

Be very careful choosing which starting materials you use!

There is some arithmetic involved but numbers have been chosen which can be approximated.

Please state any reasonable assumptions made.

Some data in the table are fictitious and should not be used for experimental purposes!

Mol. Mass b.p. (oC) m.p. (oC) Solubility (g/mL)

4-(2-chloroisopropyl)phenyl acetate


phenylacetate


FeCl3 161 454 123 EtOH: Incompatible

CH2Cl2: 0.01 CHCl3: 6 Water: Incompatible

C(CH3)2Cl2

2,2-dichloropropane 85 69 -35 Miscible with EtOH, CH2Cl2, CHCl3 and water

CH3CO2H acetic acid

60 118 16 Miscible with EtOH, CH2Cl2, CHCl3 and water

Answer

Dissolve 4-(2-chloroisopropyl)phenyl acetate (214 g), phenylacetate (150 g,10% excess) and iron(III) chloride

(at least 320 g) in chloroform (at least 100 mL). At least two equivalents of catalyst is required because of the

presence of carbonyl groups. Heating should not be required since the chloride is on a tertiary carbon.

When TLC indicates the reaction is done, the reaction mixture is washed with 10% HCl (3 × 100 mL) to remove

some of the excess starting material and the catalyst. We assume that the diester is soluble in the chloroform.

Drying is not necessary since the following step is a hydrolysis (water is added). The chloroform is then

removed by distillation since it is not soluble with the aqueous acid used for the hydrolysis step. The residual

material in the distillation flask is the crude diester product along with excess phenylacetate.

A strong aqueous acid such as 1 M HCl (aq) (100 mL) is added to the mixture and then refluxed. Once again

the hydrolysis can be monitored by TLC. Conveniently bisphenol A has low solubility in water and is likely to

precipitate out. The acetic acid and phenol (from the hydrolysis of phenyl acetate) are water soluble and

therefore should not interfere with the product. The final step is to isolate the bisphenol A by a vacuum filtration.


(b) (1 Mark) After you isolated the product from your synthesis, you obtained the IR spectrum shown

below. What conclusions about your product can be drawn from this spectrum?

(c) (2 Marks) In preparing to do your reaction, you checked whether TLC could separate the compounds

involved. The plate on the left shows a mixture of 4-(2-chloroisopropyl)phenyl acetate and bisphenol A,

eluted with hexane. Complete the blank plate to show how the same mixture of compounds would appear if

it had been eluted with ethyl acetate. Be sure to label the spots. No explanation is required.

1) There’s evidence of an alcohol present.

2) The carbonyl group at 1765 cm-1 is a sign that there’s incomplete ester

hydrolysis.


6. (6 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C8H16O2. 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) (1.0 Mark) What functional group(s) does this compound contain? Indicate the specific evidence for

your conclusion.

(c) (1.5 Marks) Between 0.88 and 1.25 ppm in the 1H NMR spectrum there are three very prominent

signals. What structures do each of these indicate?

(d) (1 Mark) There is a 1H NMR multiplet integrating for 1 hydrogen at 5.01 ppm. What type of

environment is this hydrogen in? HINT: This is a bit further downfield than we typically expect for this type

of hydrogen.

(e) (1 Mark) The two 1-hydrogen multiplets at 2.3 and 1.67 ppm are both attached to the same carbon

atom. Why do they have different chemical shifts?

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

Structure for C8H16O2

Unsaturation no. = 1

Must contain an ESTER: IR C=O stretch at ca. 1730, no OH. 13C NMR ca. 178

ppm says carboxyl-type carbonyl, not aldehyde or ketone.

6H doublet at ca. 1.25: two identical CH3 groups both with 1 neighbour. Probably CH(CH3)2

3H doublet at ca. 1.1: one CH3 group with 1 neighbor

3H triplet at ca. 0.88: a CH3 group with an adjacent CH2 group.

At ca 5 ppm we might think of vinylic C-H but this is not possible since our only unsaturation is a

C=O. This must be a C-H next to Oxygen, from the alcohol portion of our ester.

If the two protons of a CH2 group are not identical it must be because there is a stereogenic centre

somewhere in the molecule that makes them diastereotopic. See Klein Section 16.4.


Spectra for Question 6

IR

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

1H NMR

1H

m

1H

m

1H

m

6H

d

3H

tr

3H

d

1H

m

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