chem 2220 midterm 1 - university of...

Name: __________________________ Student Number: _____________________

University of Manitoba - Department of Chemistry

CHEM 2220 - Introductory Organic Chemistry II - Term Test 1

Thursday, February 12, 2015; 7-9 PM

This is a 2-hour test, marked out of 50 points total. Part marks are available on all questions.

Put all answers in the spaces provided. If more space is required you may use the backs of the

exam pages but be sure to indicate that you have done so. A spectroscopic data sheet is attached

at the end of the exam.

QUESTION MARKS

1. Mechanism

(6 Marks)

2. Mechanism

(4 Marks)

3. Mechanism

(4 Marks)

4. Mechanism

(8 Marks)

5. Reactions and Products

(22 Marks)

6. Spectra and Structures

(6 Marks)

TOTAL (50 Marks)

CHEM 2220 Test #1 of 9 Feb 12, 2015

1. (6 MARKS) You have learned that epoxides will react with acids such as HBr by nucleophilic

ring opening. Draw mechanisms and provide a brief explanation for the different regiochemical

outcomes in the two reactions shown below.

CHEM 2220 Test #1 of 9 Feb 12, 2015

2. (4 MARKS) 1,2,3,4,5,6-Hexachlorocyclohexane (C6H6Cl6) has many

stereoisomers, including the pesticide Lindane. One of these stereoisomers

undergoes E2 elimination thousands of times more slowly than the other

stereoisomers under identical conditions. Identify which stereoisomer this is and

briefly explain why it is so resistant to E2 elimination.

3. (4 MARKS) Normally, when 1,3-dienes react with HX species the 1,2-addition product is

formed under kinetic conditions while the 1,4-addition product is thermodynamically favoured.

However, when 1,4-dimethylcyclohepta-1,3-diene is treated with HBr at elevated temperature, the

1,2-product predominates rather than the 1,4-adduct! Briefly explain this result.

CHEM 2220 Test #1 of 9 Feb 12, 2015

4. (8 MARKS) When 1,2-diols are treated with strong Brønsted acids, they can undergo the

Pinacol Rearrangement. An example is shown below. Write a stepwise mechanism to explain

how each of the two products is formed in this reaction. You do not have to explain the relative

amounts of the products, just write the mechanism.

CHEM 2220 Test #1 of 9 Feb 12, 2015

5. (22 MARKS) Provide the necessary reagents/solvents or starting materials or major products to

correctly complete the following reactions. Mechanisms are NOT required. Show relative

product stereochemistry (wedge and dash bonds) where appropriate – if a racemic product is

formed, simply indicate “+/−” or “racemic”.

a. (2 Marks)

b. (2 Marks)

c. (2 Marks)

d. (2 Marks)

e. (2 Marks)

CHEM 2220 Test #1 of 9 Feb 12, 2015

f. (2 Marks)

g. (4 Marks)

h. (6 Marks)

CHEM 2220 Test #1 of 9 Feb 12, 2015

6. (6 MARKS Total) The IR and NMR spectra of an unknown organic compound A having the

formula C6H12O are shown on the next page. Based on these data, answer the following questions

about compound A.

a. (1 Mark) What is the degree of unsaturation in compound A?

b. (2 Marks) There are two functional groups in A. What are they? What specific

spectroscopic evidence identifies each one?

c. (1 Mark) In the 1H NMR spectrum the signals between 5 ppm and 6 ppm integrate for a

total of 3 hydrogens. What does this fact tell you about the structure of A?

d. (1 Mark) In the 1H NMR spectrum there is a multiplet signal between 1.3 ppm and 1.6 ppm

integrating for a total of 4 hydrogens. What does this combined set of signals suggest about

the structure?

e. (1 Mark) What is the structure of compound A?

CHEM 2220 Test #1 of 9 Feb 12, 2015

IR

C6H12O

13C NMR

C6H12O

1H NMR

C6H12O

m,

1H

tr,

3H

m,

4H

m,

1H

m,

1H

s,

1H

m,

1H

of 9

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


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








QUESTION MARKS

1. Mechanism

(6 Marks)

2. Mechanism

(4 Marks)

3. Mechanism

(4 Marks)

4. Mechanism

(8 Marks)


(22 Marks)


(6 Marks)

TOTAL (50 Marks)

CHEM 2220 Test #1 Answers Page 2 of 9 Feb 12, 2015









5.

The two products arise because there are two possible groups that can participate in a 1,2-shift. Note that the loss of water, the 1,2-shift and the formation of the C=O bond are probably actually more or less concerted, but they are written here as separate steps for clarity.


(22 MARKS) Provide the necessary reagents/solvents or starting materials or major products to




a. (2 Marks)

b. (2 Marks)

c. (2 Marks)

d. (2 Marks)

e. (2 Marks)


f. (2 Marks)

g. (4 Marks)

h. (6 Marks)

9-BBN could be used instead of BH3. B2H6 is an acceptable equivalent to BH3, as are BH3·SMe2 or BH3·THF complexes.

Note trans relationship of ester groups in Diels-Alder product requires E-alkene as dienophile.




about compound A.








the structure?


Unsaturation number is 1.

Alcohol – broad IR band at ca 3400 cm-1. Alkene – 13C NMR signals at ca 113 and ca

142 ppm, and 1H NMR signals between 5 and 6 ppm.

There are 3 vinylic protons and only 1 alkene group so this must be a terminal alkene

RCH=CH2.

There are probably 2 CH2 groups and they are likely right next to each other since they

are very similar chemical shift and show a lot of coupling.

ANSWER KEY








QUESTION MARKS

1. Mechanism

(6 Marks)

2. Mechanism

(4 Marks)

3. Mechanism

(4 Marks)

4. Mechanism

(8 Marks)


(22 Marks)


(6 Marks)

TOTAL (50 Marks)






5.

The two products arise because there are two possible groups that can participate in a 1,2-shift. Note that the loss of water, the 1,2-shift and the formation of the C=O bond are probably actually more or less concerted, but they are written here as separate steps for clarity.


(22 MARKS) Provide the necessary reagents/solvents or starting materials or major products to




a. (2 Marks)

b. (2 Marks)

c. (2 Marks)

d. (2 Marks)

e. (2 Marks)


f. (2 Marks)

g. (4 Marks)

h. (6 Marks)

9-BBN could be used instead of BH3. B2H6 is an acceptable equivalent to BH3, as are BH3·SMe2 or BH3·THF complexes.

Note trans relationship of ester groups in Diels-Alder product requires E-alkene as dienophile.




about compound A.








the structure?


Unsaturation number is 1.

Alcohol – broad IR band at ca 3400 cm-1. Alkene – 13C NMR signals at ca 113 and ca

142 ppm, and 1H NMR signals between 5 and 6 ppm.

There are 3 vinylic protons and only 1 alkene group so this must be a terminal alkene

RCH=CH2.

There are probably 2 CH2 groups and they are likely right next to each other since they

are very similar chemical shift and show a lot of coupling.


IR

C6H12O

13C NMR

C6H12O

1H NMR

C6H12O

m,

1H

tr,

3H

m,

4H

m,

1H

m,

1H

s,

1H

m,

1H

of 9

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


C H

1.5 – 2.5 C OH


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

chem 2220 midterm 1 - university of...

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