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Chemistry 351

Chemistry 243 Mid-term Test Answers and Comments

There were two different tests. Test A had a white cover paperand Test B had a blue

cover paper.

Class Performance

The average grade was 60%. This is an acceptable average for a second year class but I

know that you can do better.

The class performance is outlined in the below chart.

0

5

10

15

20

25

90-10080-8970-7960-6950-5940-49

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Answers for test B:

Part A:

1c, 2b, 3c, 4a, 5c, 6b, 7c, 8d, 9b, 10c, 11d, 12d, 13c, 14d, 15d, 16a, 17c, 18b, 19a, 20d,21d, 22a, 23c, 24c, 25b, 26d, 27b, 28d, 29c, 30a.

Part B

1) a) Draw the ring-flipped chair conformation. The isopropyl group will now be in anequatorial position and the hydroxyl group will be in an axial position. See

chapter 2.10 and 2.11 of the textbook.

b) cis-2-isopropyl-cyclohexanol. See chapter 2.7 of the textbook.

c) See chapter 2.5 of the textbookfor a discussion of Newman projection

HH

CH3H

Cl

H

CH3H

HH

Cl

H

Staggered Gauche

Most Stable

d) The methyl groups remain eclipsed because the double bond does not allowrotation. See chapter 3.2 of the textbook for a discussion of alkene bond rotation

and chapter 3.4 for the E,Z designation rules.

C C

H

H3C

H

CH3

2) a) Find the following functional groups: carboxylic acid, ether, phenyl group or

arene, amide, amine and alcohol. See chapter 2.1 of the textbook.

b) Z or E configuration is acceptable. See chapter 3.1 of the textbook.

H3C CH3

CH3Br

c) The molecule is chiral because it has a mirror image that is not identical. Because

this is potentially a mesocompound, stating that there are chiral centres is not

sufficient to explain the chirality of this molecule. See chapter 6.8 of the

textbook.Br

Br

H3C

CH3

HH

d) 2,4,6-octatriene andpara-bromotoluene orpara-methyl-bromobenzene. See

chapter 3.1 and 5.3 of the textbook.

3) a) Draw a two step reaction where the first step is the movement of electrons from

the double bond to the proton Lewis acid of HBr to produce a tertiary cation. In

the second step, the bromide anion nucleophile moves electrons towards thecarbocation to create a bond. Be sure to follow the Markovnikov rule. See

chapter 3.7, 3.9, 4.1 and 4.2 of the textbook.

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b) The reaction coordinate should have two steps and show an intermediate. The

first energy barrier should be higher than the second. See chapter 3.8 and 3.9 ofthe textbook.

c) Your answer should include a discussion of the bromonium ion. See chapter 4.5

of the textbook.

4) The phenol ring is brominated in the orthoandparapositions, the benzoic acid

ring is not. The oxygen directly attached to the phenol ring has lone pairs which

donate electrons to the ring, activating it and favouring orthoandpara

substitution. The carbonyl group attached to the benzoic acid ring is a metadirecting deactivator. This ring is much less reactive for bromine that the

activated ring. The products that you draw should be orthoandparabrominated

on the phenol ring. See problem 3 of the sample test that was handed out one

week before this test.

5) a) There are three stereocentres. See chapter 6.1 and 6.6 of the textbook.

N

CH3

b) A mesocompound has more than one stereocentre but has an internal mirror plane

so the molecule is not chiral. See chapter 6.8 of the textbook.

CH3

Br

H3C

BrHH

6) a) Use Freidel-Krafts reactions to alkylate the benzene ring. Be aware of the order

so that you place an ortho/para director on the ring first. See chapter 5.6, 5.7 and

5.8 of the textbook.

Cl

AlCl3 AlCl3

H3C

O

Cl

OH3C

O,P-director

M-director

b) The immediate product would be an enol. You must recognize that an enol will

alwaus tautomerize to a ketone. See chapter 4.13 of the textbook.

H2C C

CH3

OH

H3CC CH3

O

Initial enol product Ketone product

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Chemistry 351

Answers for test A:

Part A:

1b, 2d, 3c, 4b, 5a, 6a, 7b, 8d, 9d, 10c, 11a, 12c, 13a, 14d, 15d, 16c, 17b, 18d, 19b, 20d,21c, 22d, 23a, 24c, 25d, 26a, 27a, 28c, 29a, 30a.

Part B

1) The phenol ring is brominated in the orthoandparapositions, the benzoic acidring is not. The oxygen directly attached to the phenol ring has lone pairs which

donate electrons to the ring, activating it and favouring orthoandpara

substitution. The carbonyl group attached to the benzoic acid ring is a meta

directing deactivator. This ring is much less reactive for bromine that theactivated ring. The products that you draw should be orthoandparabrominated

on the phenol ring. See problem 3 of the sample test that was handed out one

week before this test.

2) a) There are four stereocentres. See chapter 6.1 and 6.6 of the textbook.

O

NH2O CH3

OHO

b) A mesocompound has more than one stereocentre but has an internal mirror plane

so the molecule is not chiral. See chapter 6.8 of the textbook.

CH3

Br

H3C

BrHH

3) a) Use an acetylide anion and an alkyl bromide. See chapter 4.13 of the textbook.

Br

HC C +

CHC

+ Br

b) The immediate product would be an enol. You must recognize that an enol will

alwaus tautomerize to a ketone. See chapter 4.13 of the textbook

H2C C

OH

H3C C

O

Initial enol product Ketone product

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4) a) The methyl groups remain eclipsed because he double bond does not allow

rotation. See chapter 3.2 of the textbook for a discussion of alkene bond rotationand chapter 3.4 for the E,Z designation rules.

C C

H

H3C

H

CH3

b) See chapter 2.5 of the textbookfor a discussion of Newman projection

HH

CH3H

Cl

H

CH3H

HH

Cl

H

Staggered Gauche

Most Stable

c) Draw a chair conformation with equatorial methyl groups in the 1 and 2 positions.See chapter 2.10 and 2.11 of the textbook.

d) 1,2-dimethylcyclohexane. See chapter 2.7 of the textbook.

5) a) Find the following functional groups: 3 ketones or carbonyl groups, alkene or

double bond, alcohol and an ester. See chapter 2.1 of the textbook.

b) Z or E configuration is acceptable. See chapter 3.1 of the textbook.

H3C CH3

CH3Br

c) The molecule is not chiral because it has a mirror image that is identical. There is

an internal plane of symmetry within the molecule. See chapter 6.8 of the

textbook.

CH3

Br

H3C

BrHH

d) 2,4,6-octatriene andpara-bromotoluene orpara-methyl-bromobenzene. See

chapter 3.1 and 5.3 of the textbook.

6) a) Draw a two step reaction where the first step is the movement of electrons from

the double bond to the bromine Lewis acid of Br2to produce a tertiary cation. In

the second step, the bromide anion nucleophile moves electrons towards the

carbocation to create a bond. The bromonium ion should be an intermediate.Indicate trans addition which results in a product where both bromine atoms are

axial. Now recognize that you must ring flip the chair conformation to get bothbromines in the equatorial positions. See chapter 4.5, 2.10 and 2.11 of the

textbook.

b) The reaction coordinate should have two steps and show an intermediate. The

first energy barrier should be higher than the second. See chapter 3.8 and 3.9 of

the textbook.