Chemistry 12
Solutions Manual Part A
Unit 1 Organic Chemistry
Answers to Unit 1 Preparation Questions
(Student textbook pages 4-5)
1. e
2. e
3. In addition to the equipment listed, include location of
main gas shut-off valve, fume hood (if equipped), exits
to hallway, location of lab benches or other work area.
4. Student answer will include
• Fire extinguisher type (A, B, C, D, ABC).
• Face the fire with a clear path to an exit available
behind you.
• A fire extinguisher lasts for about 30 s. If the fire
cannot be extinguished in this time, leave the area
and activate a fire alarm.
• Pull the pin from the top of the extinguisher to
release the tamper seal, activating the extinguisher.
• Aim the nozzle or nose of the extinguisher toward
the base or bottom of the fire. You want to attack the
fire at the source. Remember, this is one time you
want to aim low.
• Squeeze the handle down toward the extinguisher.
Depressing the handle releases the extinguishing
element.
• Sweep across the fire in a side to side motion until the
fire appears to be out.
• Watch the fire closely. If there seem to be any signs of
the fire reigniting, repeat the last 4 steps.
5. Emphasize possibility of spills and splashing of
chemicals onto hands and clothing. Also the possibility
of losing track of the identity of the contents in a
beaker or flask and the mixing, by mistake, of clear,
colourless solutions (such as vinegar and solution
of baking soda) that will produce a gas, causing
splattering. Also, some common kitchen chemicals
such as oven cleaners and drain cleaners are dangerous
and safety gear should be used at home.
6. Mixing of paper waste can cover broken glass which
presents a danger to the person disposing of this waste,
risking possible cuts and exposure to infection.
7. d
8. a. Wear protective eyewear, apron, thermal gloves;
know location of fire extinguisher, fire blanket, tie
back long hair, do not wear loose clothing.
b. Read label on the cylinder; be aware of reading
on pressure gauge, hold cylinder with both hands,
pointing the outlet away from your body; open valve
slowly.
c. Wear protective eyewear, apron, and protective
gloves.
d. Wear protective eyewear, apron, and protective
gloves. Know the location of eyewash station, tie
back long hair, do not wear loose clothing.
e. Wear protective eyewear, apron, protective gloves,
tie back long hair, do not wear loose clothing.
9. a. poisonous material causing immediate and serious
toxic effects
b. flammable and combustible
c. corrosive
d. compressed gas
10. a. The attraction between two atoms that results from
the sharing of electrons.
b. A compound in which the atoms are held together
by covalent bonds.
c. The electrostatic attraction between a negatively
charged ion and a positively charged ion.
d. The temperature at which a pure substance changes
from the solid state to the liquid state.
e. The temperature at which a pure substance changes
from the liquid state to the gaseous state.
11. b
12. d
13. Melting involves separating molecules from their
fixed position in the solid state. Polar molecules
have a definite shape with areas of fixed negative and
positive charge. The electrical attraction between
these oppositely charged areas on different molecules
(dipole-dipole) must be overcome for melting to occur.
This takes more energy than that which is needed
to overcome relatively weak attractions (dispersion
forces) between non-polar molecules.
14. Cl2, non-polar covalent molecule < CH3OH, polar
covalent molecule < K2O, ionic compound
Unit 1 Part A • MHR 1
adopting the symmetrical, tetrahedral arrangement of
its bonds.
15. Heat and light are given off
16. No. Hazardous (causing injury to health) products
come from combustion of a hydrocarbon is CO(g),
which does not form with excess O2.
CH3
5.
H2C
17. hydrocarbon and oxygen
18. a. Incomplete combustion occurs when there is
insufficient oxygen to react with all of the reactants
(such as hydrocarbons). Products other than carbon
dioxide and water are carbon (soot) and carbon
monoxide. Less heat is given off, and it is less
efficient than complete combustion.
b. Carbon monoxide is clear, colourless, and odourless.
It can accumulate and overcome a person, causing
death.
(Student textbook page 11)
1. Up to and including the 1800s, the term organic
was used to describe matter that came from living
organisms that contained a “vital energy,” and inorganic
was used to describe matter that came from non-living
material. The laboratory synthesis of compounds once
thought to be produced only by living organisms led
to the modern definition of an organic compound
as one in which carbon atoms are bonded to each
other, to hydrogen atoms, and sometimes to a few
specific elements, usually oxygen, nitrogen, sulfur, or
phosphorus.
2. The key to carbon’s ability to bond with several atoms
is found in its atomic structure. A carbon atom has
four valence electrons, and a half-filled outer shell of
electrons. It has an intermediate electronegativity and
is much more likely to share electrons than to gain or
lose enough electrons to form ions. Its four valence
electrons can be shared with up to four other atoms.
This leads to the potential for the formation of a wide
variety of molecules.
3. Any molecule that contains C–C or C–H are
considered organic, and molecules or ions such as CO2
and CO32– that contain no C–C or C–H bonds are
considered inorganic.
4. Atomic systems arrange themselves to minimize their
potential energy. For methane, this occurs if the four
bonds are as far apart as possible (because the electron
pairs repel each other), which the molecule achieves by
2
MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
CH2
CH3
CH2
CH2
CH2
CH2
CH3
6. There are nine constitutional isomers of heptane.
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
H
H
Chapter 1 Structure and Physical
Properties of Organic Compounds
Answers to Learning Check Questions
CH3
H
H
C
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
H
C
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H
C
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
C
H
H
H
C
H
H
H
C
C
H
H
H
H
C
C
H
H
H
H
C
C
C
C
H
H
C
H
H
H
H
H
H
H
C
C
H
H
C
C
C
H
H
H
C
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
C
H
H
H
C
H
H
H
H
C
C
C
C
C
H
H
H
H
H
H
H
H
H
H
C
C
C
C
H
H
H
C
H
H
H
H
H
H
C
H
H
H
H H
C H
H
C
C
C
C
H
H
H
H
H
C
H
H
(Student textbook page 19)
7. Examples include methane (CH4) found in natural gas
to heat homes, propane (C3H8) used in barbeques and
as a heating fuel,c12_techart_c1-lc_ans6
butane (C4H8) found in lighters and
portable burners, octane (C8H18) found in gasoline,
Art is) too
wide to fit in column width.
waxes (C20H
42 found in candles, and tars (C40H82)
used in paving.
8. In a saturated hydrocarbon, each carbon atom
is bonded to as many other atoms as possible.
Unsaturated hydrocarbons have at least one multiple
bond and are therefore not bonded to as many other
atoms as possible, or they are not completely saturated
with hydrogen atoms.
9. A homologous series is a set of molecules in which
each member differs from the next by an additional
specific structural unit. The alkane series can be
represented by a general formula, CnH2n+2. Ethane,
C2H6, differs from the next member of the series,
propane, C3H8, by –CH2.
10. 3 carbon atoms: C3H8; 7 carbon atoms: C7H16;
9 carbon atoms: C9H20; 12 carbon atoms: C12H26
11. All reasonable answers should show that an atom or
group of atoms on an alkane has been substituted
in place of a hydrogen atom on the parent chain of
carbons.
15. a. Alkanes that are liquids at room temperature have
between 5 and 16 carbon atoms inclusive.
b. Alkanes that are gases at room temperature have
between 1 and 4 carbon atoms inclusive.
16. CnH2n
17. They are similar in that they have the same number
and sequence of carbon atoms as the corresponding
alkane. They are different in that there are fewer
hydrogen atoms in an alkene, there is a double bond in
an alkene, and the shape is different.
18.
For example
H
H
H
H
H
C
C
C
H
H
H
H
H
H
C
C
C
H
H
C
H
H
H3C
CH
CH
H2C
CH
CH2
CH3
CH3
CH3
H
H3C
C
CH2
(Student textbook page 32)
19.
H3C
CH
CH
CH2
CH2
CH3
H2C
CH
H3C
CH2
CH
CH
CH2
CH3
H3C
C
CH2
CH2
CH2
CH3
H
Substituent group
EmpericalHformula
12. The root is pent-, the prefix is 2-methyl, and the suffix is
C4H
-ane. 10
Condensed
structural
formula
(Student
textbook
page 24)
Emperical
formula
13. Empirical
formula
H3C CH2 CH2
Emperical
formula
C4HC4H
10
H3C
C
CH2
CH
CH3
H3C
CH3
H3C
CH
CH2
CH
CH2
H3C
LineLine
structural
H
H formula
Hformula
H
structural
Expanded Structural formula
Expanded Structural formula
CH3CH2CH2CH3
14. They are
insoluble in water but soluble in non-polar
CH3CH2CH2CH3
solvents.
CH
C
CH2
10
Structural
H Hformula
H Hformula
Expanded
Structural
H C
H C
H C
H H
H C
CH3CH2CH2CH3
H H
C H
C H
C H
C H
CH2
CH3
CH
CH
CH3
H3C
H3C
CH3
Expanded
molecular formula
C4H
10
Structural
formula
structural formula
CHCondensed
3CH2CH2CH3
H Hstructural
H H formula
Condensed
Condensed
structural
H3C CH
CHformula
CH3
2
2
H C C C C H
H3C CH2 CH2 CH3
H H H H
Structural
formula
Structural
formula
CH2
CH2
HC
CH3
CH3
H3C
H3C
C
C
CH3
CH3 CH3
H3C
C
CH
CH2
CH3
20. alkenes < alkanes < alkynes
21. The linear structure of alkynes and the nature of the
triple bond allow them to attract one another more
strongly than corresponding alkanes and alkenes.
22. Testosterone and estrogen are two examples of steroid
hormones containing cyclic hydrocarbons.
23. Straight-chain alkanes have the general formula
CnH2n+2. Cyclic alkanes have the general formula
CnH2n. Alkenes have the general formula CnH2n.
Alkynes have the general formula CnH2n-2.
24. At least three carbon atoms are necessary and the
molecular formula is C3H6.
Line structural formula
Line structural formula
Unit 1 Part A • MHR 3
(Student textbook page 36)
25. Cyclic hydrocarbons have higher boiling points and
melting points than straight-chain hydrocarbons. There
is a greater difference in terms of melting point.
26. Aromatic hydrocarbons contain a benzene ring;
aliphatic hydrocarbons do not contain a benzene ring.
27. Benzene is more stable than cyclohexene (therefore,
less reactive), and it has fewer hydrogen atoms.
28. All the bonds in benzene are identical and have an
intermediate length compared with single and double
bonds.
Answers to Section 1.1 Review Questions
(Student textbook page 14)
1. An organic compound is one in which carbon atoms
are bonded to one another, to hydrogen and a few
other non-metal elements (O, N, S P, halogens)
2. The carbon atom is bound only to oxygen atoms.
There are no C–C or C–H bonds.
3. If the bond angle around carbon atoms was 90o, there
would be fewer possible compounds because of greater
repulsion between pairs of electrons. Also biological
compounds would be restricted sheet or cubic shapes.
29. Electrons in the “double bonds” of a benzene ring are
in fact shared by all six carbon atoms. They are not
“localized” to a bond between two carbon atoms.
H
4
MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
C
H H H
H
benzene ring
Figure 1.10 (Student textbook page 13): Triple bonds are
linear and rigid and each carbon atom has only one atom
attached.
Figure 1.13 (Student textbook page 15): They all contain
only carbon and hydrogen atoms. All contain only single
bonds. Except for methane, the carbon atoms are bonded
to other carbon atoms. Each compound has one more CH2
unit than the one before.
Figure 1.15 (Student textbook page 22): The compound
on the left is unsaturated because the carbon atoms are
not bonded to the maximum number of atoms possible,
owing to the double bond. The compound on the right
is saturated because the carbon atoms are bonded to the
maximum number of atoms; all of the bonds are single
bonds.
Figure 1.16 (Student textbook page 22): These two alkenes
are isomers. The double bond can join an end carbon atom
to another carbon atom, or the double bond can join the
two middle carbon atoms.
Figure 1.25 (Student textbook page 46): Possible response:
The CFCs act as catalysts and thus are not used up in the
reactions. The CFCs take a long time to naturally degrade
in the atmosphere. Also, older products, such as old
refrigerators, may still emit CFCs.
Figure 1.36 (Student textbook page 70): The citric acid
protonates the amines in the fish. That is, the acid adds a
hydrogen ion to the amine, making it a soluble salt which is
not volatile.
Figure 1.43 (Student textbook page 80): It is all three.
Glucosamine contains hydroxyl side groups, an amino side
group, and a formyl side group.
H
C
30.
Answers to Caption Questions
H
H
Lewis Structure
4. a.
H
H
C
C
H
HH
C
C
H
H
H
H
H
C
C
H
H
C
C
H
H
H
H
H
C
H
H
H
C
C
H
C
H
H
H
H
H
H
C
H
C
C
C
H
H
H
H
b. C2H2Cl2
Cl
H
Cl
C
C
Cl
H
C
C
Cl
H
H
H
Cl
H
C
C
5. a. Rotation occurs around a single bond and a trans
“isomer” could easily rotate into a cis “isomer” for
the same molecule
F
H
H
C
C
H
Cl
H
H
H
Cl
C
C
F
H
H
b. Each carbon atom in a triple bond is attached to
only one other atom 180o to each other.
Cl
C
C
F
6. In the list of words, “trans” means across the nation,
continent, or Atlantic respectively. The attached
identical atoms or groups in the trans form are across
the double bond from one another.
Cl
7. The wrist, the thumb side, the fingers and flat edge on
the baby finger side.
8. Venn diagram should show:
Diastereomers only:
• carbon atoms joined by double bond
• non-mirror images, no free rotation about the double
bond
• cis and trans forms have different physical properties
Diastereomers and enantiomers:
• carbon atoms
Enantiomers only:
• mirror images of one another
• single bonds
• four different atoms bonded to carbon
• same physical properties
9. The physical properties would be similar as their basic
structure is the same except for the rotation of plane
polarized light. Their chemical properties would also
be similar. However, in living systems, the enzymes
that catalyze chemical reactions recognize and bind to
only one of the two enantiomers.
10. Looking at the two carbon atoms in the double bond,
the one on the left is bonded to two identical CH3
groups. The groups must differ in order to form cis or
trans isomers.
11. a. The ring structure of the pentane causes the carbon
atoms to form a rigid plane. Notice that, in the cis
isomer, the two fluorine atoms are above the plane
of the cyclic carbon atoms and in the trans isomer,
the two fluorine atoms are on opposite sides of the
plane.
cis- 1, 2 difluoropentane
H
H
H
H
C
trans- 1, 2 difluoropentane
H
C
F
H
F
C
C
C
H
H
H
F
C
C
H
H
H
C
C
C
F
H
H
H
H
b. In the linear structures, a double bond prevents
rotation around the bond, thus preventing the
atoms attached to the carbons atoms of the bond. In
the ring structure, the ring itself prevents rotation
around the single bonds, preventing an attached
atom from moving from one side of the ring to the
other.
12. The boiling point depends upon the intermolecular
attractions between molecules. This is affected by
the surface area of each isomer. The more spherical
the shape of the isomer, the less surface area is in
contact between molecules and less energy is required
to separate them. These isomers will have a lower
boiling point than the more linear shaped isomers.
The difference in chemical reactivity may be less
notable for hydrocarbons since the bond strengths will
be much the same regardless of their configuration.
For constitutional isomers having functional groups
attached, there can be a drastic difference in chemical
reactivity.
13.
+
H
H
N
H
O
C
N
–
H
Answers to Section 1.2 Review Questions
(Student textbook page 41)
1. A homologous series is a specific series of compounds
in which each member differs from the next by an
additional specific structural unit. For example, C2H6
and C3H8 differ by CH2
2. ethane C2H6
H
H
C
H
H
C
H
H
3. prefix: 2-methyl; root: hept; suffix: ane
Each name has three parts, the prefix, the root, and
the suffix. The root indicates the number of carbon
atoms in the longest continuous chain of carbon atoms.
The prefix indicates the position(s) and name(s) of
any branches attached to the main chain. The suffix
indicates the series to which the compound belongs.
4. a. CnH2n+2
b. CnH2n
c. CnH2n–2
5. Examples are:
• octane, major component of gasoline,
CH3CH2CH2CH2CH2CH2CH2CH3
• butane, lighter fluid, CH3CH2CH2CH3
• nonane, component of camp stove fuel,
CH3CH2CH2CH2CH2CH2CH2CH2CH3
• toluene, used in paint thinner, C6H5(CH3)
Unit 1 Part A • MHR 5
6. Both groups are made up of the elements carbon and
hydrogen. Saturated hydrocarbons have single bonds
between carbon atoms so that each carbon is bonded
to four atoms, the maximum number for carbon.
Unsaturated hydrocarbons have at least one double
bond or triple bond between two carbon atoms.
Comparing the boiling points of alkanes, alkenes and
alkynes with similar number of carbon atoms, the
alkenes have the lowest and the alkynes the highest.
7. a. 3-methylheptane
b. 2-methylpropene
c. 2,3-dimethylcyclohexene
d. 2-ethyl-1,3-dimethylbenzene
8. a. CH
C CH CH CH
3
3
CH3
b.
HC
H3CH2C
C
C
CH3
CH3
CH2CH2CH3
CH
C
H3C
c.
CH2
CH2CH3
CH3
CH
CH
CH2
CH3
CH3
CH3
CH2
CH3
CH
CH2
CH
CH2
Answers to Section 1.3 Review Questions
(Student textbook page 82)
1. Alcohols can be used for solvents, antifreezes,
fuels, antiseptics. Only one, ethanol, can be used in
beverages. They can also be used as starting materials
for the synthesis of industrial compounds.
CH3
CH3
d.
12. a. Hormones such as estrogen and testosterone have
cyclic rings.
b. Flavours such as vanilla are aromatics.
13 a. There is only one spot for the triple bond which is
between carbons one and two because carbon atoms
are always numbered so that the double bond is
nearest C1.
b. There is only one double bond and multiple bonds
in cyclic compounds always begin at the number
one carbon.
14. The electrons in the second bond of double bonds are
shared equally among all the carbon atoms and are
termed delocalized. If there were alternating single and
double bonds, the molecule would have unequal bond
lengths. The lengths of all of the bonds in benzene are
all the same and between the lengths of typical single
and double bonds.
CH2
CH2
CH3
9. a.
2. Unlike propane, which would be a gas at room
temperature, propan-1-ol would be a liquid at room
temperature and would only combust with the oxygen
in the air on the surface of the liquid.
3. Ethanol is more polar than tetrachloroethene and
would not dissolve non-polar molecules as well.
Ethanol would not evaporate as well and would
therefore take more energy to recover.
4. Comparing Aldehydes and Ketones
Aldehydes
Ketones
Structure
formyl group at end of
carbon chain
carbonyl group on a nonterminal carbon atom
Naming
root: indicates number
of carbon atoms in
longest chain
suffix: –al
root: indicates number of
carbon atoms in longest
chain
suffix: position number of
carbonyl group, –one
Physical
properties
• small – soluble in
water
• polar
• < 15 carbon atoms
– liquid
• > 15 carbon atoms
– waxy solids
• small
– soluble in water
• polar
• < 15 carbon atoms
– liquid
• > 15 carbon atoms
– waxy solids
b.
c.
d.
10. Test their boiling points:
hex-1-ene < hexane < hex-1-yne < cyclohexane
11. CnH2n; The alkenes with one double bond also have
this general formula.
6
MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
5. Butter, cheese, and vomit may contain partially
digested fats.
6. The carboxyl group, –COOH, found in carboxylic acids
would conduct electric current and turn litmus red.
7. a. Propan-1-ol and ethanoic acid
b. Ethanol and propan-1-amine
8. a. One would expect alcohols, primary and secondary
amines, carboxylic acids, and primary and
secondary amides to form hydrogen bonds with
identical molecules.
b. They would be more soluble in polar solvents such
as water and have higher melting and boiling points
as compared to similar sized molecules with other
functional groups.
9. a. Three ethers could have this formula:
1-methoxypropane, ethoxyethane,
2-methoxypropane. Ethers have the general formula
R–O–R’.
b. Three alcohols could have this formula: butan-1-ol,
butan-2-ol and 2-methylpropan-1-ol.
c. The ethers would have lower melting points and
boiling points than the alcohols. The alcohols would
be more soluble in polar solvents such as water.
10 a. hexan–3–ol
b. 3-methylpentane-2,2-diol
c. 1-2-dichloro-1-fluoro-2-methylpropane
11 a. OH
b.
OH
HO
c.
Answers to Practice Problems
For full solutions to Practice Problems, see Part B
of this Solutions Manual.
(Student textbook page 19)
1. 2-methylpropane
2. 2,2-dimethylpentane
3. 5-ethyl-3,4-dimethylnonane
4. 3,3,5-trimethylheptane
5. 2,2,5-trimethylhexane
6. 2-methylbutane
7. dimethylpropane
8. 2,2,4,4-tetramethylhexane
9. pentane
10. 2,3-dimethylpentane
11. 3,3-dimethylheptane
(Student textbook page 21)
12.
CH3
H3C
13.
H3C
CH2
C
CH2
CH2
CH3
CH3
14.
CH3
C
CH
CH2
CH2
CH2
CH2
CH3
CH3 CH2
CH2
O
O
e.
CH3
H2C
OH
d.
CH2
CH3
CH3
O
HC
CH3
15. (12)
NH2
(13)
(14)
Unit 1 Part A • MHR 7
16. pentane
17. 4-ethyloctane
(Student textbook page 27)
35.
H3C CH
CH CH2
18. 3,3-diethyl-4-methylhexane
36.
19.
H
H
HH C HH
H
C
C
C
C
H
H
H
H
CH2
CH2
H3C
CH
C
CH3
CH2
37.
H
CH3
CH2
CH2
CH3
CH3
H3C
CH
C
CH
CH2
CH3
H3C
20.
H
H
H
C
H C H
H C H
C
C
C
H
H
H
H
H
H
H
H
C
C
C
C
C
C
H
H
H
H
H
H C HH C H
H
H C HH C H
C
21.
H
H
C
38.
H
H
H C H
H H C H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H H C H
H
H C H
H
H
H
H
H
H
H
C
C
H3C
H
CH3
CH3
CH
CH
CH
C
CH2
H2C
CH3
CH
CH
CH3
CH2
CH
CH
CH2
39.
22. 4-ethyl-4-propylheptane
(Student textbook page 26)
23. pent-2-ene
c12_techart_c1-pp_ans21
24. 3-methylbut-1-ene
25. 4,4-dimethylhex-2-ene
26. 2-ethylpent-1-ene
27. 3,4-diethylhex-2-ene
28. but-1-ene
29. 5,6-dimethylhept-3-ene
30. 4-ethyl-3-methylhex-2-ene
40. hex-2-ene
H3C
CH2
41. 5-ethyloct-2-ene
H3C
CH2
33. 4-methylpent-2-ene
34. 2-ethyl-3-methylpent-1-ene
8
MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
HC
CH2
42. 2-methylbut-2- ene
CH3
31. 2,5-dimethyloct-3-ene
32. propene
CH2
H3C
CH
C
CH3
CH2
CH3
CH3
CH3
43. a.
H
H C HH
H
C
C
H
b.
H
C
d.
H
C
C
H
H
52. but-1-yne
53. 4-methylpent-2-yne
H
H
H C H
HH C HH
H
C
C
C
H
H
C
H
C
51. but-1-yne
H
54. 3-ethyl-5-methyl-3-propylhex-1-yne
(Student textbook page 34)
55. cyclopentane
H
H
C H C HH
H
H
56. 1-ethyl-3-methylcyclobutane
57. 4-methylcycloheptene
58. 3-methyl-5-propylcyclopentene
c12_techart_c1-pp_ans50d
59. 3-cyclopentyloctane
44. 3-ethyl-3-propyl-4-methylhex-1-ene
60.
CH3
61.
CH3
CH3
(Student textbook page 30)
45. hex-3-yne
62.
CH3
46. 4-methylpent-1-yne
63.
47. 5-ethyloct-2-yne
48. 4,4,5-trimethylhex-1-yne
49. a. H3C
C
C
b.
CH3
CH3
HC
C
CH2
CH
64.
CH
C
C
C
CH2
d.
H3C
CH3
HC
CH2
C
C
CH
CH2
c.
CH2
CH2
CH3
68. 2,4-diphenyloctane
HC
69.
CH2
CH3
H3C
50. a.
b.
CH
67. 1-ethyl-3-propylbenzene
CH3
CH2
CH3
66. 1,2,4-trimethylbenzene
CH3
CH3
CH3
CH2
(Student textbook page 38)
65. methylbenzene, historically known as toluene
CH3
CH3
H2C
CH3
CH3
CH3
c.
CH2
CH3
CH2
70.
CH2
CH3
CH2
CH2
CH3
CH3
H3C
CH3
Unit 1 Part A • MHR 9
71.
H3C
CH2
CH2
84.
CH3
OH
CH2
72.
CH2
CH3
H2C
CH3
(Student textbook page 49)
85. 2-fluorobutane
CH2
CH3
86. 3-bromo-3-methylpentane
H3C
87. 3-chloro-5-methylhexane
CH3
73.
CH3
CH2
CH3
H2C
CH
CH2
CH2
CH3
88. 1,3-dichloro-2-fluorobutane
89. 1,3-dibromo-2-chlorocyclohexane
90.
CH2
CH2
CH
H2C
CH2
CH3
l
91.
CH2
CH3
74. 1-ethyl-4-methylbenzene
H2C
CH2
F
Cl
92.
CH3 Br
CH3
CH3
C
CH
CH2
CH3
CH3
93.
H2C
Cl
CH3
H3C
(Student textbook page 45)
75. butan-2-ol
CH
HC
CH
Br
Br
94. 1,4-difluoro-2-propylcycloheptane
76. pentane-2,3-diol
F
77. propan-2-ol
78. 3-methylbutane-1,3-diol
79. 6-phenyl-2-propylheptan-1-ol
80.
81.
F
OH
H2C
95. a. 2-chlorobutane
b. 3-bromo-4-chlorohexane
c. 1,3-dichlorocyclopentane
d. 2-chloro-3,3-dimethylbutane
CH3
OH
H2C
CH2
CH3
82.
(Student textbook page 52)
96. propanal
OH
HO
CH2
83.
CH2
H3C
H3C
CH2
CH
CH2
HC
CH3
CH3
OH
CH
HC
97. 2-methylbutanal
98. 3-ethyl-4-methylhexanal
CH3
10 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
99. 2,4-dibromopentanal
CH3
100.
C
CH2
H
101.
110.
CH3
O
CH
CH
O 3
H3C
CH3
H
C
CH3
O
C
CH2
111.
C
H3C
O
H
102.
O
CH3
H3C
HC
C
CH3
112. The ketone group lies on a carbon attached to two
other carbons
H3and
C therefore
C
CH2would never have a
numerical assignment of 1. This compound does not
exist.
O
H
103.
CH2
113. The correct name
CH3 is pentan-2-one. Numbering of
the longest carbon chain gives the ketone group the
H3number.
C C
CH2
lowest
Cl
H2C
C
O
H
104.
CH2
H3C
CH2
CH3
C
CH2
H2C
CH3
H
CH2
C
O
105. a. The correct name is ethanal. The aldehyde group
should be placed on the number 1 carbon (unless
there are other higher priority functional groups
also present such as carboxylic acids). It is not
necessary to include a number in this case as it is
assumed the aldehyde is on the first carbon.
b. The correct name is 5-methylheptanal. The longest
chain would include the ethyl group, and the
methyl group would be a side branch.
c. Since carbon only forms four bonds, an aldehyde
could not be inside a cyclic hydrocarbon as well
as having hydrogen and double-bonded oxygen.
Therefore, this compound does not exist.
d. Since the terminal carbon would need to have a
hydrogen and a double-bonded oxygen, and be
joined to the hydrocarbon chain, there would be no
room for a fluorine to bond. This compound does
not exist.
114. The correct name is 5-methyloctane-3,4-dione.
Numbering should begin with giving the ketone
3
groups the lowest possible numbers. TheCH
longest
continuous chain containing these
H3groups
C C is eight
CH2
carbons long.
115. Since carbon forms only four bonds and the linkage
between carbons in the benzene ring is considered
the intermediate of a single and double bond, the
carbon atom with the ketone group would not be able
to form a double bond with oxygen. This compound
does not exist.
(Student textbook page 61)
116. propanoic acid
117. 4-ethylhexanoic acid
118.
O
CH2
H3C
119.
HO
CH2
C
CH2
C
CH2
CH2
HC
OH
CH3
CH2
CH2
CH2
CH3
O
120.
(Student textbook page 56)
106. butan-2-one
CH3
O
C
CH2
CH
C
CH2
CH2
CH
CH2
CH3
HO
107. 2-methylpentan-3-one
108. 3-ethyl-4-methyl-hexan-2-one
109.
O
121.
OH
HO
O
CH2
C
O
Unit 1 Part A • MHR 11
122. The correct name is hexanoic acid. The carboxylic
acid gets the lowest number on the longest
hydrocarbon and is assumed to be on carbon 1.
123. Owing to carbon only being able to form four bonds,
a carboxylic acid functional group would always be at
the end of a hydrocarbon chain. This compound does
not exist.
124. The correct name is 3-ethylheptanoic acid.
Numbering should begin with giving the carboxylic
acid groups the lowest possible numbers. The longest
continuous chain containing these groups is seven
carbons long.
125. Since carbon forms only four bonds, a carboxylic acid
group could never be inside a cyclic hydrocarbon.
This compound does not exist.
(Student textbook page 69)
136. 1-ethoxypropane
137. 2-ethoxypropane
138. ethoxycyclohexane
139.
H3C
O
CH2
CH3
140.
CH3
H3C
CH2
CH2
141.
HC
O
CH2
H3C
CH2
O
HC
CH2
CH3
CH3
CH
CH2
CH3
CH3
142.
(Student textbook page 65)
126. methylethanoate
CH3
O
HC
CH2
CH2
CH2
CH2
CH3
127. ethylmethanoate
143. The correct name is methoxyethane. The root name
should be the longest hydrocarbon chain.
128. butyl 3-chlorobutanoate
129.
O
CH2
H3C
130.
O
C
CH3
CH2
CH3
O
H3C
CH2
O
131.
C
CH2
O
CH2
O
H3C
132.
CH2
CH2
HC
C
CH3
O
O
133.
146. The correct name is ethoxybenzene. It is assumed that
the R′ group is first attached to the number 1 carbon
of the benzene ring and it is not necessary to number
it.
148. pentan-3-amine
O
149. N-ethylbutan-2-amine
F
150. N-methyl-N-propylhexan-2-amine
151.
O
H3C
NH2
152.
O
135.
145. The correct name is 2-methoxypentane. The
numbering of the root name should be made to give
the longest possible chain to which the R′ group is
attached.
(Student textbook page 74)
147. ethanamine
O
134.
144. The correct name is 2-propoxybutane. The
numbering of the root name should give the
ether linkage the lowest number possible on its
hydrocarbon chain.
H3C
O
153.
O
12 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
CH2
H2C
NH2
CH2
CH2
NH
CH2
CH2
CH2
CH2
HC
CH2
CH3
NH2
CH3
Answers to Chapter 1 Review Questions
154.
H3C
CH2
CH2
155.
CH2
CH
H3C
N
CH2
CH2
CH3
CH3
(Student textbook pages 87-91)
1. b
2. c
NH2
3. a
4. d
156.
5. e
NH
6. d
7. b
157.
8. d
N
9. e
10. e
158.
11. a
N
12. c
13. e
14. b
(Student textbook page 79)
159. methanamide
15. Carbon can form four bonds as it has four unpaired
electrons. It also has intermediate electronegativity
which prevents it from forming singular ions. This
allows it forms covalent bonds which can continue in
chains.
160. 3-ethylhexanamide
161. N-propylpentanamide
162. N,N-dimethylbutanamide
163.
H 3C
O
16. Isomers are molecules with the same molecular
formula but their atoms are in a different arrangement
NH2
17. Stereoisomers have the same molecular formula
and connectivity but different three-dimensional
arrangement of their atoms in space. Constitutional
isomers are molecules that have the same molecular
formula but have different connections between the
atoms.
C
164.
O
H3C
CH
C
NH2
H3C
165.
O
H3C
CH2
CH2
CH2
CH2
C
N
H3C
CH2
CH3
CH2
166.
O
H3C
CH2
CH
CH3
C
N
CH2
H2C
CH3
167. propanamide
168. N-propyl-2,2-dimethylbutanamide
169. N,N-dimethyl-2-propylhexanamide
CH2
CH3
18. Molecules with two different atoms or groups on both
sides of a double bond can form cis/trans isomers.
Cycloalkanes with two side-groups can also form
cis/trans isomers.
19. Enantiomers have identical structures. Their physical
and chemical properties are essentially the same.
However they do affect plane polarized light differently
and may react with certain other enantiomeric
molecules differently. As well, enzymes are specific for
one form of the isomer.
20. Saturated hydrocarbons have each carbon atom
bonded to as many other atoms as possible.
Unsaturated hydrocarbons contain at least one double
or one triple bond so each carbon atom has less than
the maximum atoms bonded to it.
Unit 1 Part A • MHR 13
21. a. CnH2n+2
b. CnH2n
c. CnH2n-2
d. CnH2n
22. They are all non-polar and therefore do not dissolve
well in polar solvents such as water.
23. They are similar in that they are six-carbon rings. The
difference is that the six carbons in cyclohexane are
single bonded, whereas the carbons in benzene have
bonds that are intermediate between single and double.
The formula for cyclohexane is C6H12 whereas the
formula for benzene is C6H6.
24. Ethanol dissolves in polar solvents; ethane is soluble in
polar solvents. Ethanol has higher melting and boiling
points compared to ethane. At room temperature,
ethanol is a liquid and ethane is a gas.
25. Both aldehydes and ketones have carbon atoms double
bonded to an oxygen atom (carbonyl groups). In
aldehydes the carbon in the C=O bond is the terminal
carbon of a chain and, therefore, has a formyl group,
whereas in ketones the C=O is in the main carbon
chain with a carbon atom bonded to it on each side
32. Boiling points could distinguish the two as C2H6 would
be a gas at room temperature whereas C8H18 would be
a liquid.
33. a. correct name is 3-ethyl-2-methylpentane
b. correct name is 3-ethylhex-1-ene
34. The formula for hexane is C6H14, cyclohexane is C6H12
and benzene is C6H6
35. Some examples could be: paradichlorbenzene,
C6H5Cl2, used as a moth repellant, phenylacetone,
C6H5CH2 –CO–CH3, used to make amphetamines,
benzyl alcohol, C6H5CH2OH, used to make perfumes,
paints epoxy resins.
36. There are five constitutional isomers of C4H4 as
shown below. Constitutional isomers can have single,
double, or triple bonds as long as the number of
atoms present allow for them. The but-2-ene could
have diastereoiosmers but they are not constitutional
isomers.
H2C
H2C
CH
CH3
H2C
CH2
H2C
CH2
CH3
CH
CH
27. An amide is the product of a reaction between a
carboxyl group, –COOH, and an amine, –NH2.
CH2
CH
CH2
28. These aromatic compounds were found in naturally
occurring plants and the original names were often
taken from those sources.
CH2
26. An ester is the product of the reaction between
carboxyl, –COOH, and hydroxyl, –OH, functional
groups.
29. a. CH4 —
A carbon atom cannot bond to more than
­­
four atoms.
b. C2H6 — Alkanes must have 2n+2 hydrogen atoms.
c. hexane C6H14, or cyclohexane C6H12 As above.
d. CH4 — Pure alkanes do not contain oxygen atoms.
30. Enantiomers, or optical isomers, are nonsuperimposable mirror images of each other. Four
different atoms or functional groups must be bonded
to the central carbon atom to form enantiomers. The
mirror images of CH2BrF would be superimposable
31. CH2Br2 can form cis/trans isomers if the two bromine
atoms are not on the same carbon atom. The molecule
C2H2Br2Cl2 would have single bonds between the
carbon atoms that can rotate freely. The molecule C2HI
would have a triple bond between the carbons and
would have a linear arrangement, with the H and I
atoms 180° to one another.
14 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
CH3
CH3
CH3
C
CH3
37. a. Volatile organic compounds are organic compounds
that have a vapour pressure high enough to
evaporate and produce part per billion levels in the
atmosphere.
b. There are many beneficial uses: fuels, solvents,
refrigerants, anaesthetics, etc. Their drawbacks
include: contributing to global warming, creation of
smog, destroying the ozone layer, and they are often
are toxic at elevated levels.
c. Examples of naturally occurring VOC’s are natural
gas, methane, emitted from decomposition of
animals as well isoprene, 2-methylbuta-1,3-diene
emitted from forests.
2-methylbuta-1,3-diene CH2=C(CH3)–CH=CH2
Examples of anthropogenic sources are
tetrachlorethane, low molecular weight petroleum
derivatives (eg. hexanes, octanes) and formaldehyde
(methanal).
38. The molecule can form an enantiomer. There are four
different groups (methyl, hydroxyl, carboxyl, and
hydrogen) around the second carbon atom and can
therefore form a different mirror image.
39. Design an experiment to determine the boiling point of
each compound. Ethane would have low boiling point,
be a gas at room temperature and be poorly soluble in
a polar solvent such as water as ethane is non-polar.
Ethanol would have a higher boiling point, be a liquid
at room temperature at be soluble in water due to the
polar alcohol group. Ethanoic acid would be soluble
in water and have an even higher boiling point than
ethanol due to the two very electronegative oxygen
atoms in the carboxyl functional group. Ethanoic acid
could also conduct electric current.
40. a. hexane-3,4-diol
b. 2-chloro-3,4-dimethylpentane
c. 3-ethyl-6-methyldecan-2-one
d. 3,3-dimethylpentan-2-one
41. Report should follow the guidelines for report writing,
using proper grammar, and spelling, and identification
of sources of information. Some inorganic greenhouse
gases are CO2, N2O (nitrous oxide), water, and O3
(ozone). Examples of organic greenhouse gases
are CH4 (methane), trichlorofluoromethane and
dichlorodifluoroethane. Inorganic molecules such as
carbon dioxide and water are much more abundant in
the atmosphere and the production of CO2 by humans
is much more prevalent.
42. a.
Boiling Point vs Number of Carbon Atoms
Boiling Point (°C)
200
100
Decane
Nonane
Hexane
Pentane
0
–100
–200
Octane
Heptane
c. It should be mainly soluble in non-polar solvents
due to the long hydrocarbon chain and insoluble in
water.
d. It should form constitutional (structural) as well as
cis/trans diastereomers. There is no chiral centre
(carbon atom bonded to four different atoms or
groups) for enantiomers (optical isomers)
44. a. 2-methylbutanoic acid
b. 7-methyl-7-phenylnonan-3-one
c. ethylpentanoate
d. N-methyl-N-propylbutanamine
45. a. From lowest to highest boiling points the molecules
would be: ethane (-89°C), methoxymethane
(-23°C), ethanamine (17°C), ethanoic acid (118oC),
ethanamide (222°C).
b. The molecules of these four compounds are
about the same size and the amount of dispersion
forces between molecules will be about the same.
All except ethane are polar molecules and will
experience dipole-dipole attractions between
molecules. The greatest factor affecting the boiling
point would be the amount of hydrogen bonding
between molecules. Methoxymethane does not form
hydrogen bonds and has the lowest boiling point.
Ethanamine can form hydrogen bonds between
molecules however nitrogen is not as electronegative
as the oxygen atoms in ethanoic acid so the
degree of hydrogen bonding is less in ethanamine.
Ethanamine has a lower boiling point than ethanoic
acid. Ethanamide has both electronegative oxygen
and nitrogen atoms and the highest amount of
hydrogen bonding would be expected between
molecules of this compound.
c. Ethanoic acid and ethanamide are liquids at room
temperature and the other are gases.
Butane
Propane
Ethane
Methane
b. C11H24 about 200°C; C12H26 about 215°C
43. a. Aromatic (phenyl), alkene and alcohol.
b. It should have a high boiling point (eg. over 400°C)
due to the long hydrocarbon chain and polar alcohol
groups. It c12_techart_c1-cr_ans42a
would be a liquid due to the unsaturation.
Unit 1 Part A • MHR 15
46.
Name
Structure
Saccharin
Functional Groups
O
OH
O
HO
OH
Mannitol
300X sweeter than sucrose
bitter; USDA has
removed carcinogen
warnings
haloalkane, alcohol,
ethers
600x sweeter than sucrose; stable at
hot and cold temperatures; good for
baking
non-cited
alcohols
doesn’t cause tooth decay; does not
affect insulin levels; can be extracted
from plant products; medicinal uses;
not hygroscopic
amino group, ester
linkage, amide linkage,
carboxyl group
160 - 220 times sweeter than sucrose;
does not cause tooth decay; does not
affect insulin levels; is metabolized like
other amino acids
O
O
Cl
Cl
Cl
O
O
OH
OH
OH OH OH
OH OH OH
Aspartame
47. a.
H
H
H
H
C
H
H3C
C
H
CH2
C
Drawbacks (Cons)
ketone, aromatic,
amine, (sulfoxyl)
NH
S
Sucralose
Benefits (Pros)
H
C
O
C
H
harmful to people
with rare genetic
disease, PKU
(phenylketoneurea)
48. a. NOTE: The name is invalid and should be
3,4-dimethylheptane
3, 4-dimethylheptane
H
H
CH
CH2
CH
b.
OH
OH
b. Several constitutional isomers can form by changing
the location of the double bond and the hydroxyl
group. Many more could be formed because it could
form carboxylic acids, ketones, and aldehydes.
The double bond allows for the formation of
diasteriomers.
O
O
O
NH
OH NH2
O
16 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
c.
49. a. 1,3-dichloro-2-fluoro-2-methylpentane
b. butanal
c. 3-amino-3-hydroxybutan-2-one
50. a. methanal
b. methylbenzene
c. 1,2-dimethylbenzene
d. phenylamine
e. trichloromethane
51. Answers should include the key terms and concepts
shown on page 86. Graphic organizer could be a main
idea web, a spider map, or a concept map.
52. a. Herbicides are used to kill unwanted plants for
aesthetic purposes and to increase crop yields in
agriculture.
b. The molecule shown is glyophosate or
N-phosphonomethyl amino acetic acid. Reasons
would be that it has no chlorine substituents
expected in the name of 2,4-D and the presence
of amino and carboxylic acid groups (as well as a
phosphorus atom) present in the IUPAC name for
glyophosate.
c. General consensus would be that glyophosates
would be less harmful to the environment due
shorter persistence and fewer medical side effects to
this date.
d. Some alternatives around the home are to use
mulch, plant native plants that would compete
better with weeds, pull weeds by hand and use less
controversial household chemicals such as acetic
acid (vinegar), boiling water, rubbing alcohol or
corn gluten to remove undesirable plants. Other
reasonable alternatives are possible.
53. a. Putrescine would have the name butan-1,4-diamine
and cadaverine would be pentan-1,5-diamine.
b. Rotting flesh smells putrid and another term for an
animal corpse is a cadaver. The suffix “ine” indicates
an amino group.
c. Substances such as citric acid found in citric fruits or
other carboxylic acids would react with the slightly
basic amine group and form salts which are nonvolatile.
54. a. N,N-diethyl-3-methylbenzamide
b. The common name, DEET, comes from the letters
“dee” to represent diethyl, and the “t” in toluene. It is
simpler for everyday use.
c. It would dissolve in non-polar or slightly polar
solvents.
d. WHMIS symbols would poisonous and infectious
causing other toxic effects (if inhaled or long time
exposure); poisonous and infectious with immediate
side effects (if ingested). It is also flammable.
e.
Risks
Benefits
Poisonous if taken internally
Personal comfort by
preventing painful and
distracting parasites from
biting
When sold as an aerosol, must
be used with caution and
disposed of in a safe manner.
Controls insects that carry
disease
Insects can develop immunity
to the compound.
Persists in the environment
affecting non-harmful,
beneficial insects
f. Possible alternatives would be: wearing light
clothing, covering exposed skin, restricting outdoor
activities during peak insect hours, removing
environments that breed insects (standing water) or
attract certain insects (CO2, certain perfumes), using
naturally derived insect repellents (e.g., cinnamon),
55. Reduce fuel spillage, take public transit, have
automobiles tuned up and emissions systems in good
working order, reduce leaks of heating, cooking and
transportation fuels, reduce unnecessary mechanized
travel.
56. a. The difference in boiling point is due to the
difference in the intermolecular attraction
between molecules. The –OH group in methanol
is very polar, which allows for the formation of
hydrogen bonds with neighbouring molecules.
Methanal is slightly polar and can form dipoledipole interactions but has no hydrogen bonding.
Since there is less attraction between molecules of
methanal, this compound has a lower boiling point.
b. Formaldehyde is an aqueous solution of methanal.
57. a. Saturated fats have higher melting points than
unsaturated and are therefore solids at room
temperature whereas unsaturated fats are usually
liquid.
Unit 1 Part A • MHR 17
b. Cis fats are unsaturated and have the hydrocarbon
chains on the same side of the double bonds whereas
trans unsaturated fats have them on opposite sides.
cis form
Answers to Chapter 1 Self-Assessment Questions
(Student textbook pages 92-3)
1. a
2. c
3. e
4. e
5. d
6. c
trans form
7. a
c. The unsaturated fats are softer and less likely to form
a hard layer lining the arteries which leads to strokes
and heart attacks. Also the sites of double bonds in
unsaturated fats are more reactive than the single
bonds between carbons in the saturated fats.
58. a. adrenaline: C9H13NO3; amphetamine: C9H13N
b. Both have a phenyl group, C6H5, and an amine,
NH2 group, adrenaline has –OH groups that are not
c12_techart_c1-cr_ans57
present in amphetamines.
c. Epinephrine (adrenaline) constricts blood vessels
and dilates air passages, which reverses the effects of
an anaphylactic shock.
d. Amphetamines increase alertness and concentration.
8. c
9. c
10. e
11. The molecules are not constitutional isomers as
they are the same molecule just rotated and drawn
differently.
12.
Two pairs are possible
cis-hex-2-ene
59. a.
trans-hex-2-ene
and
Limonene
cis-hex-3-ene
CH3
HC
H2C
H
H2C
C
C
C
CH2
CH2
chiral/optical centre carbon
CH3
b. One would expect it to be found in citrus fruits.
c. The R-(+) limonene smells of oranges and the S-(–)
limonene smells of lemony turpentine.
d. It can be used as an industrial solvent because it is
less volatile and non-toxic than chlorofluorocarbons
and methyl ethyl ketone.
e. As it can be extracted from orange peels and is thus
a renewable resource. It is also biodegradable.
18 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
trans-hex-3-ene
13. a. The root must be the longest continuous chain. The
correct name is 2-methylbutane.
b. Numbering of the carbon atoms in the main chain
of an alkane must begin at the end that will give the
side groups the lowest possible numbers. The correct
name is 2,3-dimethylpentane.
c. This molecule is named correctly.
d. The numbering of the carbon atoms in the main
chain of an alkene must begin at the end nearest
the double bond. The correct name is 5-ethyl-6methylhept-3-ene.
14. a. Solid alkanes could be found in waxes, asphalt and
greases. The molecules would be over 18 carbon
atoms long.
b. Liquid alkanes could be found in gasoline,
lubricating oils and heating fuels such as kerosene.
The molecules would be between 5–22 carbon atoms
long.
c. Gaseous alkanes could be found in natural gas,
propane and butane used for heating and cooking.
The molecules would be 1–4 carbon atoms long.
b.
CH3
CH3
eg. cyclohexane
3-methylpent-1-yne
c. Example:eg.3-methylpent-1-yne;
C6H10
d. cyclohexane
d. Example: cyclohexene; C6H10
e. benzene C H
e. Example: benzene;
6 6
16. Benzene is not an alkane because the carbon atoms
are bonded to only three other atoms. Each carbon
atom is bonded to only one hydrogen atom. The ring
in the centre of the benzene diagram represents the
six-delocalized electrons that are shared equally among
atoms. This resonance hybrid has equal bond lengths
halfway between a single and double bond.
17. a. 3-methylbutanoic acid
b. N,N-dimethylpropan-2-amine
18. a.
CH2
c.
CH3
C
NH
CH
CH2
CH3
CH2
CH3
CH2
CH3
CH2
CH2
CH3
19. Both esters and amides have a carbonyl group (C=O)
but esters have an extra oxygen bonded to this carbon
whereas amides have a nitrogen.
They both can be thought of as the product between
a carboxylic acid with another molecule. The other
molecule needed to form an ester is an alcohol whereas
the other molecule needed to form an amide is an
amine. The alkyl group that was derived from the
carboxylic acid would get the prefix and the alcohol
and amine derivatives would get the suffix. Esters end
in “oate” and amides end in “amide.”
Primary and secondary amides will have high melting
and boiling points and will often be solids because
they can hydrogen bond with one another. Esters are
polar but no hydrogen bonding can occur. The melting
points are therefore lower than the amides of similar
size. At room temperature many are liquids. The longer
chain esters may be waxy solids.
20. a. They IUPAC name would be ethanedioic acid.
b. It would be a solid at room temperature due to its
ability to form multiple hydrogen bonds with other
oxalic acid molecules.
c. It would be soluble in polar solvents such as water
due to its ability to form hydrogen bonds with the
solvent
21.
propan-1-ol
propan-2-ol
methoxyethane
O
O
C
O
O
HO
CH3
CH
CH3
eg. 2-methylpentane
15. a. Example:
2-methylpentane; C6H14
b. Example: cyclohexane; C6H12
CH2
22. No. Cyclopentane has the chemical formula C5H10
while pentane has the chemical formula C5H12.
23. Use BLM A-32 Presentation Rubric or BLM A-36
Multimedia Presentation to assess answers.
24. a.
OH
CH3
CH
CH3
Unit 1 Part A • MHR 19
b. The –OH group in the alcohol can hydrogen bond
with the water whereas the non-polar propyl group
can join with the non-polar molecules
c. Keeping the gas tank as full as possible and having a
proper gas cap will prevent humid air from entering
the tank.
25. a. The source for most of the hydrocarbons we use are
from fossil fuels such as crude oil, natural gas and
coal.
b. Some problems associated with the source are oil
spills, acid rain, explosions and the fact that it is
non-renewable, accumulation of CO2 which is a
greenhouse gas associated with global warming.
c. Other sources could be using plants and bacteria to
produce biodiesel and ethanol.
(Student textbook page 107)
7. A substitution reaction occurs when the reactants are
an alkane with a substituent of a halogen or a hydroxyl
group, and the other reactant is H−X, or OH−. A
condensation reaction can occur when the reactants
are a carboxylic acid and an ammonia or amine. An
esterification reaction occurs when the reactants are a
carboxylic acid and an alcohol.
8.
condensation reactions
two molecules combine to form one
larger molecule and one very small
molecule which is usually water
esterification reaction
Chapter 2 Reactions of
Organic Compounds
a special case of a condensation
reaction in which one reactant is
an alcohol and the other is a
carboxylic acid and the products
are an ester and a water molecule
Answers to Learning Check Questions
(Student textbook page 102)
1. Both are examples of reactions of organic compounds.
In an addition reaction, atoms are added to an organic
compound at the site of a double or triple bond.
In elimination reactions, atoms are removed from
an organic molecule and a double bond is formed.
Elimination reactions are the opposite of addition
reactions.
2. They act as catalysts.
3. a. More than one product is possible with an
asymmetric molecule.
b. The major product occurs when the hydrogen atom
is removed from the carbon atom that has the most
carbon-carbon bonds.
4. The hydrogen atoms of the small molecule will attach
to the carbon of a double bond that is already bonded
to the most hydrogen atoms. The rule is used when two
products can be formed from an addition reaction.
5. No, if there is a limited amount of X–X, then a
substituted alkene will be produced.
6. Carbons can only have up to four bonds, and alkanes
are already fully saturated. Therefore, the carbons
cannot accept any more bonding partners. Alkenes
and alkynes are unsaturated and can break a double
or triple bond in order to bond with another element/
compound.
20 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
9. They are reactions that form large biomolecules such as
proteins, which are essential to living things.
10. acetylsalicylic acid, artificial flavours, and artificial
aromas
11. Nine esters: Each alcohol will react with each of
the carboxylic acids. The ester names are methyl
methanoate, methyl ethanoate, methyl propanoate,
ethyl methanoate, ethyl ethanoate, ethyl propanoate,
propyl methanoate, propyl ethanoate, and propyl
propanoate.
12. a. Esterification: The product is an ester.
b. condensation: The product is an amide.
c. substitution: A halogen could replace an hydroxyl
group on the alcohol.
(Student textbook page 120)
13. A polymer is a large, long-chain molecule with
repeating units of small molecules called monomers.
Some uses are plastics, adhesives, and chewing gum.
14.
• Addition polymerization only—Monomers are
combined by addition reactions. For example, the
monomers could be alkenes. In the polymer, the
former double bonds have become single bonds.
• Addition and condensation polymerization—In both
types of reactions, small molecules link to form long
chains of very large molecules.
• Condensation polymerization—Monomers have two
functional groups which can undergo condensation
reactions with groups on the other monomers. For
example, the monomers might have both a carboxyl
group of one end and an alcohol group on the other
end. The carboxyl group of one monomer reacts with
the alcohol group on another monomer to form an
ester linkage. The result is a polyester.
15. polyurethane
16. Addition polymer: There are fewer double bonds in the
polymer than in the monomers. No molecules, other
than the polymer, are formed.
Answers to Caption Questions
Figure 2.3 (Student textbook page 97): The carbon atoms
in the product are bonded to more atoms than the carbon
atoms in the organic reactant.
Figure 2.14 (Student textbook page 116): Possible
responses: train cars linked together to form a long train,
linking building blocks together
Answers to Section 2.1 Review Questions
(Student textbook page 115)
1. a. oxidation
b. esterification
c. elimination NB: First product should be CH2=CH2.
17. amino acids
2. a. ethylhexanoate + water
b. methanoic acid
c. 2-bromopropane + water
d. bromobenzene + hydrobromic acid
e. methanoic acid
18. A polymer containing cross-links is stronger and less
flexible than a straight chain polymer.
3. Substitution reaction:
haloalkane + OH− → alcohol + halide ion
Condensation polymer: A small molecule (usually
water) is released with the formation of each
monomer-monomer linkage.
(Student textbook page 124)
19. The petroleum is separated into classes of
hydrocarbons according to size in a process called
fractional distillation. A variety of organic reactions
are used to convert the hydrocarbons into compounds
needed as starting materials for the production of
compounds that are useful such as plastics.
20. The petrochemicals are very useful in industry.
21. Ethene is reacted with chlorine to make
dichloroethane. This is then cracked to produce HCl
and chloroethene. Chloroethene is reacted in an
addition polymerization reaction to make PVC.
22. Advantages: PVC is very useful for industrial products
such as windows, piping, and doors.
Disadvantage: The monomer, vinyl chloride, is a
carcinogen.
23. Dioxins are chlorinated aromatic hydrocarbons that are
toxic by-products from the production and disposal by
incineration of PVC.
24. Recycle, Reduce usage, Reuse plastics; purchase
degradable plastics
Oxidation reaction:
alcohol + [O] → aldehyde
Oxidation reaction:
aldehyde + [O] → carboxylic acid
4. Addition reactions and elimination reactions can
result in more than one product if the reactants are
asymmetric. In addition reactions, the H atom will
bond to the carbon with the most H atoms already
bonded to it. In elimination reactions, the H atom will
be removed from the carbon with the most carboncarbon bonds.
5. a. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(ℓ)
b. 2C4H10(ℓ) + 13O2(g) → 8CO2(g) + 10H2O(ℓ)
c. C13H28(ℓ) + 20O2(g) → 13CO2(g) + 14H2O(ℓ)
6. a. CH2=CH2−CH3 + H2O
b. CH3− HC(Cl)−HC(Cl)−CH3
c. major product CH3−CH(Br)−CH3
minor product CH2(Br) −CH2−CH3
7. It is a hydrolysis reaction.
CH3− CH(CH3)−CH2−COOH + HO−CH2−CH3
8. a. CH3OH
b. CH3−CO−CH3
Unit 1 Part A • MHR 21
9. major product: 2-methyl-2-bromooctane,
H3C−C(Br)(CH3)−CH2−(CH2)4−CH3
5. a. polyamide
O
minor product: 2-methyl-3-bromooctane,
H3C−CH(CH3)−CH2(Br) −(CH2)4−CH3
HO
(Student textbook page 127)
1. a. Natural polymers are found in living things whereas
synthetic polymers are manufactured by chemical
processes outside of living things. Both are long
chains made up of similar or identical monomer
units.
b. Three examples of naturally occurring polymers are
starch (a polysaccharide), proteins, and DNA. Three
examples of synthetic polymers are polystyrene,
polyvinylchloride, polyethylene.
c. The starch monomer is glucose. Protein monomers
are amino acids. DNA monomers are nucleotides,
which consist of a nitrogen base, a five-carbon sugar,
and a phosphate group. The polystyrene monomer is
styrene (phenylethene). The monomer of polyvinyl
chloride is vinyl chloride (chloroethene). The
monomer of polyethylene is ethylene (ethene).
2. Addition and condensation reactions.
3. a. The hydroxyl group (−OH) of an alcohol and the
carboxyl group (−COOH) group of a carboxylic
acid react to form an ester bond.
b. The carboxyl group (−COOH) of a carboxylic acid
and an amino group (−NH2) of an amine react to
form an amide bond.
4. a. (i) During the production of synthetic polymers,
harmful by-products are often produced. (ii) Many
synthetic polymers remain in the environment
for long periods of time. (iii) The incineration of
many synthetic polymers produces harmful, toxic
products.
b. Synthetic polymers are used to mass produce a wide
variety of items that are durable and long lasting
and things that could not be made from natural
reactants.
c. Possible answer: Yes. Our technological society
has become dependent on many of the synthetic
polymers for variety of applications. However, we
need to find ways to minimize risks.
22 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
(CH2)4
C
C
OH
H2N
(CH2)6
NH2
CH
NH2
b. polyamide
10. Carbon monoxide, is an odourless, colourless,
poisonous gas, is a product of incomplete combustion.
Answers to Section 2.2 Review Questions
O
O
C
HO
O
CH2
C
OH
H2N
CH2
CH2
c. polyester
O
HO
CH2
CH2
OH
HO
O
C
CH2
OH
C
6. a. addition
...
CH2
CH2
CH2
CH2
...
CH2
b. addition
...
CH2
CH
CH2
CH
CH2
...
CH
c. condensation
O
...
C
O
(CH2)3
C
O
NH
CH
NH
C
O
(CH2)3
C
NH
CH3
CH
NH
...
CH3
d. esterification
...O
O
O
C
C
O
CH2
O
O
O
C
C
7. Example: crude oil fractional distillation ethane
ethene polymerization polyethylene
O
CH2
cracking at 800ºC
Answers to Practice Problems
For full solutions to Practice Problems, see Part B
of this Solutions Manual.
(Student textbook pages 99-100)
1. H3C—CH2—CH2—CH3
2. a. H3C−CH2 −C(Br)=C(Br)−CH3
b. H3C−CH2 −C(Br)2−C(Br)2−CH3
3. The product is iodoethane. Only one product is
possible because if iodine (I) is added to carbon 1 or
carbon 2, the same product is formed.
4. CH3−CH(Cl)−CH(Cl)−CH2−CH3
...
5. most abundant
11.
methylcyclohexene + hydrobromic →
acid
CH
CH2
OH
CH
CH3
CH3
CH
CH
3
CH2
CH
CH3
(Student textbook page 102)
OH
12.
CH3
CH
CH2
CH2
OH
CH2
CH3
CH
CH
3
CH2
CH2
CH2
CH3
CH3
CH
CH
CH2
CH3
CH3
Br
CH3
CH
CH2
CH
CH3
CH3
7. a.
Br
H
C
C
C
H
H
H
H + HOCH2CH3 + NaCl
14. In addition to the HOCH2CH3 and NaBr which
will always be products, the original 3-bromo-2methylpentane could form three different products.
These products are 2-methylpent-2-ene, cis-2methylpent-3-ene, and trans-2-methylpent-3-ene.
15. Water will always be a product. The reactant,
2-methylpentan-3-ol can be converted to any or the
three products, 2-methylpent-2-ene, cis-2-methylpent3-ene, or trans-2-methylpent-3-ene.
16. Sodium bromide and ethanol will always be products.
The reactant, 3-methyl-2-iodibutane can be converted
into 2-methylbut-2-ene or 3-methylbut-1-ene.
b. most abundant
17. The major product will be 2,4-dimethylcyclohexene.
CH3
ICH3
18. The major product will be 1-ethylcyclopentene.
I
19. The possible reactants are butan-1-ol or butan -2-ol.
least abundant
CH3
20. The possible reactants are
CH3
CH3
I
I
8. a.
CH2 + H2O
13.
6. neither is preferred over the other
CH3
C
CH3
H
Br
Br
→
CH3
CH3
least abundant
CH
CH
x
CH3
or
OH
CH2
CH2
CH3
x
CH3
CH2
C
CH2
CH2
CH3
CH3
b.
CH2
Cl
CH3
9. a. H2C=CH—CH2—CH3, or
H3C—CH=CH—CH3, or
HC≡C—CH2—CH3, or H3C—C≡C—CH3
b. CH2=CH—CH2—CH3
10.
HBr
+
OH
CH3
CH
O
CH2
CH3
X represents one of four possible elements: F, Cl, Br,
or I.
21. The major product would be cyclobutene.
22. Note: NaOH should be deleted; reactant should be
+ NaOCH2CH3.
The final products would be but-1-yne,
HC≡C−CH2CH3, and but-2-yne, H3C−C≡C−CH3.
SO3H
Unit 1 Part A • MHR 23
(Student textbook page 104)
23. This reaction is not specific and, depending on the
amount of chlorine present, a chlorine atom could
replace any or all of the hydrogen atoms.
40. methanol and salicylic acid
CH3
24. There are nine possible products.
25.
CH3
CH
O
CH3
CH3
CH2
CH2
C
CH2
OH + HO
C
CH2
CH3 + Cl−
CH3
OH
A base such as NaOCH2CH3 would give an elimination
reaction but a hydroxide ion results in a substitution
reaction.
H23C CHC2
CH
CCH2OH + HO
CH2
CH2
(Student textbook page 113)
43. oxidation
44. reduction
45. reduction
30. Note: FeBr2 should be FeBr3
46. oxidation
47. addition and reduction
48. esterification and condensation
31. CH3CH2CH3 + Br2
49. CH4 + 2OCH
2 → CO2 + 2H2O + energy
32.
3
CH3
CH
50. C4H8 + 6O2 → 4CO2 + 4H2O + energy
H3C C
CH2
51. C3H8 + 5O2 → 3CO2 + 4H2O + energy
CH2Cl + OH−
CH3
52. C6H12O6 + 6O2 →6CO2 + 6H2O + energy
(Student textbook page 108-9)
33. condensation reaction
(Student textbook page 121)
53. addition
34. esterification reaction
54. addition
35. esterification reaction
55. condensation
36. esterification reaction
56. polyamide
37. condensation reaction
H2N
38. ethanoic acid and octanol
O
CH3
C
OH + HO
CH2
(CH2)6
CH3
CH2
C
HO
CH2
NH2
CH3 O
C 2)C6
C H3(CH
CH2 OH
C
57. polyester
O
CH2
CH2
O
39. butanoic acid and methanol
CH3
CH
CH3
42. butanoic acid + propan-1-ol
CH3 O
28. 2-hydroxypropane + Cl−
CH
CH2
OH
CH3
27. bromobenzene + HBr
CH2
H
H3C C
CH2 acid + propan-1,2-diol
41. 3,3-dimethylheptanoic
CH3 + H2O
26. bromobenzene + H2O
CH3
C
OH + HO
CH3
Cl
29.
OH
O
HO
OH + HO
CH3
C
CH2
CH3
OH
O H3C
C
CH2
O
HO
24 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
CH2
CH2
C
OH
CH2
CH3
58. polyester
13. e
HO
14. a
OH
O
O
HO
15. a.
CH2
C
C
OH
C
C
CH2BrCH3 + NaOCH2CH3 →
NaBr + HOCH2CH3 + CH2=CH2
n CH2=CH2 → - - - —CH2—CH2—CH2—CH2— - - -
60.
CH2
O
O
C
C
O
O
C
C
O
O
CH2
CH2
O
O
CH2
CH2
O
Z
C
C
C
...
CH2
CH
CH2
CN
CH
CH2
CN
CH
...
62.
O
...
O
CH2
C
O
O
CH2
C
CH2
C
R
C
(Student textbook pages 137-41)
1. c
2. d
3. a
OH + H
N
R
ammonia
or amine
C
N
Z
+ HOH
amide
O
...
R
Answers to Chapter 2 Review Questions
Y
water
Look for an amide and water molecule as products.
e.
O
O
+
O
carboxylic acid
CN
C
CH2 O
C O
CH2
Two compounds react to form two different
compounds and the carbon atoms are bonded to the
same number of atoms in the product and reactant.
d.
C
O
61.
C
The carbon atoms in the organic product are bonded
to fewer atoms than were the carbon atoms in the
organic reactant.
c.
Y
Z →
C
Z + A
CO Y + A
carry out addition polymerization on the ethene.
O
Y
+
Z
The carbon atoms in the products are bonded to
more atoms than the carbon atoms in the reactants.
b. Y
Z
59. First, treat the 1-bromoethane with NaOCH2CH3 to
eliminate the bromine atom and a hydrogen atom
resulting in ethene.
Then
C
Y
O
C
H2SO4
OH + HO
carboxylic acid
R
alcohol
C
O
+ HOH
ester
water
Look for an ester and water molecule as products.
f.
OH
C
O
H
+
[O]
C
4. d
5. d
6. a
7. c
8. b
9. b
10. b
11. a
alcohol
oxidizing agent
O
C
aldehyde
aldehyde or ketone
O
+
H
[O]
oxidizing agent
C
OH
carboxylic acid
Look for carbon having more bonds to oxygen or fewer
CHEM12_2.052A
bonds to hydrogen.
12. d
Unit 1 Part A • MHR 25
g.
OH
O
C
+
aldehyde or ketone
C
C
alkene
[H]
reducing agent
+
[H]
reducing agent
C
H
alcohol
H
H
C
C
alkane
Look for carbon having fewer bonds to oxygen or
more bonds to hydrogen.
h. hydrocarbon + O2 → CO2 + water + energy.
Look for a hydrocarbon chain reacting with oxygen
to form carbon dioxide and water.
16. Addition and elimination reactions are the opposite
of each other. In addition reaction a double or triple
bond between carbon atoms is lost and two new
single carbon bonds with new atoms are formed.
With elimination reactions two single bonds are lost
CHEM12_2.069A
and a double or triple
bond between carbon atoms
are formed while two atoms are eliminated from the
reactant.
17. During an addition reaction use Markovnikov’s
rule and add the hydrogen atom to the carbon atom
already bonded to the most hydrogen atoms. During
an elimination reaction remove the hydrogen from the
carbon with the most bonds to other carbon atoms.
18. The reactions result in a large mixture of products.
There are other ways to obtain the desired product.
19. Both reactions involve two large molecules coming
together to form a large molecule and a small molecule
(water). The difference is that esterification reactions
always occur between an alcohol and a carboxylic acid.
An esterification reaction is a special case of a more
general condensation reaction.
20. A molecule is broken apart by adding the hydroxyl
group from a water molecule to one side of a bond and
the hydrogen atom of the water molecule to the other
side of the bond.
21. When naming polymers add poly in front of the name
of the monomer unit. In this case the monomer is
called propylene (propene).
22. Polymers are long-chain molecules made with
repeating units of small molecules and these natural
molecules contain repeating units of small molecules.
For example, proteins have repeating units of amino
acids
26 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
23. a. substitution
b. elimination
c. addition
d. substitution
e. addition
f. elimination
24. a. oxidation,
b. reduction,
c. oxidation
25. a. esterification,
b. condensation,
c. hydrolysis
26. polyvinylacetate
27. polyesters – polymers in which the monomers are
connected by ester bonds. Typically, one reactant
monomer has an alcohol group on both ends and the
other reactant monomer has a carboxyl group on both
ends.
polyesters and polyamides – Both are made through
condensation polymerization reactions using two
different monomers. Both use a carboxylic acid
monomer as a reactant. Both reactions also produce
a small molecule such as H2O. Both are durable
products.
polyamides – polymers in which the monomers are
connected by amide linkages. One of the reactant
monomers has an amine group on both ends and the
other has a carboxyl group of both ends.
28. a. addition,
b. condensation,
c. addition
29. a. elimination,
b. addition,
c. combustion,
d. substitution
30. a. butan-2-ol
b. major product: 2-bromo-2-methylbutane
minor product: 2-bromo-3-methylbutane
c. propene
d. major product: 3-methylpent-2-ene
minor product: 3-methylpent-1-ene
31. a. propan-1-ol
b. N-ethylpropanamide
32. a. Both of the compounds have an amine group and
a carboxyl group. Any two molecules could react
to form an amide bond. Therefore, although they
are shown to alternate in the figure below, the two
compounds will not necessarily alternate in the
polymer.
O
O
...
N
...
N
N
N
H
H
N.
O
O
. .N
O
O
.
..
O
b.
N
H
N
...
...
O
...
...
O
F
F
...
F
F
F
F
F
F
C
C
C
C
C
C
C
C
F
F
F
F
F
F
F
F
F
F
F
C
C
C
C
C
FC
F
F
F
F
F
F
F
c.
O
...
N
F
...
O
O
O
.C. . NH CCH2
...
C
33. a.
O
CH2
HO
CH2
C
C
NH
O
O
O
C
2
C
C
CH
CH3
b.
CH2
CH
CH2
CH3
b. Br2(aq) WHMIS: Corrosive. Harmful if inhaled.
Causes respiratory tract irritation and possible
burns. Causes eye and skin irritation and possible
burns. Wear gloves when making the bromine water.
Work in a fume hood.
c. addition
d. H3C–CH(Br)−CH(Br)−CH3
37.
• heat an alcohol in the presence of H2SO4
• react a haloalkane with a strong base such as sodium
ethoxide NaOCH2CH3
38. a. hex-3-ene + HCl
F
F
F
F
b. 3-methylbutanoic acid + propan-2-ol
.
.
.
C C
c. 2-methylpropan-2-ol + HCl
...
C C
d. 2,5,-dimethylhex-2,4-diene
F
F
F
F
e. methane + chlorine
f. 2,3-dimethybutan-2-ol → 2,3-dimethylbutanal
O
O
O
39.
O
O
...
C NH CH
CH
NH
• monomer CH3−CH=CH−CH3
...
C
C NH CH2 CH2 NH
CH2
NH
CH2 NH
...
C• polybutene −[−CH(CH
C NH CH
NH
CH2
CH2 NH
CH
NH
2
2
)−CH(CH
)−]
−
3
3
n
• solvent alkylethers, tertiary alcohols
OH
• acid catalyst
WHMIS: 2-butene is a gas at room temperature, highly
flammable, explosive with air mixture, inhalation can
cause dizziness and unconsciousness, evaporation from
skin can lead to frostbite, used face shield
2
40. Major product: 2-bromopropane, CH3—CHBr—CH3
c.
O
NH2
CH2
CH
NH2
+
HO
C
Minor product: 1-bromopropane, CH2Br—CH2—CH3
O
CH2
C
OH
CH
34. A polyamide because the functional groups that react
to form proteins are a carboxyl group and an amino
group which form amide linkages.
35. a. 2C2H6 + 7O2 → 4CO2 + 6H2O
b. 2C2H5OH + 6O2 →4CO2 + 6H2O
36. a. • use 2 fats and 2 oils
• add equal portions of each fat and oil to separate
test tubes
• added a measured volume of bromine water to
each test tube
• shake gently or warm in a water bath while
looking for a decrease in the intensity of the
reddish colour of bromine water.
41. The product, 1-bromopropane, gives the impression
that the reaction is a substitution reaction because a
bromine atom is seen in place of a hydrogen atom.
However, if it was a properly written equation, there
would have to be an HBr in the products. Also, this
method of substitution with alkanes is nonspecific and
there would also be many other products.
42. 2 glucose → maltose + water
43. a. 2-phenylacetic acid + ethanol
b. butyric acid + ethanol
c. ethanoic acid + ethanol
44. Without the alcohol present, the esterification reaction
shown will slow down and the reverse hydrolysis
reaction will be favoured.
45. Degradable plastics are good because they break down
over time when exposed to environmental conditions.
It may be difficult to use across the world due to cost of
manufacturing
Unit 1 Part A • MHR 27
46. Hex-1-ene + HBr →
2-bromohexane + 1-bromohexane
proteins – monomers are amino acids joined through
Brcondensation
reactions with amide linkages between
Br
the amino acids; include various functions of proteins
CHCH
2 2 CH
2 2 CH
2 2 CH
2 2 CH
3 +
3 3 CH
2 2 CH
2 2 CH
2 2 CH
3 3
CH CHCH
CH
CH
CH
HBr→ →CHCH
CH
CH
CH
CH
CH
3 +HBr
in
theCH
human
body
such
as building
of muscle tissue,
Br
Br
Br Br
forming enzymes, hormones. DNA—monomers are
→ CH
CH2 2 CH
CH3 3 ++ HBr
HBr →
CH3 3 CH
CH
CH
nucleotides, forms through condensation reactions
CH2 2 CH
CH2 2 CH
CH2 2 CH
CH
3 CH
+ 3+
CH
CH
CHCH
2 2 CH
2 the
2 2 CH
2 found
2 in
3nuclei,
CH
CH
CH
CH
2 CH
2 CH
2 cell
3
of
nucleotides,
codes for
Br → 3-bromohexane
hex-3-ene +Br
HBr
CH2 2
CH
amino acid sequence to make a protein, controls cell
development.
Br Br
CH2 2 CH
CH2 2 CH
CH2 2 CH
CH2 2 CH
CH2 2 CH
CH3 3
++CH
51. The wax candle is made up of very large alkanes.
CHCH
3 3 CH
2 2 CH
2 2 CH
3 +
3 3 CH
2 2 CH
2 2 CH
2 2 CH
3 3
CH
CH CHCH CHCH
CH
HBr→ →CHCH
CH
CH CHCH
CH
CH
3 +HBr
Incomplete
combustion
of these alkanes
results in
Br
Br
the formation of the black soot. The black residue on
the watch glass is soot which consists of incompletely
→ CH
CH3 3 ++ HBr
HBr →
CH3 3 CH
CH2 2 CH
CH CH
CH2 2 CH
CH2 2 CH
CH3 3
CH
burned hydrocarbons.
When HBr is added to hex-1-ene the Br can be added
52.
to either carbon 1 or 2. Carbon 2 is preferred because it
• PET is polyethylene terephthalate, a thermoplastic,
is bonded to the most carbon atoms.
polyester
When HBr is added to hex-3-ene the Br can be added
• monomer is ethylene terephthalate made from an
to either carbon 3 or 4. However, because the molecule
esterification reaction between ethylene glycol and
is symmetrical, and carbon atoms are always numbered
terephthalic acid
such that the substituent has the lowest number,
• repeating unit in the polymer is C10H8O4
adding to either carbon 3 or 4 results in the same
• modification of product is accomplished through
product, 3-bromohexane.
a copolymerization reaction e.g. if cyclohexane
47. Some reactions can be categorized as two different
dimethanol replaces ethylene glycol or if isophthalic
reaction names because they fit both descriptions. For
acid replaces terephthalatic acid, a softer product is
example, the reduction of an alkene to an alkane can
obtained
also be thought of as an addition reaction (hydrogen
53. a. One estimate is 400 years.
adding across the double bond)
b. Plastic bags are not part of curbside recycling
48. The concept map should include the following types
programs likely because of expense to get rid them.
of reactions, with an example of each: addition,
Individuals are reusing more and more.
elimination, substitution, condensation, esterification,
c. Much less plastic waste goes into landfill. It is also
hydrolysis, oxidation, reduction, combustion (complete
more economical.
and incomplete), as well as addition and condensation
d.
Cleanliness and contamination can be a problem
polymerization. The students should create categories
especially if the bag is used for carrying food.
such as “opposites” with examples of hydrolysis and
condensation, or oxidation and reduction. They could
group reactions with respect to the functional groups
involved.
49. Answer can focus on such information as: plastics are
made up of long chains so they are flexible so they can
be molded into a variety of shapes but may also be hard
and rigid and have uses where durability is important.
They are used for many everyday products such as
water bottles and plastic shopping bags, piping, conduit
and their low density and high durability leads to many
industrial and construction uses.
50. Refer to Appendix A page 712 SE format for Spider
Organizer. The central concept should be the fact
that both are polymers found in living systems.
Other ideas from brainstorming would include:
28 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
54. Answers can include: recycling of plastic bottles and
other plastics in cafeteria; use of dishes rather than
use plastics plates when possible; reuse plastic garbage
bags; refrigerate left overs in reusable containers rather
than covering with plastic wrap, use refillable bottle for
water; …and more
55. Students should consult Appendix A, pages 712-14
if they are unsure of which organizer to choose.
Alternatively, they could prepare a short audio report,
animation, or other type of electronic presentation that
covers the topics outlined on page 136 of the student
textbook.
56.
OH + n
n HO
61.
O
CH3
CH3
Cl
CH3
O
Cl
•R
eact the 2-methylprop-1-ene-1,3-diol with HBr
to produce 2-bromo-2-methylpropan-1,2-diol, as
shown below. There will also be some 1-bromo2-methylpropan-1,2-diol which will have to be
removed.
• React carbonic acid with the 2-bromo-2methylpropan-1,2-diol to get the final product, also
shown below. The second reaction is shown with the
compounds shaped so it is clear that the two OH
groups on the carbonic acid will react with the two
alcohol groups on the 2-bromo-2-methylpropan-1,2diol.
2n HCl
O
O
CH3
n
BPA mimics human hormones and can have negative
neurological effects especially if ingested at a young
age. BPA is useful because it makes Polycarbonate
plastic, which is clear and nearly shatter-proof.
57. Note: Assume that the starting material is
bromomethane and the product is methyl octanoate.
HO
CH
C
Convert bromomethane to methanol through a
substitution reaction. Oxidize octan-1-ol to an
aldehyde then to the carboxylic acid, octanoic acid.
Esterification reaction between methanol and octanoic
acid will produce methyl octanoate
C
O
HO
CH3OH + C7H15COOH →
HO
H2O + CH3-O–CO−C7H15
•C
arry out an elimination reaction by treating
bromoethane with NaOCH2CH3 to produce ethene.
• Carry out a reduction reaction on propan-2-one to
produce propan-2-ol.
• Carry out an elimination reaction by treating
propan-2-ol with with sulfuric acid to produce
propene.
• Carry out an addition polymerization with the ethene
and propene.
60. a. As the particles become small enough, animals
including fish and birds can ingest the plastic. This
cannot be digested and can interfere with normal
functions in their bodies. Death can result from
this or from blockage of airways. Also, further
degradation can release harmful chemicals.
b. Reduce, Reuse, Recycle.
CH2
CH2
OH
Br
C7H15CH=O + [O] → C7H15COOH
59.
→
CH3
HO
C8H17OH + [O] → C7H15CH=O → + [O] →
C7H15COOH
58. There is cross linking between strands specifically the
formation of hydrogen bonds between N−H and O=C
As well there is strength from the aromatic stacking
interactions.
OH + HBr
CH3
CH3−Br + OH− → CH3OH + Br−
CH2
C
OH
+
CH2
CH3
O
OH
C
→ 2H2O +
CH2
O
C
+
CH2
CH3
C
O
CH2
Br
Br
62. You can use the alcohol of the carboxylic acid you
need with the same R group and then oxidize it with
an oxidizing agent. You could also find an alkene
with a C1–C2 double bond and perform an addition
reaction to make it into an alcohol which you can then
oxidize
63. For example:
alanine
phenylalanine
O
H
N
CH
H
CH3
C
serine
O
N
CH
H
CH2
C
O
N
CH
H
CH2
C
OH
OH
64. • O
xidize the original oct-6-ene-1-ol to oct-6-eneoic
acid
• Carry out an addition reaction with water to convert
the oct-6-eneoicacid to 6-hydroxyoctanoic acid.
Unit 1 Part A • MHR 29
• Carry out a condensation polymerization reaction to
product the poly-6-hydroxyoctanoic acid.
OH
O
+ [O] →
OH
O
+ H2O
OH
OH
O
n
12. a. 2-methylpropan-2-yl 2,2-dimethylbutanoate +
water,
b. N-propyl-3-chlorobutanamide + water
O
OH
13. cycloheptene + H2O → cycloheptanol + [O] →
cycloheptanone
OH → nH2O +
14. a. addition polymer
CH2=CH—C≡N
b. condensation polymer
OH
O
...
11. Amino acids would be found in muscle-building
supplements because they are the monomers of
proteins which help athletes build muscle. (In reality,
only exercise can build muscle.)
O
O
O
O
O ...
65. Use different coloured paper clips. For example
nylon 6,6. The monomers are adipyl chloride and
hexamethylene diamine can be represented by different
colours or sizes of paper clips.
66. a.
H
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
H
b. 2C8H18(ℓ) + 25O2(g) → 16CO2(g) + 18H2O(ℓ)
c. CO2, a major greenhouse gas that is linked to
global warming and acid precipitation, is constantly
added to the environment. Combustion is usually
incomplete, thus adding soot and carbon monoxide
to the atmosphere. Also, the octane has additives
that can further pollute the air.
Answers to Chapter 2 Self-Assessment Questions
(Student textbook pages 142-3)
1. d
2. c
3. b
4. d
5. a
6. e
7. b
8. d
9. b
10. c
30 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
O
HO
O
OH
HO
HO
15. Both cellulose and starch are polymers with glucose as
the monomer. However they differ in how the glucose
monomers are linked together. Cellulose has beta
linkages and starch has alpha linkages. Because of the
beta linkages humans cannot digest cellulose; we do
not have the enzymes that recognize beta linkages.
Cellulose fibers are used to compose wood, paper,
cotton and flax. Starch is used as an energy storage unit
for plants.
16. • reduce the carboxylic acid to an aldehyde
•reduce the aldehyde to an alcohol
• an elimination reaction of the alcohol to make an
alkene
• add the alkene to itself to perform an addition
polymerization reaction
17. a. 2-chloroheptane and 3-chloroheptane; In the
reactant, the two carbon atoms involved in the
double bond are bonded to an equal number of H
atoms and an equal number of C atoms. Therefore,
neither product will be preferred.
b. 2-methylhept-3-ene and 6-methylhept-3-ene
Carbon atoms 3 and 5 are bonded to the same
number of other carbon atoms as well as the same
number of hydrogen atoms. Therefore, neither
product is preferred.
18. Essay can focus on common uses such as furniture,
cooking utensils, car parts, fabrics, toys, footwear,
bags, phones, or computers. If we depended on natural
materials for these uses, many applications would
not be possible. Also there would not be enough
natural products to make items to support the world
population.
19. (CH3)2 CHCH=CHCH3 + 9O2 → 6CO2 + 6H2O
20. Polyamide. The −COOH of one amino acid reacts with
NH2 of another amino acid to form amide linkages.
21. The number of C−O bonds decrease or the number of
C−H bonds increase
22. a. CH3CH2C(Cl)=CH(Cl)
and CH3CH2C(Cl)2–CH(Cl)2,
b. cyclopentene + water
c. HO-CH2CH2CH3 + Cl23. butanone + [H] → butan-2-ol
butan-2-ol + HCl → 2-chlorobutane + water
24. a. hexyl propanoate + water,
b. cyclobutanol
25. HOOC−CH2−CH2−CH2−CH2−OH with H2SO4 as
a catalyst
The −OH group on one end of the molecule will react
with the −COOH group on the other end to form the
cyclic compound.
Answers to Unit 1 Review Questions
(Student textbook pages 147-53)
1. c
2. c
3. b
4. b
5. a
6. d
7. b
8. c
9. b
10. e
11. a
12. e
13. a
14. c
15. a. alkene
b. alkane
c. alkene or cycloalkane
d. alkyne
16 a. These are not constitutional (structural) isomers but
the same molecule rotated 180 degrees.
b. These are constitutional (structural) isomers, as the
longest chain containing the double bond is five
carbons long, whereas the second has the longest
chain containing the double bond is only four
carbons long.
17. Both types of stereoisomers have their atoms bonded
in the same sequence on the carbon chain. Cis/trans
(Z/E) diastereomers (geometric isomers) have two
unique atoms or groups arranged on the same side
(cis/Z) or on opposite sides (trans/E) of the double
bonds. They have different physical but similar
chemical properties. Enantiomers (optical isomers)
are non-superimposable mirror images of each other
and don’t require a double bond. They have essentially
the same physical and chemical properties except how
the rotate plane-polarized light and react with other
enantiomers and with enzymes.
18 a. substitution
b. addition
c. condensation
d. hydrolysis
e. elimination
f. combustion
19. A polymer is a long-chained molecule made up of
repeating units. A monomer is the unit which repeats
in a polymer. Glucose is a monomer and starch is one
of its polymers.
20. Primary amides would be expected to be more soluble
in water as they have the greatest number of N–H
bonds that allows for hydrogen bonding with H2O.
21. They are both made through condensation
polymerization reactions. Polyamides contain amide
linkages, –CO–NH– while polyesters contain ester
linkages, –CO–O–.
22. The reactant in an addition reaction is an alkene which
must contain a double bond between two carbon
atoms. When a small molecule such as HX is added to
an asymmetric alkene, two products can form because
the X could be added to either of the carbon atoms
involved in the double bond. Markovnikov’s rule is
used to predict which product is more abundant. Also,
two different products can form when each double
bond has two different atoms or groups of atoms
single bonded to it. The products will be cis and trans
isomers.
23. Petrochemicals are products derived from petroleum.
Basic hydrocarbons, such as ethene and propene, are
converted into plastics and other synthetic materials.
24. a. chloroprene,
b. methyl methacrylate
Unit 1 Part A • MHR 31
low solubility in water. Propan-1-ol would both be
soluble in water because both can enter into hydrogen
bonding. The –COOH group in propanoic acid makes
this molecule more polar than the corresponding
alcohol and it will have the higher boiling point will
be higher than the alcohol. Propanoic acid will turn
litmus red.
25. condensation polymerization
26. a. addition
b. condensation
c. combustion
27. a. oxidation
b. reduction
28. a. hydrolysis
b. esterification and condensation
33. Esterification (cyclic) and polymerization
O
29. a. N-butylethanamide
b. cyclopentylbutanoate
30. a. The longest carbon chain is five carbon atoms long.
The correct name is pentane.
b. Numbering of the side groups should give the lowest
places on the longest chain for alkanes. The correct
name would be 2,3-dimethylpentane.
c. Side groups on an alkane should be placed in
alphabetic order excluding multiplying prefixes. The
correct name would be 3-ethyl-2,2-dimethylnonane.
d. The benzene group has the greatest number of
carbon atoms and therefore would receive the root
name. The correct name would be 1-pentylbenzene.
31. a. Aldehydes are always at the end of a carbon chain.
The molecule would be a ketone and have the name
propanone.
b. Numbering of a cycloalkane would place the
halogens in alphabetic order giving the first in the
alphabet the lowest number. The correct name
would be 1-bromo-2-chlorocyclobutane.
c. The alcohol group would receive the lowest number
in the longest chain. The longest carbon chain that
does not contain multiple bonds is nine carbons
long. The correct name would be 7-methylnonan-3ol.
d. The longest carbon chain in an ether gets the root
name. The correct name would be ethoxybutane.
e. The longest unsaturated chain receives the root
name. Numbering gives the multiple bond the
lowest number. The correct name is 2-ethylpent-1ene.
f. The longest carbon chain in the root name is five
carbons long. The correct name would be methyl-3methylpentanoate
32. Use difference in their boiling points, solubility in
water and reaction to litmus to distinguish between
these molecules. Propyne would have the weakest
intermolecular forces (dispersion forces) between
molecules and therefore the lowest boiling point and
be a gas at room temperature. The gas would have a
32 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
O
H2SO4
OH
HO
C
CH2
CH2
CH2
CH2
O
n
O
H2SO4
OH
HO
+
H2O
CH2
n H2O +
O
...
O
C
CH2
CH2
CH2
CH2
CH2
O
O
C
CH2
CH2
CH2
CH2
CH2
...
34.
Convert ethane to ethene by cracking.
CH3—CH3 800PtºC CH2=CH2 + H2
Convert ethene into ethanol by addition of H2O.
CH2=CH2 + H2O → CH3—CH2OH
Convert ethanol to ethoxy ethane by dehydration.
This reaction requires strong sulfuric acid as a
catalyst and high temperatures.
H SO
2CH3—CH2OH heat CH3—CH2—O—CH2—CH3
+ H2O
35. a. propan-1-ol + ethanol
b. 3-chloro-2,2-dimethylpentane
c. N-methyl-N-(2-methylpropyl)ethanamide
•
•
•
2
4
36. a. A is ethane; C is ethanol; B is ethyl ethanoate ; D is
ethanoic acid
b. ethane CH3—CH3; ethanol CH3—CH2—OH; ethyl
ethanoate CH3—CH2—CO—O—CH2—CH3;
ethanoic acid CH3— COOH
37. a. Both molecules have aromatic (benzene) rings and
halogens.
b. Both names contain the name phenyl which is
benzene ring, the prefix bi- which indicates that
there are two benzene rings, and the name of a
halogen.
c. The molecules would be expected to be liquids or
solids due to the large number of carbon atoms and
highly electronegative halogen atoms there would be
significant dispersion forces between molecules.
d. These molecules would not be soluble in water
but be soluble in non-polar substances such as
hydrocarbons (oils & fats for example).
e. They would be expected to have other similar
physical properties such as melting point boiling
point density. They may have similar physiological
effects since they can dissolve and accumulate in fat
tissue
38. an ether
39. Carry out a substitution reaction of benzene with
bromine.
benzene + 2Br2 →1,3-dibromobenzene + HBr
40. a. 7-methylocta-2,6-dien-1-ol
b. It would be a volatile liquid at room temperature. It
would be slightly soluble in polar solvents and very
soluble in non-polar solvents.
c. Ethanol would be a suitable solvent as it would
evaporate quickly allowing the geraniol to slowly
volatilize.
41 a. Amino acids have both amino, –NH2, and carboxyl,
–COOH, functional groups.
b. Glycine would be named 2-aminoethanoic acid and
alanine would be 2-amino-propanoic acid.
c. Glycine would be more soluble in water than alanine
as it has fewer alkyl groups.
42. a.
O
b. It should be a gas at room temperature, be soluble in
non-polar solvents and of low solubility in water.
c. Its extreme flammability would be one of the reasons
for not being used in operating rooms.
43.
• monomer CH3—CH=CH—CH3
• polybutene –[—CH(CH3)—CH(CH3)—]n –
• solvent alkylethers, tertiary alcohols
• acid catalyst
WHMIS: 2-butene is a gas at room temperature, highly
flammable, explosive with air mixture, inhalation can
cause dizziness and unconsciousness, evaporation from
skin can lead to frostbite, use face shield
44. a. butanal, butanoic acid
b. 4-methylhexan-3-ol
45. a. poly-1,1-dichloroethene;
...
CH2
CCl2
CH2
CCl2
CCl2
CH2
CH2
...
CCl2
b. trade name is Dacron
...
O
O
O
C
C
O
CH2
CH2
O
O
O
C
C
O
CH2
CH2
...
46. a.
• use 2 fats and 2 oils
• a dd equal portions of each fat and oil to separate
test tubes
• a dded a measured volume of bromine water to each
test tube
• s hake gently or warm in a water bath while looking
for a decrease in the intensity of the reddish colour
of bromine water.
b. Br2(aq) WHMIS: Corrosive. Harmful if inhaled.
Causes respiratory tract irritation and possible
burns. Causes eye and skin irritation and possible
burns. Wear gloves when making the bromine water.
Work in a fume hood.
c. addition
d. H3C—CH(Br)—CH(Br)—CH3
47. Ethanol is the only one of the three that will mix with
water. Adding water to each will indicate which one
is ethanol. Hex-2-ene can be separated from benzene
by the addition of Br2(aq). The bromine water will
discolour as it reacts with the hex-2-ene but will not
react with benzene.
48. a. 3-hydroxypropanoic acid
b. 4-aminobutanal
49. butane; 3-methylpentane; 2-chloropentane;
2-iodopentane; 1-bromohexane
50.
• Oxidize the ethanol all the way to ethanoic acid.
• Carry out an addition reaction using HCl with the
non-1-en-5-amine to form 2-chloronon-5-amine.
• Carry out a condensation reaction between the
ethanoic acid with the 2-chloronon-5-amine in the
presence of an acid catalyst to form the amide bond.
51. The teacher might suggest to the students that they
start this question early and carry a list of the classes of
compounds for several days or a week. As they are in
different locations, they can look for examples. It would
be interesting for them to look for a certain category of
used (e.g. foods, medications, cleaning solutions) that a
particular hydrocarbon or derivative fits into.
Unit 1 Part A • MHR 33
‑52.
58. Students have 58 terms to consider and link;
organization will most likely reflect the section
organization of the two chapters. Answers should
include additional explanatory words to show
meaningful relationships between the words and
clusters of words.
H
C
H
Cl
H
The chlorine atom is above the plane of the paper, one
hydrogen atom is below and two hydrogen atoms are in
the plane of the paper.
59. Some ideas to expand on: The disposal of items and
products made of plastic leads to very large amounts
of waste that is harmful to the environment. Since they
take a very long time to degrade, they take up space
in land disposal sites. No one wants garbage sites near
where they live so there are not many places to put this
waste. There is a large mass of garbage in the ocean
which is mostly made up of plastics. This interferes
with marine life and can cause the death of animal
life in the ocean. Reuse and recycling are part of the
solution to this problem.
53. The simplest hydrocarbon that can form cis and trans
isomers is 2-butene.
H
H
HH
3C
C
C
HCH3
C
C
cis-2-butene
CH3
H3C
Cl
54.
Br Cl
C F
Br
CCH3F
CH3
CH3
H
C
H3H
C
H3C
C
H3
CH
C -2-butene
C
trans
H
Cl
F ClC
F
Br
60. All aspects of our everyday life have uses of organic
compounds. The answer should cover a wide variety of
uses and abuses of e.g. pharmaceuticals, plastics, food
additives, fuels.
CCH3Br
CH3
The diagrams show that 1-bromo-1-chloro-1fluoroethane can form non-superimposable mirror
images of each other. Therefore, there can be
enantiomers of 1-bromo-1-chloro-1-fluoroethane.
55. Sample answer: Cotton requires large amounts of water
to grow and a large amount of pesticides are used on
the plant. Polyester uses 10 times more energy than
cotton during production and produces 4 times more
carbon dioxide
61. One possible flowchart is:
Organic compounds
Addition
Elimination
Substitution
Esterification
56.
• complete combustion – excess of O2(g) available;
products of complete combustion of hydrocarbons
are only carbon dioxide and water.
• complete and incomplete combustion - starting
material is a hydrocarbon; CO2 and H2O are always
produced
• incomplete combustion – insufficient O2(g) available;
less heat given off; in addition to CO2 and H2O, CO
and C are also products.
Incomplete combustion is dangerous because CO
is poisonous, colourless, odourless, gas that is
undetectable to our senses.
57. Life in the 1800s would centre around providing
necessities rather than conveniences. Discuss
availability of choices in modes of travel, clothing,
furniture, type of work.
34 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
Functional groups
Reactions between organic
compounds and the resulting
products are controlled by the
types of functional groups present.
Reactions of
organic compounds
Synthetic polymers
Petrochemicals
Organic compounds
are defined by the
funtional group(s)
that they contain.
Polymerization
reactions(addition
and condensation)
produce very large
molecules
Natural polymers
Biological molecules
and their functions
62. One possible flowchart is illustrated below:
Crude Oil
Fractional distillation
Cracking
and reforming
Ethene (ethylene)
oxidation by
addition of
oxygen gas with
a silver catalyst
Ethylene oxide
Xylene
oxidation by
addition of
potassium
permanganate
Terephthalic acid
Polymerization reactions
(addition and condensation)
produce very large molecules
Ethylene glycol
condensation
polymerization
Terephthalic acid
63. With respect to solubility, the statement is too general
to agree with. The general statement that matches
this is that “like dissolves like” Non-polar solvents
dissolve non-polar compounds and polar solvents
dissolve polar compounds. Water is a polar solvent.
Hydrocarbons are non-polar and have low solubility
in water while organic compounds with a polar part
on the molecule will dissolve in water. The ability to
conduct an electric current depends on the presence
of ions. Few organic compounds will form ions in
solution; even carboxylic acids ionize only slightly. The
statement is generally true regarding conductivity.
64. a. Formic acid is the simplest of the carboxylic acids,
HCOOH. The IUPAC name is methanoic acid.
b. Human use is to use formic acid as an antibacterial
agent and preservative with animal feed; it can be
reduced to formaldehyde, a preservative for animal
specimens, it can act as a reducing agent to reduce
metals from their solutions; it is used in the tanning
industry; it is used to synthesize esters.
c. Stinging insects such as bees and beetles will secrete
formic acid.
d. It is a defense mechanism to ward off predators.
65. a. Teflon has a very low coefficient of friction and the
Teflon polymer is non-reactive to compounds found
in foods.
b. At high temperatures, above 240oC, it is known that
Teflon breaks down and gives off toxic particles and
gases.
c. A safety note should be included cautioning users
to not heat this pan above normal temperatures
used for cooking. Do not place in an oven or on a
barbeque.
66. a. Any molecule that has a large hydrocarbon chain
with polar groups such as alcohols or carboxylic
acids on the end would suffice.
b. Emulsifiers make heterogeneous mixtures such
with oil and water appear to be homogeneous. For
example lecithin in egg yolks helps to prevent the
separation of oil and vinegar mixtures in certain
salad dressings.
67. Some examples would be pharmacist, medical doctor,
researchers in the cosmetics, pesticides and polymer
industries, food and drug analyst, petrochemical
engineers, hazardous waste handlers as well as a
myriad of others. Training for most of these programs
would involve a four year undergraduate degree as well
as possible post graduate degrees. To become a medical
doctor would require at least six years. Only certain
schools would offer pharmacy and medical programs
68.
H3C
CH2
CH2
CH2
O
CH2
CH2
OH
b. One would expect them to be considered flammable
and possibly toxic in large doses.
c. The sorbitan molecule contains alcohol, ether, ester
and large alkyl functional groups. This combination
would make one end of the molecule soluble in
water and the other end soluble in the oil.
d. Skimmers recapture the oil which could be reused
but require much energy and effort. Burning
removes it from the ecosystem but creates
greenhouse gases and harms organisms on the
surface. The dispersants reduce the concentration of
the oil but just spread it and when used underwater
may damage the ecosystem deeper than just on
the surface. Other benefits and repercussions are
possible.
Unit 1 Part A • MHR 35
e. Reduce one’s use of fossil fuels such as using public
transit, recycling plastics, encouraging governments
companies to have safeguards in place are just some
of the many answers students could give.
69. The chains of this polymer have –NH2 and –C=O
groups attached, which allows hydrogen bonding to
occur. This increases the strength.
70. The greater the intermolecular attractions between
molecules the greater the boiling point. The boiling
points of benzene and cyclohexane are higher than for
hexane because the structure of these two compounds
gives a greater surface area. This results in greater
intermolecular attractions between their molecules.
71. (1) It is similar to a condensation reaction
(2) reduction,
(3) polymerization
72. The tar-like substance is most likely a long chain
carbon compound that is non-polar. Vegetable oil
would likely be the best since it has fairly long carbon
chains and the esters that make it up are only slightly
polar.
73. This is oxidation because a carbon atom in the product
is bonded to fewer hydrogen atoms.
77. a. With so many organic compounds widely used,
solvents have become specific for various classes
of compounds. The advantage is that there are no
organic compounds for which a solvent cannot
be found so that any compound can be cleaned of
a surface. The disadvantage is that many organic
solvents are volatile and are hazardous to our health
if inhaled. Disposal of solvents is another problem
since they can persist in the environment for long
periods of time.
b. Research should focus on absorption of solvents
through the skin and inhalation.
c. Results can be organized, for example, with respect
type of health effect, seriousness of health effect,
type of solvent
d. Greater care could be taken to avoid using more
of the solvent than necessary. Research the type of
solvent that will dissolve a particular compound and
use the one that has the least effect on health and the
environment.
Answers to Unit 1 Self-Assessment Questions
(Student textbook pages 154-5)
1. b
74 a. Ethanol or grain alcohol, C2H5OH, is the alcohol
present in wine, spirits and beer.
b. Due to their ability to form hydrogen bonds,
alcohols would be liquids or solids at room
temperature.
c. Due to the toxic and combustible nature one
would expect the WHMIS symbols for poison and
flammable to be on containers used in the workplace
2. e
75. Rusting is an oxidation of a metal. The oxidation
and reduction discussed in this chapter have organic
compounds as reactants.
9. b
76. Student answers will likely include some of the
following points. The public often believes that
“natural” is safer and better than “synthetic,” even if
the compounds are identical. This is evident in natural
and artificial flavourings, and with natural and artificial
sweeteners. The public is prepared to pay a premium
for “organic” or “natural” products, without knowing or
understanding the differences between them. In order
for consumers to make informed product choices, they
have to thoroughly research the differences between
the products. Many manufactured products have strict
guidelines around labelling, while there are few if any
guidelines around “natural” products.
36 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
3. b
4. b
5. e
6. b
7. d
8. b
10. a
11. There are five structural isomers, one of which has two
diasteroisomers.
H
H
H
H
H
C
C
C
C
H
H
cis
or
trans H
H
H
H
H
H
C
C
C
C
H
14. a. Vitamin C could form enantiomers as it has two
carbon atoms that are bonded to four different
atoms or groups. These two carbon atoms are circled
in the figure below.
H
H
H
H
H
C
C
H
H
C
C
H
H
H
HO
HO
HO
H
C
H
C
C
C
H
H
12. a. CH
2
H
H
C
H
C
C
C
H
H
H
CH
H
b.
O
CH2
CH2
CH3
CH3
c.
O
CH3
CH2
CH2
CH
CH2
C
OH
CH2
CH2
CH3
e.
CH3
alcohol to carboxylic acid: oxidation
O
C
H
C
C
H
CH3
H
O
C
C
C
H
O
H H
CH2
CH2
H H
H
C
C
H
C
C
16. a. fuels
b. beverages, antiseptics
c. pharmaceuticals
d. sour tasting foods such as vinegar
e. artificial and natural flavourings
These are just some examples of the many places
students may find these substances
haloalkane to alcohol: substitution
H
O
H
OH
17. alkane to haloalkane: substitution
CH3
d.
O
15 a. 3-ethyl-2,2-dimethyloctane
b. 1-ethyl-2-fluorobenzene
c. butan-2-ol
d. 3-phenylnonane
CH
CH
O
b. Due to the large number of hydroxyl groups and
oxygen atoms, which could hydrogen bond with
other molecules, it would have a high melting point
and be a solid at room temperature. Due to the large
number of polar hydroxyl groups it would be soluble
in water.
c. Yes as fats are non-polar and vitamin C molecules
polar, the vitamin C would not dissolve in the fats.
It is water soluble and would be quickly eliminated
from the body.
H H
H
H
H
H
H
13. Some benefits would be better medicines, dyes, foods,
cosmetics, transportation, and agricultural yields.
Some drawbacks would be increased reliance on
hydrocarbons with the resulting environmental effects
such as oil spills and emission of VOCs and ozone
destroying CFCs. The release of toxins such as dioxins
as well as drugs and pesticides that have had negative
side effects on humans and the environment could also
be mentioned
CH2
CH2
CH3
Unit 1 Part A • MHR 37
C
H
18. Molecules must have double bonds to undergo
addition polymerization or they must have two
functional groups which can react to combine and
release a small molecule to undergo condensation
polymerization.
19. The first polymer is an addition polymer and the
second is a condensation polymer (polyamide). The
polyamide would be stronger because its chains can
hydrogen bond to each other.
20. Recycle bags, have a no plastic bag rule, buy degradable
garbage bags. Have recycle containers available for soft
drink and water bottles.
21. a. 3-methylpentan-2-ol + HCl
b. 3-methylpentan-3-ol and H2SO4 as a catalyst
22.
O
O
HO
HO
OH
CH2
CH2
OH
38 MHR • Chemistry 12 Solutions Manual 978-0-07-106042-4
23.
• Oxidize the ethanol first to ethanal and then to
ethanoic acid.
• Carry out an addition reaction to add water to the
double bond of the but-3-en-2-amine producing
3-aminobutan-2-ol.
• Combine the ethanoic acid and the 3-aminobutan2-ol and use sulfuric acid as a catalyst to carry out
condensation reactions. The carboxyl group of one
molecule of ethanoic acid will react with the amino
group of the 3-aminobutan-2-ol, and the carboxyl
group of a second molecule of ethanoic acid will
react with the alcohol group on the other end of the
3-aminobutan-2-ol to form the desired compound.
24. a. major product: 3-bromo-3-methylpentane
minor product: 2-bromo-3-methylpentane
b. 3-methylbutanoic acid and ethanol
c. 2-methylpentan-3-one
25. Small carbon chains ignite too early and explode
instead of burn, causing incomplete combustion. This
is also damaging to the engine. Carbon chains with
more than 12 carbon atoms would not vaporize well
enough and would therefore not undergo complete
combustion. Soot would build up in the engine and
cause damage.