Chapter 9
Hydrocarbons
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 332
1. Problem
Heptane has 7 carbon atoms. What is the chemical formula of heptane?
What Is Required?
You need to establish the chemical formula of heptane.
What Is Given?
Heptane has 7 carbon atoms.
Plan Your Strategy
Heptane is an alkane, meaning it is a saturated hydrocarbon. In these, there are 2
hydrogen atoms for every middle C, and 3 H atoms for the two outer C atoms.
Multiply the number of middle carbon atoms by 2H and the outer carbons by 3H
and add the results for the total number of H atoms in the formula.
Act on Your Strategy
(5 × 2 H atoms) + (2 × 3 H atoms) = 16 H atoms. Therefore, the formula should
be C7H16.
Check Your Solution
The formula of heptane can be found in most chemistry books and on the Internet.
Make a search to verify your answer.
2. Problem
Nonane has 9 carbon atoms. What is the chemical formula?
What Is Required?
You need to establish the chemical formula of nonane.
What Is Given?
Nonane has 9 carbon atoms.
Plan Your Strategy
Nonane is an alkane, meaning it is a saturated hydrocarbon. In these, there are 2
hydrogen atoms for every middle C, and 3 H atoms for the two outer C atoms.
Multiply the number of middle carbon atoms by 2H and the outer carbons by 3H
and add the results for the total number of H atoms in the formula.
Act on Your Strategy
(7 × 2 H atoms) + (2 × 3 H atoms) = 20 H atoms. Therefore, the formula should
be C9H20.
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Check Your Solution
The formula of nonane can be found in most chemistry books and on the Internet.
Make a search to verify your answer.
3. Problem
An alkane has 4 carbon atoms. How many hydrogen atoms does it have?
What Is Required?
You need to find the number of hydrogen atoms in the alkane.
What Is Given?
The alkane has 4 carbon atoms.
Plan Your Strategy
An alkane is a saturated hydrocarbon. In these, there are 2 hydrogen atoms for every
middle C, and 3 H atoms for the two outer C atoms. Multiply the number of middle
carbon atoms by 2H and the outer carbons by 3H and add the results for the total
number of H atoms in the formula.
Act on Your Strategy
(2 × 2 H atoms) + (2 × 3 H atoms) = 10 H atoms.
Check Your Solution
You can check by searching in books or on the Internet if the alkane C4H10. really
does exist. It does, and is in fact butane.
4. Problem
Candle wax contains an alkane with 52 hydrogen atoms. How many carbon atoms
does this alkane have?
What Is Required?
You need to find the number of carbon atoms in the candle wax alkane.
What Is Given?
The number of hydrogen atoms is 52.
Plan Your Strategy
An alkane is a saturated hydrocarbon. In these, there are 2 hydrogen atoms for every
middle C, and 3 H atoms for the two outer C atoms. Subtract 6 H atoms from the
total number given, and reserve these 6 for the 2 outer carbons. Divide the difference
by 2 to get the number of middle carbons.
Act on Your Strategy
Subtracting from the 2 outer C atoms: 52 H atoms − 6 H atoms = 46 H atoms
Middle C atoms: 46 H atoms / 2 H atoms per C atom = 23 C atoms.
23 middle C atoms + 2 outer C atoms = 25 C atoms
Check Your Solution
You can check by searching in books or on the Internet if the candle wax alkane
C25H52 really does exist. It does, and the unbranched alkane is called pentacosane.
Solutions for Practice Problems
Student Textbook pages 336–337
5. Problem
Name each compound.
CH3
(a)
CH3
CH
CH2
CH3
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(b)
CH3
CH3
C
CH3
CH3
(c)
CH3
CH
CH2
CH2
CH3
CH
CH
CH2
CH3
CH3
CH3
(d)
CH3
CH3
CH2
CH3
CH2
CH3
C
CH2
CH3
CH3
(e)
CH3
CH2
CH3
CH2
C
CH2
CH3
CH2
C
CH2
CH3
CH2
CH2
CH3
What Is Required?
You have to name the compounds listed.
What Is Given?
The structural formula for each compound is given.
Plan Your Strategy
For each structure, follow the following steps.
Step 1 Find the parent chain, or the longest continuous chain. The number of carbons in the parent chain forms the root of the main alkane name.
Step 2 Identify any branches present. The end closest to the branch is the lower-numbered end of the main parent chain. Number the parent from this end across.
Step 3 If there are more than one type of branch, remember that the numbering of
the branch must reflect its lowest possible number in the chain.
Step 4 Identify the location of the branch relative to the main chain. Number the
branch by the number of main chain C it is attached to. Change the -ane suffix of the branch to -yl.
Step 5 If more than one of the same type of branch alkane is present, name them
once using the appropriate number prefixes (di- for two, tri- for 3, tetra- for 4,
and so on).
Step 6 Put the prefix + suffix + root name together for the name of the compound, in
alphabetical order of branches.
Act on Your Strategy
(a) The main chain has 4 carbons, so it will be butane.
The methane branch is attached to C2 of the main root, so it is a 2-methyl
branch.
The name of the structure is 2-methylbutane.
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(b) The main branch has 3 carbons, so it with be propane. The two methane branch-
es are both on the C2 of the main chain, so they will be 2, 2-dimethyl branches
(di- because there is two of them). The name of the structure is 2, 2-dimethylpropane.
(c) The longest main branch has 6 carbons, so it will be hexane. There are 2 ethane
groups attached to the root and one methane group. Since the ethane groups must
have the lowest numbering possible, numbering of the chain will go from left to
right, so that they are at C2 and C4 (if numbering was right to left, they would be
at positions C3 and C5 which ere not the lowest possible numbers). The methane
is at C5. The name is therefore 2,4-diethyl-5-methylhexane.
(d) The main branch has 6 carbons, so it is a hexane. Two methane groups are
attached to the C2 of the main branch as well as C4, so these will be 2,2,4,4tetramethyl branches (tetra- because there are 4 of them). The name is 2,2,4,4tetramethylhexane.
(e) There are 8 main carbons in the root so it is an octane. There are 3 methane
parts, so in order to give it their lowest number, numbering of the chain will proceed from right to left. The methane groups are thus two at C2 and one at C4, so
they will be names 2,2,4-trimethyl branches. A propyl branch is attached to the
C4 main carbon, so it will be called 4-propyl. The name, taken together in alphabetical order, is 2,2,4-trimethyl-4-propyloctane.
Check Your Solution
Try searching for these names on the Internet and in chemistry books and see if the
structures and names match.
6. 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.
7. Problem
Name each compound
(a)
(b)
(c)
Solution
In a line structural formula, the end of the line, and a point at which lines meet,
represent a carbon atom. Sufficient hydrogen atoms are assumed to be around each
carbon atom to fulfill the valence requirements for carbon. There are only single
carbon-carbon bonds shown so each example in this question is an alkane or an
alkane derivative.
(a) There are 7 carbon atoms and 16 hydrogen atoms. The compound is heptane.
(b) The longest chain has 5 carbons and the compound will be a derivative of
pentane. On carbons C2 and C3, there is a methyl group. The compound is
2,3-dimethylpentane.
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(c) The longest chain has 6 carbon atoms and the compound will be a derivative of
hexane. On carbon C4 there is an ethyl group and on carbons C2 and C3 there
are methyl groups. The compound is 4-ethyl-2,3-dimethylhexane.
Check Your Solution
Recheck the number of carbons in the longest chain and verify that the branches are
on the lowest possible numbered carbon.
Solutions for Practice Problems
Student Textbook pages 338–339
8. 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
(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
CH
CH2
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
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
Solution
(a) CH3
➀
CH
➁
C
➂
CH2
➃
CH3
CH2
➄
CH3
➅
The propyl group should be part of the main chain. The correct name is
3-methyl-2-hexene.
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(b)
➆
CH3
CH2
➅
CH3
CH2
CH
CH
CH
➃
➂
➁
➄
CH3
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
➄
CH2
➃
CH2
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.
10. Problem
Draw a line structural diagram for each alkane.
3-ethyl-3,4-dimethyloctane
2,3,4-trimethylhexane
5-ethyl-3,3-dimethylheptane
4-butyl-6-ethyl-2,5-dimethylnonane
(a)
(b)
(c)
(d)
Solution
In a line structural formula, the end of the line, and a point at which lines meet,
represent a carbon atom. Sufficient hydrogen atoms are assumed to be around each
carbon atom to fulfill the valence requirements for carbon.
(a) The root and suffix is octane. On this chain of 8 carbon atoms there will be an
ethyl group, C2H5, on C3 and methyl groups, CH3 on C3 and C4.
(b) The root and suffix is hexane. On this chain of 6 carbons, there will be methyl
groups, CH3, on C2, C3, and C4.
(c) The root and suffix is heptane. On this chain of 7 carbon atoms there will be an
ethyl group, C2H5, on C5 and two methyl groups, CH3, on C3.
(d) The root and suffix is nonane. On this chain of 9 carbon atoms there will be a
butyl group, C4H9, on C4, an ethyl group, C2H5, on C6 and methyl groups,
CH3, on C2 and C5.
Check Your Solution
Confirm that the number of carbons in the longest chain matches the root of the
name and that the alkyl groups are attached to the proper numbered carbon atom in
the longest chain.
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11. Problem
One way to assess how well you have learned a new skill is to identify mistakes.
Examine the following compounds and their names. Identify any mistakes, and correct the names.
(a) 4-ethyl-2-methypentane
CH3
CH3
CH
CH2
CH
CH3
CH2
CH3
(b) 4,5-methylhexane
CH3
CH3
CH2
CH2
CH
CH
CH3
CH2
(c) 3-methyl-3-ethylpentane
CH3
CH3
CH2
C
CH2
CH3
CH2
CH2
CH3
What Is Required?
You have to verify if the names for the given structural diagrams are correct.
What Is Given?
The structural diagram is given with a chemical name.
Plan Your Strategy
For each diagram, use the rules for naming alkanes from structural diagrams.
Step 1 Find the parent chain, or the longest continuous chain. The number of carbons in the parent chain forms the root of the main alkane name.
Step 2 Identify any branches present. The end closest to the branch, or which has the
most branches, is the lower-numbered end of the main parent chain. Number
the parent from this end across.
Step 3 Identify the location of the branch relative to the main chain. Number the
branch by the number of main chain C it is attached to. Change the -ane suffix of the branch to -yl.
Step 4 If more than one of the same type of branch alkane is present, name them
once using the appropriate number prefixes (di- for two, tri- for 3, tetra- for 4,
and so on)
Act on Your Strategy
(a) 2,4-dimethylhexane → the longest chain should be 6 carbons, not 5, and two
methyl branches therefore occur at C2 and C4.
(b) 2,3-dimethylhexane → numbering on the main chain should occur at the end
closest to substitution, or the end with the most branches.
(c) 3-ethyl-3-methylhexane → the longest chain is 6 carbons, not 5.
Chapter 9 Hydrocarbons • MHR
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Check Your Solution
Try searching for these names on the Internet and in chemistry books and see if the
structures and names match. Also, try drawing the structures for the incorrect names
given, using the drawing rules for alkanes, and see how the structures differ.
Solutions for Practice Problems
Student Textbook pages 343–344
12. Problem
The following equation shows the combustion of 3-ethyl-2,5-dimethylheptane:
C11H24 + 17O2 → 11CO2 + 12H2O
(a) Does this equation show complete or incomplete combustion ?
(b) Draw the structural formula for 3-ethyl-2,5-dimethylheptane.
What Is Required?
(a) You have to decide if the combustion is complete or incomplete.
(b) You have to draw the structural formula for the alkane.
What Is Given?
The balanced chemical equation is given.
Plan Your Strategy
(a) A complete combustion will produce only carbon dioxide and water. An incom-
plete combustion will produce other by-products.
(b) Follow the steps below.
Step 1 The root chain is heptane, so it will have 6 carbons.
Step 2 There will be a ethane group on C3 and two methane groups on C2 and
C5. Add the methane groups first at these two positions (you can arbitrarily choosing either end as C1).
Step 3 Add the ethane group last to C3.
Act on Your Strategy
(a) Since only CO2 and H2O are in the equation, it shows complete combustion.
(b)
CH3
CH3
CH3
CH
CH
CH2
CH
CH2
CH3
C2H5
3-ethyl-2,5-dimethylheptane
Check Your Solution
For (b), work backward. Use the naming rules for an alkane to name the structure
you have drawn. The answer should match.
13. Problem
(a) Write a balanced equation for the complete combustion of pentane, C5H12.
(b) Write a balanced equation for the complete combustion of octane, C8H18.
(c) Write two possible balanced equation for the incomplete combustion of ethane,
C2H6.
What Is Required?
You need to write balanced equations for the alkanes listed.
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What Is Given?
The given compounds are the reactants. You know that O2 is a reactant too.
Equations (a) and (b) are complete combustion reactions. Equation (c) is an incomplete combustion.
Plan Your Strategy
For pentane and octane, the complete combustion will give rise to the products CO2
and H2O only.
Step 1 Write the equation.
Step 2 Balance the C atoms first.
Step 3 Balance the H atoms next.
Step 4 Balance the O atoms last.
For ethane, the reaction is incomplete, so the possible products are carbon, carbon
monoxide, carbon dioxide, and water. Since water is the only hydrogen-containing
product, you should balance the H atoms first, the C atoms next, and the O atoms
last.
Act on Your Strategy
(a) C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g)
(b) C8H18(g) + 25O2(g) → 8CO2(g) + 9H2O(g)
(c) (i) C2H6(g) + 2O2(g) → C(s) + CO(g) + 3H2O(g)
(ii) C2H6(g) + 3O2(g) → CO(g) + CO2(g) + 3H2O(g)
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equation.
14. Problem
The flame of a butane lighter is usually yellow, indicating incomplete combustion of
the gas. Write a balanced chemical equation for the incomplete combustion of butane
in a butane lighter. Use the condensed structural formula for butane.
What Is Required?
You need to write the balanced equation of the incomplete combustion of butane.
What Is Given?
You have to check the formula for butane, which should be C4H10. You know butane
and O2 are the reactants in the equation.
Plan Your Strategy
(a) For the incomplete reaction, the possible products are carbon, carbon monoxide,
carbon dioxide, and water. Since water is the only hydrogen-containing product,
you should balance the H atoms first, the C atoms next, and the O atoms last.
Act on Your Strategy
CH3 −CH2 −CH2 −CH3 + 4 O2 → 2 C + CO + CO2 + 5 H2O
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equation.
15. Problem
The paraffin wax in a candle burns with a yellow flame. If it had sufficient oxygen it
would burn with a blue flame, it would burn rapidly and release a lot of energy. It
might even be dangerous! Write the balanced chemical equation for the complete
combustion of candle wax, C25H52(s).
What Is Required?
You need to write a balanced equation for the complete combustion of candle wax.
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What Is Given?
the formula for candle wax is C25H52. You know C25H52 and O2 are the reactants in
the equation
Plan Your Strategy
The complete combustion will give rise to the products CO2 and H2O only. Balance
first by C, then by H, and finally by O.
Act on Your Strategy
C25H52(s) + 38O2(g) → 25CO2(g) + 26H2O(g)
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equation.
16. Problem
4-Propyldecane burns to give solid carbon, water vapour, carbon monoxide, and carbon dioxide.
(a) Draw the structural formula for 4-propyldecane.
(b) Write two different balanced equations for the reaction described in this problem.
(c) Name the type of combustion. Explain.
What Is Required?
(a) You need to draw the structural formula for 4-propyldecane.
(b) You need to write two possible balanced equations for this reaction.
(c) You have to explain whether the react was complete or not.
What Is Given?
The reactants are 4-propyldecane and O2.
Plan Your Strategy
(a) From the name given, you know the root chain has 10 C atoms. A propyl group
is on C4. Since there is only one branch, you can arbitrarily choose either end of
the chain as C1 and place the propyl branch at C4.
(b) Count the number of C and H atoms in the structural diagram in (a). This gives
the formula for the molecule. Since the problem asks for 2 balanced equations,
you can guess it is an incomplete one, so the possible products are C, CO, CO2,
and H2O. Since water is the only molecule with H atoms in it, start by balancing
the H, then by C, and finally by O.
Act on Your Strategy
(a)
C3H7
CH3
CH2
CH2
CH
CH2
CH2
CH2
CH2
CH2
CH3
4-propyldecane
(b) C13H28(l) + 15O2(g) → 3C(s) + 4CO(g) + 6CO2(g) + 14H2O(g)
C13H28(l) + 12O2(g) → 5C(s) + 6CO(g) + 2CO2(g) + 14H2O(g)
(c) In this case, incomplete combustion has occurred as both solid carbon product
and carbon monoxide gas have been formed as well.
Check Your Solution
The same number of C, H, and O atoms must appear on both sides of the equations.
Check the structure by working backward, using the naming rules of alkanes to name
the structure drawn.
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Solutions for Practice Problems
Student Textbook page 348
17. Problem
Name each hydrocarbon.
(a) CH3
(b) CH3
CH2
CH2
CH
CH2
CH CH2
CH2 C
CH3
CH CH3
CH2
CH2
CH3
(c)
CH2
CH3
CH
CH3
CH
C
CH
CH2
CH2
CH3
CH2
CH3
What is Required?
You have to name the alkenes given.
What Is Given?
The full structural diagrams of the alkenes are given.
Plan Your Strategy
Follow the following steps for all the structures.
Step 1 Find the chain that contains the double bond. This is the main root chain,
and it need not be the longest chain possible. The number of C atoms in the
main chain will determine the main alkene name.
Step 2 Number the main chain, starting from the end closest to the double bond.
Identify the C number at the double bond; this will be the alkene number.
Step 3 Identify the branches on the main chain and number there position relative to
the main chain C atom it is attached to. The suffix of the branch should be
changed to -yl.
Step 4 If more than one of the same type of branch is present, name them once using
the appropriate number prefixes (di- for two, tri- for 3, tetra- for 4, and so
on).
Step 5 Put the prefix + suffix + root name together for the name of the compound, in
alphabetical order of branches.
Act on Your Strategy
(a) The main double-bonded chain has 6 C atoms. The double-bond is on C3, so it
is simply 3-hexene.
(b) The main double-bonded chain has 7 C atoms, with the double bond at C2. It
will be a 2-heptene. The propane branch is on C3, so it will be the 3-propyl
branch. The final name is 3-propyl-2-heptene.
(c) The main double-bonded chain has 8 C atoms, with the double bond on C4. It
will be 4-octene. Numbering will start from left to right. On the C2 and C3,
there is one methane on each; these will be 2,3-dimethy branches. An ethane
group is on C4, so it will be the 4-ethyl branch. Putting these together alphabetically, we get 4-ethyl-2,3-dimethyl-4-octene.
Check Your Solution
Try searching for these names on the Internet and in chemistry books and see if the
structures and names match.
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18. Problem
Draw a condensed structural diagram for each compound.
(a) 2-methyl-1-butene
(b) 5-ethyl-3,4,6-trimethyl-2-octene
What Is Required?
You need to draw the structural diagrams of the given alkenes.
What Is Given?
The branched name of the alkene is given.
Plan Your Strategy
For each of the alkenes, follow the following steps.
Step 1 Identify the root chain in the name, which is usually the last name with the
suffix -ene.
Step 2 Draw the main chain first, preferably as a straight chain, of linked C atoms
only. The number before the root chain name is the number of the C with the
double bond.
Step 3 Number the rest of the main chain carbons according to the position of the
double bond.
Step 4 Add the branches to the main chain. The number in front of the -yl branch
alkane identifies the main chain C it must be attached to. Add the branches to
the main branch first.
Step 5 Finish off the diagram by adding the H atoms to the main branch C.
Remember that each main branch carbon should be attached to 4 other
atoms, be it C or H or both. In the case of the double-bonded carbons, they
only have 2 other atom attachments available each, one being a C atom and
one a H atom.
Act on Your Strategy
(a) 2-methyl-1-butene
CH3
CH2
C
CH2
CH3
(b) 5-ethyl-3,4,5-trimethyl-2-octene
CH3 CH3 CH3
CH3
CH
C
CH
C
CH2
CH2
CH3
CH2
CH3
Check Your Solution
Try searching for these names on the Internet and in chemistry books and see if the
structures and names match. Work backward and try to rename the structures you
have drawn using the naming rules for alkenes.
19. Problem
You have seen that alkenes, such as C6H12, can have isomers. Draw condensed structural formulas for the isomers of C4H8. Then name the isomers.
What Is Required?
You need to draw the isomers of the alkene C4H8.
What Is Given?
You are given the formula of C4H8.
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Plan Your Strategy
Isomers have the same number of C and H atoms, but in a completely different
attachment to one another. Set out the four C atoms and add the double bond in different areas of the links. You can even create branches along the root chain. the
important facts to remember are:
1. Each C atom in the double bond is only allowed to form 2 more bonds with
another atom, be it a C or H atom.
2. Those C atoms not part of the double bond can form 4 bonds with other atoms.
3. Any “free-end” C atom is usually a methyl group, unless it is double (or triple)
bonded to the C2 atom.
4. The total number of C atoms must be 4, and the total number of H atoms must
be 8 for C4H8.
Act on Your Strategy
(a) 2-methyl-1-propene
CH3
CH2
C
CH3
(b) 1-butene
CH2
CH
CH2
CH3
Check Your Solution
Check other chemistry textbooks or the Internet to see if your answers are correct.
See also Student Textbook page 348 for the geometric isomers of 2-butene, as well as
the Electronic Learning Partner for more about geometric isomers.
Solutions for Practice Problems
Student Textbook page 348
20. Draw and name the cis-trans isomers for C5H10.
What Is Required?
You have to draw the geometric isomers of C5H10.
What Is Given?
The formula of C5H10 is given.
Plan Your Strategy
Step 1 First draw the five C atoms as a straight chain. You will see that there are 4
links in which a double bond can be placed. Repeat this drawing 3 more
times.
Step 2 Draw a double bond into each of the 4 diagrams of the chain, one for each
main chain position of the link.
Step 3 Add the H atoms to the C atoms. Remember that each double-bonded C
atom can only have two other atom attachments.
Step 4 For the double bond being at the centre links; you can actually have two geometric isomers depending on how the H atoms are positioned relative to the
C atoms.
Chapter 9 Hydrocarbons • MHR
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Act on Your Strategy
(a) cis-2-pentene
CH
CH
CH3
CH2
CH3
(b) trans-2-pentene
CH3
CH2
CH
CH
CH3
Check Your Solution
Try searching for pentene geometric isomers on the Internet and in other chemistry
books and see if the structures and names match.
21. Problem
Why can 1-butene not have cis-trans isomers? Use a structural diagram to explain.
What Is Required?
You need to explain why 1-butene will not generate geometric isomers.
What Is Given?
The double bond position of the butene is given.
Plan Your Strategy
Repeat the steps given in Problem 10 above. It was noted that two of the 3 drawn
structures would be identical. These are the 1-butene structures.
Act on Your Strategy
H
H
H
H
C
C
C
C
H
H
H
H
Therefore, no matter from which end of the chain you had started the double bond,
the result is identical.
Check Your Solution
If available, use ball and stick models to check the results. Wine gums and toothpicks
can be used if model kits are not available. You will see that no matter how you twist
the C atoms around the double bond, the configuration is the same. there are no true
isomers of 1-butene.
22. Problem
Like other isomers, two cis-trans isomers have the same atomic weight. They also
yield the same elements when decomposed. How might you distinguish between two
such isomers in the lab?
What Is Required?
You need to explain how two isomers can be distinguished in a lab.
What Is Given?
The isomers are geometric only.
Chapter 9 Hydrocarbons • MHR
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Plan Your Strategy
All isomers have different physical and chemical properties, including cis-trans
isomers. These can be investigated in the laboratory.
Act on Your Strategy
The simplest way to determine between cis and trans isomers in the lab is to evaluate
their boiling points and compare to reference data.
Check Your Strategy
Review some organic chemistry reference books or Internet sources for laboratory
data on cis-trans isomers, particularly for boiling points. Also see the Electronic
Learning Partner for more on cis-trans isomers.
23. Problem
C6H12 has four possible pairs of cis-trans isomer. Draw and name all four pairs.
What Is Required?
You have to draw the 4 geometric isomers of C6H12.
What Is Given?
The formula of C6H12 is given.
Plan Your Strategy
Step 1 First draw the six C atoms as a straight chain. You will see that there are 5
links in which a double bond can be placed. Repeat this drawing 4 more
times.
Step 2 Draw a double bond into each of the 5 diagrams of the chain, one for each
main chain position of the link.
Step 3 Add the H atoms to the C atoms. Remember that each double-bonded C
atom can only have two other atom attachments.
Step 4 For the double bond at the centre links; you can actually have two geometric
isomers depending on how the H atoms are positioned relative to the C
atoms.
Act on Your Strategy
(a) cis-3-hexene
CH
CH
CH2
CH2
CH3
CH3
(b) trans-3-hexene
CH3
CH2
CH
CH
CH2
CH3
(c) cis-2-hexene
CH
CH3
CH
CH2
CH2
CH3
Chapter 9 Hydrocarbons • MHR
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(d) trans-2-hexene
CH3
CH2
CH2
CH
CH
CH3
Check Your Strategy
Review some organic chemistry reference books or Internet sources on cis-trans
isomers of C6H12. Also see the Electronic Learning Partner for more on cis-trans
isomers.
Solutions for Practice Problems
Student Textbook pages 353–354
24. 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.
25. 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.
(a) CH2
CHCH2CH2CH2CH3 + HBr →
Br
CH3CHCH2CH2CH2CH3 + Br
CH2CH2CH2CH2CH2CH3
Chapter 9 Hydrocarbons • MHR
178
CH3
(b)
CH3
C
CH
CH2
CH3 + HOH →
CH3
CH3
C
CH3
CH2
CH2
CH3 + CH3
CH
CH
CH2
CH3
OH
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.
(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.
26. Problem
Draw the major product of each reaction.
CH3CHCH2 + Br2 →
CH2 CH2 + HOH →
CH2 CHCH2CH3 + HBr →
(CH3)2C 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.
CH3CH2OH
(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
27. Problem
For each reaction, name and draw the reactants that are needed to produce the
given product.
(a) ? + ? → CH3CH(Cl)CH3
(b) ? + ? → Br CH2CH2Br
OH
(c) ? + HOH → CH3CH2CCH2CH3
CH3
Chapter 9 Hydrocarbons • MHR
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(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 CH3CH(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.
CH3CHCH2 + HCl
(b) This is an addition reaction. The reactants are ethene and bromine, shown below.
CH2CH2 + Br2
(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 pages 354–355
28. Problem
Name each alkyne.
(a)
CH3
CH3
C
C
C
CH3
CH3
(b)
CH3
CH2
CH3
CH3
CH
CH2
C
C
CH
CH2
CH2
CH3
What Is Required?
You have to name the unsaturated hydrocarbons listed.
What Is Given?
The structural diagrams of the alkynes are given.
Plan Your Strategy
Alkynes are named in the same way as alkenes.
Chapter 9 Hydrocarbons • MHR
180
Step 1 Find the chain that contains the triple bond. This is the main root chain, and
it need not be the longest chain possible. The number of C atoms in the main
chain will determine the main alkyne name.
Step 2 Number the main chain, starting from the end closest to the trile bond.
Identify the C number at the triple bond; this will be the alkyne number.
Step 3 Identify the branches on the main chain and number there position relative to
the main chain C atom it is attached to. The suffix of the branch should be
changed to -yl.
Step 4 If more than one of the same type of branch is present, name them once using
the appropriate number prefixes (di- for two, tri- for 3, tetra- for 4, and so
on).
Step 5 Put the prefix + suffix + root name together for the name of the compound, in
alphabetical order of branches.
Act on Your Strategy
(a) The main chain has 5 carbons so it is a pentyne; the triple bond is at C2, so it is a
2-pentyne. There are 2 methane branches at the C4 position; these will be 4,4dimethyl branches. The final name is 4,4-dimethyl-2-pentyne.
(b) The main chain with the triple bond has 6 C atoms, with the bond being at position C1, so it is 1-hexyne. There is an ethane group and a propane group attached
to C3, so these will be 3-ethyl and 3-propyl branches, respectively. A methane
branch is attached to C5, so it is a 5-methyl branch. Taken together, the name of
the alkyne is 3-ethyl-5-methyl-3-propyl-1-hexyne.
Check Your Solutions
Look up other organic chemistry books or Internet sources to find these structures
and see if the names and structures match.
29. Problem
Draw a condensed structural diagram for each compound.
2-pentyne
4,5-dimethyl-2-heptyne
3-ethyl-4-methyl-1-hexyne
2,5,7-trimethyl-3-octyne
(a)
(b)
(c)
(d)
What Is Required?
You have to draw the condensed structural diagram of the alkynes listed.
What Is Given?
The branched names of the alkynes are given.
Plan Your Strategy
For each of the alkynes, follow the following steps.
Step 1 Identify the root chain in the name, which is usually the last name with the
suffix -yne.
Step 2 Draw the main chain first, preferably as a straight chain, of linked C atoms
only. The number before the root chain name is the number of the C with the
triple bond.
Step 3 Number the rest of the main chain carbons according to the position of the
triple bond.
Step 4 Add the branches to the main chain. The number in front of the -yl branch
alkane identifies the main chain C it must be attached to. Add the branches to
the main branch first.
Chapter 9 Hydrocarbons • MHR
181
Step 5 Finish off the diagram by adding the H atoms to the main branch C.
Remember that each main branch carbon should be attached to 4 other
atoms, be it C or H or both. In the case of the triple-bonded carbons, they
only have 1 other atom attachment available each.
Act on Your Strategy
(a) 2-pentyne
CH3
C
C
CH2
CH3
(b) 4,5-dimethyl-2-heptyne
CH3
C
C
CH3
CH3
CH
CH
CH2
CH3
(c) 3-ethyl-4-methyl-1-hexyne
CH3
CH2
CH3
CH3
CH
CH2
C
C
CH
CH2
(d) 2,5,7-trimethyl-3-octyne
CH3
CH3
CH
CH3
C
C
CH
CH3
CH2
CH
CH3
Check Your Solutions
Look up other organic chemistry books or Internet sources to find these names and
see if the names and structures match.
Solutions for Practice Problems
Student Textbook page 358
30. Problem
Name each compound.
(a) CH3
(b)
CH2
CH3
H3C
H3C
CH3
(c)
CH3
(d) CH3
CH2
(e)
CH3
CH2
CH3
CH3
Chapter 9 Hydrocarbons • MHR
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(f)
CH2CH3
CH3
CH3
(g) CH3CH2
CH2CH3
(h)
CH3
CH2CH2CH2CH2CH3
What Is Required?
You have to name each compound drawn.
What Is Given?
The condensed structural diagram for each cyclic hydrocarbon is given.
Plan Your Strategy
The following rules can be applied to naming cyclic compounds.
Rule 1 When naming cyclic compounds, all the C in the ring are treated as equal, so
any carbon can be used as C1.
Rule 2 When there are branches attached, the C atom with the branch is assigned
the lowest number.
Rule 3 If there are more than two branches attached, number in one direction (clockwise or counterclockwise) according to alphabetical listing of the branches;
that is, an ethyl group will usually start first, followed by methyl, and so on.
Rule 4 Wherever a double bond occurs in the ring, it must be assign the lowest 2 C
numbers (that is C1 and C2 on either side of the bond). Then number either
clockwise or counterclockwise to the nearest branch, in such as way that the
branches have the lowest possible number in the ring.
Act on Your Strategy
(a) The ring has 5 C atoms so it is a cyclopentane. There is an ethane branch and a
methane branch attached to the ring. Name the ethane first, so positioning it as
C1. It will be the 1-ethyl branch. The methane group is clockwise to the ethane
group, so number the rest of the C atoms in a clockwise direction. Methane is
thus the 3-methyl branch. Taken together, the name is 1-ethyl-3-methylcyclopentane.
(b) The ring has 6 C atoms, so it is a cyclohexane. There are methane branches
attached consecutively to 4 of its C atoms. Choose as C1 either one of the 2 carbon atoms with methane attached, but which does not have both C neighbours
with branches. Count towards the neighbouring branched carbons in the ring
(you can see that no matter which C1 you choose, the clockwise or counterclockwise counting makes not difference to the numbering of other branches). This
structure is therefore 1,2,3,4-tetramethylcyclohexane.
(c) The ring has 4 C atoms so it is a cyclobutane. There is only one methane branch
so this will be position C1. The name is 1-methylcyclobutane.
(d) The ring has 5 C atoms, but one double bond, so it is a cyclopentene. A propane
branch and a methane branch is attached to it. Choose the carbon at the double
bond, which is closer to methane as C2 since it should count alphabetically.
Counting counterclockwise, the structure is therefore named 3-methyl-5-propyl1-cyclopentene.
Chapter 9 Hydrocarbons • MHR
183
(e) The ring has 8 carbons with a double bond, so it is a 1-cyclooctene. Counting
clockwise from C1 to the methane branch at C4, the final name is 4-methyl-1cyclooctene.
(f) The ring has 9 C atoms and one double bond, so it is a 1-nonene. Counting
counterclockwise from the double bond to the nearest branch, we find a methane
at C3 and C4 and an ethane branch at C5. Considered alphabetically, the final
name is 5-ethyl-3,4-dimethyl-1-nonene.
(g) The ring has 6 carbon atoms with a double bond so it is a 1-cyclohexene. There
are two ethane groups attached, and the lowest numbering of these is achieved if
the counting is done counterclockwise from the double bond, giving the branches
at C3 and C5. The name is 3,5-diethyl-1-cyclohexene.
(h) The ring has 5 C atoms so it is a cyclopentane. There is a methane branch and a
pentane branch attached, so by alphabetical default, the methane branch will be
assigned as the C1 position. Counting clockwise, the name is 1-methyl-2-pentylcyclopentane.
Check Your Solutions
Look up other organic chemistry books or Internet sources to find these names and
see if the names and structures match.
31. Problem
Draw a condensed structural diagram for each compound.
1,2,4-trimethylcycloheptane
2-ethyl-3-propyl-1-cyclobutene
3-methyl-2-cyclopentene
cyclopentene
1,3-ethyl-2-methylcyclopentane
4-butyl-3-methyl-1-cyclohexene
1-1-dimethylcyclopentane
1,2,3,4,5,6-hexamethylcyclohexane
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
What Is Required?
You need to draw the condensed structural formula of the compounds given.
What Is Given?
The branched names of the cyclic hydrocarbons are given.
Plan Your Strategy
The following rules can be applied to drawing cyclic hydrocarbons.
Rule 1 The ring structure and number of carbons is given by the last cyclo-prefixed
word in the name. The presence of a double bond is indicated by the suffix ene and the number in front of cyclo-.
Rule 2 Where there is a double bond, its C atoms always have the lowest numbers.
This can be placed first in any position in the ring.
Rule 3 The numbers and -yl groups in the front of the root name are the positions
and names of the branches. They always have the lowest possible number
position around the ring, and so place relative to any double bonds in this
way.
Chapter 9 Hydrocarbons • MHR
184
Act on Your Strategy
(a) 1,2,4-trimethylcycloheptane
CH3
CH3
CH3
(b) 2-ethyl-3-propyl-1-cyclobutene
CH2 CH3
CH2CH2CH3
(c) 3-methyl-2-cyclopentene
CH3
(d) cyclopentene
(e) 1,3-ethyl-2-methylcyclopentane
CH3
CH2CH3
CH2CH3
(f) 4-butyl-3-methyl-1-cyclohexene
CH2CH2CH2CH3
CH3
(g) 1-1-dimethylcyclopentane
H3C
CH3
(h) 1,2,3,4,5,6-hexamethylcyclohexane
H3C
H3C
H3C
CH3
CH3
CH3
Check Your Solutions
Look up other organic chemistry books or Internet sources to find these names and
see if the names and structures match.
Chapter 9 Hydrocarbons • MHR
185
Solutions for Practice Problems
Student Textbook page 361
32. Problem
Name the following aromatic compound.
CH3
H 3C
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.
33. 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
34. Problem
Give another name for the compound in question 11(a).
Solution
The compound can also be called meta-ethylmethylbenzene.
35. 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
CH3
CH2CH2CH3
CH3
CH3
CH3
CH3
CH3
Chapter 9 Hydrocarbons • MHR
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