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CHEMISTRY 12

Chapter 1

Classifying
Organic Compounds
Note to Teacher: You will notice that there are two different formats for the Sample
Problems in the student textbook. Where appropriate, the Sample Problem contains the
full set of steps: Problem, What Is Required, What Is Given, Plan Your Strategy, Act on
Your Strategy, and Check Your Solution. Where a shorter solution is appropriate, the
Sample Problem contains only two steps: Problem and Solution. Where relevant, a
Check Your Solution step is also included in the shorter Sample Problems.

Solutions for Practice Problems
Student Textbook page 10
1. Problem

Predict and sketch the three-dimensional shape of each single-bonded atom.
(b) C in CH4

(a) C and O in CH3OH

Solution
(a) The carbon atom has four single bonds, so it will have a tetrahedral shape.
The oxygen atom has two single bonds, so it will have a bent shape.
H



H

OH
around the carbon atom

O



H



C
H

CH3

around the oxygen atom

(b) the carbon atom has four single bonds, so it will have a tetrahedral shape.

H
C
H

H
H

2. Problem

Predict and sketch the three-dimensional shape of each multiple-bonded molecule.
(a) HC CH
(b) H2C O
Solution
(a) Each carbon atom has one triple bond and one single bond. The shape around

each carbon atom is linear, so the shape of this molecule is linear.
H

C

C

H

(b) The carbon atom has one double bond and two single bonds, so the shape around

the carbon atom (and the shape of the molecule) will be trigonal planar.
O
C
H

H
Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

3. Problem

Identify any polar bonds that are present in each molecule in questions 1 and 2.
Solution
The molecule in question 1(a) has two polar bonds: C O and O H. The molecule in question 1(b) has only C H bonds, which are usually considered to be
non-polar.
The molecule in question 2(a) has only C C and C H bonds, which are usually
considered to be non-polar.
The molecule in question 2(b) has one polar bond: C O.
4. Problem

For each molecule in questions 1 and 2, predict whether the molecule as a whole is
polar or non-polar.
Solution
For question 1(a):
Step 1 The molecule has polar bonds.
Step 2 There are two polar bonds: C O and O H
Step 3 Because there is a bent shape around the oxygen atom, the polar bonds do
not counteract each other. The molecule is polar.
For question 1(b):
There are no polar bonds, so the molecule is non-polar.

Step 1

For question 2(a):
Step 1 There are no polar bonds, so the molecule is non-polar.
For question 2(b):
The molecule has a polar bond.
Since there is only one polar bond, the molecule is polar.

Step 1
Step 2

Solutions for Practice Problems
Student Textbook pages 16–17
5. Problem

Name each hydrocarbon.
(a) H3C

CH3

(e)

(b) H2C

CH

(f)

H2C

CH

CH3

CH3

(c)

CH3

C

CH3

(g)

CH

CH2

CH3

(d)

Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

Solution
(a) Step 1
Step 2

There are only two carbon atoms, so the root is -eth-.
The name of this compound is ethane.

(b) Step 1
Step 2

There are four carbon atoms in a ring. The root is -cyclobut-.
This is an alkene, so the name of the compound is cyclobutene.

There are five carbon atoms in the main chain, so the root is -pent-.
There is a double carbon-carbon bond, so the suffix is -ene.
Number the compound from the left to give the lowest position number
to the double bond.
Step 4 There is a methyl group on the second carbon, so the prefix is 2-methyl-.
Step 5 The full name is 2-methyl-2-pentene.

(c) Step 1
Step 2
Step 3

There are six carbon atoms in the longest chain, so the root is -hex-.
The compound is an alkane, so the suffix is -ane.
Number from the left to give the branch the lowest possible
position number.
Step 4 There is a methyl group on the third carbon atom, so the prefix
is 3-methyl-.
Step 5 The full name is 3-methylhexane.

(d) Step 1
Step 2
Step 3

(e) Step 1
Step 2
Step 3
Step 4

Step 5

There are seven carbon atoms in the longest chain that contains the
double bond. The root is -hept-.
The compound is an alkene, so the suffix is -ene.
Number from the left to give the lowest possible position number to the
double bond.
There is a methyl group on the second carbon atom, and an ethyl
group on the third carbon atom. In alphabetical order, the prefix is
3-ethyl-2-methyl-.
The full name is 3-ethyl-2-methyl-2-heptene.

There are six carbon atoms in the main ring, so the root is -cyclohex-.
The compound is an alkane, so the suffix is -ane.
Start numbering at a branch, so that the two branches have the lowest
possible position numbers.
Step 4 There are two methyl groups, so the prefix is 1,2-dimethyl-.
Step 5 The full name is 1,2-dimethylcyclohexane.

(f) Step 1
Step 2
Step 3

There are five carbon atoms in the main chain, so the root is -pent-.
There is a triple bond, so the suffix is -yne.
Number from the left to give the lowest possible position number to the
triple bond.
Step 4 There are no branches.
Step 5 The full name is 2-pentyne.

(g) Step 1
Step 2
Step 3

6. Problem

Draw a condensed structural diagram for each hydrocarbon.
(a) propane
(b) 4-ethyl-3-methylheptane
(c) 3-methyl-2,4,6-octatriene
Solution
(a) Step 1
Step 2
Step 3
Step 4

The root is -prop-, so there are three carbon atoms in the main chain.
The compound is an alkane, so all the bonds are single.
There are no branches.
CH3 CH2 CH3
Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

(b) Step 1
Step 2
Step 3

The root is -hept- so there are seven carbon atoms in the main chain.
There are no double or triple bonds.
The ethyl group is attached to carbon 4. The methyl group is attached
to carbon 3.
CH3

Step 4

CH3

CH2

CH2

CH

CH

CH2

CH2

CH3

CH3

The root is -oct- so there are eight carbon atoms in the main chain.
There are three double bonds, between carbons 2 and 3, 4 and 5,
and 6 and 7.
Step 3 There is a methyl branch at carbon 3.

(c) Step 1
Step 2

CH3

Step 4

CH3

CH

C

CH

CH

CH

CH

CH3

7. Problem

Identify any errors in the name of each hydrocarbon.
(a) 2,2,3-dimethylbutane
(b) 2,4-diethyloctane
(c) 3-methyl-4,5-diethyl-2-nonyne

Solution
(a) There are three methyl groups, so the name should be 2,2,3-trimethylbutane.
(b) If you draw this compound, you can see that the main chain has more than eight

carbon atoms (the ethyl group on carbon 2 should be counted as part of the main
chain). The correct name is 5-ethyl-3-methylnonane.
(c) If you draw this compound, you will see that the third carbon atom forms more
than four covalent bonds. This is not possible. One solution would be to change
the name to 3-methyl-4,5-diethyl-2-nonene.
8. Problem

Correct any errors so that each name matches the structure beside it.
(a) 4-hexyne
(b) 2,5-hexene

CH3
CH3

CH2
C

CH

CH
C

C

C

CH2

CH3

CH3

Solution
(a) This compound has a double bond, not a triple bond. Also, the bond is located at
carbon 3. The correct name is 3-hexene.
(b) This compound has triple bonds, not double bonds. Also, since there are two
triple bonds, the prefix -di- should be used. Finally, the triple bonds are located at
carbons 2 and 4. The correct name is 2,4-hexadiyne.
9. Problem

Use each incorrect name to draw the corresponding hydrocarbon. Examine your drawing, and rename the hydrocarbon correctly.
(a) 3-propyl-2-butene
(b) 1,3-dimethyl-4-hexene
(c) 4-methylpentane

Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

Solution
(a) CH3

C

CH





CH3



CH2


CH2


CH3


The propyl group should be part of the main chain. The correct name is
3-methyl-2-hexene.
(b)



CH3
CH2


CH3
CH2

CH





CH


CH3

CH




The first methyl group should be part of the main chain. Also, you should
number in the opposite direction to give a lower position number to the double
bond. The correct name is 4-methyl-2-heptene.
CH3

(c)

CH3

CH2



CH2





CH


CH3


You should number in the opposite direction to give the lowest possible position
number for the methyl group. The correct name is 2-methylpentane.

Solutions for Practice Problems
Student Textbook page 19
10. Problem

Name the following aromatic compound.
CH3

H3C

CH3

Solution
Step 1 Start numbering at one of the branches. Since they are identical and spaced
evenly, it doesn’t matter which one.
Step 2 There are methyl groups at carbons 1, 3, and 5.
Step 3 The name is 1,3,5-trimethylbenzene.
11. Problem

Draw a structural diagram for each aromatic compound.
(a) 1-ethyl-3-methylbenzene
(b) 2-ethyl-1,4-dimethylbenzene
(c) para-dichlorobenzene (Hint: Chloro refers to the chlorine atom, Cl.)

Solution
(a)

CH2

CH3

(b)

(c)

CH3

Cl

CH2CH3
CH3
CH3

Cl

12. Problem

Give another name for the compound in question 11(a).
Solution
The compound can also be called meta-ethylmethylbenzene.
Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

13. Problem

Draw and name three aromatic isomers with the molecular formula C10H14 .
(Isomers are compounds that have the same molecular formula, but different structures. See the Concepts and Skills Review for a review of structural isomers.)
Solution
Aside from the six carbons in the benzene ring, there are an additional four carbon
atoms that exist as branches. Three possible isomers are 1,2,3,4-tetramethylbenzene,
1,2,3,5-tetramethylbenzene, and 1-methyl-4-propylbenzene, shown below. There are
many other possible isomers.
CH3

CH3
CH3

CH2CH2CH3

CH3

CH3

CH3
CH3

CH3

CH3

Solutions for Practice Problems
Student Textbook pages 26–27
14. Problem

Name each alcohol. Identify it as primary, secondary, or tertiary.
OH
(a) CH3

CH2

CH2

OH

(d) CH3

CH

CH

CH2

CH3

OH
(b)

OH

(e)

OH

OH

(c)

Solution
The main chain has three carbon atoms. The name of the parent
alkane is propane.
Step 2 Replacing -e with -ol gives propanol.
Step 3 The OH group is at carbon 1, giving 1-propanol. This is a
primary alcohol.

(a) Step 1

There are four carbon atoms in the main chain, so the parent alkane
is butane.
Step 2 The base name is butanol.
Step 3 A position number is needed, so the compound is 2-butanol. This is a
secondary alcohol.

(b) Step 1

There are four carbons in the main ring, so the parent alkane
is cyclobutane.
Step 2 The name of the compound is cyclobutanol. No position number is
needed. This is a secondary alcohol.

(c) Step 1

(d) Step 1
Step 2

There are five carbons in the main chain, so the parent alkane is pentane.
There are two OH groups, so the base name is pentanediol.

Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

Step 3

(e) Step 1
Step 2
Step 3

Step 4
Step 5

The OH groups are located on carbon 2 and carbon 3 of the main
chain. The compound name is 2,3-pentanediol. Both OH groups
are secondary.
There are seven carbons in the main chain, so the parent alkane
is heptane.
The base name is heptanol.
The OH group is located on carbon 1, giving 1-heptanol. (Note that
an OH group is always given priority over alkyl groups such as a
methyl group.)
There are methyl groups at carbon 2 and carbon 4. The prefix
is 2,4-dimethyl-.
The full name is 2,4-dimethyl-1-heptanol. This is a primary alcohol.

15. Problem

Draw each alcohol.
(a) methanol
(b) 2-propanol
(c) 2,2-butanediol

(d) 3-ethyl-4-methyl-1-octanol
(e) 2,4-dimethyl-1-cyclopentanol

Solution
(a) There is only one carbon atom in this molecule.

CH3

OH

(b) There are three carbon atoms in the main chain. The OH group is at

carbon 2.
OH
CH3

CH

CH3

(c) There are four carbon atoms in the main chain. There are two OH groups,

both located at carbon 2.
OH
CH3

C

CH2

CH3

OH
(d) There are eight carbons in the main chain. The OH group is at carbon 1, and

there is an ethyl group at carbon 3 and a methyl group at carbon 4.
CH2
HO

CH2

CH2

CH

CH3
CH

CH2

CH2

CH2

CH3

CH3
(e) There are five carbons, and the compound is ring-shaped. The OH group is at

carbon 1, and there are two methyl groups at carbons 2 and 4.
OH
CH3
CH3

Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

16. Problem

Identify any errors in each name. Give the correct name for the alcohol.
(a) 1,3-heptanol

OH
CH2

HO

CH2

CH

CH2

CH3

(b) 3-ethyl-4-ethyl-1-decanol

OH
(c) 1,2-dimethyl-3-butanol

CH3
CH2

CH

CH

CH3

CH3

OH

Solution
(a) There are five carbon atoms, so the root should be -pent-. Also, there are two

OH groups. The correct name is 1,3-pentanediol.

(b) The two ethyl groups should be grouped together, so the correct name

is 3,4-diethyl-1-decanol.
(c) There are five carbons in the main chain so the root should be -pent-, not -but-.

The OH group should be given the lowest possible position number, so the
correct name is 3-methyl-2-pentanol.

17. Problem

Sketch a three-dimensional diagram of methanol. Hint: Recall that the shape around
an oxygen atom is bent.
Solution
Note to teacher: If students complete a three-dimensional diagram of methanol
successfully, have them attempt a similar diagram of ethanol, as shown below.
H
C

H

C

H
C
H

H

H
H

H





O



• •





O

H
methanol

H

ethanol

Solutions for Practice Problems
Student Textbook page 28
18. Problem

Draw a condensed structural diagram for each alkyl halide.
(a) bromoethane
(b) 2,3,4-triiodo-3-methylheptane

Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

Solution
(a) There are two carbon atoms in the main chain. No position number is needed
because the two carbon atoms are indistinguishable.
CH3 CH2 Br
(b) There are seven carbon atoms in the main chain. There is one methyl group, at
carbon 3, and three iodine atoms, at carbons 2,3, and 4.

CH3

I

CH3 I

CH

C

CH

CH2

CH2

CH3

I
19. Problem

F

Name the alkyl halide at the right.
Then draw a condensed structural
diagram to represent it.

F

Solution
There are six carbon atoms in the main ring, and the compound is a cyclic alkane.
The parent alkane is cyclohexane. There are two fluorine atoms at positions 1 and 2.
The name is 1,2-difluorocyclohexane.
CH2
CH2

CH

F

CH2

CH

F

CH2
20. Problem

Draw and name an alkyl halide that has three carbon atoms and one iodine atom.
Solution
1-iodopropane
CH3 CH2 CH2 I
21. Problem

Draw and name a second, different alkyl halide that matches the description in the
previous question.
Solution
2-iodopropane
CH3

CH

CH3

I

Solutions for Practice Problems
Student Textbook page 30
22. Problem

Use the IUPAC system to name each ether.
(a) H3C

O

CH3

(c) CH3

CH2

CH2

CH2

O

CH3

CH3
(b) H3C

O

CH

CH3

Chapter 1 Classifying Organic Compounds • MHR

9

CHEMISTRY 12

Solution
The longest alkyl group is based on methane.
The alkoxy group is also based on methane. The prefix is 1-methoxy, or
just methoxy. (No 2-methoxy position is possible.)
Step 3 The full name is methoxymethane.

(a) Step 1
Step 2

(b) Step 1
Step 2
Step 3

The longest alkyl group is based on propane.
The prefix is 2-methoxy.
The full name is 2-methoxypropane.

(c) Step 1
Step 2
Step 3

The longest alkyl group is based on butane.
The prefix is 1-methoxy.
The full name is 1-methoxybutane.

23. Problem

Give the common name for each ether.
(a) H3C

O

CH2CH3

(b) H3C

O

CH3

Solution
(a) Step 1
Step 2

The two alkyl groups are methyl and ethyl.
The full name is ethyl methyl ether.

(b) Step 1
Step 2

The two alkyl groups are both methyl.
The full name is dimethyl ether.

24. Problem

Draw each ether.
(a) 1-methoxypropane
(b) 3-ethoxy-4-methylheptane
(c) tert-butyl methyl ether
Solution
(a) The main alkyl group has three carbon atoms. The alkoxy group has one carbon

atom, and is attached to the first carbon of the larger group.
CH3 O CH2 CH2 CH3
(b) The parent alkane is 4-methylheptane. Draw the parent alkane, then add an
ethoxy group at the third carbon atom of the main chain.
CH2

CH3

O
CH3

CH2

CH

CH

CH2

CH2

CH3

CH3
(c) There are two alkyl groups: the tert-butyl group, and the methyl group. These two

groups are connected by an oxygen atom.
CH3
CH3

C

O

CH3

CH3
25. Problem

Sketch diagrams of an ether and an alcohol with the same number of carbon atoms.
Generally speaking, would you expect an ether or an alcohol to be more soluble in
water? Explain your reasoning.

Chapter 1 Classifying Organic Compounds • MHR

10

CHEMISTRY 12

Solution
The alcohol will be more soluble in water because it has an O H bond, and can
form hydrogen bonds with water molecules. The ether can accept hydrogen bonds
from water molecules, but cannot reciprocate them, as shown below.
O
H

H
CH3CH2CH2

O

O

H

H
O

H

CH3CH2

H

O
CH3

H
ether

alcohol

Solutions for Practice Problems
Student Textbook page 32
26. Problem

Name each amine.
(a) CH3

NH2

(c) CH3

CH2

CH

CH3

NH2

CH3
(b) C(CH3)3CH2

N

CH2CH3

(d)

N

CH3

H

Solution
(a) Step 1
Step 2

The parent alkane is methane.
The name of the compound is methanamine. No position number
is needed.

The largest group has three carbon atoms in the main chain, and two
methyl groups on the second carbon atom. (This may be easier to see
if you draw a full structural diagram of the alkyl branch.) The parent
alkane is 2,2-dimethylpropane.
Step 2 The nitrogen atom is attached at carbon 1, so the base name
is 2,2-dimethyl-1-propanamine.
Step 3 An ethyl group is attached to the nitrogen atom. The corresponding
prefix is N-ethyl.
Step 4 The full name is N-ethyl-2,2-dimethyl-1-propanamine.

(b) Step 1

(c) Step 1
Step 2

The parent alkane is butane.
The nitrogen atom is attached at carbon 2, numbering from the right, so
the full name is 2-butanamine.

The parent alkane is cyclopentane.
No position number is needed, since there are no other branches on the
ring. The base name is cyclopentanamine.
Step 3 The prefix is N,N-dimethyl-.
Step 4 The full name is N,N-dimethylcyclopentanamine.

(d) Step 1
Step 2

27. Problem

Draw a condensed structural diagram for each amine.
(a) 2-pentanamine
Chapter 1 Classifying Organic Compounds • MHR

11

CHEMISTRY 12

(b) cyclohexanamine
(c) N-methyl-1-butanamine
(d) N,N-diethyl-3-heptanamine

Solution
NH2

(a)

CH3

CH2

CH

CH2

CH3

CH2

(b)

CH2

CH

CH2

CH2

NH2

CH2
(c) CH3 CH2 CH2 CH2 NH CH3
(d) CH3

CH2

CH2

CH

CH2

CH2

CH3

N
CH3

CH2

CH2

CH3

28. Problem

Classify each amine in the previous question as primary, secondary, or tertiary.
Solution
(a) primary

(b) primary

(c) secondary

(d) tertiary

29. Problem

Draw and name all the isomers with the molecular formula C4H11N.
Solution
NH

NH2

NH2
1-butanamine

N-methyl-2-propanamine

NH

2-butanamine

N

NH2

N-ethylethanamine

2-methyl-1-propanamine

NH

N,N-dimethylethanamine

NH2

N-methyl-1-propanamine

2-methyl-2-propanamine

Solutions for Practice Problems
Student Textbook page 36
30. Problem

Name each aldehyde or ketone.
O

O
(a) HC

CH2

CH2

CH3

(b)

Chapter 1 Classifying Organic Compounds • MHR

12

CHEMISTRY 12

O
(c) CH3

CH

C

O
CH2

CH2

CH3

(d) CH3

CH2

CH3

CH

CH

CH2CH3

O
(e)

Solution
(a) Step 1
Step 2

The parent alkane is butane.
The compound’s name is butanal.

(b) Step 1
Step 2
Step 3

The parent alkane is octane.
Changing the suffix gives octanone.
The carbonyl group is on the third carbon from the right, so the full
name is 3-octanone.

(c) Step 1
Step 2
Step 3

The parent alkane is 2-methylhexane.
Changing the suffix gives 2-methylhexanone.
The carbonyl group is on carbon 3, so the full name
is 2-methyl-3-hexanone.

The parent alkane is 2-ethylbutane. (Note that the main chain must
contain the carbonyl group.)
Step 2 The full name is 2-ethylbutanal.

(d) Step 1

(e) Step 1
Step 2
Step 3

The parent alkane is 4-ethyl-3-methylheptane.
Replacing the suffix gives 4-ethyl-3-methylheptanone.
The carbonyl group is at carbon 2, so the full name
is 4-ethyl-3-methyl-2-heptanone.

31. Problem

Draw a condensed structural diagram for each aldehyde or ketone.
(a) 2-propylpentanal
(b) cyclohexanone
(c) 4-ethyl-3,5-dimethyloctanal
Solution
O
CH

(a) HC

CH2

CH2

CH2

CH2

CH3

CH3

O

(b)

C
CH2

CH2

CH2

CH2
CH2

CH2

O
(c) HC

CH2

CH
CH3

CH

CH3
CH

CH2

CH2

CH3

CH3
Chapter 1 Classifying Organic Compounds • MHR

13

CHEMISTRY 12

32. Problem

Is a compound with a C O bond and the molecular formula C2H4O an aldehyde
or a ketone? Explain.
Solution
This compound must be an aldehyde. There are only two carbon atoms, so it is not
possible for the carbonyl group to have an alkyl group on each side.
33. Problem

Draw and name five ketones and aldehydes with the molecular formula C6H12O.
Solution
O
O

O

3-hexanone

2-hexanone

hexanal

O
O
4-methyl-2-pentanone

2-methylpentanal

Solutions for Practice Problems
Student Textbook page 40
34. Problem

Name each carboxylic acid.
O
(a) HO

C

CH3 O
CH2

CH3

(b) CH3

CH2

C

C

OH

CH3
O
(c)

OH

Solution
(a) Step 1
Step 2

The parent alkane is propane.
Replacing the suffix gives propanoic acid.

(b) Step 1
Step 2
Step 3

The parent alkane is butane.
Replacing the suffix gives butanoic acid.
The two methyl branches are both at carbon 2, since the carboxyl group
is carbon 1. The full name is 2,2-dimethylbutanoic acid.

The parent alkane is hexane. (Note: Even though a longer carbon chain is
possible, the main chain must include the carboxyl group.)
Step 2 Replacing the suffix gives hexanoic acid.
Step 3 There is an ethyl group at carbon 2; and two methyl groups at carbon 4
and carbon 5. The full name is 2-ethyl-4,5-dimethylhexanoic acid.

(c) Step 1

Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

35. Problem

Draw a condensed structural diagram for each carboxylic acid.
(a) hexanoic acid
(b) 3-propyloctanoic acid
(c) 3,4-diethyl-2,3,5-trimethylheptanoic acid

Solution
O
(a) CH3

CH2

CH2

CH2

CH2

C

OH
O

(b) CH3

CH2

(c) HO

CH2

CH2

CH2

O

CH3 CH3 CH2

CH3

C

CH

CH

C

CH

CH2

CH

CH2

CH2

CH2

CH2

CH3

C

OH

CH3

CH3 CH3

36. Problem

Draw a line structural diagram for each compound in question 35.
Solution
O
(a)

(c)

O
(b)

OH

OH

HO
O

37. Problem

Draw and name two carboxylic acids with the molecular formula C4H8O2.
Solution
O

O

OH

OH
butanoic acid

2-methylpropanoic acid

Solutions for Practice Problems
Student Textbook page 45
38. Problem

Name each ester.
O

O
(a) CH3CH2

O

(b) CH3CH2CH2C

CH

O

CH3

O
(c) CH3CH2CH2CH2C

O

CH2CH2CH2CH2CH3

Chapter 1 Classifying Organic Compounds • MHR

15

CHEMISTRY 12

Solution
The parent acid is methanoic acid. (Note that this is not the largest
group; rather, it is the group that contains the C O bond.)
Step 2 Replacing the suffix gives methanoate.
Step 3 An ethyl group is attached to the oxygen atom.
Step 4 The full name is ethyl methanoate.

(a) Step 1

(b) Step 1
Step 2
Step 3
Step 4

The parent acid is butanoic acid.
Replacing the suffix gives butanoate.
A methyl group is attached to the oxygen atom.
The full name is methyl butanoate.

(c) Step 1
Step 2
Step 3
Step 4

The parent acid is pentanoic acid.
Replacing the suffix gives pentanoate.
A pentyl group is attached to the oxygen atom.
The full name is pentyl pentanoate, or n-pentyl pentanoate.

39. Problem

For each ester in the previous question, name the carboxylic acid and the alcohol that
are needed to synthesize it.
Solution
(a) methanoic acid/ethanol
(b) butanoic acid/methanol
(c) pentanoic acid/1-pentanol
40. Problem

Draw each ester.
(a) methyl pentanoate
(b) heptyl methanoate
(c) butyl ethanoate

(d) propyl octanoate
(e) ethyl 3,3-dimethylbutanoate

Solution
O
O

(a)

(d)

O

O

O
(b)

O

(e)

O

O

O
(c)

O

41. Problem

Write the molecular formula of each ester in the previous question. Which esters are
isomers of each other?
Solution
(a) C6H12O2
(b) C8H16O2
(c) C6H12O2
Isomers: (a) and (c); (b) and (e)

(d) C11H22O2
(e) C8H16O2

Chapter 1 Classifying Organic Compounds • MHR

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CHEMISTRY 12

42. Problem

Draw and name five ester isomers that have the molecular formula C5H10O2 .
Solution
O

O

O
O

O

O

methyl butanoate

ethyl propanoate

methyl 2-methylpropanoate

O
O
O
O

butyl methanoate
(n-butyl methanoate)

isopropyl ethanoate

Solutions for Practice Problems
Student Textbook page 48
43. Problem

Name each amide.
O
NH2

(a) CH3CH2CH2C

O
(b) H3C

NH

N

(c)

O

CCH2CH2CH2CH2CH3

Solution
(a) Step 1
Step 2
Step 3

The parent acid is butanoic acid.
Replacing the suffix gives butanamide.
The compound is a primary amide, so no prefix is needed. The full name
is butanamide.

(b) Step 1
Step 2
Step 3
Step 4

The parent acid is hexanoic acid.
Replacing the suffix gives hexanamide.
This is a secondary amide. The prefix is N-methyl-.
The full name is N-methylhexanamide.

(c) Step 1
Step 2
Step 3
Step 4

The parent acid is 3-methylheptanoic acid.
Replacing the suffix gives 3-methyl heptanamide.
The compound is a tertiary amide. The prefix is N,N-diethyl-.
The full name is N,N-diethyl-3-methylheptanamide.

44. Problem

Draw each amide.
nonanamide
N-methyloctanamide
N-ethyl-N-propylpropanamide
N-ethyl-2,4,6-trimethyldecanamide

(a)
(b)
(c)
(d)

Chapter 1 Classifying Organic Compounds • MHR

17

CHEMISTRY 12

Solution
NH2

(a)

N

(c)

O

O
O

O
(b)

NH

(d)

NH

45. Problem

Name each amide.
(a) CH3CONH2
(b) CH3CH2CH2CH2CH2CH2CONHCH3
(c) (CH3)2CHCON(CH3)2

Solution
(a) Step 1
Step 2

The parent acid is ethanoic acid.
Replacing the suffix gives ethanamide. This is the name of
the compound.

(b) Step 1
Step 2
Step 3
Step 4

The parent acid is heptanoic acid.
Replacing the suffix gives heptanamide.
This is a secondary amide. The prefix is N-methyl-.
The full name is N-methylheptanamide.

(c) Step 1
Step 2
Step 3
Step 4

The parent acid is 2-methylpropanoic acid.
Replacing the suffix gives 2-methylpropanamide.
This is a tertiary amide. The prefix is N,N-dimethyl-.
The full name is N,N-dimethyl-2-methylpropanamide.

46. Problem

Draw a line structural diagram for each amide in the previous question.
Solution
O
(a)

N

(c)

NH2

O
O
(b)

NH

Chapter 1 Classifying Organic Compounds • MHR

18

CHEMISTRY 12

Chapter 2

Reactions of Organic
Compounds
Solutions for Practice Problems
Student Textbook page 63
1. Problem

Identify each reaction as an addition, substitution, or elimination reaction.
OH

(a)

CH3

CH2

CH

CH3 + HBr → CH3

CH2

CH3

(b)

CH3
(c)

Br

CH

CH

CH

CH3 + Cl2 → CH3

CH3 + HOH

CH
Cl

Cl

CH3

CH

CH

CH

CH3

Cl



Solution
(a) No double bond is broken or formed. Two reactants form two products. The

OH group in the organic reactant is replaced with a Br atom. This is a
substitution reaction.
(b) A double bond becomes a single bond. Two reactants become one product. The
organic product has more non-carbon atoms bonded to carbon (14 in total) than
the organic reactant (12 in total). This is an addition reaction.
(c) A double bond is formed. One reactant forms two products (HCl is eliminated
from the reactant). This is an elimination reaction.
2. Problem

Identify each reaction as an oxidation or a reduction. The oxidizing and reducing
agents are not shown.
CH3

(a)

CH3

CH

CH

CH2

O

CH3

CH → CH3

CH

CH2CH3
(b) CH3CH2

CH3
C

H

O
CH

CH2

C

OH

CH2CH3
CH3

→ CH3CH2CH2CHCH3

C
CH3

Chapter 2 Reactions of Organic Compounds • MHR

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CHEMISTRY 12

(c)

O

OH


O

OH

Solution
(a) The reactant has 16 C H bonds and one double C O bond. The product has

15 C H bonds, one double C O bond, and one C O bond. This is an
oxidation, since the product has gained a C O bond and lost a C H bond.
(b) The reactant has 12 C H bonds, while the product has 14 C H bonds. The
product has gained C H bonds, so this is a reduction. (Note that this is also an
addition reaction. The other reactant, H2 , is not shown.)
(c) The reactant has two C O bonds, while the product has two C O bonds.
Also, although they are not shown, two new C H bonds have been formed at
the site of each double bond. Since the product has lost C O bonds and gained
C H bonds, this is a reduction.
3. Problem

Classify each reaction in two different ways: for example, as oxidation and as an
elimination reaction.
H

(a) CH3

C

+ H2 → CH3CH2CH2CH2CH3

C

H

CH2CH3
O

O

(b)

CH3

C

OH + HO

CH3

CH3

C

O

CH3 + HOH

Solution
(a) A double bond becomes a single bond so this is an addition reaction. Also, the

product has gained two C H bonds, so this is a reduction.
(b) Two organic molecules combine to form a large organic molecule, producing

water as a second product. This is a condensation reaction. Also, the OH group
on the carboxylic acid is replaced with the methoxy OCH3 group, so this is a
substitution reaction.

4. Problem

Identify the type of reaction.
OH

(a)

CH3

CH

CH2 + HOH → CH3

CH

CH3

OH

(b)

CH3

CH2

CH

O
CH2

CH3 → CH3

CH2

C

CH2

CH3

Solution
(a) A double bond becomes a single bond. This is an addition reaction.
(b) The product loses a C H bond, and gains a C O bond, so this is an oxidation.

Chapter 2 Reactions of Organic Compounds • MHR

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CHEMISTRY 12

Solutions for Practice Problems
Student Textbook pages 67–68
5. Problem

Draw the reactants and products of the following reaction.
3-ethyl-2-heptene + HOH → 3-ethyl-3-heptanol + 3-ethyl-2-heptanol
Use Markovnikov’s rule to predict which of the two products will form in the greater
amount.
Solution
CH2
CH3

CH

C

CH3
CH2

CH2
CH2

CH2

CH3 + H2O → CH3

CH2

C

CH2
CH2

CH2

CH2

CH3

OH
3–ethyl–3–heptanol



CH3

CH

CH2

CH3

CH

CH2

CH2

CH2

CH3

OH
3–ethyl–2–heptanol

The OH group goes to the third carbon atom, which is the more substituted carbon
atom, that is, the carbon atom that is bonded to the largest number of other carbon
atoms. Therefore, the main product is 3-ethyl-3-heptanol.
6. Problem

Name the reactants and products of each reaction. Use Markovnikov’s rule to predict
which of the two products will form in the greater amount.
CHCH2CH2CH2CH3 + HBr →

(a) CH2

Br
CH3CHCH2CH2CH2CH3 + Br

CH2CH2CH2CH2CH2CH3

CH3

(b)

CH3

C

CH

CH2

CH3 + HOH →

CH3
CH3

C
OH

CH3
CH2

CH2

CH3 + CH3

CH

CH

CH2

CH3

OH

Solution

(a) 1-hexene + hydrobromic acid → 2-bromohexane + 1-bromohexane

The bromine atom will go to the second carbon atom, since that atom is bonded
to more carbon atoms than the first atom. The major product will be
2-bromohexane.

Chapter 2 Reactions of Organic Compounds • MHR

21

CHEMISTRY 12
(b) 2-methyl-2-pentene + water → 2-methyl-2-pentanol + 2-methyl-3-pentanol

The OH group will go to the second, more substituted carbon atom.
2-methyl-2-pentanol is the major product.

7. Problem

Draw the major product of each reaction.
CH3CH CH2 + Br2 →
CH2 CH2 + HOH →
CH2 CHCH2CH3 + HBr →
(CH3)2 C CHCH2CH2CH3 + HCl →

(a)
(b)
(c)
(d)

Solution
(a) Only one product is possible, since one of the reactants is symmetrical.
The product is 1,2-dibromopropane.
CH3

CH

CH2

Br

Br

(b) Only one product is possible, since one of the reactants is symmetrical.

CH3 CH2 OH
(c) Both reactants are asymmetrical, so two products are possible. The bromine atom
will add to the second carbon atom, so the major product is 2-bromobutane.
Br
CH3

CH

CH2

CH3

(d) Both reactants are asymmetrical, so two products are possible. The Cl atom will

add to the second carbon atom, which has more bonds to other carbon atoms
than the third carbon atom. The major product is 2-chloro-2-methylhexane.
CH3
CH3

C

CH2

CH2

CH2

CH3

Cl
8. Problem

For each reaction, name and draw the reactants that are needed to produce the
given product.
(a) ? + ? → CH3CH(Cl)CH3
(b) ? + ? → Br CH2CH2 Br

OH
(c) ? + HOH → CH3CH2CCH2CH3

CH3
(d) CH2 CHCH3 + ? → CH3CH2CH3

Solution
(a) This is probably an addition reaction, since water is not shown as a second

product (in which case, it could be a substitution reaction of CH3 CH(OH)CH3
and HCl). The organic reactant in an addition reaction must have a double bond.
The two reactants are propene and hydrochloric acid, shown below.
CH3 CH CH2 + HCl
(b) This is an addition reaction. The reactants are ethene and bromine, shown below.
CH2 CH2 + Br2

Chapter 2 Reactions of Organic Compounds • MHR

22

CHEMISTRY 12

(c) This is an addition reaction. There are two possible answers to this question.

Both answers are given below.
CH3

CH2

C

CH

CH3

CH3
3-methyl-2-pentene

CH3

CH2

C

CH2

CH3

CH2
2-ethyl-1-butene
(d) This is an addition reaction. The missing reactant is hydrogen, H2 . (Note: This

particular reaction requires a catalyst, so the actual reactant would be a reducing
agent such as H2 /Pt, where the platinum is a catalyst.)

Solutions for Practice Problems
Student Textbook page 73
9. Problem

Name each type of reaction.
(a) 1-propanol + HCl → 1-chloropropane + H2O
(b) 1-butanol + [O] → butanal
(c) CH3CH2CH2Cl + NaOH → CH3CH2CH2OH + NaCl
OH

(d)

CH3

CH

CH2

CH2

CH3

H2SO4

CH3 CH

CH

CH2

CH3 + H2O

Solution

(a) The OH group of the alcohol is replaced with a Cl atom. This is a substitution

reaction.
(b) An oxidizing agent is present, so it is an oxidation. Even if the oxidizing agent

were not given, you know that butanal has one more C O bond and one less
C H bond than 1-butanol.
(c) The Cl atom in 1-chloropropane is replaced with an OH group, so this is a
substitution reaction.
(d) A double bond is formed by heating an alcohol in acid, so this is an elimination
reaction. Water is produced as a second product.
10. Problem

Draw the structures of the reactants and products in parts (a) and (b) of question 9.
Solution
(a) CH3

CH2

CH2

OH + HCl

CH3

CH2

CH2

Cl + H2O
O

(b) CH3

CH2

CH2

CH2

OH + [O]

CH3

CH2

CH2

CH

11. Problem

Name each type of reaction.
(a) CH3

CH2

CH2

CH2

OH

(b) CH3

CH2

CH2

CH2

CH2

H2SO4

OH + [O] → (i) + [O] → (ii)

Chapter 2 Reactions of Organic Compounds • MHR

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CHEMISTRY 12

OH
(c) CH3

CH

CH

CH3 + HBr



CH3
CH3
(d) CH3

CH

CH2

Br + NaOH →

OH
(e) CH3

CH2

CH

CH3 + [O] →

OH
+ HCl →

(f)

Solution
(a) An alcohol is heated in the presence of sulfuric acid. These are the conditions

required for an elimination reaction.
(b) An oxidizing agent is present, and the reactant is not a tertiary alcohol. An
(c)
(d)
(e)
(f)

oxidation will occur.
A substitution reaction will occur.
A substitution reaction will occur.
An oxidation will occur.
A substitution reaction will occur.

12. Problem

Draw and name the products of each reaction in question 11.
Solution

(a) 1-butene + water

CH3

CH2

CH

CH2 + H2O

(b) (i) pentanal

O
CH3

CH2

CH2

CH2

CH

(ii) pentanoic acid

O
CH3

CH2

CH2

CH2

C

OH

(c) 2-bromo-3-methylbutane + water

Br
CH3

CH

CH

CH3

+

H2O

CH3
(d) 2-methyl-1-propanol + sodium bromide

CH3
CH3

CH

CH2

OH

+

NaBr

(e) butanone

O
CH3

CH2

C

CH3

Chapter 2 Reactions of Organic Compounds • MHR

24

CHEMISTRY 12
(f) chlorocyclohexane + water

Cl
+ H2O

13. Problem

Is the following reaction possible? Why or why not?
CH3
CH3

CH2

CH2

C

OH

+



[O]

CH3

Solution
No, this reaction cannot occur. Tertiary alcohols have no hydrogen atom available to
be removed, so they do not react with ordinary oxidizing agents.

Solutions for Practice Problems
Student Textbook page 78
14. Problem

Identify each type of reaction. (Oxidizing and reducing agents are not shown.)
O
(a)

OH


O

(b) (CH3)3C

C

OH
C(CH3)3 → (CH3)3C

CH

C(CH3)3

(c) 2-pentanone → 2-pentanol
(d) hexanoic acid + ethanol
ethyl hexanoate + water
(e) butyl methanoate + water
methanoic acid + 1-butanol

Solution
(a) The product has lost a C O bond and gained a C H bond (at the carbon
bonded to the oxygen atom). This is a reduction.
(b) A ketone is reduced to an alcohol. This is a reduction.
(c) This is a reduction.
(d) This is an esterification (condensation) reaction.
(e) This is a hydrolysis reaction.
15. Problem
(a) Name the reactants and products of the first two reactions in question 14.
(b) Draw the reactants and products of the last three reactions in question 14.

Solution
(a) cyclobutanone, cyclobutanol; 2,2,4,4-tetramethyl-3-pentanone,

2, 2, 4, 4-tetramethyl-3-pentanol
O
(b)

CH3

CH2

CH2

C

2-pentanone

OH
CH3

CH3

CH2

CH2

CH

CH3

2-pentanol

Chapter 2 Reactions of Organic Compounds • MHR

25

CHEMISTRY 12

O
CH3

CH2

CH2

CH2

CH2

C

OH

CH3

hexanoic acid

CH2

OH

ethanol

O
CH3

CH2

CH2

CH2

CH2

C

O

CH2

CH3

H

ethyl hexanoate

O

water

H

O
H

C

O

CH2

CH2

CH2

CH3

H

butyl methanoate

O

water

H

O
H

C

OH

CH3

CH2

CH2

methanoic acid

CH2

OH

1-butanol

16. Problem

Name and draw the product(s) of each reaction.
hexanal + [O] →
octanal + [H] →
propanoic acid + methanol

propyl ethanoate + water

3-hexanone + [H] →
2-propanol + 3-methylpentanoic acid


(a)
(b)
(c)
(d)
(e)
(f)

Solution
(a) hexanoic acid

CH3

CH2

O
CH2

CH2

CH2

C

OH

CH2

CH2

CH2

CH2

+

H2O

(b) 1-octanol

CH3

CH2

CH2

CH2

OH

(c) methyl propanoate + water

O
CH3

CH2

C

O

CH3

(d) 1-propanol + ethanoic acid

O
CH3

CH2

CH2

OH

+

CH2

CH2

CH3

C

OH

(e) 3-hexanol

OH
CH3

CH2

CH

CH3

(f) 2-propyl 3-methylpentanoate (isopropyl 3-methylpentanoate)

O

CH3
CH3

CH2

CH

CH2

C

CH3
O

CH
CH3

Chapter 2 Reactions of Organic Compounds • MHR

26

CHEMISTRY 12

17. Problem

Name and draw the reactant(s) in each reaction.

(a) ? + [H] → CH3CH2CH(CH3)CH2OH
(b) ? + [O] →? + [O] → CH3CH2CH2CH2COOH

O

(c) ? + ? →

O

Solution
(a) 2-methylbutanal
CH3

CH2

CH3

O

CH

CH

(b) 1-pentanol; pentanal

CH3

CH2

CH2

CH2

CH2

OH

O
CH3

CH2

CH2

CH2

CH

(c) 1-propanol + 3-methylhexanoic acid

OH

+

HO
O

Solutions for Practice Problems
Student Textbook page 84
18. Problem

A monomer called methylmethacrylate polymerizes to form an addition polymer that
is used to make bowling balls. What is the name of this polymer?
Solution
The name of the polymer is based on the name of the monomer, so the polymer is
polymethylmethacrylate (commonly known as PMMA).
19. Problem

Classify each polymerization reaction as an addition or condensation polymerization
reaction.

(a) nHO



O

O

C

C

O

O

C

C

CN

(b)

nCH2

OH + nHO

CH2

CH2

O

CH2

CN

CH → …

CH2

nHO

CH2

C

O



CN

CH

CH2

CH



O

O

(c)

OH → …

OH →

CH2

O

CH2

C

O
O

CH2

C



Chapter 2 Reactions of Organic Compounds • MHR

27

CHEMISTRY 12

Solution
(a) Ester bonds are formed between monomers, so this is a condensation
polymerization.
(b) The reactant has a double bond, while the product does not. Therefore, this is an
addition polymerization.
(c) Ester bonds are formed between monomers, so this is a condensation
polymerization.
20. Problem

Draw the product of each polymerization reaction. Include at least two linkages for
each product.
O
(a) nHO

OH + nHO

CH2CH2CH2

O

C

OH →

C

CH2

CH3
CH →

(b) nH2C

O

O
CH2NH2 + nHO

(c) nH2NCH2

OH →

C(CH2)6C

Solution
O

(a) …

O

O

CH2CH2CH2
CH3

(b) …

CH2

CH

O

C

CH2

CH3
CH2

NHCH2

C

O

CH2CH2CH2

O

C

O
CH2

C



CH3

CH

CH2NH



CH

CH2

O

O

(c) …

O

O
NHCH2

C(CH2)6C

CH2NH

O

C(CH2)6C



21. Problem

Classify each polymer as an addition polymer (formed by addition polymerization) or
a condensation polymer (formed by condensation polymerization). Then classify each
condensation polymer as either a polyester or a nylon (polyamide).
(a) …

CH2

CH

CH2



CH

Br

Br
O

(b) …

NH

CH2

NH

O
CH2CH2

C

O
(c) …

O

CH2CH2

O

C

O

NH

NH

CH2

O
CH2CH2

O
O

(d) …

C

C

O
O

CH2CH2
O

O

CCH2C

O



O

C



O

CCH2C



Chapter 2 Reactions of Organic Compounds • MHR

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CHEMISTRY 12

Solution
(a) The polymer has a backbone of carbon atoms, with no ester or amide bonds, so it
is an addition polymer.
(b) The polymer has amide bonds in the main chain, so it was formed by condensation polymerization. It is a condensation polymer, and it is a polyamide, or nylon.
(c) The polymer has ester bonds in the main chain, so it is a condensation polymer
and a polyester.
(d) The polymer has ester bonds in the main chain, so it is a condensation polymer
and a polyester.
22. Problem

Draw the structure of the repeating unit for each polymer in question 21. Then draw
the structure of the monomer(s) used to prepare each polymer.
Solution
(a) repeating unit:



CH2



CH
Br

monomer:

CH2

CH
Br

(b) repeating unit:

O


C

NH

CH2

NH

O
CH2CH2

C



O
monomers:

CH2

H2N

NH2

HO

O
CH2CH2

C

C

OH

O
(c) repeating unit:



O

CH2CH2



C

O
monomer:

HO

CH2CH2

C

OH
O

O

(d) repeating unit:



O

O



CCH2C
O

monomers:

HO

OH

HO

O

CCH2C

OH

Chapter 2 Reactions of Organic Compounds • MHR

29

SCH4U Chem 12 Chapter 3

CHEMISTRY 12

Chapter 3

Atoms, Electrons, and
Periodic Trends
Solutions for Practice Problems
Student Textbook page 136
1. Problem

What are the allowed values for l in each of the following cases?

(a) n = 5
(b) n = 1

Solution
For any value of the quantum number n, the allowed values of the quantum number
l are integers ranging from 0 to (n − 1).
(a) n = 5, ∴(n − 1) = 4, and l can have values 0, 1, 2, 3, or 4.
(b) n = 1, ∴(n − 1) = 0, and l can have only the value 0.
Check Your Solution
For any value of n, there are n possible values of l. When n = 5, there were five
allowed values of l. When n = 1, there was one allowed value for l. The answer is
correct.
2. Problem

What are the allowed values for ml for an electron with the following quantum
numbers?
(a) l = 4
(b) l = 0
Solution
For any value of the quantum number l, the allowed values of the quantum number
ml are integers ranging from –l…0…+l.
(a) For l = 4, ml can have the values –4, –3, –2, –1, 0, +1, +2, +3, and +4.
(b) For l = 0, ml can only have the value 0.
Check Your Solution
For any value of l there can be (2l + 1) values of ml . For l = 4, there are
2(4) + 1 = 9 values for ml . For l = 0, there is 2(0) + 1 = 1 value for ml .
3. Problem

What are the names, ml values, and total number of orbitals described by the
following quantum numbers?
(a) n = 2, l = 0
(b) n = 4, l = 3

Chapter 3 Atoms, Electrons, and Periodic Trends • MHR

30

CHEMISTRY 12

Solution
The type of orbital is determined by the value of n and l as n(l ). The orbital is s, for
l = 0, p for l = 1, d for l = 2 and f for l = 3. You can find the possible values for
ml from l. The total number of orbitals is given by the total number of ml values.
(a) For l = 0, ml can have only the value 0. Since n = 2 and l = 0, the quantum
numbers represent a 2s orbital. Since there is only one allowed value for ml , there
is one 2s orbital.
(b) For l = 3, ml can have allowed values of –3, –2, –1, 0, +1, +2, and +3. Since
n = 4 and l = 3, the quantum numbers represent a 4f orbital. Since there are
seven allowed values for ml , there are seven 4f orbitals.
Check Your Solution
Since the total number of orbitals for any given l value is (2l + 1), there should be
2(0) + 1 = 1 orbital when l = 0, and 2(3) + 1 = 7 orbitals when l = 3. This answer
is correct.
4. Problem

Determine the n, l, and possible ml values for an electron in the 2p orbital.
Solution
The type of orbital takes its name from the value of the quantum numbers n and l.
A p orbital corresponds to l = 1. Since it is a 2p orbital, n = 2. You can find the
possible values for ml from l. For l = 1, the allowed values that are possible for ml
are –1, 0, and +1.
Check Your Solution
For any value of l, there can be (2l + 1) values of ml . For l = 1, there are
2(1) + 1 = 3 values for ml . This answer is correct.
5. Problem

Which of the following are allowable sets of quantum numbers for an atomic orbital?
Explain your answer in each case.
(a) n = 4, l = 4, ml = 0
(b) n = 3, l = 2, ml = 1
(c) n = 2, l = 0, ml = 0
(d) n = 5, l = 3, ml = −4
Solution
For any value of the quantum number n, the allowed values of the quantum number
l are integers ranging from 0 to (n − 1). For any value of the quantum number l, the
allowed values of the quantum number ml are integers ranging from –l …0…+l.
Apply these criteria to each case.
(a) For n = 4, the allowed values for l are 0, 1, 2, or 3. l = 4 is not an allowed value.
It is allowable for ml to have the value 0. This combination of quantum numbers
is not allowable.
(b) For n = 3, the allowed values for l are 0, 1, or 2. For the value of l = 1, ml can
have the value 1. This combination of quantum numbers is allowable.
(c) For n = 2, the allowed values for l are 0 or 1. For the value of l = 0, ml can have
the value 0. This combination of quantum numbers is allowable.
(d) For n = 5, the allowed values for l are 0, 1, 2, 3, or 4. For l = 3, ml cannot have
the value –4. This combination of quantum numbers is not allowable.
Check Your Solution
In (b) and (c), the criteria for allowed quantum numbers are met.

Chapter 3 Atoms, Electrons, and Periodic Trends • MHR

31

CHEMISTRY 12

Solutions for Practice Problems
Student Textbook pages 145–146
6. Problem

Use the aufbau principle to write complete electron configurations and complete
orbital diagrams for atoms of the following elements: sodium, magnesium,
aluminum, silicon, phosphorus, sulfur, chlorine, and argon (atomic numbers 11
through 18).
Solution
(See solution for Problem 7)
7. Problem

Write condensed electron configurations for atoms of these same elements.
Solution (Problem 6 and Problem 7)
The aufbau principle is a process of building up the ground state electron configuration of atoms, by adding one electron to the lowest available energy level in order of
increasing atomic number. To build up to element 18, the order in which orbitals
are filled increases: 1s<2s<2p<3s<3p. The Pauli Exclusion Principle specifies that
no two electrons in an atom can have the same set of four quantum numbers. This
leads to the conclusion that an s orbital can hold a maximum of two electrons and
the three p orbitals at any energy level can hold, in total, a maximum of six electrons.
Electrons in the same orbital are shown to have opposite spin, ↑↓, a consequence
of the Pauli Exclusion Principle that is referred to as Hund’s Rule. A condensed
electron configuration represents the electrons in the filled inner core by the symbol
of the noble gas to which it corresponds, followed by the electron configuration of
the valence shell.

18

1s 22s 22p63s 23p6




















































[Ne]3s 23p4
[Ne]3s 23p5
[Ne]3s 23p6



Ar


→ → →

→ →




1s 22s 22p63s 23p5

[Ne]3s 23p3



17

[Ne]3s 23p2



Cl

[Ne]3s 23p1



1s 22s 22p63s 23p4

[Ne]3s 2



16

[Ne]3s1



S

Condensed
configuration

3p



1s 22s 22p63s 23p3



15



P



1s 22s 22p63s 23p2



14



Si



1s 22s 22p63s 23p1



13



Al



1s 22s 22p63s 2



12



Mg

3s


→ →
→ → →

1s 22s 22p63s1

2p

→ → → → → → →

11

2s

→ → → → → → → →
→ → → → → → → →
→ → → → → → → →

Na

1s

→ → → → → → → →

Z

Orbital diagram

→ → → → → → → →

Element

Electron
configuration

Check Your Solution
The number of electrons increases in order of increasing atomic number, and the
sequence predicted by the aufbau principle is followed. In any orbital containing
two electrons, the electrons are of opposite spin.
8. Problem

Make a rough sketch of the periodic table for elements 1 through 18, including
the following information: group number, period number, atomic number, atomic
symbol, and condensed electron configuration.

Chapter 3 Atoms, Electrons, and Periodic Trends • MHR

32

CHEMISTRY 12

Solution
1
IA

2
IIA

13
IIIA

14
IVA

15
VA

16
VIA

17
VIIA

18
VIIIA

1

1
H
1.01
1s1

2
He
4.003
1s 2

2

3
Li
6.941
[He]2s1

4
Be
9.012
[He]2s 2

5
B
10.81
[He]2s 22p1

6
C
12.01
[He]2s 22p2

7
N
14.01
[He]2s 22p3

8
O
16.00
[He]2s 22p4

9
F
19.00
[He]2s 22p5

10
Ne
20.18
[He]2s 22p6

3

11
Na
22.99
[Ne]3s1

12
Mg
24.13
[Ne]3s 2

13
Al
26.98
[Ne]3s 23p1

14
Si
28.09
[Ne]3s 23p2

15
P
30.97
[Ne]3s 23p3

16
S
32.07
[Ne]3s 23p4

17
Cl
35.45
[Ne]3s 23p5

18
Ar
39.95
[Ne]3s 23p6

Check Your Solution
The arrangement of elements 1–18 follows a regular pattern. The number of energy
levels in any horizontal row corresponds to the period number, and the number of
electrons in the valence shell corresponds to the group number (or to the last digit
of the group number).
9. Problem

A general electron configuration for atoms belonging to any element of group 1
(IA) is ns 1 , where n is the quantum number for the outermost occupied energy level.
Based on the patterns you can observe so far for elements 1 to 18, predict the general
electron configuration for the outermost occupied energy levels of groups 2 (IIA),
13 (IIIA), 14 (IVA), 15 (VA), 16 (VIA), 17 (VIIA), and 18 (VIIIA).
Solution
Referring to the solution given for problem 8, it can be observed that within any
group, the electrons in the valence shell are found in the same type of orbitals, and
the number of valence electrons is equal to the group number or last digit of the
group number. Using these criteria, the pattern should be the same for all elements
within the group: 2 (IIA): ns 2 ; 13 (IIIA): ns 2np1; 14 (IVA): ns 2np2; 15 (VA): ns 2np3;
16 (VIA): ns 2np4; 17 (VIIA): ns 2np5; 18 (VIIIA): ns 2np6.
Check Your Solution
The number of the electrons in the valence shell of elements within a group follows
the expected pattern — the number of electrons is equal to the number of the group
or the last digit of the group number. The link between the orbital notation and the
group number has been applied correctly.

Solutions for Practice Problems
Student Textbook page 150
10. Problem

Without looking at a periodic table, identify the group number, period number, and
block of an atom that has the following electron configuration.
(a) [Ne]3s 1
(b) [He]2s 2
(c) [Kr]5s 24d 105p 5

Chapter 3 Atoms, Electrons, and Periodic Trends • MHR

33

CHEMISTRY 12

Solution
(a) This element corresponds to the general notation ns 1 , which is an s block element,
since there are only s electrons in the valence shell. n = 3, so the element is in
period 3. Since there is only one valence electron, the element is in group 1.
(b) This element corresponds to the general notation ns 2 , which is an s block element,
since there are only s electrons in the valence shell. n = 2, so the element is in
period 2. Since there are two valence electrons, the element is in group 2.
(c) This element corresponds to the general notation ns 2np5, which is a p block
element, since there are s and p electrons in the valence shell. n = 5, so the
element is in period 5. Since there are seven valence electrons, the element is
in group 17.
Check Your Solution
The link between n value and period, and the match between the valence electrons
for each element and the general notation for elements in that group have been
correctly applied.
11. Problem

Use the aufbau principle to write the complete electron configurations for the atom
of the element that fits the following descriptions:
(a) group 2 (IIA) element in period 4
(b) noble gas in period 6
(c) group 12 (IIB) element in period 4
(d) group 16 (IVB) element in period 2
Solution
(a) Group 2 elements have the general notation ns 2 , and for an element in period 4,

n = 4. Therefore, the valence electrons are represented by 4s 2 . Also, an element in
period 4 will have an inner core that corresponds to the noble gas in period 3 of
the periodic table, which is argon. The element is Ca, and its complete electron
configuration is 1s 22s 22p63s 23p64s 2 .
(b) The noble gas in period 6 will have 6 energy levels and have the general notation
6s 26p6. Follow the sequence outlined in the aufbau principle until you reach
6p6 . This should be the noble gas radon. Its complete electron configuration is
1s 22s 22p63s 23p64s 23d 104p65s 24d 105p66s 24f 145d 106p6
(c) Group 12 elements have the general notation ns 2(n − 1)d 10 , and for an element
in period 4, n = 4. Also, an element in period 4 will have an inner core that
corresponds to the noble gas in period 3 of the periodic table, which is argon.
The condensed electronic configuration will be [Ar]4s 23d 10 . The element is Zn,
and its complete electron configuration is 1s 22s 22p63s 23p64s 23d 10 .
(d) Group 16 elements have the general notation ns 2p 4 , and for an element in
period 2, n = 2. Also, an element in period 2 will have an inner core that
corresponds to the noble gas in period 1 of the periodic table, which is helium.
The element is O, and its complete electron configuration is 1s 22s 22p4 .
Check Your Solution
A check of the complete electron configurations shows that the link between n and
period number, the sequence predicted by the aufbau principle, and the match
between the valence electrons for each element and the general notation for elements
in that group are correctly followed.

Chapter 3 Atoms, Electrons, and Periodic Trends • MHR

34


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