CHAPTER 9: ANSWERS TO SELECTED PROBLEMS
SAMPLE PROBLEMS (“Try it yourself”)
CH3
9.1
CH3
CH3
CH CH CH3
and
CH3
CH CH2 CH2
OH
OH
9.2
a) Ethylene gives only one product, because the two sides of the ethylene molecule are
identical.
OH
OH
CH2 CH3
and
CH3 CH2
are the same molecule
b) Cis-2-hexene gives two different products, as shown below.
OH
OH
CH3 CH CH2 CH2 CH2 CH3
9.3
and
CH3 CH2 CH CH2 CH2 CH3
2-butanol
OH
OH
9.4
CH3 CH2 CH CH2 CH2 CH2 CH2 CH2 CH3
9.5
Cyclohexanol has the higher boiling poing, because it can form hydrogen bonds while
cyclohexene cannot. (The two molecules are roughly the same size, so we can ignore the size in
this question.)
9.6
3-pentanol is more soluble than propane, because the alcohol group in 3-pentanol can
form hydrogen bonds. Propane does not contain any atoms that can form hydrogen bonds.
9.7
Compound X is the least soluble, followed by compound Y, followed by compound Z
(the most soluble).
CH3
9.8
CH2 CH
C
CH3
CH3
CH3
9.9
CH3 CH2
C
CH3
CH
CH3
and
CH3 CH2
CH CH CH2
9.10 This compound does not contain a phenol group, because the hydroxyl group is not
directly bonded to the aromatic ring.
9.11 Compound Y is the least soluble, followed by compound X, followed by compound Z
(the most soluble).
SECTION PROBLEMS
Section 9.1
OH
9.1
CH3
OH CH3
a) CH3 CH CH2 CH CH3
and
CH3 CH2 CH CH CH3
OH
OH
and
b) CH2 CH2 CH2 CH3
OH
c) CH3 CH2 CH CH2 CH2 CH3
CH3
CH CH2 CH3
(Only one product is formed, because the
two sides of cis-3-hexene are identical.)
OH
d)
OH
CH2 CH3
and
CH CH3
OH
OH
e)
and
CH2 CH3
CH2 CH3
9.2
In part b, the second product will predominate. In part d, the first product will
predominate. In part e, the second product will predominate.
9.3
Alkenes b and e form only one product when they react with water, because the two sides
of the molecule are identical (when you divide the molecules in half through the C=C bond.)
CH3 CH2 CH
CH
CH2 CH3
CH3
Section 9.2
9.4
Most biological reactions are linked in a sequence. If the enzyme does not select the
correct product, the next enzyme cannot use the product in the next reaction, so the incorrect
product goes to waste.
O
9.5
HO
C
OH
CH2
O
C
CH2
C
OH
C
OH
O
Section 9.3
9.6
a) 1-butanol
b) cyclohexanol
c) 4-octanol
OH
9.7
d) 1-pentanol
OH
a) CH3 CH2 CH CH2 CH3
b) CH3 CH CH2 CH2 CH2 CH2 CH2 CH2 CH3
OH
c)
d) CH3 OH
9.8
a) This name is incorrect, because it does not tell where the alcohol group is located.
b) This name is correct.
c) This name is correct.
d) This name is incorrect, because the principal chain must be numbered from the end
that is closest to the functional group. (The correct name would be 2-pentanol.)
Section 9.4
9.9
O
CH3
δ+
H
δ–
O
H
CH2
CH2
CH3
9.10 a) 3-pentanol has a higher boiling point than pentane, because it contains a functional
group that can form hydrogen bonds.
b) The second compound has the higher boiling point, because it contains two alcohol
groups, both of which can form hydrogen bonds.
9.11
a) 2-butanol has the higher solubility, because it has the smaller hydrophobic region.
b) 2-butanol has the higher solubility, because it contains a group that can form hydrogen
bonds while hexane does not.
c) Ethanol has the higher solubility, because it contains a group that can form hydrogen
bonds while ethyne does not.
9.12 a) The solution is basic, because KOH dissociates into K+ and OH– ions, and OH– ion is a
strong base.
b) This solution is neutral, because CH3OH (methanol) does not dissociate into ions.
c) This solution is basic, because Ba(OH)2 dissociates into Ba2+ and OH– ions.
Section 9.5
9.13
a) CH2 CH2 (This is the only product.)
and
CH3 CH CH CH2 CH3
b) CH2 CH CH2 CH2 CH3
(Both the cis and the trans isomers of the second alkene will be formed.)
CH3
CH3
c) CH3 C CH CH2 CH3 and
CH3 CH CH CH CH3
(Both the cis and the trans isomers of the second alkene will be formed.)
CH3
d) CH3 C
(This is the only product, but both the cis and the
trans isomers will be formed.)
CH CH CH3
CH3
e) Methanol cannot be dehydrated, because it does not have a second carbon atom
adjacent to the functional group.
f)
(This is the only product.)
CH3
9.14
CH3
OH
CH3
CH CH2 CH CH2 CH2 CH2 CH3
and
CH3
OH
CH CH2 CH2 CH CH2 CH2 CH3
9.15 In a dehydration reaction, there must be at least one hydrogen atom directly bonded to the
carbon adjacent to the functional group. In this molecule, the adjacent carbon is not bonded to a
hydrogen atom.
OH
C
CH2
This carbon atom is not bonded to a
hydrogen atom.
Section 9.6
9.16
SH
OH
OH
OH
CH3
thiol
alcohol
phenol
alcohol
9.17 a) Phenol has a higher solubility than benzene, because the hydroxyl group in phenol can
form hydrogen bonds.
b) The second molecule (1-propanol) has a higher solubility than the first (a thiol),
because alcohol groups can form hydrogen bonds, while thiol groups cannot.
9.18
Compounds b and d can be dehydrated. Phenols and thiols cannot be dehydrated.
CUMULATIVE PROBLEMS (Odd-numbered problems only)
9.19
a) There are many possible answers. Here are two:
OH
OH
(cyclopropanol)
(cyclohexanol)
b) There are three possible answers:
SH
SH
CH2 CH2 CH3
SH
CH3 CH CH3
c) There are many possible answers. Here are two:
OH
CH3
C
OH
CH3
CH3
CH2 CH2 CH2 CH CH3
CH3
d) There are many possible answers. You can either move the functional group to a
different location, or convert the principal chain (pentane) into a branched chain.
OH
CH3 CH CH2 CH2 CH3
Here we have moved the
functional group from the #1
carbon to the #2 carbon.
OH
CH3
CH2 CH2 CH CH3
Here we have changed the
unbranched carbon chain
into a branched chain.
OH
9.21
a) CH3 CH2
(This is the only product.)
b) CH3
CH CH2
(This is the only product.)
c) CH3
CH CH2
(This is the only product.)
OH
d)
CH2 CH2 CH2
e) CH3 CH C
CH3
OH
and
and
CH2 CH CH3
CH2 CH CH CH3
CH3
CH3
CH3
f) CH2 CH C
CH3
(This is the only product.)
CH3
g) This alcohol cannot be dehydrated, because there is no hydrogen atom on the carbon
adjacent to the functional group.
OH
h) CH3 CH2 CH CH2 CH2 CH3
i)
(This is the only product.)
(This is the only product
OH
OH
j)
CH3
and
CH3
k) There are three products of this dehydration.
C
CH2 CH2 CH2 CH3
C
CH3
CH CH2 CH2 CH3
CH3
C
CH2 CH2 CH2 CH3
CH2
OH
9.23
d) The main product is
CH2 CH CH3
OH
j) The main product is
9.25
a) methanol
e) cyclohexanol
CH3
b) 2-butanol
a) CH3
d) 3-octanol
OH
OH
9.27
c) phenol
CH2
b) CH3 CH CH2 CH2 CH3
OH
OH
c)
d) CH3
CH CH3
OH
e) CH2 CH2 CH2 CH2 CH2 CH3
9.29
a) C3H8O
b) C3H6O
9.31 a) The first molecule is more soluble, because it contains a functional group that can form
hydrogen bonds (the alcohol group), while the second molecule does not.
b) The first molecule is more soluble, for the same reason as part a. (The thiol group
does not form hydrogen bonds.)
c) The second molecule is more soluble, for the same reason as part a.
d) The first molecule is more soluble. Both contain an alcohol group, but the first
molecule has the shorter hydrocarbon chain.
e) The second molecule is more soluble, because it contains a functional group that can
form hydrogen bonds, while the first molecule does not. (The chain length only matters when
both molecules contain hydrogen-bonding groups.)
f) The second molecule is more soluble, because it contains two functional groups that
can form hydrogen bonds, while the first molecule only contains one.
9.33 a) Compound C has the lowest solubility in water, because it cannot form hydrogen
bonds.
b) Compound B can be dehydrated. Thiols and phenols cannot be dehydrated.
9.35 Enzymes speed up reactions, so they occur at a rate that is useful to the cell. When more
than one product is possible, enzymes also select the correct product, ensuring that the cell does
not waste its resources.
9.37
H
O
δ+
H
δ–
O
Here, water is
the donor and
methanol is
the acceptor.
CH3
δ+
H
δ– H
O
H
Here, methanol
is the donor and
water is the
acceptor.
9.39
Isomers have the same molecular formula, but different structures (and different names).
a) These two molecules are isomers.
b) These two molecules are not isomers; they are two different ways to draw the same
compound (2-butanol).
c) These two molecules are not isomers, because they have different molecular formulas.
The second molecule has five carbon atoms, while the first molecule has only four.
d) These two molecules are isomers.
e) These two molecules are isomers. (Isomers do not have to have the same functional
group—they only need to have the same molecular formula.)
f) These two molecules are not isomers. The first molecule has 12 hydrogen atoms,
while the second has only 10.
9.41
Reaction #2 has two possible products, so the enzyme must select the correct product.
9.43
There are four possible products.
OH
CH3
OH
CH CH2 CH2 CH CH3
OH
CH3
OH
CH3
OH
CH2 CH CH2 CH CH3
OH
CH CH2 CH2 CH2 CH2
OH
CH3
OH
CH2 CH CH2 CH2 CH2
9.45 a) The two alkenes are 1-pentene and 2-pentene. (Actually, there are three alkenes, since
both the cis and trans isomers of 2-pentene would work.) Here are their structures:
CH2 CH CH2 CH2 CH3
CH3
CH CH CH2 CH3
(1-pentene)
(2-pentene)
b) The best choice would be 1-pentene. According to Markovnikov’s Rule, the hydration
of 1-pentene would produce 2-pentanol, with only a trace of 1-pentanol. If you used 2-pentene,
you would get roughly equal amounts of 2-pentanol and 3-pentanol.
9.47 There are many possible answers for each part of this question. I’ll give two examples
for each part.
OH
a)
CH3
OH
CH2
and
CH3 CH2 CH CH2 CH3
OH
b)
CH3
CH2
CH
OH
CH3
and
CH3
OH
OH
c) CH3 CH2 C
CH2 CH2 CH3
and
C
CH2 CH3
CH3
CH3
OH
d) CH3 OH
and
CH2
9.49 Ethanol molecules can form hydrogen bonds with one another, because they contain an
–O–H group. Molecules of dimethyl ether, in contrast, cannot form hydrogen bonds with one
another, because they do not contain a hydrogen atom that is directly bonded to oxygen.
Therefore, the attraction between ethanol molecules is much stronger than the attraction between
dimethyl ether molecules, giving ethanol a much higher boiling point.
9.51
alcohol
CH3
HO
phenol
OH
C
CH
alkyne
9.53
CH3
H
alkene
C
H
C
CH2 SH
thiol
9.55 Ethylene glycol has a very high tendency to form hydrogen bonds. The attraction
between ethylene glycol molecules is so strong that hexane molecules (which cannot form
hydrogen bonds) are forced out of the area between the ethylene glycol molecules. This is
analogous to the reason that water will not mix with hexane.
Ethanol has a lower tendency to form hydrogen bonds, so hexane molecules are not
forced out of the area between ethanol molecules. As a result, ethanol and hexane mix with one
another.
9.57 You need 6.4 g of ethanol (the calculator answer is 6.44952 g). The isotonic
concentration is 0.28 M, so we need to use enough ethanol to make a 0.28 M solution. First,
calculate the number of moles of ethanol in 500 mL of this solution (0.14 moles). Then use the
formula weight of ethanol (46.068) to convert moles to grams.
9.59 You make 14 g of ethanol (the calculator answer is 13.9590047 g). The molecular
formula of ethene is C2H4, and the molecular formula of ethanol is C2H6O. The balanced
equation for the hydration reaction is C2H4 + H2O → C2H6O. The formula weights of ethene
and ethanol are 28.052 and 46.068 respectively, so we get 46.068 g of ethanol when we start
with 28.052 g of ethene. Using this relationship as a conversion factor gives you the final
answer.
9.61 a) The molecular formula of 2-butanol is C4H10O, and the formula weight is 74.12, so one
mole of 2-butanol weighs 74.12 g. Therefore, two moles weighs 148.24 g.
b) The volume is 185 mL (the calculator answer is 184.837905 mL). From part a, you
know that the mass is 148.24 g. Use the density as a conversion factor to convert grams into
milliliters (you need to turn the density upside down).
9.63
a) You need 4 g of isopropyl alcohol.
b) You need 1.8 g of isopropyl alcohol (the calculator answer is 1.80282 g). The
molecular formula of isopropyl alcohol (2-propanol) is C3H8O, so the formula weight is 60.094.
Therefore, one mole of isopropyl alcohol weighs 60.094 g. Use this, the volume of the solution,
and the molarity to calculate the mass of isopropyl alcohol. (Review molarity in Chapter 5 if you
have forgotten how to do this.)
9.65
C2H6O + 3 O2 → 2 CO2 + 3 H2O
9.67 Sodium ion is positively charged, so it will be attracted to the negatively-charged oxygen
atoms in the surrounding methanol molecules.
H
H
O
CH3
CH3
O
+
Na
H
O
H
CH3
O
CH3