chem 2220 midterm 1 - university of manitobahultin/chem2220/archive/...chem 2220 test #1 answers...

Name: __________________________ Student Number: _____________________

University of Manitoba - Department of Chemistry

CHEM 2220 - Introductory Organic Chemistry II – MIDTERM TEST

Take-Home Test, Wed. March 1 2017 5:00 pm – Fri. March 3 2017 5:00 pm,

This is a TAKE HOME test. It is intended to be done in 2 hours, CLOSED-BOOK. It will be

marked out of 50 points total. Part marks are available on all questions.

The test is delivered electronically – this version is for later study purposes. Questions are

identical to the e-version.

A spectroscopic data sheet is attached at the end of the exam.

QUESTION MARKS

1. Mechanism

(4 Marks)

2. Mechanism

(4 Marks)

3. Mechanism

(6 Marks)

4. Mechanism

(6 Marks)

5. Reactions and Products

(24 Marks)

6. Spectra and Structures

(6 Marks)

TOTAL (50 Marks)

CHEM 2220 Test #1 of 10 Feb 11, 2016

1. (4 MARKS) When cyclohexene is treated with dilute HCl in acetic acid (AcOH), the major

product is cyclohexyl acetate, while cyclohexyl chloride is only the minor product. When the

reaction is performed with the addition of Me4N+Cl− but otherwise identical conditions, two

differences are observed: 1) the major product becomes cyclohexyl chloride; 2) the rate of the

reaction increases significantly. Briefly explain the reasons for these two effects.

2. (4 MARKS) Briefly explain using mechanistic arguments why the following reaction gives

exclusively the product isomer shown.

CHEM 2220 Test #1 of 10 Feb 11, 2016

3. (6 MARKS) When the tricyclic compound shown was treated with strong acid, the product was

perhaps surprising. Write out a stepwise mechanism to account for this outcome. Notice the

stereochemistry of the starting material and the product.

CHEM 2220 Test #1 of 10 Feb 11, 2016

4. (6 MARKS) The following reaction is similar to the process by which Nature builds complex

structures like steroids. Provide a detailed stepwise mechanism for this reaction.

CHEM 2220 Test #1 of 10 Feb 11, 2016

5. (24 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)

CHEM 2220 Test #1 of 10 Feb 11, 2016

e. (2 Marks)

f. (4 Marks)

g. (4 Marks)

CHEM 2220 Test #1 of 10 Feb 11, 2016

h. (6 Marks)

CHEM 2220 Test #1 of 10 Feb 11, 2016

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

formula C6H10O2 are shown on the next page. Based on these data, answer the following questions

about compound A.

a. (1 Mark) Based on the molecular formula, what is the degree of unsaturation in A?

b. (1 Mark) The 13C NMR signal at 168.5 ppm is characteristic of a particular functional

group. Specifically, what is this functional group?

c. (1 Mark) What can be concluded from the 1H NMR singlet at 3.73 ppm?

d. (2 Marks) The 1H NMR signal at 6.86 ppm is labelled as a multiplet on the spectrum

because there is coupling to two different types of protons. A close-up of this signal is

shown here, revealing more detail. What conclusion can be drawn from the splitting pattern

of this signal?

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

CHEM 2220 Test #1 of 10 Feb 11, 2016

IR

C6H10O2

13C NMR

C6H10O2

1H NMR

C6H10O2

m,

3H m,

1H

s,

3H

m,

3H

of 10

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

University of Manitoba - Department of Chemistry

CHEM 2220 - Introductory Organic Chemistry II – MIDTERM TEST

Take-Home Test, Wed. March 1 2017 5:00 pm – Fri. March 3 2017 5:00 pm,

This is a TAKE HOME test. It is intended to be done in 2 hours, CLOSED-BOOK. It will be

marked out of 50 points total. Part marks are available on all questions.

The test is delivered electronically – this version is for later study purposes. Questions are

identical to the e-version.

A spectroscopic data sheet is attached at the end of the exam.

QUESTION MARKS

1. Mechanism

(4 Marks)

2. Mechanism

(4 Marks)

3. Mechanism

(6 Marks)

4. Mechanism

(6 Marks)

5. Reactions and Products

(24 Marks)

6. Spectra and Structures

(6 Marks)

TOTAL (50 Marks)

CHEM 2220 Test #1 ANSWERS Page 2 of 10 Mar 3, 2016

1. (4 MARKS) When cyclohexene is treated with dilute HCl in acetic acid (AcOH), the major

product is cyclohexyl acetate, while cyclohexyl chloride is only the minor product. When the

reaction is performed with the addition of Me4N+Cl− but otherwise identical conditions, two

differences are observed: 1) the major product becomes cyclohexyl chloride; 2) the rate of the

reaction increases significantly. Briefly explain the reasons for these two effects.

2. (4 MARKS) Briefly explain using mechanistic arguments why the following reaction gives

exclusively the product isomer shown.

In the simple reaction, the major product is the acetate because the concentration of acetic acid (solvent) is much higher than that of chloride ion (dilute HCl) and the second product-determining step is bimolecular. That is, its rate depends on the concentrations of both components, so reaction of acetate will be faster. When tetrabutylammonium chloride is added, this increases the concentration of chloride ion, so the reaction of chloride with the cation becomes faster.

When this alkene has one aryl ring and one proton on each end, but the aryl rings are not the same. Thus, when it is protonated, two possible cations can form. Note that while both cations are adjacent to a phenyl ring (benzylic) the one on the left gains stabilization from the electron-donating methoxy group. The one on the right is also delocalized, but lacks this extra stabilization. This means that the preferred cation is the one next to the methoxyphenyl group and this will lead to the observed product.


3. (6 MARKS) When the tricyclic compound shown was treated with strong acid, the product was

perhaps surprising. Write out a stepwise mechanism to account for this outcome. Notice the

stereochemistry of the starting material and the product.

There are two ways we could represent the attack of a carboxylic acid on an electrophile such as a cation. In the mechanism above, I drew the OH group acting as the nucleophile, but it could also be drawn with the C=O group as nucleophile. Either is acceptable here. Later in the course, we will talk more about this. There is good evidence to suggest that the C=O should be the nucleophile in most reactions, not the OH!

This is a series of 1,2-shifts towards a lower-energy cation, and then attack by the carboxyl group. Recall that these 1,2-shifts occur across one face of the molecule, via 3-membered transition states, so each group remains on the same side of the molecule that it started out on.


4. (6 MARKS) The following reaction is similar to the process by which Nature builds complex

structures like steroids. Provide a detailed stepwise mechanism for this reaction.

I drew H3O+ and the participation of water as a base explicitly, but I didn’t expect students necessarily to show this. You could just write H+ in the first step, and show the proton leaving in the last step without explicitly showing what the base is.


5. (24 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)

f. (4 Marks)

g. (4 Marks)


h. (6 Marks)


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

formula C6H10O2 are shown on the next page. Based on these data, answer the following questions

about compound A.

a. (1 Mark) Based on the molecular formula, what is the degree of unsaturation in A?

b. (1 Mark) The 13C NMR signal at 168.5 ppm is characteristic of a particular functional

group. Specifically, what is this functional group?

c. (1 Mark) What can be concluded from the 1H NMR singlet at 3.73 ppm?

d. (2 Marks) The 1H NMR signal at 6.86 ppm is labelled as a multiplet on the spectrum

because there is coupling to two different types of protons. A close-up of this signal is

shown here, revealing more detail. What conclusion can be drawn from the splitting pattern

of this signal?

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

2 degrees of unsaturation.

IR at 1717 cm-1 says CARBONYL; 13C NMR at 168.5 ppm says “CARBOXYL-TYPE” (not aldehyde or ketone). Can’t be COOH (no OH in IR); Can’t be amide, acid chloride etc (formula), MUST BE ESTER.

1H NMR 3-proton singlet at 3.73 ppm is indicative of OCH3. This is a METHYL ESTER. alkene

This is clearly a “quartet of quartets”, which means

it must be coupled to TWO DIFFERENT CH3

GROUPS.

We have a methyl ester and an alkene with ONLY ONE H on it. The alkene H is coupled to TWO DIFFERENT CH3 groups. We can see the two CH3 groups as multiplets at ~1.7 – 1.9 ppm in the 1H NMR. These are all the atoms present in the molecular formula. How can we assemble them? Obviously the ester and the two methyls must be attached to the three remaining locations on the alkene – there are three ways to do this.


IR

C6H10O2

13C NMR

C6H10O2

1H NMR

C6H10O2

m,

3H m,

1H

s,

3H

m,

3H

chem 2220 midterm 1 - university of manitobahultin/chem2220/archive/...chem 2220 test #1 answers...

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