chem 2220 term test 2 - university of manitobahultin/chem2220/archive… · · 2015-03-26......

Name: __________________________ Student Number: _____________________

University of Manitoba - Department of Chemistry

CHEM 2220 - Introductory Organic Chemistry II - Term Test 2

Thursday, March 15, 2012; 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

(5 Marks)

2. Mechanism

(5 Marks)

3. Mechanism

(8 Marks)

4. Reactions and Products

(22 Marks)


(5 Marks)

6. Spectra and Structures

(5 Marks)

TOTAL (50 Marks)

Important Note: We will NOT accept papers written in pencil

back for re-marking after they have been returned to you.

Please do not ask!

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

CHEM 2220 Test #2 of 9 March 15, 2012

1. (5 MARKS) Draw a reasonable stepwise mechanism for the following reaction.

2. (5 MARKS) When aniline (aminobenzene) is treated with fuming sulfuric acid, electrophilic

aromatic substitution occurs at the meta position even though we expect an amino group to be an

ortho, para director. Briefly explain why this is so.


3. (8 MARKS) The Benzoin Addition Reaction was discovered in the 1830s, but its mechanism

was only determined in 1903. In its classic form, two equivalents of benzaldehyde are heated with

aqueous potassium cyanide to form benzoin (2-hydroxy-2-phenylacetophenone). Suggest a

stepwise mechanism for this reaction. HINT: the CN group in a nitrile behaves as a good electron-

withdrawing group.


4. (22 MARKS TOTAL) Provide the necessary reagents/solvents or starting materials or major

products to correctly complete the following reactions. Mechanisms are NOT required. Give

product stereochemistry where appropriate.

a. (2 Marks)

b. (2 Marks)

c. (2 Marks)

d. (6 Marks)


e. (2 Marks)

f. (4 Marks)

g. (4 Marks)


5. (5 MARKS) We have seen that acetylene is a very useful starting material for organic

synthesis. Using acetylene as the starting compound, propose a sequence of 3 reactions to produce

(E)-oct-3-ene-1-ol. For each reaction, give the necessary reagent(s) and conditions, as well as the

product formed. You may use any additional organic compound having four or fewer carbon

atoms, as well as any reagents or solvents you may require.


6. (5 MARKS TOTAL) The spectra of an unknown organic compound A having the formula C9H7NO2 are shown on the next page. Answer the following questions about this compound.

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

b. (1 MARK) What functional group(s) is(are) present? Indicate the specific signal(s) in

the spectra that support your answer.

c. (1 MARK) What does the 13

C NMR spectrum tell you about compound A? Be

specific.

d. (2 MARKS) What is the structure of compound A?


IR

1H NMR

13C NMR


Spectroscopy “Crib Sheet” for 2.222 – 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

University of Manitoba - Department of Chemistry

CHEM 2220 - Introductory Organic Chemistry II - Term Test 2

Thursday, March 15, 2012; 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

(5 Marks)

2. Mechanism

(5 Marks)

3. Mechanism

(8 Marks)


(22 Marks)


(5 Marks)

6. Spectra and Structures

(5 Marks)

TOTAL (50 Marks)

Important Note: We will NOT accept papers written in pencil

back for re-marking after they have been returned to you.

Please do not ask!

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

CHEM 2220 Test #2 ANSWERS Page 2 of 9 March 15, 2012

1. (5 MARKS) Draw a reasonable stepwise mechanism for the following reaction.

2. (5 MARKS) When aniline (aminobenzene) is treated with fuming sulfuric acid, electrophilic

aromatic substitution occurs at the meta position even though we expect an amino group to be an

ortho, para director. Briefly explain why this is so.

This is problem 20.65a from Klein. Note that students could just write “H+” instead of showing

H3O+. It was not required to draw water explicitly as the base when “losing a proton”, and

students could draw “forward arrows” instead of “equilibrium arrows” if they wanted.

When aniline is put into strong acid, the amino group is immediately protonated. Ammonium

ions are strong electron withdrawing groups, and since the lone pair is tied up now, they

cannot stabilize the Wheland complexes formed from ortho or para addition. The ammonium

group behaves as a deactivating meta director, similar to a nitro group.

This is based on material on page 882 and Table 19.1 in Klein, and is similar to problem 19.83.


3. (8 MARKS) The Benzoin Addition Reaction was discovered in the 1830s, but its mechanism

was only determined in 1903. In its classic form, two equivalents of benzaldehyde are heated with

aqueous potassium cyanide to form benzoin (2-hydroxy-2-phenylacetophenone). Suggest a

stepwise mechanism for this reaction. HINT: the CN group in a nitrile behaves as a good electron-

withdrawing group.

This is based on the formation of cyanohydrins. Because the CN group is a good EWG, the

adjacent C-H bond becomes much more acidic than usual and can be deprotonated (reversibly)

by hydroxide ion. The resulting carbanion then attacks another molecule of benzaldehyde.

Note that only relatively small amounts of the anion are present at any given time, as the

equilibrium constants for the first few steps do not favour the products. However, the later

steps do favour their products, and the final equilibrium is extremely favourable, so the overall

reaction proceeds smoothly. Students did not have to give any explanatory details: the

question only required a mechanism.


4. (22 MARKS TOTAL) Provide the necessary reagents/solvents or starting materials or major

products to correctly complete the following reactions. Mechanisms are NOT required. Give

product stereochemistry where appropriate.

a. (2 Marks)

b. (2 Marks)

c. (2 Marks)

d. (6 Marks)

The first step must form the “Hoffmann” product, so a sterically bulky base for E2

elimination is the key point.

In the second step, it might be tempting to think of ozonolysis but note that the number of

carbons in the product is the same as in the starting material. There is no cleavage therefore.

Anti-Markovnikov hydration followed by PCC oxidation is thus the only answer.


e. (2 Marks)

f. (4 Marks)

g. (4 Marks)

The molecule might look scary but this is just Grignard addition to a ketone. This

reaction is actually part of the synthesis of the estrogen-modulating drug Raloxifene,

which is prescribed for treatment of osteoporosis in post-menopausal womene.


5. (5 MARKS) We have seen that acetylene is a very useful starting material for organic

synthesis. Using acetylene as the starting compound, propose a sequence of 3 reactions to produce

(E)-oct-3-ene-1-ol. For each reaction, give the necessary reagent(s) and conditions, as well as the

product formed. You may use any additional organic compound having four or fewer carbon

atoms, as well as any reagents or solvents you may require.

This is similar to many examples in Chapter 10 (Alkynes) and Chapter 12

(Synthesis) in Klein.


6. (5 MARKS TOTAL) The spectra of an unknown organic compound A having the formula

C9H7NO2 are shown on the next page. Answer the following questions about this compound.

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

b. (1 MARK) What functional group(s) is(are) present? Indicate the specific signal(s) in

the spectra that support your answer.

c. (1 MARK) What does the 13

C NMR spectrum tell you about compound A? Be

specific.

d. (2 MARKS) What is the structure of compound A?

Unsaturation number is SEVEN!

There is a C=O group present (IR, ~1710 cm-1

). From the 13

C NMR (see below) it is

clearly a carboxyl-family carbonyl, but the lack of any IR signal for an OH shows

that it is NOT a carboxylic acid. The 1H NMR has no signals around 4 ppm so it is

not an ester. If it were a nitrile there would be an IR band at around 2250 cm-1

,

which there is not. It is probably some kind of amide therefore.

The signal at ~168 ppm is a carboxyl-family carbonyl carbon. The signal at ~24 ppm

is likely a CH3 group. There are only 3 other signals, which are in the aromatic

region. This also tells us there are no alkenes or alkynes present. Since there are 9

carbons in total, this molecule is highly symmetrical! The aromatic ring accounts for

4 unsaturations, leaving 3 more which must be two carbonyls plus one more ring.

Note integral in 1H NMR is 4:3.

Four aryl protons means

disubstituted aromatic ring. The

methyl group at about 3 ppm is

consistent with N-CH3.

Solving this problem was all about symmetry. The 13

C NMR showed only 1 kind of

carbonyl and 1 kind of aliphatic, with 3 kinds of aryl carbons. The key was to ask

yourself “how many ways can 9 carbons be sorted into 5 kinds? Furthermore, since

the 1H NMR told you that there was only 1 methyl, you could deduce that it had to lie

on the symmetry plane. Finally, the symmetry strongly suggested that there were

actually 2 carbonyls, both identical. These points all should have suggested an ortho-

disubstituted aromatic compound as shown above.


IR

1H NMR

13C NMR

chem 2220 term test 2 - university of manitobahultin/chem2220/archive… · · 2015-03-26......

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