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CHEM 2220 Final Exam 2012R of 14

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)

OHPBr3 Ph3P

Product is ionic - show bothcation and anion parts!

nBuLi, THFthen add

H

O

(6 Marks)


(g) (2 Marks)

(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (6 Marks)


2. (a) (7 MARKS) Synthesis. 1-(Hydroxymethyl)cyclohexene can be prepared in four steps starting from diethyl heptanedioate. Provide the reagents and reaction conditions required for each of the four necessary reactions, as well as the intermediate products formed at each step.

OO

OEtEtO

OH

Reaction 1

Reaction 2

Reaction 3

Reaction 4

(b) (3 Marks) Cyclohexene compounds can often be made using the Diels-Alder reaction. Would the Diels-Alder method be suitable for preparing 1-(hydroxymethyl)cyclohexene? Assuming that the required starting materials can be obtained, briefly explain why or why not.


3. (8 MARKS) Mechanism. In his 1982 synthesis of the enediandric acids A-D, K.C. Nicolaou used the following reaction. Actually, in this single step there are two separate processes that occur sequentially. Identify the separate processes, and explain the stereochemical outcome of this remarkable transformation. Notice that this process occurs at an unusually low temperature for this type of reaction.

CH2Cl2/MeOH25 oC

H

H

OH

OHHO OH(racemic)


4. (12 MARKS Total) Mechanism. Many polycyclic natural products including steroids are biosynthesized from acyclic polyenes such as squalene, in a two-stage process catalyzed by two enzymes.

This concept was developed by W.S. Johnson and co-workers as a route to synthesize steroids in the laboratory (see Johnson, W.S. Angew. Chem. Int. Ed. Engl. 1976, 15, 9-17).

(a) (5 MARKS) Johnson’s team studied the following reaction as a model process. Suggest a stepwise mechanism for this reaction. (Nitromethane was used as a polar but non-nucleophilic solvent.)


(b) (7 MARKS) Johnson also explored the possibility of making tricyclic compounds using this type of reaction. The following reaction formed a mixture of products, but the ring structure was identical in all of them. Write a mechanism for the formation of the two products shown.


5. (20 MARKS TOTAL) Mechanism grab-bag!

(a) (4 MARKS) The Darzens reaction is closely related to chemistry discussed in CHEM 2220. Write a mechanism for this reaction.

(b) (4 MARKS) In 1964 Paul de Mayo of the University of Western Ontario reported the following reaction. Write a stepwise mechanism to explain this transformation.


(c) (4 MARKS) In mildly acidic conditions, glutaraldehyde (pentanedial) exists in equilibrium with a cyclic hemiacetal. Provide a mechanism for the conversion of glutaraldehyde to this cyclic structure.

(d) (8 MARKS) The Overman Rearrangement is a two-step procedure that provides a very convenient access to allylic amines. An example of this rearrangement is shown. Write a mechanism for each step of this two-step sequence, and show the structure of the intermediate product A. (Assume an aqueous workup at the end of the first step. In the second step, xylene is simply a high-boiling solvent).


6. Lab Questions (15 MARKS Total) Rookie chemist Neville Roundbottom was asked to synthesize about 122 g of Alcohol B from the reduction of Acid A using lithium aluminium hydride. Lithium aluminium hydride (LiAlH4) is a much more reactive reducing agent than sodium borohydride and will react with water and alcohol. Neville’s proposed synthetic procedure is shown below.

Most data and procedures in this question are fictitious and should not be used for experimental purposes.

Acid A Alcohol B LiAlH4 Diethyl ether MW (g/mol) 136.15 122.14 38.39 74.12 MP (oC) 52 20 N/A -116 BP (oC) 120 195 N/A 35 Solubilities (g/mL)

Water – 0.3 Ethanol - 6 Diethyl ether - 3


Water – N/A Ethanol – N/A Diethyl ether – 10

Water - 0.03 Ethanol – miscible

Densities 1.76 0.89 N/A 0.88 Neville’s Proposed Procedure

Dissolve Acid A (10 g) in diethyl ether (100 mL). Cool the solution on an ice bath. Slowly add LiAlH4 (8 g). After 30 minutes of stirring, slowly add ice cold water (50 mL) to the solution. Remove and discard the top (aqueous) layer. The organic layer is extracted three times with 50 mL aliquots of 1 M HCl. The solvent in the retained aqueous layer is removed by distillation and the liquid product is isolated from the distillation flask (round bottom flask).

(a) (9 MARKS) There are several problems with this proposed experiment. Identify as many problems with Neville’s procedure as you can, and indicate how you would modify the procedure to make it work. Your goal is to accomplish the synthesis of the required amount of Alcohol B in the most pure form possible. You are free to use any reagents and chemicals you wish. Assume that the reaction produces a 50% yield. State all assumptions.


(b) (3 MARKS) After implementing your recommendations, Neville collected his product and then brought you the TLC plate shown in Figure 1. Unfortunately Neville had used up all the starting material so he could not use it as a reference spot. However, he found a sample of pure Alcohol B and used that as the reference. Help Neville interpret this TLC plate by explaining whether the product has been made and comment on the purity of the product.

(c) (3 MARKS) Neville performed the reaction again and examined his product by TLC on silica gel. The plate is shown on the left in Figure 2.

Using the blank TLC plate template on the right in Figure 2, draw a diagram of how the plate should look if the plate substrate (absorbent) was modified to the LESS POLAR cellulose but using the same eluent. The spots from left to right are starting material, co-spot, and product.

Eluent: ethyl acetate:hexanes (1:9)Silica plate

Solvent front

Origin

Alcohol Bcospot

Productfrom the reaction

Figure 1: TLC of Neville's first product.

Figure 2: TLC for Neville's second attempt.


7. (5 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C7H16O. The IR, 1H NMR and 13C NMR spectra of this compound are shown on the next page. Answer the following questions about this compound. (a) (1 MARK) What is the unsaturation number for this compound?

(b) (1 MARK) What functional group(s) is(are) present? What specific spectroscopic evidence supports your answer?

(c) (1 MARK) Briefly explain the signals in the 1H NMR spectrum between 0.80 – 0.88 ppm.

(d) (2 MARKS) Draw the structure of this compound in the box below.

Structure for C7H16O

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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

HH

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 RCNRCCR

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 2012R

Friday April 20, 2012 1:30 pm – 4:30 pm Frank Kennedy Brown 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 (10 Marks)

Question 3 – Mechanism (8 Marks)

Question 4 – Mechanism (12 Marks)

Question 5 – Mechanism grab-bag (20 Marks)

Question 6 – Laboratory (15 Marks)

Question 7 – Spectroscopy (5 Marks)

TOTAL: (100 Marks)

http://www.wpclipart.com/science/atoms_molecules/periodic_tables/periodic_table_of_elements_BW.png

CHEM 2220 Final Exam 2012R ANSWER KEY Page 2 of 15

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) (6 Marks)


(g) (2 Marks)

(h) (2 Marks)

(i) (2 Marks)

(j) (2 Marks)

(k) (6 Marks)


2. (a) (7 MARKS) Synthesis. 1-(Hydroxymethyl)cyclohexene can be prepared in four steps starting from

diethyl heptanedioate. Provide the reagents and reaction conditions required for each of the four necessary

reactions, as well as the intermediate products formed at each step.


(b) (3 Marks) Cyclohexene compounds can often be made using the Diels-Alder reaction. Would the

Diels-Alder method be suitable for preparing 1-(hydroxymethyl)cyclohexene? Assuming that the required

starting materials can be obtained, briefly explain why or why not.

The Diels-Alder reaction generally requires an electron-withdrawing group on

the dienophile. The cyclohexene product will have the double bond between

the carbons that were C2 and C3 of the diene. The diene and dienophile

needed to make 1-(hydroxymethyl)cyclohexene in the simplest way do not fit

these requirements, so the DA reaction is probably not appropriate.


3. (8 MARKS) Mechanism. In his 1982 synthesis of the enediandric acids A-D, K.C. Nicolaou used the following

reaction. Actually, in this single step there are two separate processes that occur sequentially. Identify the

separate processes, and explain the stereochemical outcome of this remarkable transformation. Notice that

this process occurs at an unusually low temperature for this type of reaction.


4. (12 MARKS Total) Mechanism. Many polycyclic natural products including steroids are biosynthesized from

acyclic polyenes such as squalene, in a two-stage process catalyzed by two enzymes.

This concept was developed by W.S. Johnson and co-workers as a route to synthesize steroids in the

laboratory (see Johnson, W.S. Angew. Chem. Int. Ed. Engl. 1976, 15, 9-17).

(a) (5 MARKS) Johnson’s team studied the following reaction as a model process. Suggest a stepwise

mechanism for this reaction. (Nitromethane was used as a polar but non-nucleophilic solvent.)


(b) (7 MARKS) Johnson also explored the possibility of making tricyclic compounds using this type of

reaction. The following reaction formed a mixture of products, but the ring structure was identical in all of

them. Write a mechanism for the formation of the two products shown.


5. (20 MARKS TOTAL) Mechanism grab-bag!

(a) (4 MARKS) The Darzens reaction is closely related to chemistry discussed in CHEM 2220. Write a

mechanism for this reaction.

(b) (4 MARKS) In 1964 Paul de Mayo of the University of Western Ontario reported the following reaction.

Write a stepwise mechanism to explain this transformation.


(c) (4 MARKS) In mildly acidic conditions, glutaraldehyde (pentanedial) exists in equilibrium with a cyclic

hemiacetal. Provide a mechanism for the conversion of glutaraldehyde to this cyclic structure.

(d) (8 MARKS) The Overman Rearrangement is a two-step procedure that provides a very convenient

access to allylic amines. An example of this rearrangement is shown. Write a mechanism for each step of

this two-step sequence, and show the structure of the intermediate product A. (Assume an aqueous

workup at the end of the first step. In the second step, xylene is simply a high-boiling solvent).


6. Lab Questions (15 MARKS Total) Rookie chemist Neville Roundbottom was asked to synthesize about 122 g

of Alcohol B from the reduction of Acid A using lithium aluminium hydride. Lithium aluminium hydride (LiAlH4) is

a much more reactive reducing agent than sodium borohydride and will react with water and alcohol. Neville’s

proposed synthetic procedure is shown below.

Most data and procedures in this question are fictitious and should not be used for experimental purposes.

Acid A Alcohol B LiAlH4 Diethyl ether

MW (g/mol) 136.15 122.14 38.39 74.12

MP (oC) 52 20 N/A -116

BP (oC) 120 195 N/A 35

Solubilities (g/mL)



Water – N/A Ethanol – N/A Diethyl ether – 10

Water - 0.03 Ethanol – miscible

Densities 1.76 0.89 N/A 0.88

Neville’s Proposed Procedure

Dissolve Acid A (10 g) in diethyl ether (100 mL). Cool the solution on an ice bath. Slowly add LiAlH4 (8

g). After 30 minutes of stirring, slowly add ice cold water (50 mL) to the solution. Remove and discard

the top (aqueous) layer. The organic layer is extracted three times with 50 mL aliquots of 1 M HCl. The

solvent in the retained aqueous layer is removed by distillation and the liquid product is isolated from

the distillation flask (round bottom flask).

(a) (9 MARKS) There are several problems with this proposed experiment. Identify as many problems with

Neville’s procedure as you can, and indicate how you would modify the procedure to make it work. Your

goal is to accomplish the synthesis of the required amount of Alcohol B in the most pure form possible. You

are free to use any reagents and chemicals you wish. Assume that the reaction produces a 50% yield.

State all assumptions.

Here are some key points that could be part of your answer:

1) The amount of Acid A is incorrect. Since we want 122 g of Alcohol A (1 mole) and we know that

the reaction produces product at 50% yield, then we require 2 moles of starting material, or 272 g

of Acid A.

2) To dissolve Acid A, notice that the solubility in diethyl ether is 3 g/mL, we’ll require a minimum of

90 mL to dissolve all of Acid A.

3) Reduction of an acid requires 3 hydride equivalents: 1 to neutralize the acidic proton plus 2 more

to add to the carbonyl. A minimum of 28.8 g of LiAlH4 is required (0.75 moles) assuming the

reagent delivers all four hydrides. In practice one would probably add at least 1 mole, and

probably more just to be sure.

4) As mentioned by Dr.Hultin in class, the LiAlH4 reduction reaction will have to be refluxed (note

that THF is usually used as the solvent since it has a higher boiling point than diethyl ether) and

the reaction should be monitored by TLC to determine its completion.

5) After water is added to quench the reaction, the ether will be the top layer and not the bottom.

6) A WASHING should be done and not an EXTRACTION using aliquots of 1 M NaOH to remove

the residual acid. Note that the product is in the ether layer.

7) Dry the ether layer with a drying agent potassium carbonate would be a good choice to remove

any final traces of acid.

8) The product can be isolated by distilling off the solvent and saving the liquid in the round bottom

flask.


(b) (3 MARKS) After implementing your recommendations, Neville

collected his product and then brought you the TLC plate shown in

Figure 1. Unfortunately Neville had used up all the starting

material so he could not use it as a reference spot. However, he

found a sample of pure Alcohol B and used that as the reference.

Help Neville interpret this TLC plate by explaining whether the

product has been made and comment on the purity of the product.

(c) (3 MARKS) Neville performed the reaction again

and examined his product by TLC on silica gel. The

plate is shown on the left in Figure 2.

Using the blank TLC plate template on the right in

Figure 2, draw a diagram of how the plate should

look if the plate substrate (absorbent) was modified

to the LESS POLAR cellulose but using the same

eluent. The spots from left to right are starting

material, co-spot, and product.

Figure 1: TLC of Neville's first product.

Figure 2: TLC for Neville's second attempt.

There are two spots in the product lane. The one with

the higher Rf corresponds to the product while the

bigger spot with the lower Rf corresponds to the

starting material. Therefore the TLC suggests that very

little product was made in the reaction. The product, be

it the acid or alcohol is considered impure since there

are two spots in the TLC.

The samples will travel to higher Rf values since the plate is less

polar. Alcohol A will still have a higher Rf than Acid A.


7. (5 MARKS Total) Spectroscopy. The molecular formula of an unknown organic compound is C7H16O. The

IR, 1H NMR and

13C NMR spectra of this compound are shown on the next page. Answer the following

questions about this compound.

(a) (1 MARK) What is the unsaturation number for this compound?

(b) (1 MARK) What functional group(s) is(are) present? What specific spectroscopic evidence supports

your answer?

(c) (1 MARK) Briefly explain the signals in the 1H NMR spectrum between 0.80 – 0.88 ppm.

(d) (2 MARKS) Draw the structure of this compound in the box below.

Structure for C7H16O

Unsaturation number is ZERO.

There is an alcohol because the IR shows a broad band at around 3400 cm-1

.

There are 9 protons in this region. One signal is a 3-proton singlet suggesting a CH3 group with no

neighbour protons. The other signal integrates for 6 protons and at first might appear to be a triplet.

However, the peaks of a triplet must be in a 1:2:1 ratio and this is not the case. It looks more like this

is actually two different signals on top of one another, perhaps a 3-proton triplet with a 3-proton

singlet on top.

The 1-proton quartet at about 3.5 ppm tells us that the CH next to the OH group is also

next to a CH3 and nothing else. We see that CH3 as the doublet at 1.1 ppm.

Our analysis for part c above also tells us that there must be two methyls with no

neighbours, and one methyl with a CH2 next to it.


Spectra for Question 7

IR

1H NMR

13C NMR

NB: all signals are single lines.

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

student no p age 1 o - university of manitobahultin/chem2220/archive/2012/final... · chem 2220...

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