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) HCCH
(b) H2CO
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
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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: CO and OH. The molecule in question 1(b) has only CH bonds, which are usually considered to be
non-polar.
The molecule in question 2(a) has only CC and CH bonds, which are usually
considered to be non-polar.
The molecule in question 2(b) has one polar bond: CO.
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: CO and OH
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)
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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.
CH3CH2CH3
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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
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CHEMISTRY 12
Solution
(a) CH3
C
CH
➀
➁
CH3
➂
CH2
➃
CH2
➄
CH3
➅
The propyl group should be part of the main chain. The correct name is
3-methyl-2-hexene.
(b)
➆
CH3
CH2
➅
CH3
CH2
CH
➄
➃
CH
➂
CH3
CH
➁
➀
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
➄
CH2
➃
➂
CH
➁
CH3
➀
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.
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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.
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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
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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
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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.
CH3CH2Br
(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
CH3CH2CH2I
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
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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.
CH3OCH2CH2CH3
(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.
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CHEMISTRY 12
Solution
The alcohol will be more soluble in water because it has an OH 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
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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) CH3CH2CH2CH2NHCH3
(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)
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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
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CHEMISTRY 12
32. Problem
Is a compound with a CO 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
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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
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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 CO 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
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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
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