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Answers
Chapter 1 Matter, Energy, and Measurement
1.1 multiplication (a) 4.69 3 105 (b) 2.8 3 10215;
division (a) 2.00 3 1018 (b) 1.37 3 105
1.2 (a) 147°F (b) 8.3°C
1.3 109 kg
1.4 13.8 km
1.5 743 mph
1.6 78.5 g
1.7 2.43 g/mL
1.8 1.016 g/mL
1.9 4.8 3 103 cal 5 48 kcal
1.10 46°C
1.11 0.0430 cal/g # deg
1.13 (a) Matter is anything that has mass and takes up
space. (b) Chemistry is the science that studies matter.
1.15 Dr. X’s claim that the extract cured diabetes would be
classified as (c) a hypothesis. No evidence had been provided
to prove or disprove the claim.
1.17 (a) 3.51 3 1021 (b) 6.021 3 102 (c) 1.28 3 1024
(d) 6.28122 3 105
1.19 (a) 6.65 3 1017 (b) 1.2 3 101 (c) 3.9 3 10216
(d) 3.5 3 10223
1.21 (a) 1.3 3 105 (b) 9.40 3 104 (c) 5.139 3 1023
1.23 4.45 3 106
1.25 (a) 2 (b) 5 (c) 5 (d) 5 (e) ambiguous, better to
write as 3.21 3 104 (three significant figures) or 32100. (five
significant figures) (f) 3 (g) 2
1.27 (a) 92 (b) 7.3 (c) 0.68 (d) 0.0032 (e) 5.9
1.29 (a) 1.53 (b) 2.2 (c) 0.00048
1.31 330 min 5 5.6 h
1.33 (a) 20 mm (b) 1 inch (c) 1 mile
1.35 Weight would change slightly. Mass is independent of
location, but weight is a force exerted by a body influenced by
gravity. The influence of the Earth’s gravity decreases with
increasing distance from sea level.
1.37 (a) 77°F, 298 K (b) 104°F, 313 K (c) 482°F, 523 K,
(d) 2459°F, 0 K
1.39 (a) 0.0964 L (b) 27.5 cm (c) 4.57 3 104 g (d) 4.75 m
(e) 21.64 mL (f) 3.29 3 103 cc (g) 44 mL (h) 0.711 kg
(i) 63.7 cc (j) 7.3 3 104 mg (k) 8.34 3 104 mm (l) 0.361 g
1.41 50 mi/h
1.43 solids and liquids
1.45 No, melting is a physical change.
1.47 bottom: manganese; top: sodium acetate; middle:
calcium chloride
1.49 1.023 g/mL
1.51 water
1.53 One should raise the temperature of water to 4°C.
During this temperature change, the density of the crystals
decreases, while the density of water increases. This brings
the less dense crystals to the surface of the more dense
water.
1.55 The motion of the wheels of the car generates kinetic
energy, which is stored in your battery as potential energy.
1.57 0.34 cal/g # C°
1.59 334 mg
1.61 The body shivers. Further temperature lowering results
in unconsciousness and then death.
1.63 Methanol, because its higher specific heat allows it to
retain the heat longer.
1.65 0.732
1.67 kinetic: (b), (d), (e); potential: (a), (c)
1.69 the European car
1.71 kinetic energy
1.73 The largest is 41 g. The smallest is 4.1310 3 1028 kg.
1.75 10.9 h
1.77 The heavy water. When converting the specific heat
given in J/g # °C to cal/g # °C, one finds that the specific heat
of heavy water is 1.008 cal/g # °C, which is somewhat greater
than that of ordinary water.
1.79 (a) 1.57 g/mL (b) 1.25 g/mL
1.81 two
1.83 60 J would raise the temperature by 4.5°C; thus the final
temperature will be 24.5°C.
1.85 Number (b), 4.38, has three significant figures. Number (a), 0.00000001, has only one significant figure. The zeros
merely indicate the location of the decimal point.
1.87 To do this calculation you need a conversion factor from
kilometers to miles. Table 1.3 gives 1 mile 5 1.609 km
95 km 3
1 mile
~ 59 km
1.609 km
If you use the other possible conversion factor
95 km 3
1.609 km 153 km2
~
mi
mi
Both the numbers and the units are incorrect.
1.89 In photosynthesis, the radiant energy of sunlight is converted to chemical energy in the sugars produced.
1.91 Converting 30°C from the Celsius to Fahrenheit temperature scales gives 86°F. You are most likely to be wearing
a T-shirt and shorts.
1.93 Cells that have been exposed to several cycles of freezing and thawing will have expanded quite a bit. The expansion
process tends to break open the cells to make their contents
available for fractionation and further study.
1.95 We use the specific heat of water and the information
that a liter of water weighs 1000. grams.
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Answers
Amount of heat 5 SH 3 m 3 (T2 2 T1)
Amount of heat 5
1.00 cal
1000. grams
3 2.000 L 3
3 4.85C
gC
L
Amount of heat 5 9.70 3 103 calories
1.97 Determining the amount of substance and its effectiveness can be done together. You separate the components of the
original material and, in the process, determine its amount.
One possible way is to weigh amounts of recovered material.
You would then test the substance to see whether the individual compound produces the predicted results.
1.99
4.85 3 103 calories
Chapter 2 Atoms
2.1 (a) NaClO3 (b) AlF3
2.2 (a) The mass number is 31. (b) The mass number is 222.
2.3 (a) The element is phosphorus (P); its symbol is 31
15 P.
Rn.
(b) The element is radon (Rn); its symbol is 222
86
2.4 (a) The atomic number of mercury (Hg) is 80; that of lead
(Pb) is 82.
(b) An atom of Hg has 80 protons; an atom of Pb has
82 protons.
(c) The mass number of this isotope of Hg is 200; the mass
number of this isotope of Pb is 202.
202
(d) The symbols of these isotopes are 200
80 Hg and 82 Pb.
2.5 The atomic number of iodine (I) is 53. The number
of neutrons in each isotope is 72 for iodine-125 and 78 for
iodine-131. The symbols for these two isotopes are 125
53 I and
131
53 I, respectively.
2.6 Lithium-7 is the more abundant isotope.
2.7 The element is aluminum (Al). Its Lewis dot structure is
?
AlC
2.9 (a) F (b) T (c) T (d) T (e) F (f ) T
(g) T (h) T (i) F (j) F (k) T (l) F
2.11 (a) Oxygen (b) Lead (c) Calcium
(d) Sodium (e) Carbon (f) Titanium (g) Sulfur
(h) Iron (i) Hydrogen (j) Potassium (k) Silver (l) Gold
2.13 (a) Americium (b) Berkelium (c) Californium
(d) Dubnium (e) Europium (f) Francium
(g) Gallium (h) Germanium (i) Hafnium
(j) Hassium (k) Holmium (l) Lutetium
(m) Magnesium (n) Polonium o) Rhenium
(p) Ruthenium (q) Scandium (r) Strontium
(s) Ytterbium, Yttrium, Erbium (t) Thulium
2.15 (a) K2O (b) Na3PO4 (c) Li NO3
2.17 (a) The law of conservation of mass states that matter
can neither be created nor destroyed. According to Dalton’s
theory, matter is made up of indestructible atoms and a chemical reaction just changes the attachments among atoms and
does not destroy the atoms themselves.
(b) The law of constant composition states that any compound
is always made up of elements in the same proportion by mass.
Dalton’s theory explains that this is because molecules consist
of tightly bound groups of atoms, each of which has a particular mass. Therefore, each element in a compound always constitutes a fixed proportion of the total mass.
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2.19 No. CO and CO2 are different compounds, and each
obeys the law of constant composition for that particular compound.
2.21 (a) F (b) T (c) T (d) F (e) T (f) T (g) T (h) T
(i) F (j) F (k) T (l) F (m) T (n) F (o) T (p) T (q) T
(r) F (s) T (t) F
2.23 The statement is true in the sense that the number of
protons (the atomic number) determines the identity of the
element.
2.25 (a) The element with 22 protons is titanium (Ti)
(b) The element with 76 protons is osmium (Os)
(c) The element with 34 protons is selenium (Se)
(d) The element with 94 protons is plutonium (Pu).
2.27 Each would still be the same element, because the
number of protons has not changed.
(a) The element has 21 protons and is scandium (Sc).
(b) The element has 22 protons and is titanium (Ti).
(c) The element has 47 protons and is silver (Ag).
(d) The element has 18 protons and is argon (Ar).
2.29 Radon (Rn) has an atomic number of 86, so each isotope has
86 protons. The number of neutrons is mass number 2 atomic
number.
(a) Radon-210 has 210 2 86 5 124 neutrons
(b) Radon-218 has 218 2 86 5 132 neutrons
(c) Radon-222 has 222 2 86 5 136 neutrons
2.31 Tin-120, Tin-121, and Tin-124
2.33 (a) An ion is an atom that has either gained or lost one
or more electrons.
(b) Isotopes are atoms with the same number of protons in their
nuclei but which differ with respect to the number of neutrons.
2.35 Rounded to three significant figures, the calculated
value is 12.0 amu. The value given in the Periodic Table is
12.011 amu.
98.90
1.10
3 12.000 amu 1
3 13.000 amu 5 12.011 amu
100
100
2.37 Carbon-11 has 6 protons, 6 electrons, and 5 neutrons.
2.39 Americium-241 (Am) has atomic number 95. This isotope
has 95 protons, 95 electrons, and 241 2 95 5 146 neutrons.
2.41 (a) T (b) F (c) F (d) F (e) F (f) T (g) T
(h) T (i) T
2.43 (a) Groups 2A, 3B, 4B, 5B, 6B, 7B, 8B, 1B, and 2B contain only metals. Note that Group 1A contains one nonmetal,
hydrogen.
(b) No groups contain only metalloids
(c) Only Groups 7A and 8A contain only nonmetals.
2.45 Elements in the same Group of the Periodic Table should
have similar properties.
As, N, and P I and F Ne and He Mg, Ca, and Ba K and Li
2.47 (a) Aluminum . silicon (b) Arsenic . phosphorus
(c) Gallium . germanium (d) Gallium . aluminum
2.49 (a) T (b) T (c) T (d) F (e) T (f) F (g) T
(h) T (i) F (j) T (k) T (l) T (m) T (n) T (o) F
(p) F (q) T (r) T (s) T (t) F
2.51 The group number tells the number of electrons in the
valence shell of an element in the group.
2.53 (a) Li(3): 1s22s1 (b) Ne(10): 1s22s22p6 (c) Be(4): 1s22s2
(d) C(6): 1s22s22p2 (e) Mg(12): 1s22s22p63s2
2.55 (a) He(2): 1s2 (b) Na(11): 1s22s22p63s1
(c) Cl(17): 1s22s22p63s23p5 (d) P(15): 1s22s22p63s23p3
(e) H(1): 1s1
11/22/08 1:17:29 AM
Answers
2.57 In (a), (b), and (c), the outer-shell electron configurations
are the same. The only difference is the number of the valence
shell being filled.
2.59 The element might be any Group 2A element, all of
which have two valence electrons. It might also be helium
(in Group 8A).
2.61 (a) T (b) T (c) T (d) F (e) F (f) T (g) F (h) T
2.63 (a) T (b) T (c) T (d) T (e) F (f) T
2.65 (a) Fact: The atomic radius of an anion is always larger
than that of the atom from which it is derived. For anions, the
nuclear charge is unchanged, but an added electron introduces
new repulsions and the electron cloud swells because of the
increased electron-to-electron repulsions.
(b) Fact: The atomic radius of a cation is always smaller than
that of the atom from which it is derived. When an electron is
removed from an atom, the nuclear charge remains the same
but fewer electrons are repelling each other. Consequently,
the positive nucleus attracts the remaining elections more
strongly causing the electrons to contract more toward the
nucleus.
2.67 Here are ground-state electron configurations for each
O, O1, and N, N1.
One of these
electrons is lost
O 1s2 2s2 2px2 2py1 2pz1
O+ 1s2 2s2 2px1 2py1 2pz1 + e-
This electron is lost
N 1s2 2s2 2px1 2py1 2pz1
O+ 1s2 2s2 2px1 2py1
+ e-
The electron removed from O is one of the paired electrons in
the doubly occupied 2px orbital, whereas the electron removed
from N is an electron from the singly occupied 2pz orbital.
There is some repulsion between the two paired electrons in
the case of oxygen, which means that it is easier to remove an
electron from O than it is an electron from the singly occupied
2pz orbital for nitrogen.
2.69 Sulfur and iron are essential components of proteins,
and calcium is a major component of bones and teeth.
2.71 Because the 2H/1H ratio on Mars is five times larger
than that on Earth, the atomic weight of hydrogen on Mars
would be greater than that on Earth.
2.73 Bronze is an alloy of copper and tin.
2.75 (a) 1s (b) 2s, 2p (c) 3s, 3p, 3d (d) 4s, 4p, 4d, 4f
2.77 (a) Fact: atomic radius decreases in going across a
period in the Periodic Table. Although the principal quantum
number of the outermost orbital remains the same, as each
successive electron is added, the nuclear charge also increases
by the addition of one proton. The resulting increased attraction between the nucleus and electrons is somewhat stronger
than the increasing repulsion between electrons, which causes
the atomic radius to decrease.
(b) Fact: Energy is required to remove an electron from an
atom. Energy is required to overcome the force of attraction
of the positively charged nucleus for the negatively charged
electron.
2.79 (a) Group 3A elements have three electrons in their
valence shell. If we let n indicate the principal energy level,
then Group 3A elements have the outer-shell electron
configuration ns2, np1.
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(b) Group 7A, the halogen, have an outer-shell electron
configuration ns2, np5.
(c) Group 5A elements have the outer-shell electron configuration ns2, np3.
2.81 (a) Carbon-12 has 6 protons and 6 neutrons. Neutrons
contribute 50% of its mass.
(b) Calcium-40 has 20 protons and 20 neutrons. Neutrons
contribute 50% of its mass.
(c) Iron-55 has 26 protons and 29 neutrons. Neutrons contribute 53% of its mass.
(d) Bromine-79 has 35 protons and 44 neutrons. Neutrons
contribute 56% of its mass.
(e) Platinum-195 has 78 protons and 117 neutrons. Neutrons
contribute 60% of its mass.
(f) Uranium-238 has 92 protons and 146 neutrons. Neutrons
contribute 61% of its mass.
2.83 (a) P (b) K (c) Na (d) N (e) Br (f) Ag (g) Ca
(h) C (i) Sn (j) Zn
2.85 (a) Silicon is in Group 4A. It has four outer-shell electrons.
(b) Bromine is in Group 7A. It has seven outer-shell electrons.
(c) Phosphorus is in Group 5A. It has five outer-shell electrons.
(d) Potassium is in Group 1A. It has one outer-shell electron.
(e) Helium is in Group 8A. It has two outer-shell electrons.
(f) Calcium is in Group 2A. It has two outer-shell electrons.
(g) Krypton is in Group 8A. It has eight outer-shell electrons.
(h) Lead is in Group 4A. It has four outer-shell electrons.
(i) Selenium is in Group 6A. It has six outer-shell electrons.
(j) Oxygen is in Group 6A. It has six outer-shell electrons.
2.87 (a) An electron has a charge of 21, a proton has a charge
of 11, and a neutron has no charge.
(b) An electron has a mass of 0.0005 amu; a proton and a neutron each has mass of 1 amu.
2.89 The atomic number of this element is 54, which means
that it is xenon (Xe). This isotope has 54 protons, 54 electrons,
and 131 2 54 5 77 neutrons.
2.91 (a) Predict that its ionization energy will be smaller
than that of asatine (At) but greater than that of radium (Ra).
2.93 As illustrated by the answer to Problem 2.81, the neutron to proton ratio of an element generally increases as atomic
number increases. We can make the following generalizations.
As illustrated by the answer to Problem 2.81, the neutron to
proton ratio of the elements generally increases as atomic
number increases.
For light elements (H through Ca), the stable isotopes usually
have equal numbers of protons and neutrons.
Beyond calcium (Ca), the neutron/proton ratio becomes
increasingly greater than 1.
CHAPTER 3 Chemical Bonds
3.1 By losing two electrons, Mg acquires a complete octet. By
gaining two electrons, sulfur acquires a complete octet.
(a) Mg (12 electrons): 1s22s22p63s2 h Mg21 (10 electrons):
1s22s22p6
(b) S (16 electrons): 1s22s22p63s23p4 h S22 (18 electrons):
1s22s22p63s23p6
3.2 Each pair of elements is in the same column (group) of
the Periodic Table and electronegativity increases from bottom
to top within a column. Therefore
(a) Li . K (b) N . P (c) C . Si
3.3 (a) KCl (b) CaF2 (c) Fe2O3
3.4 (a) magnesium oxide (b) barium iodide
(c) potassium chloride
11/22/08 1:17:30 AM
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Answers
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3.5 (a) MgCl2 (b) Al2O3 (c) LiI
3.6 (a) iron(II) oxide, ferrous oxide
(b) iron(III) oxide, ferric oxide
3.7 (a) potassium hydrogen phosphate
(b) aluminum sulfate
(c) iron(II) carbonate, ferrous carbonate
3.8 (a) SiH (2.5 2 2.1 5 0.4); nonpolar covalent
(b) PiH (2.1 2 2.1 5 0.0); nonpolar covalent
(c) CiF (4.0 2 2.5 5 1.5); polar covalent
(d) CiCl (3.0 2 2.5 5 0.5); polar covalent
d⫹
␦⫹
d⫺
3.9 (a) C 9 N
3.10
␦⫺
␦⫹
(b) N 9 O
H
H
H
(b)
H
H
109.5°
109.5° O
(c) H
␦⫺
Cl 109.5°
C
Cl
109.5°
H
C
O
120°
3.16 (a) H2S contains no polar bonds and is a nonpolar
molecule.
(b) H 9 C 9 Cl
H
H
H
O
H
(a) H 9 C 9 C 9 H
109.5°
C9O
(a)
(c) C 9 Cl
H
109.5°
H
(a)
H
S
H
H
Nonpolar
(c) H 9 C # N
3.11
(b) HCN contains a polar CiN bond and is a polar molecule.
H
H
H
(a) H 9 C 9 C 9 H
H
H
H
H
2 single bonds
and
1 double bond
C"C"C
H
(d)
H
2 double bonds
H!C#C!H
1 single bond
and
1 triple bond
(a)
H
C
(c)
H
H
H
C
H
H
Nonpolar
a ⫺
H ! C ! OC
Li: 1s22s1 h Li1: 1s2 1 e2
⫹
(b) An oxygen atom has the electron configuration 1s22s2p4.
When O gains two electrons, it forms O22, which has the
electron configuration 1s22s22p6. This configuration is the
same as that of neon, the noble gas nearest oxygen in atomic
number.
a
H!C"O
O: 1s22s22p4 1 2e2 h O22: 1s22s22p6 (complete ocet)
A ⫺
COC
⫹
(c)
(c) C2H6 contains no polar bonds and is not a polar molecule.
COC
A ⫺
A ⫺
COC
(b)
Nonpolar
3.17 (a) F (b) T (c) F (d) T (e) T (f) F (g) T
(h) F (i) F
3.19 (a) A lithium atom has the electron configuration 1s22s1.
When Li loses its single 2s electron, it forms Li1, which has the
electron configuration 1s2. This configuration is the same as
that of helium, the noble gas nearest Li in atomic number.
3.12 (a) Nitrogen dioxide (b) Phosphorus tribromide
(c) Sulfur dichloride (d) Boron trifluoride
3.13
(b) H9C#N
H
H
4 single bonds
(c)
H
C"C
(b)
E ! CH3
CH3 ! C " O
3.14 (a) A valid pair of contributing structures.
(b) Not a valid pair. The contributing structure on the right
has 10 electrons in the valence shell of carbon and thus
violates the octet rule. The valence shell of carbon consists of
one s orbital and three p orbitals, which can hold a maximum
of eight valence electrons, hence the octet rule.
3.15 Given are three-dimensional structures showing all
unshared electron pairs.
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3.21 (a) Mg21 (b) F2 (c) Al31 (d) S22 (e) K1 (f) Br2
3.23 The stable ions are: (a) I2, (c) Na1, and (d) S22
3.25 Being intermediate in electronegativity, carbon and
silicon are reluctant to accept electrons from a metal or lose
electrons to a halogen to form ionic bonds. Instead, carbon and
silicon share electrons in nonpolar covalent and polar covalent
bonds.
3.27 (a) T (b) T (c) F (d) T (e) F (f) T (g) F (h) T
(i) T (j) F (k) F (l) F (m) T (n) T
3.29 (a) T (b) T (c) T (d) T (e) F (g) T (h) T (i) F
(j) F (k) T (l) F (m) T (n) T (o) F
11/24/08 11:54:27 AM
Answers
3.31 Electronegativity generally increases in going from left
to right across a row of the Periodic Table because the number
of positive charges in the nucleus of each successive element
in the row increases going from left to right. The increasing
nuclear charge exerts a stronger and stronger pull on the
valence electrons.
3.33 Electrons are shifted toward the more electronegative
atom. (a) Cl (b) O (c) O (d) Cl (e) negligible
(f) negligible (g) O
3.35 (a) CiCl polar covalent (b) CiLi polar covalent
(c) CiN polar covalent
3.37 (a) T (b) F (c) T (d) T (e) T (f) F (g) F
(h) F (i) F
3.39 (a) NaBr (b) Na2O (c) AlCl3 (d) BaCl2 (e) MgO
3.41 Sodium chloride in the solid state forms a lattice in
which each Na1 ion is surrounded by six Cl2 ions and each
Cl2 ion is surrounded by six Na1 ions.
3.43 (a) Fe(OH)3 (b) BaCl2 (c) Ca3(PO4)2 (d) NaMnO4
3.45 (a) The formula (NH4)2PO4 is incorrect. The correct
formula is (NH4)3PO4
(b) The formula Ba2CO3 is incorrect. The correct formula is
BaCO3
(c) The formula Al2S3 is correct.
(d) The formula MgS is correct.
3.47 (a) T (b) F (c) T (d) F (e) F (f) F (g) T (h) F
(i) T (j) F (k) F
3.49 The formula of potassium nitrite is KNO2.
3.51 (a) Na1, Br2 (b) Fe21, SO322 (c) Mg21, PO432
(d) K1, H2PO42 (e) Na1, HCO32 (f) Ba21, NO32
3.53 (a) KBr (b) CaO (c) HgO (d) Cu3(PO4)2
(e) Li2SO4 (f) Fe2S3
3.55 (a) F (b) F (c) F (d) T (e) T (f) T (g) T (h) F
(i) T (j) T (k) F (l) T (m) F (n) T
3.57 (a) A single bond results when one electron pair is
shared between two atoms.
(b) A double bond results when two electron pairs are shared
between two atoms.
(c) A triple bond results when three electron pairs are shared
between two atoms.
H
3.59
(a)
H!C!H
(b)
H!C#C!H
H
H
C"C
H
E
H
(c)
CFC
a
B ! FC
(d)
H
E
CFC
A CClC
A
CClC
H
C"O
a
(e)
H
(f )
3.61 The total number of valence electrons for each
compound is:
(a) NH3 has 8 (b) C3H6 has 18 (c) C2H4O2 has 24
(d) C2H6O has 20 (e) CCl4 has 32 (f) HNO2 has 18
(g) CCl2F2 has 32 (h) O2 has 12
3.63 (a) A bromine atom has seven electrons in its valence shell.
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(b) A bromine molecule has two bromine atoms bonded by a
single covalent bond.
(c) A bromide ion is a bromine atom that has gained one electron
in its valence shell; it has a complete octet and a charge of 21.
(a)
a
CBrD
(b)
a ! BrC
a
CBr
(c)
a CBrC
3.65 Hydrogen has the electron configuration 1s1. Hydrogen’s
valence shell has only one s orbital, which can hold only two
electrons.
3.67 Nitrogen has five valence electrons. By sharing three
more electrons with other atoms or another atom, nitrogen can
achieve the outer-shell electron configuration of neon, the noble
gas nearest to it in atomic number. The three shared pairs of
electrons may be in the form of three single bonds, one double
bond and one single bond, or one triple bond. With these combinations, there is one unshared pair of electrons on nitrogen.
3.69 Oxygen has six valence electrons. By sharing two electrons with another atom or other atoms, oxygen can achieve
the outer-shell electron configuration of neon, the noble gas
nearest it in atomic number. The two shared pairs of electrons
may be in the form of a double bond or two single bonds. With
either of these configurations, there are two unshared pairs of
electrons on oxygen.
3.71 O61 has a charge too concentrated for a small ion.
Furthermore, it would take an excessive amount of energy to
remove all six electrons.
3.73 (a) BF3 does not obey the octet rule because in this compound, boron has only six electrons in its valence shell.
(b) CF2 does not obey the octet rule because in this compound,
carbon has only 6 electrons in its valence shell.
(c) BeF2 does not obey the octet rule because in this compound,
beryllium has only 4 electrons in its valence shell.
(d) C2H4, ethylene, obeys the octet rule. In this compound, each
carbon has a double bond to the other carbon and single bonds
to two hydrogen atoms, giving each carbon a complete octet.
(e) CH3 does not obey the octet rule. In this compound, carbon
has a single bond to three hydrogens, which gives carbon only
seven electrons in its valence shell.
(f) N2 obeys the octet rule. Each nitrogen has one triple bond
and one unshared pair of electrons and, therefore, eight electrons in its valence shell.
(g) NO does not obey the octet rule. This compound has 11
valence electrons, and any Lewis structure drawn for it will
show either oxygen or nitrogen with only 7 electrons in its
valence shell.
3.75 (a) Sulfur dioxide (b) Sulfur trioxide (c) Phosphorus
trichloride (d) Carbon disulfide
3.77 (a) A Lewis structure for ozone must show 18 valence
electrons.
(b, c) Note that each contributing structure has one positive
and one negative charge.
a ! C ! C ! ClC
a
CCl
CClC
E CClC
E
■
O
O
O
O
O
O
(d) The central oxygen atom is surrounded by three regions of
electron density. Therefore, predict an OiOiO bond angle of 120°.
(e) The given contributing structure is not acceptable because
it places 10 electrons in the valence shell of the central oxygen atom. Oxygen is a second-period element, and orbitals
available to it for covalent bonding are the single 2s orbital and
11/22/08 1:17:35 AM
A13
Answers
■
three 2p obitals. These orbitals can hold only eight electrons
(the octet rule).
3.79 (a) T (b) F (c) T (d) F (e) T (f) T (g) F (h) T
(i) T (j) T (k) F (l) T (m) T
3.81 (a) H2O has 8 valence electrons and H2O2 has 14 valence
electrons.
(b)
H9O9H
H9O9O9H
Water
Hydrogen peroxide
(c) Each oxygen in each molecule is surrounded by four regions
of electron density. Therefore, predict all bond angles to be 109.5°.
3.83 Shape of each molecule and approximate bond angles
about its central atom:
H
(a)
H
H!C!H
H!P!H
(b)
F!C!F
(c)
H
H
F
Tetrahedral
(109.5°)
Pyramidal
(109.5°)
Tetrahedral
(109.5°)
O
(d) O " S " O
Cl
(e) O " S " O
(f )
F
Pyramidal
(109.5°)
Pyramidal
(109.5°)
3.85
(a) T
3.87
(a)
(b) T
(c) F
(d) T
(e) T
(f) T
3.103 Sodium fluoride, NaF, and stannous fluoride, SnF2, are
used as fluoride sources in fluoridated toothpaste and dental gels.
3.105 Sodium iodide, NaI, is used as a source of iodide in
table salt.
3.107 Magnesium hydroxide, Mg(OH)2, and aluminum
hydroxide, Al(OH)3.
3.109 (a) calcium phosphate
(b) magnesium hydroxide
(c) potassium chloride, potassium
iodide
(d) Iron oxide
(e) calcium phosphate
(f) zinc sulfate
(g) manganese sulfate
(h) titanium dioxide
(i) silicon dioxide
(j) cupric sulfate
(k) calcium borate
(l) sodium molybdate
(m) chromium chloride
(n) potassium iodide
(o) sodium selenate
(p) vanadyl sulfate
(q) nickel sulfate
(r) stannic sulfate
Cl
H
C"C
H
(g) T
(h) T
a ! B ! FC
a
CF
E
CFC
(b) The FiBiF bond angles are 120°
(c) BF3 has three polar bonds, but because of its geometry it is
a nonpolar molecule.
3.89 No, molecular dipoles are the result of the sum of the
direction and magnitude of individual polar bonds.
3.91 The individual polar CiCl bonds in CCl4 act in equal
but opposite directions canceling each other’s effect on the
molecular dipole.
3.93 Sodium iodide, NaI, is used as an iodide source in table
salt.
3.95 Potassium permanganate, KMnO4, is used as an external antiseptic.
3.97 Nitric oxide, NO, is rapidly oxidized by oxygen in the air
to nitrogen dioxide, which then dissolves in rain water to form
nitric acid, HNO3.
3.99 The compounds are (a) silane, SiH4, (b) phosphine, PH3,
and (c) hydrogen sulfide, H2S.
3.101 Predict a shape like that created by putting together
the bases of two square-based pyramids. This shape is called
octahedral because it has eight faces.
91123_00_Ans_pA8-A63 pp2.indd 13
90°
F
F
F
(g) H 9 N 9 H (h) Cl ! P ! Cl
Cl
F
S
Tetrahedral
(109.5°)
H
90°
3.111 (a) Cd(II) (b) Cr(III) (c) Ti(IV) (d) Mn(II)
(e) Co(III) (f) Fe(III)
3.113 (a) Following is a Lewis structure for vinyl chloride.
F ! C ! Cl
Trigonal planar
(120°)
Bent
(120°)
F
F
␦⫺
␦⫹
H
(b) Predict that all bond angles are 120°.
(c) Vinyl chloride has a polar CiCl bond, is a polar molecule,
and has a dipole.
3.115 (a) Incorrect. The left carbon has five bonds.
(b) Incorrect. The carbon in the middle has only three bonds.
(c) Incorrect. The second carbon from the right has only three
bonds, and the oxygen on the right has only one bond.
(d) Incorrect. Fluorine has two bonds.
(e) Correct
(f) Incorrect. The second carbon from the left has five bonds.
3.117 LiAlH4 or Li1 AlH42.
Chapter 4 Chemical Reactions
4.1 (a) ibuprofen, C13H18O2 5 206.1 amu
(b) Ba3(PO4)2 5 601 amu
4.2 1500. g H2O is 83.3 mol H2O.
4.3 2.84 mol Na2S is 222 g Na2S.
4.4 In 2.5 mol of glucose, there are 15 mol of C atoms, 30 mol
of H atoms, and 15 mol of O atoms.
4.5 0.062 g CuNO3 contains 4.9 3 1024 mol Cu1.
4.6 235 g H2O contains 7.86 3 1024 molecules H2O.
4.7 The balanced equation is
6CO2(g) ⫹ 6H2O(ℓ)
photosynthesis
C6H12O6(aq) ⫹ 6O2(g)
4.8 The balanced equation is
2C 6H 14 1 g 2 1 19O 2 1 g 2 h 12CO 2 1 g 2 1 14H 2O 1 g 2
4.9 The balanced equation is
3K 2C 2O 4 1 aq 2 1 Ca 3 1 AsO 4 2 2 1 s 2 h
2K 3AsO 4 1 aq 2 1 3CaC 2O 4 1 s 2
11/24/08 2:15:10 PM
Answers
4.10
(a) The balanced equation is:
2Al2O3(s)
electrolysis
4Al(s) ⫹ 3O2(g)
(b) It requires 51 g of alumina to prepare 27 g of aluminum.
4.11 From the balanced equation, we see that the molar
ratio of CO required to produce CH3COOH is 1:1. Therefore, it
requires 16.6 moles of CO to produce 16.6 moles of CH3COOH.
4.12 From the balanced equation, we see that the molar ratio
of ethylene to ethanol is 1:1. Therefore, 7.24 mol of ethylene
gives 7.24 mole of ethanol, which is 334 g of ethanol.
4.13 (a) H2 (1.1 mole) is in excess and C (0.50 mole) is the
limiting reagent.
(b) 8.0 g CH4 is produced.
4.14 The percent yield is 80.87%.
4.15 The net ionic equation is:
Cu21(aq) 1 S22(aq) h CuS(s).
4.16 (a) Ni21 gained two electrons so is reduced. Cr lost two
electrons so is oxidized. Ni21 is the oxidizing agent and Cr is
the reducing agent.
(b) CH2O gained hydrogens so is reduced. H2 gains oxygens
in being converted to CH3OH and so is oxidized. CH2O is the
oxidizing agent and H2 is the reducing agent.
4.17 (a) F (b) F (c) T (d) T (e) T
4.19 (a) sucrose, C12H22O11 342.3 amu
(b) glycine, C2H5NO2 75.07 amu
(c) DDT, C14H9Cl5 354.5 amu
4.21 (a) 32 g CH4 5 2.0 mol CH4
(b) 345.6 g NO 5 11.52 mol NO
(c) 184.4 g ClO2 5 2.734 mol ClO2
(d) 720. g glycerine 5 7.82 mol glycerine
4.23 (a) 18.1 mol CH2O 5 18.1 mol O atoms
(b) 0.41 mol CHBr3 5 1.2 mol Br atoms
(c) 3.5 3 103 mol Al2(SO4)3 5 4.2 3 104 mol O atoms
(d) 87 g HgO 5 0.40 mol Hg atoms
4.25 (a) 25.0 g TNT (MW 5 227 g/mol) contains
1.99 3 1023 N atoms
(b) 40 g ethanol (MW 5 46 g/mol) 5 1.0 3 1024 mol C atoms
(c) 500. mg aspirin (MW 180.2 g/mol) 5 6.68 3 1021 O atoms
(d) 2.40g NaH2PO4 (MW 120 g/mol) 5 1.20 3 1022 Na atoms
4.27 (a) 100. molecules CH2O (MW 30 g/mol) 5 4.98 3 10221
gram CH2O.
(b) 3000. molecules CH2O (MW 30 g/mol) 5 1.495 3 10219 g CH2O.
(c) 5.0 3 106 molecules CH2O 5 2.5 3 1016 grams CH2O
molecules.
(d) 2.0 3 1024 molecules CH2O 5 100 g CH2O.
4.29 3.9 mg cholesterol (MW 386.7 g/mol) 5 6.1 3 1018 molecules
cholesterol.
4.31 10 g copper (63.6 g/mol) 5 0.157 mol Cu atoms.
10 g chromium (52.0 g/mol) 5 0.192 mol Cr atoms
A 10 g sample of Cu contains 0.192 2 0.157 5 0.035 more
moles of Cr.
0.035 mole Cr 5 2.11 3 1022 atoms.
4.33 Following are the balanced equations.
(a) Hl 1 NaOH h Nal 1 H2O
(b) Ba(NO3)2 1 H2S h BaS 1 2HNO3
(c) CH4 1 2O2 h CO2 1 2H2O
(d) 2C4H10 1 13O2 h 8CO2 1 10H2O
(e) 2Fe 1 3CO2 h Fe2O3 1 3CO
4.35 CO2(g) 1 Ca(OH)2(aq) h CaCO3(s) 1 H2O 1 , 2
91123_00_Ans_pA8-A63 pp2.indd 14
■
A14
4.37 2Mg(s) 1 O2(g) h 2MgO(s)
4.39 2C(s) 1 O2(g) h 2CO(g)
4.41
2AsH3(g)
heat
2As(s) ⫹ 3H2(g)
4.43
2NaCl(aq) ⫹ 2H2O(ℓ)
electrolysis
Cl2(g) ⫹ 2NaOH(aq) ⫹ H2(g)
4.45 (a) 1 mol O2 requires 0.67 mol of N2.
(b) 0.67 mol of N2O3 are produced from 1 mol of O2.
(c) To produce 8 mol N2O3 requires 12 mol O2.
4.47 1.50 mol CHCl3 requires 319 g of Cl2.
4.49 (a) 2NaClO 2 1 aq 2 1 Cl2 1 g 2 h 2ClO 2 1 g 2 1 2NaCl 1 aq 2
(b) 5.5 kg of NaClO2 will yield 4.10 kg of ClO2.
4.51 To produce 5.1 g of glucose requires 7.5 g of CO2.
4.53 To completely react with 0.58 g of Fe2O3, we need 0.13 g C.
4.55 51.1 g of salicyclic acid.
4.57 The theoretical yield from 5.6 g of ethane is 12 g of chloroethane. The percentage yield is 68%.
4.59 (a) T (b) F (c) T (d) T (e) T (f) T (g) F (h) F
(i) T (j) F (k) T (l) F
4.61 The following chemical reactions are balanced net ionic
equations.
(a) Ag1(aq) 1 Br2(aq) h AgBr(s)
(b) Cd21(aq) 1 S22(aq) h CdS(s)
(c) 2Sc31(aq) 1 3SO422(aq) h Sc2(SO4)3(s)
(d) Sn21(aq) 1 2Fe21(aq) h Sn(s) 1 2Fe31(aq)
(e) 2K(s) 1 2H2O(ℓ) h 2K1(aq) 1 2OH2(aq) 1 H2(g)
4.63 (a) Ca3(PO4)2 will precipitate.
3Ca21(aq) 1 2PO432(aq) h Ca3(PO4)2(s)
(b) No precipitate will form (Group1 chlorides and sulfates are
soluble).
(c) BaCO3 will precipitate.
Ba21(aq) 1CO322(aq) h BaCO3(s)
(d) Fe(OH)2 will precipitate.
Fe21(aq) 1 2OH2(aq) h Fe(OH)2(s)
(e Ba(OH)2 will precipitate.
Ba21(aq) 1 2OH2(aq) h Ba(OH)2(s)
(f) Sb2S3 will precipitate.
2Sb21(aq) 1 3S22(aq) h Sb2S3(s)
(g) PbSO4 will precipitate.
Pb21(aq) 1 SO422(aq) h PbSO4(s)
4.65 The net ionic equation is
SO322(aq) 1 2H1(aq) h SO2(g) 1 H2O 1 , 2
4.67 (a) KCl (soluble: all Group 1 chlorides are soluble).
(b) NaOH (soluble: all sodium salts are soluble).
(c) BaSO4 (insoluble: most sulfates are insoluble).
(d) Na2SO4 (soluble: all sodium salts are soluble).
(e) Na2 CO3 (soluble: all sodium salts are soluble).
(f) Fe(OH)2 (insoluble: most hydroxides are insoluble).
11/24/08 12:11:17 PM
A15
■
Answers
4.69 (a) T (b) T (c) T (d) T (e) T (f) F (g) F (h) T
(i) T (j) T (k) T (l) T (m) T (n) T
4.71 (a) No, one species gains electrons and another must
lose electrons. Electrons are not destroyed but transferred from
one chemical species to another.
4.73 (a) C7H12 is oxidized (the carbons gain oxygens in going
to CO2) and O2 is reduced
(b) O2 is the oxidizing agent and C7H12 is the reducing agent.
4.75 (a) T (b) F (c) T (d) T (e) T (f) T
4.77 (a) endothermic (22.0 kcal appears as a reactant).
(b) exothermic (124 kcal appears as a product).
(c) exothermic (94.0 kcal appears as a product).
(d) endothermic (9.80 kcal appears as a reactant).
(e) exothermic (531 kcal appears as a product).
4.79 1.6 3 102 kcal of heat is evolved in burning 0.37 mol of
acetone.
4.81 Ethanol has a greater heat of combustion per gram
(7.09 kcal/g) than glucose 3.72 kcal/g).
4.83 156.0 kcal will produce 88.68 g of Fe metal.
4.85 Hydroxyapatite is composed of calcium ions, phosphate
ions, and hydroxide ions.
4.87 C2H4O is oxidized and H2O2 is reduced. H2O2 is the oxidizing agent and C2H4O is the reducing agent.
4.89 Cu1 is oxidized. The species that is oxidized during the
course of the reaction gives up an electron and is the reducing
agent. Therefore Cu1 is the reducing agent.
4.91 More than 90% of the energy needed to heat, cool, and
light our buildings, operate our cars, trucks, airplanes, shops,
and farm and factory machinery comes from the combustion of
coal, oil, and natural gas.
4.93 488 mg of aspirin (MW 180.2 g/mol) is equal to
2.71 3 1023 mol aspirin.
4.95 N2 is the limiting reagent and H2 is in excess.
4.97 4 3 1010 molecules of hemoglobin are present in a red
blood cell.
4.99 29.7 kg N2 5 1061 mol N2 and 3.31 kg H2 5
1655 mol H2
(a) From the balanced chemical equation, we see that the
two gases react in the ratio 3H2/N2. Complete reaction of
1061 mol N2 requires 3183 mol H2 but less than this number of moles of H2 is present. Therefore, H2 is the limiting
reagent.
(b) Under the balanced equation are moles of each present before reaction, moles reacting, and moles present after
complete reaction.
N2 1 3H2 h
Before rexn 1061
1655
Reacting 551
1655
After rexn 510
0
2NH3
0
0
1102
551 mol N2 5 14.3 kg of N2 remains after the reaction.
(c) 1102 moles of NH3 5 18.7 kg of NH3 is formed.
4.101 (a) Following are balanced equations for each oxidation.
C16H32O2(s) 1 23O2(g) h 16CO2 1 16H2O 1 , 2 1 238.5 kcal/mol
C6H12O6(s) 1 6O2(g) h 6CO2 1 6H2O 1 , 2 1 670 kcal/mol
(b) The heat of combustion of palmitic acid is 9.302
kcal/gram.
The heat of combustion of glucose is 3.72 kcal/gram.
(c) Palmitic acid has the greater heat of combustion per mole.
(d) Palmitic acid also has the greater heat of combustion
per gram.
91123_00_Ans_pA8-A63 pp2.indd 15
Chapter 5 Gases, Liquids, and Solids
5.1 0.41 atm
5.2 16.4 atm
5.3 0.053 atm
5.4 4.84 atm
5.5 0.422 mol Ne
5.6 9.91 g He
5.7 0.107 atm of H2O vapor
5.8 (a) Yes, there can be hydrogen bonding between water and
methanol, because a hydrogen atom on each molecule is bonded
to an electronegative oxygen atom. The OiH hydrogen can form
a hydrogen bond to an oxygen lone pair on another molecule.
(b) No, there is no polarity to a CiH bond and, therefore, it
cannot participate in a hydrogen bonding.
5.9 The heat of vaporization of water is 540 cal/g. 45.0 kcal is
sufficient to vaporize 83.3g H2O
5.10 The heat required to heat 1.0 g of iron to
melting 5 2.3 3 102 cal.
Heat (up to melting) 5 166 cal
Heat to melt 5 63.7 cal
5.11 According to the phase diagram of water (Figure 5.20), the
vapor will first condense to liquid water and then freeze to give ice.
5.13 As the volume of a gas decreases, the concentration of
gas molecules per unit of volume increases and the number
of gas molecules colliding with the walls of the container
increases. Because gas pressure results from the collisions
of gas molecules with the walls of the container, as volume
decreases, pressure increases.
5.15 The volume of a gas can be decreased by (1) increasing
the pressure on the gas or (2) lowering the temperature (cooling) of the gas. (3) The volume of the gas can be decreased by
removing some of the gas,
5.17 7.37 L
5.19 2.0 atm of CO2 gas
5.21 615 K
5.23 6.2 L of SO2 gas upon heating
5.25 The pressure read by the manometer is the difference
between the gas in the bulb and the atmospheric pressure: 833
mm Hg 2 760 mm Hg 5 73 mm Hg
5.27 2.6 atm of halothane
5.29
V1
T1
P1
V2
T2
P2
6.35 L
10°C
0.75 atm
4.6 L
0°C
1.0 atm
75.6 L
0°C
1.0 atm
88 L
35°C
735 torr
1.06 L
75°C
0.55 atm
3.2 L
0°C
0.14 atm
5.31 The volume of the balloon will be 3 3 106 L
5.33 The new temperature is 300 K.
5.35 1.87 atm
5.37 (a) 2.33 mol of gas are present
(b) No. The only information you need to know about the gas is
that it is an ideal gas.
5.39 Using the ideal gas law PV 5 nRT and n(moles) 5
mass/MW, the following equation can be derived and solved for
the molecular weight of the gas.
MW 5
1 mass 2 RT
PV
5
1 8.00g 2 1 0.0821 L ? atm ? mol21 ? K 21 2 1 273K 2
1 2.00 atm 2 1 22.4L 2
5 4.00 g/mol
11/22/08 1:17:43 AM
Answers
5.41 At constant temperature, gas density increases as pressure increases. At constant pressure, gas density decreases as
temperature increases.
5.43 (a) 24.7 mol O2 are needed to fill the chamber.
(b) 790 g of O2 are needed to fill the chamber.
5.45 5.5 L of air contains 1.16 L of O2, which, under these conditions, is 0.050 mol of O2.
0.050 mol of O2 contains 13.0 3 1022 molecules of O2.
5.47 (a) The mass of one mol of air is 28.95 grams.
(b) The density of air is 1.29 g/L.
5.49 The density of each gas is
(a) SO2 5 2.86 g/L (b) CH4 5 0.714 g/L (c) H2 5 0.0892 g/L
(d) He 5 0.179 g/L (e) CO2 5 1.96 g/L
Gas comparison: SO2 and CO2 are more dense than air; He, H2,
and CH4 are less dense than air.
5.51 The density of octane is 0.7025 g/mL
The mass of 1.00 mL of octane is 0.07025g
Using the ideal gas equation, calculate that this mass of octane
occupies 0.197 L
5.53 The density would be the same. Density of a substance
does not depend on its quantity.
5.55 (a) T (b) F (c) T (d) F
PT 5 PN 1 PO2 1 PCO 1 PH O
2
2
2
PN 5 (0.740)(1.0 atm) 5 0.740 atm (562.4 mm Hg)
2
PO 5 (0.194)(1.0 atm) 50.194 atm (147.5 mm Hg)
2
PH O 5 (0.062)(1.0 atm) 5 0.062 atm (47.1 mm Hg)
2
PCO 5 (0.004)(1.0 atm) 5 0.004 atm (3.0 mm Hg)
2
PT 5 1.00 atm (760.0 mm Hg)
5.59 (a) T (b) F (c) T (d) F (e) T (f) T (g) T (h) F
(i) T (j) T
5.61 (a) F (b) F (c) T (d) T (e) F (f) T (g) T (h) T
(i) F (j) F (k) F (l) F (m) T (n) T (o) F
5.63 Gases behave most ideally under low pressure and high
temperature to minimize non-ideal intermolecular interactions.
Therefore, choice (c) best suits these conditions.
5.65 (a) CCl4 is nonpolar; London dispersion forces
(b) CO is polar; dipole2dipole interactions.
The most polar molecule (CO) will have the largest surface
tension.
5.67 Yes. London dispersion forces range from 0.001 to 0.2
kcal/mol, whereas the lower end of dipole2dipole attractive
forces can be as low as 0.1 kcal/mol.
5.69 (a) T (b) F (c) F (d) T (e) F (f) T (g) T (h) T
(i) T (j) T (k) F (l) F (m) T (n) F (o) T (p) F
5.71 (a) T (b) T (c) T (d) F (e) T (f) F (g) T (h) F
(i) F (j) F (k) T
5.73 (a) T (b) T (c) F (d) T (e) T (f) F (g) F (h) F
(i) F (j) T (k) T
5.75 1.53 kcal is required to vaporize one mol of CF2Cl2.
5.77 The vapor pressures are approximately:
(a) 90 mm Hg
(b) 120 mm Hg
(c) 490 mm Hg
5.79 (a) HI . HBr . HCl The increasing size in this series
increases London dispersion forces.
(b) H2O2 . HCl . O2 # O2 has only weak London dispersion
intermolecular forces to overcome for boiling to occur, whereas
HCl is a polar molecule with stronger dipole2dipole attractions to overcome for boiling to occur. H2O2 has the strongest
intermolecular forces (hydrogen bonding) to overcome for
boiling to occur.
91123_00_Ans_pA8-A63 pp2.indd 16
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A16
5.81 The difference between heating water from 0°C to 37°C
and heating ice from 0°C to 37°C is the heat of fusion.
The energy required to heat 100.g of ice from 0°C to 37°C is
11700. cal
The energy required to heat 100.g of water from 0°C to 37°C is
3700 cal.
5.83 The name of the phase change is sublimation, which is
the conversion of a solid to a gas, bypassing the liquid phase.
5.85 1.00 mL of Freon-11 is 1.08 3 1022 mol of Freon-11.
Vaporizing this volume of Freon-11 from the skin will remove
6.96 3 1022 kcal.
5.87 When the temperature of a substance increases, molecular motion and hence entropy increase. Therefore, a gas at
100°C has a lower entropy than a gas at 200°C.
5.89 When a person lowers the diaphragm, the volume of the
chest cavity increases, thus lowering the pressure in the lungs
relative to atmospheric pressure. Air at atmospheric pressure
then is drawn into the lungs, beginning the breathing process.
5.91 The first tapping sound one hears is the systolic pressure, which occurs when the sphygmomanometer pressure
matches the blood pressure and the ventricle contracts pushing
blood into the arm.
5.93 When water freezes, it expands (water is one of the few
substances that expands on freezing) and will break the bottle
when the expansion exceeds the volume of the bottle.
5.95 Compressing a liquid or a solid is difficult because their
molecules or atoms are already close together and there is very
little empty space between them.
5.97 34 psi 5 2.3 atm
5.99 Aerosol cans already contain gases under pressure.
Gay-Lussac’s law predicts that the pressure inside the can will
increase as it is heated, with the potential of explosive rupturing of the can causing injury.
5.101 112 mL
5.103 Water, which forms strong intermolecular hydrogen
bonds, has the highest boiling point. Boiling points of these
three compounds are:
(a) pentane, C5H12 (36°C)
(b) chloroform, CHCl3 (61°C)
(c) water, H2O (100°C)
5.105 (a) As a gas is compressed under pressure, the molecules are forced closer together and the intermolecular forces
pull molecules together, forming a liquid.
(b) 9.1 kg of propane
(c) 2.1 3 102 moles of propane
(d) 4.6 3 103 L of propane
5.107 The density of the gas is 3.00 g/L.
Using the ideal gas law, show that
MW 5
massRT
PV
and then calculate that the molecular weight of the gas is
91.9 g/mol.
5.109 313K (40°C)
5.111 The temperature of a liquid drops during evaporation because as the molecules with higher kinetic energy
leave the liquid and enter the gas phase, the average kinetic
energy of molecules remaining in the liquid decreases. The
temperature of the liquid is directly proportional to the
average kinetic energy of molecules in the liquid phase and
as the average kinetic energy decreases, the temperature
decreases.
11/22/08 1:17:44 AM
A17
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Answers
5.113 (a) The pressure on the body at 100 feet is 3.0 atm.
(b) At 1.00 atm, PN 5 593 mm Hg (0.780 atm) and thus
2
makes up 78.0% of the gas mixture, which does not change
at a depth of 100 feet. At this depth, the total pressure on
the lungs, which is equalized by pressure of air delivered by
the SCUBA tank, is 3.0 atm and the partial pressure of N2 is
2.34 atm.
(c) At 2 atm, PO 5 158 mm Hg (0.208 atm) and thus makes up
2
20.8% of the gas mixture at 2 atm, which does not change at a
depth of 100 feet. At this depth, the total pressure on the lungs,
which is equalized by pressure of air delivered by the scuba
tank is 3.0 atm. Thus, at 100 feet, the partial pressure of O2 5
0.63 atm.
(d) As a diver ascends from 100 ft, the external pressure on
the lungs decreases and, therefore, the volume of gases in the
lungs increases. If the diver does not exhale during a rapid
ascent, the diver’s lungs could overinflate due to the expansion
of gases in the lungs, causing injury.
Chapter 6 Solutions and Colloids
6.1 To 11 g of KBr, add a quantity of water sufficient to dissolve the KBr. Following dissolution of the KBr, add water to
the 250-mL mark, stopper, and mix.
6.2 1.7 %w/v
6.3 First calculate the number of moles and mass of KCl
needed, which is 2.12mol and 158 g of KCl. To prepare the solution, place 158 g of KCl in a 2-L volumetric flask, add some
water until the solid has dissolved, and then fill the flask with
water to the 2.0-L mark.
6.4 Because the units of molarity are moles of solute/L of
solution, grams of KSCN must be converted to moles of KSCN
and mL of solution must be converted to L of solution. When
these conversions are completed, find that the concentration of
the solution is 0.0133M.
6.5 First convert grams of glucose into moles of glucose, then
convert moles of glucose into mL of solution. 10.0 g of glucose is
0.0556 mol of glucose. This mass of glucose is contained in 185
mL of the given solution.
6.6 First convert 100 gallons to liters of solution. 3.9 3 102 g
NaHSO3 must be added to the 100-gallon barrel.
6.7 Place 15.0 mL of 12.0 M HCl solution into a 300-mL volumetric flask, add some water, swirl to mix completely, and then
fill the flask with water to the 300-mL mark.
6.8 Place 0.13 mL of the 15% KOH solution into a 20-mL volumetric flask, add some water, swirl until completely dissolved,
and then fill the flask with water to the 20 mL mark.
6.9 The Na1 concentration is 0.24 ppm Na1.
6.10 215 g of CH3OH (molecular weight 32.0 g/mol) is 6.72
mol of CH3OH.
DT 5 (1.86°C/mol)(6.72 mol) 5 12.5°C The freezing point it
lowered by 12.5°C. The new freezing point is 212.5°C.
6.11 Compare the number of moles of ions or molecules in
each solution. The solution with the most ions or molecules in
solution will have the lowest freezing point.
Solution
(a) 6.2 M NaCl
(b) 2.1 M Al(NO3)3
(c) 4.3 M K2SO3
Particles in solution
2 3 6.2 M 5 12.4 M ions
4 3 2.1 M 5 8.4 M ions
3 3 4.3 M 5 12.9 M ions
Solution (c) has the highest concentration of solute particles
(ions); therefore it will have the lowest freezing point.
91123_00_Ans_pA8-A63 pp2.indd 17
6.12 The boiling point is raised by 3.50°C. The new boiling
point is 103.5°C.
6.13 The molarity of the solution prepared by dissolving
3.3 g Na3PO4 in 100 mL of water is 0.20M Na3PO4. Each formula unit of Na3PO4 dissolved in water gives 3 Na1 ions and
1 PO432 ion, for a total of 4 particles. The osmolarity of the
solution is (0.20 M)(4 ions) 5 0.80 osmol.
6.14 The osmolarity of red blood cells is 0.30 osmol.
Solution
(a) 0.1 M Na2SO4
(b) 1.0 M Na2SO4
(c) 0.2 M Na2SO4
Mol particles/L
3 3 0.1 M 5 0.30 osmol
3 3 1.0 M5 3.0 osmol
3 3 0.2 M 5 0.6 osmol
Solution (a) has the same osmolarity as red blood cells and,
therefore, is isotonic with red blood cells.
6.15 (a) T (b) T (c) T (d) T
6.17 The solvent is water.
6.19 (a) both tin and copper are solids
(b) solid solute (caffeine, flavorings) and liquid solvent (water).
(c) both CO2 and H2O (steam) are gases.
(d) gas (CO2) and liquid (ethanol) solutes in a liquid solvent
(water).
6.21 Mixtures of gases are true solutions because they mix
in all proportions, molecules are distributed uniformly, and the
component gases do not separate upon standing.
6.23 The prepared aspartic acid solution was unsaturated.
Over the two days, some of the solvent (water) evaporated
and the solution has become saturated. When water continued evaporating, the remaining water could not hold all the
dissolved solute so the excess aspartic acid precipitated as a
white solid.
6.25 (a) NaCl is an ionic solid and will be dissolved in the
water layer.
(b) Camphor is a nonpolar molecular compound and will dissolve in the nonpolar diethyl ether layer.
(c) KOH is an ionic solid and will be dissolved in the water layer.
6.27 Isopropyl alcohol would be a good first choice. The oil
base in the paint is nonpolar. Both benzene and hexane are
nonpolar solvents and may dissolve the oil-based paint, thus
destroying the painting.
6.29 The solubility of aspartic acid in water at 25°C is 0.250
g in 50.0 mL of water. The cooled solution of 0.251 g of aspartic
acid in 50.0 mL water will be supersaturated by 0.001 g of
aspartic acid.
6.31 According to Henry’s law, the solubility of a gas in a
liquid is directly proportional to pressure. A closed bottle of
a carbonated beverage is under pressure. After the bottle
is opened, the pressure is released and the carbon dioxide
becomes less soluble and escapes, leaving the contents “flat.”
6.33
(a)
(b)
1 min
3 10 6 5 1 ppm
1.0 3 10 6 min
1p
3 10 6 5 1 ppm
1.05 3 10 6 p
1 min
3 10 9 5 1 ppb
1.05 3 10 9 min
1p
3 10 9 5 1 ppm
1.05 3 10 9 p
6.35 (a) Dissolve 76 mL of ethanol in 204 mL of water (to
give 280 mL of solution).
11/22/08 1:17:45 AM
Answers
(b) Dissolve 8.0 mL of ethyl acetate in 427 mL of water (to give
435 mL to solution).
(c) Dissolve 0.13 L of benzene in 1.52 L chloroform (to give
1.65 L of solution).
6.37 (a) 4.15% w/v casein.
(b) 0.30% w/v vitamin C
(c) 1.75% w/v sucrose
6.39 (a) Place 19.5 g NH4Br in a 175 mL volumetric flask,
add some water, swirl until completely dissolved, and then fill
the flask with water to the 175 mL mark.
(b) Place 167 g of NaI into a 1.35 L volumetric flask, add some
water and swirl until completely dissolved, and then fill the
flask with water to the 1.35 L mark.
(c) Place 2.4 g of ethanol into a 330 mL volumetric flask, add
some water, swirl until completely dissolved, and then fill the
flask with water to the 330 mL mark.
6.41 0.2 M NaCl
6.43 0.509 M glucose
0.0202 M K1
7.25 3 1024 M Na1
6.45 2.5 M sucrose
6.47 The total volume of the dilution is 30.0 mL. Starting
with 5.00 mL of the stock solution, add 25.0 mL of water to
reach a final volume of 30.0 mL. Note that this is a dilution by
a factor of 6.
6.49 Place 2.1mL of the 30.0% H2O2 into a 250 mL volumetric
flask, add some water and swirl until completely mixed, and
then fill the flask with water to the 250 mL mark.
6.51 (a) 3.85 3 104 ppm Captopril
(b) 6.8 3 104 ppm Mg21
(c) 8.3 3 102 ppm Ca21
6.53 Assume the density of the lake water to be 1.00 g/mL.
The dioxin concentration is 0.01 ppb dioxin.
No, the dioxin level in the lake did not reach a dangerous level.
6.55 (a) 10 ppm Fe or 1 3 101 ppm
(b) 3 3 103 ppm Ca
(c) 2 ppm vitamin A
6.57 (a) KCl An ionic compound, very soluble in water:
a strong electrolyte
(b) Ethanol A covalent compound: a nonelectrolyte
(c) NaOH An ionic compound, very soluble in water: a strong
electrolyte
(d) HF A weak acid only partially dissociated in water: a weak
electrolyte
(e) Glucose A covalent compound, very soluble in water: a
nonelectrolyte
6.59 Water dissolves ethanol by forming hydrogen bonds with
it. The O—H group of ethanol is both a hydrogen bond acceptor
and a hydrogen bond donor.
6.61 (a) T (b) T
6.63 (a) homogeneous (b) heterogeneous (c) colloidal
(d) heterogeneous (e) colloidal (f) colloidal
6.65 As the temperature of the solution decreased, the protein
molecules must have aggregated and formed a colloidal mixture. The turbid appearance is the result of the Tyndall effect.
6.67 (a) 1.0 mol NaCl, freezing point 23.72°C.
(b) 1.0 mol MgCl2 freezing point 25.58°C.
(c) 1 mol (NH4)2CO3 freezing point 25.58°C
(d) 1 mol Al (HCO3) 3 freezing point 27.44°C
91123_00_Ans_pA8-A63 pp2.indd 18
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A18
6.69 Methanol dissolves in water but does not dissociate; it is
a nonelectrolyte. It would require 344 g of CH3OH in 1000. g of
water to lower the freezing point to 220°C.
6.71 Acetic acid, a weak acid, is only weakly dissociated in
water. KF is a strong electrolyte, completely dissociating in
water and nearly doubling the effect on freezing-point depression compared with that of acetic acid.
6.73 In each case side with greater osmolarity rises.
(a) B (b) B (c) A (d) B (e) neither (f) neither
6.75 (a) 0.39 M Na2CO3 5 0.39 M 3 3 particles/formula
unit 5 1.2 osmol
(b) 0.62 M Al(NO3)3 5 0.62 3 4 particles/formula unit 5 2.5
osmol
(c) 4.2 M LiBr 5 4.2 3 2 particles/formula unit 5 8.4 osmol
(d) 0.009 M K3PO4 5 0.009M 3 4 particles/formula unit 5
0.04 osmol
6.77 Cells in hypertonic solutions undergo crenation (shrink).
(a) 0.3% NaCl 5 0.3 osmol NaCl
(b) 0.9 M glucose 5 0.9 osmol glucose
(c) 0.9% glucose 5 0.05 osmol glucose
Solution (b) has a concentration greater than the isotonic solution so it will crenate red blood cells.
6.79 Carbon dioxide (CO2) dissolves in rainwater to form a
dilute solution of carbonic acid (H2CO3), which is a weak acid.
6.81 Nitrogen narcosis is the intoxication caused by the
increased solubility of nitrogen in the blood as a result of high
pressures as divers descend.
6.83
CaCO3(s) ⫹ H2SO4(aq)
CaSO4 ⫹ 2H2O
CaSO4(s) ⫹ CO2(g) ⫹ H2O(ℓ)
CaSO4 • 2H2O
Gypsum dihydrate
6.85 The minimum pressure required for reverse osmosis in
the desalinization of seawater exceeds 100 atm (the osmotic
pressure of seawater).
6.87 Yes, the change made a change in the tonicity. A 0.2%
NaHCO3 solution is 0.05 osmol. A 0.2% solution of KHCO3
is 0.04 osmol. This difference arises because of the difference in formula weight of NaHCO3 (84 g/mol) compared with
that of KHCO3 (100.1 g/mol). The error in replacing NaHCO3
with KHCO3 results in a hypotonic solution and an electrolyte imbalance by reducing the number of ions (osmolarity) in
solution.
6.89 When a cucumber is placed in a saline solution, the
osmolarity of the saline is greater than the water in the
cucumber, so water moves from the cucumber to the saline
solution. When a prune (a partially dehydrated plum) is
placed in the same solution, it expands because the
osmolarity in the prune is greater than the saline solution,
so water moves from the saline solution to inside the
prune.
6.91 The solubility of a gas is directly proportional to the
pressure (Henry’s law) and inversely proportional to the
temperature. The dissolved carbon dioxide formed a
saturated solution in water when bottled under 2 atm pressure. When the bottles are opened at atmospheric pressure,
the gas becomes less soluble in water. The excess carbon
11/24/08 12:34:32 PM
Answers
■
dioxide escapes through bubbles and frothing. In the
other bottle, the solution of carbon dioxide in water is
unsaturated at lower temperature and does not lose
carbon dioxide.
6.93 Methanol is more efficient at lowering the freezing
point of water. A given mass of methanol (32 g/mol) contains a
greater number of moles than the same mass of ethylene glycol
(62 g/mol).
6.95 CO (g) ⫹ H O(ℓ)
H CO (aq)
2
2
2
3
Carbonic acid
SO2(g) ⫹ H2O(ℓ)
H2SO3(aq)
Sulfurous acid
6.97 Place 39 mL of 35% HNO3 into a 300-mL volumetric
flask, add some water, swirl until completely mixed, and then
fill the flask with water to the 300 mL mark.
6.99 6 3 1023 g of pollutant.
6.101 Assume that the density of the pool water is 1.00 g/mL.
The Cl2 concentration in the pool is 355 ppm.
7.09 kg of Cl2 must be added to reach this concentration.
Chapter 7 Reaction Rates
and Chemical Equilibrium
7.1 rate of O2 formation50.022 L O2/min
7.2 rate 5 4 3 1022 mol H2O2/L # min for disappearance
of H2O2
3 H 2SO 4 4
7.3 K 5
3 SO 3 4 3 H 2O 4
3 N2 4 3 H2 43
7.4
K5
7.5
K 5 0.602 M21
7.6
K5
3 NH 3 4 2
3 CH 3COOCH 2CH 3 4 3 H 2O 4
3 CH 3COOH 4 3 HOCH 2CH 3 4
7.7 Le Chatelier’s principle predicts that adding Br2 (a product) will shift the equilibrium to the left—that is, toward the
formation of more NOBr(g).
7.8 Because oxygen’s solubility in water is exceeded, oxygen
bubbles out of the solution, driving the equilibrium toward the
right.
7.9 If the equilibrium shifts to the right with the addition of
heat, heat must have been a reactant, and the reaction is endothermic.
7.10 The equilibrium in a reaction where there is an increase
in pressure favors the side with fewer moles of gas. Therefore,
this equilibrium shifts to the right.
7.11 rate of formation of CH3I 5 57.3 3 1023 M CH3I/min
7.13 Reactions involving ions in aqueous solution of ions are
faster because they do not require bond breaking and have low
activation energies. In addition, the attractive force between
positive and negative ions provides energy to drive the reaction. Reactions between covalent compounds require the breaking of covalent bonds and have higher activation energies and,
therefore, slower reaction rates.
91123_00_Ans_pA8-A63 pp2.indd 19
7.15
Transition state
Activation energy
⫽ 14 kcal/mol
Energy
A19
Energy of
reactants
Heat of reaction
⫽ 9 kcal/mol
Energy of products
Progress of reaction
7.17 A general rule for the effect of temperature on the rate of
reaction states that for every temperature increase of 10°C, the
reaction rate doubles. In this case, a reaction temperature of
50°C would predict completion of the reaction in 1 h.
7.19 You might (a) increase the temperature, (b) increase the
concentration of reactants, or (c) add a catalyst.
7.21 A catalyst increases the rate of a reaction by providing an alternative reaction pathway with lower activation
energy.
7.23 Other examples of irreversible reactions include digesting a piece of candy, rusting of iron, exploding TNT, and the
reaction of Na or K metal with water.
7.25 (a) K 5 3 H 2O 4 2 3 O 2 4 / 3 H 2O 2 4 2
(b) K 5 3 N 2O 4 4 2 3 O 2 4 / 3 N 2O 5 4 2
(c) K 5 3 C 6H 12O 6 4 3 O 2 4 6 / 3 H 2O 4 6 3 CO 2 4 6
7.27 K 5 0.667
7.29 K 5 0.099 M
7.31 Products are favored in (b) and (c). Reactants are favored
in (a), (d), and (e).
7.33 No. The rate of reaction is independent of the energy difference between products and reactants—that is, it is independent of the heat of reaction.
7.35 The reaction reaches equilibrium quickly, but the position of equilibrium favors the reactants. It would not be a very
good industrial process unless products are constantly drawn
off to shift the equilibrium to the right.
7.37 (a) right (b) right (c) left (d) left (e) no shift
7.39 (a) Adding Br2 (a reactant) will shift the equilibrium to
the right.
(b) The equilibrium constant will remain the same.
7.41 (a) no change (b) no change (c) smaller
7.43 As temperatures increase, the rates of most chemical
processes increase. A high body temperature is dangerous
because metabolic processes (including digestion, respiration,
and biosynthesis of essential compounds) take place at a faster
rate than is safe for the body. As temperatures decrease, so
do the rates of most chemical reactions. As body temperature
decreases below normal, the vital chemical reactions will slow
to rates slower than is safe for the body.
7.45 The capsule with the tiny beads will act faster than the
solid coated-pill form. The small bead size increases the drug’s
surface area, allowing the drug to react faster and deliver its
therapeutic effects more quickly.
7.47 Assuming that there is an excess of AgCl from the
previous recipe, the recipe does not need to be changed. The
desert conditions add nothing that would affect the coating
process.
11/24/08 12:04:29 PM
Answers
7.49 At 25°C the rate is 0.70 moles per liter per second.
At 45°C the rate is 22 moles per liter per second.
7.51
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A20
7.71 When you add sodium chloride, the presence of more
chloride ions increases the concentration of one of the products
of the reaction. The equilibrium shifts to the left, increasing
the amount of solid silver chloride.
1 - uncatalyzed
reacton
Chapter 8 Acids and Bases
8.1 acid reaction: HPO422 1 H2O m PO432 1 H3O1;
base reaction: HPO422 1 H2O m H2PO42 1 OH2
8.2 (a) toward the left;
H3O Energy
2 - catalyzed
reaction
Weaker
acid
I
Weaker
base
H2O Stronger
base
HI
Stronger
acid
(b) toward the right;
Reactants
CH3COO H2S
Weaker
base
Weaker
acid
CH3COOH HS
Stronger
acid
Stronger
base
Products
Progress of reaction
Profile 2 represents the addition of a catalyst.
7.53 0.14 M
7.55 The rate of a gaseous reaction could be increased by
decreasing the volume of the container. This would increase the
number of collisions between molecules.
7.57 4 NH3 1 7O2 h NO2 1 6 H2O
7.59 The reaction with spherical molecules will proceed
more rapidly since in the case of the rod-like molecules some
collisions will be ineffective because the molecules will not
interact with the proper orientations.
7.61 Initial rate 5 0.030 moles I2 per liter per second
7.63
(a)
O
CH3COH HOCH2CH3
Initial
At equilibrium
1.00 mol
0.33 mol
1.00 mol
0.33 mol
O
CH3COCH2CH3 H2O
0 mol
0.67 mol
0 mol
0.67 mol
(b) K 5 4.1
7.65 Monitoring the disappearance of a reactant is a possible
way to determine the rate of a reaction. It will do just as well
as monitoring the formation of product because the stoichiometry of the reaction relates the concentrations of products and
reactants to one another.
7.67 Some reactions are so fast that they are over before you
can turn on a stopwatch or timer. You need specialized instruments with sophisticated electronics to follow the rates of very
fast reactions.
7.69 The rate of conversion of diamond to graphite is so slow
that it does not take place in any measurable length of time.
91123_00_Ans_pA8-A63 pp2.indd 20
8.3 pKa is 9.31
8.4 (a) ascorbic acid (b) aspirin
8.5 1.0 3 1022
8.6 (a) 2.46 (b) 7.9 3 1025, acidic
8.7 pOH 5 4, pH 5 10
8.8 0.0960 M
8.9 (a) 9.25 (b) 4.74
8.10 9.44
8.11 7.7
8.13 (a) HNO3(aq)1H2O 1 , 2 h NO32(aq)1H3O1(aq)
(b) HBr(aq)1H2O 1 , 2 h Br2(aq)1H3O1(aq)
(c) H2SO3(aq)1H2O 1 , 2 h HSO32(aq)1H3O1(aq)
(d) H2SO4(aq)1H2O 1 , 2 h HSO42(aq)1H3O1(aq)
(e) HCO32(aq)1H2O 1 , 2 h CO322(aq)1H3O1(aq)
(f) NH41(aq) 1 H2O 1 , 2 h NH3(aq)1H3O1(aq)
8.15 (a) weak (b) strong (c) weak (d) strong
(e) weak (f) weak
8.17 (a) false (b) false (c) true (d) false
(f) true (g) false (h) false
8.19 (a) A Brønsted-Lowry acid is a proton donor.
(b) A Brønsted-Lowry base is a proton acceptor.
8.21 (a) HPO422 (b) HS2 (c) CO322
(d) CH3CH2O2 (e) OH2
8.23 (a) H3O1 (b) H2PO42 (c) CH3NH31
(d) HPO422 (e) NH41
8.25 The equilibrium favors the side with the weaker
acid2weaker base combination. Equilibria (b) and (c) lie
to the left; equilibriam (a) lies to the right.
(a)
C6H5OH C2H5O
Stronger
acid
(b)
Stronger
base
C6H5O C2H5OH
Weaker
base
HCO3 H2O
H2CO3 OH
Weaker
base
Stronger
acid
Weaker
acid
Weaker
acid
Stronger
base
11/22/08 1:17:48 AM
A21
■
Answers
(c)
CH3COOH H2PO4
Weaker
acid
Weaker
base
CH3COO H3PO4
Stronger
base
Stronger
acid
8.27 (a) the pKa of a weak acid (b) the Ka of a strong acid
8.29 (a) 0.10 M HCl (b) 0.10 M H3PO4 (c) 0.010 M H2CO3
(d) 0.10 M NaH2PO4 (e) 0.10 M aspirin
8.31 Only (b) is a redox reaction. The others are acid2base
reactions.
(a) Na 2CO 3 1 2HCl h 2NaCl 1 CO 2 1 H 2O
(b) Mg 1 2HCl h MgCl2 1 H 2
(c) NaOH 1 HCl h NaCl 1 H 2O
(d) Fe 2O 3 1 6HCl h 2FeCl3 1 3H 2O
(e) NH 3 1 HCl h NH 4Cl
(f) CH 3NH 2 1 HCl h CH 3NH 3Cl
(g) NaHCO 3 1 HCl h NaCl 1 H 2O 1 CO 2
8.33 (a) 1023 M (b) 10210 M (c) 1027 M (d) 10215 M
8.35 (a) pH 5 8 (basic) (b) pH 5 10 (basic) (c) pH 5 2
(acidic) (d) pH 5 0 (acidic) (e) pH 5 7 (neutral)
8.37 (a) pH 5 8.5 (basic) (b) pH 5 1.2 (acidic)
(c) pH 5 11.1 (basic) (d) pH 5 6.3 (acidic)
8.39 (a) pOH 5 1.0, [OH2] 5 0.10 M
(b) pOH 5 2.4, [OH2] 5 4.0 3 1023 M
(c) pOH 5 2.0, [OH2] 51.0 3 1022 M
(d) pOH 5 5.6, [OH2] 5 2.5 3 1026 M
8.41 0.348 M
8.43 (a) 12 g of NaOH diluted to 400 mL of solution:
pH to be stable. In that situation, you might start with a buffer
that was initially set to have more of the conjugate base so that
it could absorb more of the H1 that you know will be produced.
8.59 No. 100 mL of 0. 1 M phosphate at pH 7.2 has a total
of 0.01 moles of weak acid and conjugate base with equimolar
amounts of each. 20 mL of 1 M NaOH has 0.02 moles of base,
so there is more total base than there is buffer to neutralize it.
This buffer would be ineffective.
8.61 (a) According to the Henderson-Hasselbalch equation,
no change in pH will be observed as long as the weak acid/conjugate base ratio remains the same.
(b) The buffer capacity increases with increasing amounts of
weak acid/conjugate base concentrations; therefore, 1.0-mol
amounts of each diluted to 1 L would have a greater buffer
capacity than 0.1 mol of each diluted to 1 L.
8.63 From the Henderson-Hasselbach equation,
pH 5 pKa 1 log(A2/HA)
A2/HA 5 10, log(A2/HA) 5 1 since 101 5 10
pH 5 pKa 1 1
8.65 When 0.10 mol of sodium acetate is added to 0.10 M HCl,
the sodium acetate completely neutralizes the HCl to acetic
acid and sodium chloride. The pH of the solution is determined
by the incomplete ionization of acetic acid.
3 CH 3COO 2 4 3 H 3O 1 4
Ka 5
[H3O1]5[CH3COO2] 5 x
3 CH 3COOH 4
"x 2 5 "K a 3 CH 3COOH 4 5 "1 1.8 3 10 2 5 2 1 0.10 2
x 5 [H3O1] 5 1.34 3 1023 M
pH 5 2log[H3O1] 5 2.9
1L sol
0.75 mol NaOH
400 mL sol a
ba
b
8.67 TRIS-H1 1 NaOH h TRIS 1 H2O 1 Na1
1000 mL sol
1 L sol
8.69 The only parameter you need to know about a buffer is its
40.0 g NaOH
pKa. Choosing a buffer involves identifying the acid form that
3 a
b 5 12 g NaOH
1 mol NaOH
has a pKa within one unit of the desired pH.
8.71
Choosing a buffer involves identifying the acid form that
(b) 12 g of Ba(OH)2 diluted to 1.0 L of solution:
has a pKa within one unit of the desired pH (a pH of 8.15). The
171.4 Ba 1 OH 2 2
0.071 mol Ba 1 OH 2 2
b 5 12 g Ba 1 OH 2 2
b a
a
TRIS buffer with a pKa 5 8.3 best fits this criteria.
1 L sol
1 mol Ba 1 OH 2 2
8.73 Mg(OH)2 is a weak base used in flame-retardant plastics.
(c) 2.81 g KOH diluted to 500 mL
8.75 (a) Respiratory acidosis is caused by hypoventilation,
(d) 49.22 g sodium acetate diluted to 2 liters
which occurs due to a variety of breathing difficulties, such as
8.45 5.66 mL
a windpipe obstruction, asthma, or pneumonia. (b) Metabolic
8.47 3.30 3 1023 mol
acidosis is caused by starvation or heavy exercise.
8.49 The point at which the observed change occurs during a
8.77 Sodium bicarbonate is the weak base form of one of the
titration. It is usually so close to the equivalence point that the
blood buffers. It tends to raise the pH of blood, which is the
difference between the two becomes insignificant.
purpose of the sprinter’s trick, so that the person can absorb
8.51 (a)
more H1 during the event. By putting NaHCO3 into the sysH 3O 1 1 CH 3COO 2 m CH 3COOH 1 H 2O 1 removal of H 3O 1 2 tem, the following reaction will occur:
(b)
HCO32 1 H1 m H2CO3. The loss of H1 means that the blood
HO 2 1 CH 3COOH m CH 3COO 2 1 H 2O 1 removal of OH 2 2
pH will rise.
8.79 (a) Benzoic acid is soluble in aqueous NaOH.
8.53 Yes, the conjugate acid becomes the weak acid and the
C 6H 5COOH 1 NaOH m C 6H 5COO 2 1 H 2O
weak base becomes the conjugate base.
pK a 5 4.19
pK a 5 15.56
8.55 The pH of a buffer can be changed by altering the weak
(b) Benzoic acid is soluble in aqueous NaHCO3.
acid/conjugate base ratio, according to the HendersonC 6H 5COOH 1 NaHCO 3 m CH 3C 6H 4O 2 1 H 2CO 3
Hasselbalch equation. The buffer capacity can be changed
pK a 5 4.19
pK a 5 6.37
without a change in pH by increasing or decreasing the
(c) Benzoic acid is soluble in aqueous Na2CO3.
amount of weak acid/conjugate base mixture while keeping
C 6H 5COOH 1 CO 32 2 m CH 3C 6H 4O 2 1 HCO 32
the ratio of the two constant.
pK a 5 4.19
pK a 5 10.25
8.57 This would occur in a couple of cases. One is very com8.81 The strength of an acid is not important to the amount
mon: You are using a buffer, such as Tris with a pKa of 8.3, but
of NaOH that would be required to hit a phenolphthalein endyou do not want the solution to have a pH of 8.3. If you wanted
point. Therefore, the more concentrated acid, the acetic acid,
a pH of 8.0, for example, you would need unequal amounts of
would require more NaOH.
the conjugate acid and base, with there being more conjugate
8.83 3.70 3 1023 M
acid. Another case might be a situation where you are perform8.85 0.9 M
1
ing a reaction that you know will generate H but you want the
91123_00_Ans_pA8-A63 pp2.indd 21
11/22/08 1:17:51 AM
Answers
8.87 Yes, a pH of 0 is possible. A 1.0 M solution of HCl has
[H3O1] 5 1.0 M. pH 5 2log[H3O1] 5 2log[1.0 M] 5 0
8.89 The qualitative relationship between acids and their
conjugate bases states that the stronger the acid, the weaker
its conjugate base. This can be quantified in the equation Kb 3
Ka 5 Kw or Kb11.0 3 10214/Ka, where Kb is the base dissociation equilibrium constant for the conjugate base, Ka is the acid
dissociation equilibruim constant for the acid, and Kw is the
ionization equilibrium constant for water.
8.91 Yes. The strength of the acid is irrelevant. Both acetic
acid and HCl have one H1 to give up, so equal moles of either
will require equal moles of NaOH to titrate to an end point.
8.93 You would need a ratio of 0.182 parts of the conjugate
base to 1 part of the conjugate acid.
8.95 An equilibrium will favor the side of the weaker acid/
weaker base. The larger the pKa value, the weaker the acid.
8.97 (a) HCOO2 1 H3O1 m HCOOH 1 H2O
(b) HCOOH 1 OH2 m HCOO2 1 H2O
8.99 (a) 0.050 mol (b) 0.0050 mol (c) 0.50 mol
8.101 According to the Henderson-Hasselbalch equation,
3 HPO 42 2 4
pH 5 7.21 1 log
3 H 2PO 42 4
9.2
223
90 Th
9.3
74
33 As
9.4
201
81 Tl
139
54 Xe
h 42He 1
h
1
0
2 1e
0
1 1e
1
h
248
96 Cm
0
e
21
1
28
10 X
h
h
51
23 V
116
51 Sb
160
55 Cs
28
9.59
10
B
5
1 0n h
11
5B
h 3Li 1 42He
1g
1
The bombarding nucleus was neon 10Ne
9.25 (a) beta emission (b) gamma emission (c) positron
emission (d) alpha emission
9.27 Gamma emission does not result in transmutation.
74
32 Ge
201
Hg
80
A22
151
63Eu
209
83 Bi
9.5 Barium-122 has decayed through five half-lives, leaving
0.31 g. 10 g h 5.0 g h 2.5 g h 1.25 g h
0.625 g h 0.31 g
9.6 The dose is 1.5 mL.
9.7 The intensity at 3.0 m is 3.3 3 1023 mCi.
9.9 Alpha rays are He21 ions (42 He) whereas protons are positively charged H1 ions (11 H).
9.11 (a) 4.0 3 1025 cm, which is visible light (blue).
(b) 3.0 cm (microwave radiation)
(c) 2.7 3 1025 cm (ultraviolet light)
(d) 2.0 3 1028 cm (X-ray)
9.13 (a) Infrared has the longest wavelength.
(b) X-rays have the highest energy.
9.14 (a) nitrogen-13 (b) phosphorus-33 (c) lithium-9
(d) calcium-39
9.17 oxygen-16
91123_00_Ans_pA8-A63 pp2.indd 22
9.23
1
1
9.57
0
2 1e
1
51
24 Cr
0
2 1e
219
88 Ra
22 4
Chapter 9 Nuclear Chemistry
h
9.21
h
239
94 Pu
3
139
53 I
151
62 Sm
9.29
HPO 4
As the concentration of H2PO42 increases, the log 3 H
2
2 PO 4 4
becomes negative, lowering the pH and becoming more acidic.
8.103 No. A buffer will have a pH equal to its pKa only if equimolar amounts of the conjugate acid and base forms are present.
If this is the basic form of Tris, then just putting any amount of
it into water will give a pH much higher than the pKa value.
8.105 (a) pH 5 7.1, 3 H 3O 1 4 5 7.9 3 10 2 8 M, basic
(b) pH 5 2.0, [H3O1] 5 1.0 3 1022 M, acidic
(c) pH 5 7.4, 3 H 3O 1 4 5 4.0 3 10 2 8 M, basic
(d) pH 5 7.0, 3 H 3O 1 4 5 1.0 3 10 2 7 M, neutral
(e) pH 5 6.6, 3 H 3O 1 4 5 2.5 3 10 2 7 M, acidic
(f) pH 5 7.4, 3 H 3O 1 4 5 4.0 3 10 2 8 M, basic
(g) pH 5 6.5, 3 H 3O 1 4 5 3.2 3 10 2 7 M, acidic
(h) pH 5 6.9, 3 H 3O 1 4 5 1.3 3 10 2 7 M, acidic
8.107 4.9 : 1, or 5 : 1 to one significant figure
9.1
9.19
■
1
1 42He h 240
Am 1 11H 1 2 0n
95
9.31 Iodine-125 decayed through approximately six halflives, with 0.31 mg remaining: 20 mg h 10 mg h 5.0 mg
h 2.5 mg h 1.25 mg h 0.625 mg h 0.31 mg.
9.33 The plutonium underwent four half-lives since the glacier deposited it. There were 16 mg of plutonium/kg at the time
of deposition.
16 mg h 8 mg h 4 mg h 2 mg h 1 mg
9.35 The rate of radioactive decay is independent of all conditions, and is a property of each specific isotope. There is no way
we can increase or decrease the rate.
9.37 (a) The iodine-131 remaining after two hours will be
8.88 3 108 counts/s. (b) After 24 days, three half-lives have
passed: 1/2 3 1/2 3 1/2 5 1/8 or 12.5% of the original amount
remains. 24.0 mCi 3 0.125 5 3.0 mCi.
9.39 Gamma radiation has the greatest penetrating power;
therefore, protection from it requires the largest amount of
shielding.
9.41 30 m
9.43 The curie (Ci) measures radiation intensity.
9.45 0.63cc
9.47 At 20 cm, the intensity would be 3 3 103Bq (8 3
1022μCi).
9.49 Person A was exposed to the larger dose of radiation and
injured more seriously.
9.51 Iodine-131 is concentrated in the thyroid and would be
expected to induce the cancer.
9.53 (a) Cobalt-60 is used for (4) cancer therapy.
(b) Thallium-201 is used in (1) heart scans and exercise stress
tests. (c) Tritium is used for (2) measuring water content
of the body. (d) Mercury-197 is used for (3) kidney scans.
9.55 The product of fusion of hydrogen-2 and hydrogen-3
nuclei is helium-4 plus a neutron and energy.
1
58
26 Fe
1
1
h 0n 1
266
109 Mt
11
5B
7
9.61 The assumption of a constant carbon-14 to carbon12 ratio rests on two assumptions: (1) that carbon-14 is
continuously generated in the upper atmosphere by the
production and decay of nitrogen-15 and (2) that carbon-14
is incorporated into carbon dioxide, CO2, and other carbon
compounds and then distributed worldwide as part of the
carbon cycle. The continual formation of carbon-14; transfer
of the isotope within the oceans, atmosphere, and biosphere;
and decay of living matter keep the supply of carbon-14
constant.
9.63 2003 2 1350 5 653 years (if the experiment was run in
2003). 653 years/5730 years 5 0.111 half-lives.
9.65 Radon-222 decays by alpha emission to polonium-218.
222
86 Rn
h
218
84 Po
4
1 2He
11/22/08 1:17:57 AM
A23
■
Answers
9.67
(b)
Decay of P-32
109.5˚
H C
0.009
0.008
Mass of P32 (mg)
120˚
H
0.010
C
C
H
H H H
0.007
0.006
10.2 Of the four alcohols with the molecular formula C4H10O,
two are 1°, one is 2°, and one is 3°. For the Lewis structures of
the 3° alcohol and one of the 1° alcohols, some CiCH3 bonds
are drawn longer to avoid crowding in the Lewis structures.
0.005
0.004
0.003
H
0.002
0.001
0.000
H
H
H
H 9 C 9 C 9 C 9 C 9 O 9 H CH3CH2CH2CH2OH
0
25
50
75
100
125
150
175
200
H
H
H
Primary (1°)
H
Time (hr)
9.69 Neon-19 decays to fluorine-19 and sodium-20 decays to
neon-20.
19
Ne
10
20
Ne
11
h
0
e
11
h
0
e
11
1
19
F
9
1
20
Ne
10
1
64
Ni
28
h
266
Ds
110
H
H
O
H
OH
H 9 C 9 C 9 C 9 C 9 H CH3CH2CHCH3
H
9.71 Both the curie and the becquerel have units of
disintegrations/second, a measure of radiation intensity.
9.73 (a) Natural sources 5 82%
(b) Diagnostic medical sources 5 11%
(c) Nuclear power plants 5 0.1%
9.75 X-rays will cause more ionization than radar waves.
X-rays have higher energy.
9.77 The decay product is neptunium-237. 1000/432 5 2.3
half-lives, so somewhat less that 25% of the original americium
will remain after 1000 years.
9.79 One sievert is equal to 100 rem. This is sufficient to
cause radiation sickness but not certain death.
9.81 (a) Radioactive elements are constantly decaying to
other elements or isotopes and these decay products are mixed
with the original sample.
(b) Beta emission results from the decay of a neutron in the
nucleus to a proton (the increase in atomic number) and an
electron (the beta particle).
9.83 Oxygen-16 is stable because it has an equal number of
protons and neutrons. The others are unstable because the
numbers of protons and neutrons are unequal. In this case,
the greater the difference in numbers of protons and neutrons, the faster the isotope decays.
9.85 The new element is darmstadtium-266.
208
Pb
82
H
1
1 6 0n
H
H
Secondary (2°)
H
H
H
H
C
H
H
CH3
H9C9C9C9O9H
H
H
CH3CHCH2OH
Primary (1°)
H
H
H
H
C
H
CH3
H 9 C 9 C 9 OH CH3COH
H
H
C
H
CH3
Tertiary (3°)
H
10.3 The three secondary (2°) amines with the molecular
formula C4H11N are
CH3
CH3CH2CH2NHCH3
CH3CHNHCH3
CH3CH2NHCH2CH3
10.4 The three ketones with the molecular formula C5H10O are
O
O
O
9.87 The intermediate nucleus is boron-11.
10
B
5
1 0n h
1
11
B
5
h 3Li 1 2He
7
11
B
5
10.1 Following are Lewis structures showing all valence electrons and with all bond angles labeled.
109.5˚
109.5˚
H H
H C
C
CH3CH2CCH2CH3 CH3CCHCH3
CH3
4
Chapter 10 Organic Chemistry
(a)
CH3CH2CH2CCH3
10.5 Following are the two carboxylic acids with the molecular formula C4H8O2. The second structure drawn for each
shows the fully condensed 2CO2H group.
O
CH3CH2CH2COH or CH3CH2CH2CO2H
and
O
a H
O
H H
91123_00_Ans_pA8-A63 pp2.indd 23
CH3CHCOH or CH3CHCO2H
CH3
CH3
11/22/08 1:18:02 AM
Answers
10.6 The four esters with the molecular formula C4H8O2 are
O
O
HCOCH2CH2CH3
H
C
HCOCHCH3
Cl
H
CH3
(1)
A24
H
(f)
Chloromethane
(2)
10.19
O
■
Following is a Lewis structure for each ion.
O
O
CH3COCH2CH3
CH3CH2COCH3
(a)
(3)
B
DCD
B
(4)
10.17
(b)
DOD
(c)
(6)
(a)
O
(4)
H
DND
(5)
O
O
H
(d)
C
O
O
10.7 (a) T (b) T (c) F (d) F
10.9 Assuming each is pure, there are no differences in chemical or physical properties.
10.11 Wöhler heated ammonium chloride and silver cyanate,
both inorganic compounds, and obtained urea, an organic compound.
10.13 The four most common elements that make up organic
compounds and the number of bonds each typically forms are
H: forms one bond
C: forms four bonds
O: forms two bonds
N: forms three bonds.
10.15 Following are Lewis dot structures for each element.
Under each is the number of electrons in its valence shell.
(a)
H
F
(7)
(b)
O
C
O
O
(c)
(d)
CH3 C
O
Cl
10.21 To use the VSEPR model to predict bond angles and the
geometry about atoms of carbon, nitrogen, and oxygen
(1) write the Lewis structure for the target molecule showing
all valence electrons; (2) determine the number of regions of
electron density around an atom of C, N, or O. (3) if you find
four regions of electron density, predict bond angles of 109.5°;
If you find three regions, predict bond angles of 120° if you find
two regions, predict bond angles of 180°.
10.23 You would find two regions of electron density around
oxygen and, therefore, predict 180° for the CiOiH bond angle.
If only two regions of electron density are shown, predict 180°
for the CiOiH bond angle
Hydrogen peroxide
(b)
H
N
N
H
H
H
Hydrazine
H
(c)
H
C
O
H
H
Methanol
H
(d)
H
C
H
(d) 9 N 9 H
H
Methanethiol
H
H
C
N
H
H
Methylamine
91123_00_Ans_pA8-A63 pp2.indd 24
C
C
H
H
O
H
(b) 9 C 9 O 9 H
(c) 9 O 9 H
H
H
(e)
H
10.25 (a) 120° about C and 109.5° about O.
(b) 109.5° about N.
(c) 120° about N.
(d) This is a molecular model of (c) and shows the 120° bond
angle about N.
10.27 A functional group is a part of an organic molecule that
undergoes a predictable set of chemical reactions.
10.29
O
O
(a) 9 C 9
S
H
H
O
(e) 9 C 9 O 9
10.31 When applied to alcohols, tertiary (3°) means that the
carbon bearing the 2OH group is bonded to three other carbon
atoms.
10.33 When applied to amines, tertiary (3°) means that the
amine nitrogen is bonded to three carbon groups.
10.35 (a) The four primary (1°) alcohols with the molecular
formula C5H12O are
11/22/08 1:18:05 AM
A25
■
Answers
CH3CH2CH2CH2CH2OH
109.5° 180°
CH3CH2CHCH2OH
CH3
CH3
CH3CCH2OH
(d) CH3 ! C # C ! CH3
CH3CHCH2CH2OH
109.5° 120°
CH3
CH3
(b) The three secondary (2°) alcohols with the molecular
formula C5H12O are
OH
OH
CH3CHCH2CH2CH3
(e) CH3 ! C ! O ! CH3
OH
CH3CH2CHCH2CH3
109.5°
O
109.5°
CH3
CH3CHCHCH3
CH3
(c) The one tertiary (3°) alcohol with the molecular formula
C5H12O is
(f) CH3 ! N ! CH3
10.43 Predict 109.5° for C-P-C bond angle.
109.5°
CH3
CH3CH2C
CH3 ! P ! CH3
OH
CH3
CH3
10.37 The eight carboxylic acids with the molecular formula
C6H12O2 are:
a five-carbon
chain with a
one-carbon
branch
a sixcarbon chain
10.45 The eight aldehydes with the molecular formula C6H12O
are below. The aldehyde functional group is written CHO.
a four-carbon
chain with
two carbons as
branches
a four-carbon
chain with
two carbons as
branches
a five-carbon
chain with a
one-carbon
branch
a sixcarbon chain
CH3
CH3
CH3CH2CH2CH2CH2CO2H
CH3CHCH2CH2CO2H
CH3
CH3CHCHCO2H
CH3CH2CH2CH2CH2CHO
CH3CHCH2CH2CHO
CH3
CH3
CH3CH2CHCH2CO2H
CH3CH2CHCO2H
CH3
CH3CH2CHCHO
CH2CH3
CH3
CH3
CH3CH2CH2CHCO2H
CH3
CH3CH2CHCH2CHO
CH2CH3
CH3
CH3CHCHCHO
CH3
CH3CH2CH2CHCHO
CH3CH2CCO2H
CH3CH2CCHO
CH3
CH3
CH3
CH3
CH3
CH3CCH2CO2H
CH3CCH2CHO
CH3
10.39 Taxol was discovered by a survey of indigenous plants
sponsored by the National Cancer Institute with the goal of
discovering new chemicals for fighting cancer.
10.41 Arrows point to atoms and show bond angles about
each atom.
109.5°
CH3
10.47 (a) nonpolar covalent (b) nonpolar covalent
(c) nonpolar covalent (d) polar covalent (e) polar covalent
(f) polar covalent (g) polar covalent (h) polar covalent
10.49 Under each formula is given the difference in electronegativity between the atoms of the most polar bond.
H
H
(a) CH3 ! CH2 ! CH2 ! OH
109.5°
d⫺
(a)
120°
O
(b) CH3 ! CH2 ! C ! H
d⫹
(b)
O ! H (3.5–2.1 ⫽ 1.4)
H
(c)
H
d⫹
H!S!C!C!N!N
H
H
H
d⫹
N! H (3.0 – 2.1 ⫽ 0.9)
91123_00_Ans_pA8-A63 pp2.indd 25
H
d⫹
N! H (3.0 – 2.1 ⫽ 0.9)
H
d⫺
(c) CH3 ! C " CH2
H!C!N!H
H
H
109.5° 120°
d⫹
d⫺
H!C!O!H
d⫺
O
H
d⫹
(d)
H!C!C!C!H
H
H
C " O (3.5 – 2.5 ⫽ 1.0)
11/22/08 1:18:07 AM
Answers
H
H
d⫹
(e)
C " Od⫺
O
d⫺
(c)
H
H
C " O (3.5 – 2.5 ⫽ 1.0) O ! H (3.5 – 2.1 ⫽ 1.4)
(a)
CH3CH2CH2C ! OH
CH3 ! CH ! C ! CH3
(d)
(e)
H
CH3
CH3
trans-1,3-Dimethylcyclohexane
H ! C ! C ! OH
11.8 In order of increasing boiling point, they are
(a) 2,2-dimethylpropane (9.5°C), 2-methylbutane (27.8°C),
pentane (36.1°C)
(b) 2,2,4-trimethylhexane, 3,3-dimethylheptane, nonane
11.9 The two chloroalkanes with their IUPAC and common
names are
CH3
OH
CH3
or
H
CH3
O
CH3
CH3
CH3CH2 ! C ! O ! CH3
(b)
H
cis-1,3-Dimethylcyclohexane
O
O
(c)
or
CH3
H
10.51 Following is one structural formula for each part. More
than one answer is possible for parts a, b and c.
O
CH2" CH ! CH2OH
Chapter 11 Alkanes
Cl
11.1 This compound is octane, and its molecular formula is C8H18.
11.2 (a) constitutional isomers (b) the same compound.
11.3 Following are structural formulas and line-angle formulas for the three constitutional isomers with the molecular
formula C5H12.
CH3
CH3CH2CH2CH2CH3
A26
CH3
CH3
d⫹
H!C!C!O!H
(f)
■
CH3
CH3CHCH2CH3
CH3CCH3
CH3
11.4 (a) 5-isopropyl-2-methyloctane. Its molecular formula is
C12H26.
(b) 4-isopropyl-4-propyloctane. Its molecular formula is C14H30.
11.5 (a) isobutylcyclopentane, C9H18
(b) sec-butylcycloheptane, C11H22
(c) 1-ethyl-1-methylcyclopropane, C6H12
11.6 The structure with the three methyl groups equatorial is
4
H3C
1 CH3
2
Cl
1-Chloropropane
(Propyl chloride)
2-Chloropropane
(Isopropyl chloride)
11.11 (a) A hydrocarbon is a compound that contains only
carbon and hydrogen.
(b) An alkane is a saturated hydrocarbon.
(c) A saturated hydrocarbon contains only CiC and CiH
single bonds.
11.13 In a line-angle formula, each line terminus and vertex
represents a carbon atom. Bonds are represented by combinations of one, two, or three parallel lines.
11.15 (a) C10H22 (b) C8H18 (c) C11H24
11.17 (a) T (b) T (c) F (d) F
11.19 None represent the same compound. There are three
sets of constitutional isomers. Compounds (a), (d), and (e) have
the molecular formula C4H8O and are one set, compounds (c)
and (f) have the molecular formula C5H10O and are a second
set, and compounds (g) and (h) have the molecular formula
C6H10O and are a third set.
11.21 (a) T (b) T (c) T
11.23 2-methylpropane and 2-methylbutane.
11.25 (a) T (b) F (c) T
11.27
CH3
11.7
Cycloalkanes (a) and (c) show cis-trans isomerism.
H3C
(a)
CH3
CH3
(a)
(b)
CH3
(c)
or
H
(d)
H
cis-1,3-Dimethylcyclopentane
H3C
CH3
CH3
H
(f)
(e)
or
H
CH3
(g)
(h)
trans-1,3-Dimethylcyclopentane
91123_00_Ans_pA8-A63 pp2.indd 26
11/22/08 1:18:09 AM
A27
■
Answers
11.29 A condensed structural formula shows only the order of
bonding of the atoms in the compound. There is no attempt in
it to show bond angles or molecular shape.
11.31 (a) F (b) F (c) T (d) T (e) F
11.33 No
11.35 Structural formulas for the six cycloalkanes with the
molecular formula C5H10 are
CH3
CH3
CH3
Cyclopentane
H3C
Methylcyclobutane
CH3
trans-1,2-Dimethylcyclopropane
H3C
1,1-Dimethylcyclopropane
CH3
cis-1,2-Dimethylcyclopropane
CH2CH3
Ethylcyclopropane
11.37 (a) T (b) F (c) F (d) F (e) T (f) T
11.39 Heptane, C7H16, has a boiling point of 98°C and a
molecular weight of 100. Its molecular weight is approximately
5.5 times that of water. Although they are considerably smaller,
water molecules associate in the liquid phase by relatively
strong hydrogen bonds whereas the much larger heptane molecules associate only by relatively weak London dispersion forces.
11.41 Alkanes are insoluble in water.
11.43 Boiling points of unbranched alkanes are related to
their surface area; the larger the surface area, the greater
the strength of dispersion forces, and the higher the boiling
point. The relative increase in molecular size per CH2 group is
greatest between CH4 and CH3CH3 and becomes progressively
smaller as the molecular weight increases. Therefore, increase
in boiling point per added CH2 group is greatest between CH4
and CH3CH3 and becomes progressively smaller for higher
alkanes.
11.45 (a) F (b) T (c) T
11.47 The heat of combustion of methane is 212 kcal/mol or
212/16 5 13.3 kcal/gram. The heat of combustion of propane is
530 kcal/mol or 530/44 5 12.0 kcal/gram. Therefore, the heat
energy per gram is greater for methane.
11.49
Cl
Cl
Cl
1-Chloropentane
2-Chloropentane
3-Chloropentane
11.51 (a) One ring contains only carbon atoms.
(b) One ring contains two nitrogen atoms.
(c) One ring contains two oxygen atoms.
11.53 Octane will produce more engine knocking than
heptane.
11.55 The Freons are a class of chlorofluorocarbons. They
were considered ideal for use as heat transfer agents in refrigeration systems because they are nontoxic, not corrosive, and
nonflammable and odorless. (c) The two Freons used for this
purpose were Freon 11(CCl3F) and Freon-12 (CCl2F2).
11.57 They are hydrofluorocarbons and hydrochlorofluorocarbons. These compounds are much more chemically reactive
in the atmosphere than the original Freons and are destroyed
before they reach the stratosphere.
91123_00_Ans_pA8-A63 pp2.indd 27
11.59 (a) The longest chain is pentane. Its IUPAC name is
2-methylpentane.
(b) The pentane chain is numbered incorrectly. Its IUPAC
name is 2-methylpentane.
(c) The longest chain is pentane. Its IUPAC name is 3-ethyl3-methylpentane.
(d) The longest chain is hexane. Its IUPAC name is 3,
4-dimethylhexane.
(e) The longest chain is heptane. Its IUPAC name is
4-methylheptane.
(f) The longest chain is octane. Its IUPAC name is 3-ethyl3-methyloctane.
(g) The ring is numbered incorrectly. Its IUPAC name is
1-ethyl-3-methylcyclohexane.
(h) The ring is numbered incorrectly. Its IUPAC name is
1-ethyl-3-methylcyclohexane.
11.61 Tetradecane is a liquid at room temperature.
11.63 On the left is the planar hexagon representation. On
the right is the more stable chair conformation of this isomer.
CH3
CH3
OH
OH
2-Isopropyl-5-methylcyclohexanol
In this chair conformation
all groups on the ring are
in equatorial positions
11.65 Following is the alternative representation of 2-deoxyD-ribose.
HO
OH
O
H
H
H
H
OH
H
Chapter 12 Alkenes and Alkynes
12.1 (a) 3,3-dimethyl-1-pentene (b) 2,3-dimethyl-2-butene
(c) 3,3-dimethyl-1-butyne
12.2 (a) trans-3,4-dimethyl-2-pentene
(b) cis-4-ethyl-3-heptene
12.3 (a) 1-isopropyl-4-methylcyclohexene
(b) cyclooctene (c) 4-tert-butylcyclohexene
12.4 Line-angle formulas for the other two heptadienes are
2
3
4
1
2
5
6
7
cis,trans-2,4-Heptadiene
6
3
7
4 5
1
cis,cis-2,4-Heptadiene
12.5 Four stereoisomers are possible (two pairs of cis-trans
isomers).
12.6
Br
(b)
(a) CH3CHCH3
Br
CH3
12.7 Propose a two-step mechanism similar to that for the
addition of HCl to propene.
11/22/08 1:18:13 AM
Answers
Step 1: Reaction of H1 with the carbon–carbon double bond
gives a 3° carbocation intermediate.
⫹
CH3 ⫹ H⫹
CH3
(a)
Step 2: Reaction of the 3° carbocation intermediate with
bromide ion completes the valence shell of carbon and gives the
product.
⫹
CH3 ⫹ Br
120°
(b)
CH2OH
120°
180°
120°
(d)
Br
⫺
CH3
12.17
CH3
(a)
(c)
CH3CCH2CH3
H3C
⫹
CH3 ⫹ H⫹
CH3
A 3° carbocation
intermediate
Step 2: Reaction of the 3° carbocation intermediate with water
completes the valence shell of carbon and gives an oxonium ion.
H
H
⫹
O9H
CH3 ⫹ CO9H
CH3
An oxonium ion
Step 3: Loss of H1 from the oxonium ion completes the reaction
and generates a new H1 catalyst.
H
HC # CCCH2CH3
(e)
CH3
CH3C " CCH2CH3
12.19 (a) 2,5-dimethyl-1-hexene
(b) 1,3-dimethylcyclopentene
(c) 2-methyl-1-butene
(d) 2-propyl-1-pentene
12.21 (a) The longest chain is four carbon atoms. The correct
name is 2-methyl-1-butene.
(b) The ring is numbered incorrectly. The correct name is
4-isopropylcyclohexene.
(c) The longest chain is six carbon atoms. The correct name is
3-methyl-2-hexene.
(d) The longest chain containing the C5C is five carbon atoms.
The correct name 2-ethyl-3-methyl-1-pentene
(e) The ring is numbered incorrectly. The correct name is 3,
3-dimethylcyclohexene.
(f) The longest chain is seven carbon atoms. The correct name
is 3-methyl-3-heptene.
12.23 Only (b) 2-hexene, (c) 3-hexene, and (e) 3-methyl2-hexene show cis-trans isomerism.
12.25 Arachidonic acid is the all cis isomer.
CO2H
H
⫹
⫹
O9H
O
CH3
CH3
⫹H
⫹
Arachidonic acid
12.27
Only parts (b) and (d) show cis-trans isomerism.
Cl
CH3
(a) CH3 9 C 9 CH 9 CH2
91123_00_Ans_pA8-A63 pp2.indd 28
(d)
CH2CH3
12.9 Propose a three-step mechanism similar to that for the
acid-catalyzed hydration of propene.
Step 1: Reaction of the carbon-carbon double bond with H1
gives a 3° carbocation intermediate.
(a) F
CH3CHC # CCH2CH3
CH3
CH2 " CCH2CH3
OH
12.11
(b)
CH3
CH3
12.10
120°
(c) HC#C9CH"CH2
12.8 The product from each acid-catalyzed hydration is the
same alcohol.
⫹
A28
12.13 A saturated hydrocarbon contains only carbon–carbon
and carbon–hydrogen single bonds. An unsaturated hydrocarbon contains one or more carbon–carbon double or triple bonds
or an aromatic ring (Chapter 13).
12.15
109.5°
A 3° carbocation
intermediate
■
H3C
Br
(b) F
(c) F
(b)
CH2Cl
CH3
(b)
Br
(d) T
(d)
CH3
12.29
CH3
CH3
Following is a structural formula for b-ocimene.
11/22/08 1:18:14 AM
A29
■
Answers
1
12.41
2
5
8
7
6
(a)
or
(b)
3
4
(c)
or
␤-Ocimene
12.31 (a) T (b) T (c) F (d) F
12.33 The four isoprene units are shown in bold.
CH3
12.43
CH3
(a) CH2 " CCH2CH3 or CH3C " CHCH3
CH3
OH
(b) CH3CHCH " CH2
Vitamin A (retinol)
12.35 In an alkene addition reaction, one bond of the carbon–
carbon double bond breaks and single bonds to two new atoms
or groups of atoms form in its place.
(c) CH2 " CHCH2CH2CH3
12.45 Reaction involves acid-catalyzed hydration of each
double bond. There are two cis-trans isomers. The structural
formula for terpin hydrate is shown below in the most stable
chair conformation with the i(CH3)2CHOH group equatorial.
OH
OH
CH3CH " CH2 ⫹ H2O
H2SO4
⫹ 2 H2O
CH3CHCH3
H2SO4
12.37
OH
Cl
! CH2CH3 ⫹ HCl
(a)
Limonene
CH2CH3
Terpin
OH
OH
! CH2CH3 ⫹ H2O
(b)
H2SO4
OH
CH2CH3
CH3
I
(c)
12.47
CH3(CH2)5CH " CH ⫹ HI
Compound A is 2-methyl-1,3-butadiene.
CH3(CH2)5CHCH3 12.49 The reagents are shown over the arrows.
Br
Cl
CH2
⫹ HCl
(d)
CH3
CH3
C ! CH3
CH3
CH3
H2/Pd
HBr
(a)
(c)
OH
Br
H2O/H2SO4
(e)
CH3 CH " CHCH2CH3 ⫹ H2O
OH
H2SO4
Br
OH
(b)
CH3 CHCH2CH2CH3 ⫹ CH3 CH2 CHCH3CH3
(f)
CH2 " CHCH2CH2CH3 ⫹ H2O
H2SO4
OH
CH3 CHCH2CH2CH3
CH3
12.39
(a)
CH3CCH2CH2CH3
Cl
91123_00_Ans_pA8-A63 pp2.indd 29
CH3
(b)
Br2
(d)
12.51 Ethylene is a natural ripening agent for fruits.
12.53 Its molecular formula is C16H30O2.
12.55 Rods are used for peripheral and night vision. Cones
function in bright light and are used for color vision.
12.57 The most common consumer items made of highdensity polyethylene (HDPE) are milk and water jugs, grocery
bags, and squeeze bottles. The most common consumer items
made of low-density polyethylene (LDPE) are packaging for
baked goods, vegetables, and other produce, as well as trash
bags. Currently, only HDPE materials are recyclable.
CH3CCH2CH2CH
OH
11/22/08 1:18:16 AM
Answers
12.59 There are five compounds with the molecular formula
C4H8. All are constitutional isomers. The only cis-trans isomers
are cis-2-butene and trans-2-butene.
Cyclobutane
Methylcyclopropane
1-Butene
cis-2Butene
trans-2Butene
■
A30
13.5 The members of each class of hydrocarbons contain
fewer hydrogens than an alkane or cycloalkane with the same
number of carbon atoms. Alternatively, each class of hydrocarbons contains one or more carbon-carbon double or triple
bonds.
13.7 No.
13.9 (a) CH4
(b) CH2 " CH2
Methane
12.61 (a) The carbon skeleton of lycopene can be divided into
eight isoprene units, here shown in bold bonds.
Ethene
(Ethylene)
H
(c)
(b) Eleven of the 13 double bonds have the possibility for
cis-trans isomerism. The double bonds at either end of the
molecule cannot show cis-trans isomerism.
12.63
HC # CH
H
H
H
H
(d)
Ethyne
(Acetylene)
H
Benzene
(a)
CH3
(b)
(c)
(d)
12.65 Each alkene hydration reaction follows Markovnikov’s
rule. H adds preferentially to carbon-3 and OH adds to carbon-4 to give 3-hexanol. Each carbon of the double bond again
has the same pattern of substitution, so 3-hexanol is the only
product.
12.67 Reagents are shown over the arrows.
Br
I
Br2
Br
HI
13.11 Benzene consists of six carbons, each surrounded by
three regions of electron density, which gives 120° for all bond
angles. The presence of only trigonal planar ring carbons
means that all ring substituents are coplanar (lie in the same
plane) and cis-trans isomerism is not possible. Cyclohexane,
on the other hand, consists of carbons each surrounded by
four regions of electron density, which gives 109.5° for all bond
angles. It is the tetrahedral nature of the ring atoms in cyclohexane that allows for up and down substituents and hence
cis-trans isomerism (because no CiC rotation is possible in
such a cyclic system.
13.13 (a) T (b) F (c) F (d) T (e) F (f) F
13.15
Br
CH3
Cl
(a)
H2O
H2SO4
CH3
(b)
H2/Pd
OH
CH2CH3
Br
CH3
NO2
O2N
12.69 Oleic acid has one double bond about which cis-trans
isomerism is possible, so 2152 isomers are possible (one pair of
cis-trans isomers).
Linoleic acid has two double bonds about which cis-trans isomerism is possible, so 2254 isomers are possible (two pairs of
cis-trans isomers).
Linolenic acid has 3 double bonds about which cis-trans isomerism is possible, so 2358 isomers are possible (four pairs of
cis-trans isomers).
Chapter 13 Benzene and Its Derivatives
13.1 (a) 2,4,6-tri-tert-butylphenol
(b) 2,4-dichloroaniline
(c) 3-nitrobenzoic acid
13.3 A saturated compound contains only single covalent
bonds. An unsaturated compound contains one or more double or
triple bonds or aromatic rings. The most common double bonds
are C5C, C5O, and C5N. The most common triple bond is C;C.
91123_00_Ans_pA8-A63 pp2.indd 30
(c)
(d)
NO2
OH
OH
Cl
(e)
(f)
CH3
CH3
13.17 Only cyclohexene will react with a solution of bromine
in dichloromethane. A solution of Br2/CH2Cl2 is reddish-purple,
whereas 1,2-dibromocyclohexane is colorless. To tell which
bottle contains which compound, place a small quantity of each
compound in a test tube and to each add a few drops of Br2/
CH2Cl2 solution. If the red color disappears, the compound is
cyclohexene, which has been converted to 1,2-dibromocyclohexane.
11/22/08 1:18:18 AM
A31
■
Answers
If the reddish-purple color remains, the compound is benzene
because in the absence of a catalyst, aromatic compounds do
not react with Br2 in dichloromethane.
Br
⫹
Br2
CH2Cl2
Br
Cyclohexene
(colorless)
Bromine
(red)
⫹
Br2
13.39 BHT participates in one of the chain propagation steps
of autooxidation. It forms a stable radical and thus terminates
the radical chain reaction.
13.41 Styrene reacts with bromine by addition to the carbon–
carbon double bond.
1,2-dibromocyclohexane
(colorless)
CH2Cl2
Br
CH " CH2
No reaction
CH CH2Br
⫹ Br2
Benzene
13.19
CH3
CH3
CH3
Br
Br
Br
2-Bromotoluene
(o-Bromotoluene)
3-Bromotoluene
(m-Bromotoluene)
4-Bromotoluene
(p-Bromotoluene)
13.21 (a) Nitration using HNO3/H2SO4 followed by sulfonation using H2SO4. The order of the steps may be reversed.
(b) Bromination using Br2/FeCl3 followed by chlorination using
Cl2 /FeCl3. The order of the steps may also be reversed.
13.23 (a) T (b) T (c) F (d) F (e) F (f) T (g) T
(h) T (i) T
13.25 Autoxidation is the reaction of a C—H group with oxygen, O2, to form a hydroperoxide CiOiOiH.
13.27 Vitamin E participates in one of the chain propagation
steps of autoxidation and forms a stable radical, which breaks
the cycle of chain propagation steps.
13.29 By definition, a carcinogen is a cancer-causing substance. The most important carcinogens present in cigarette
smoke belong to the class of compounds called polynuclear
aromatic hydrocarbons (PAHs).
13.31 Cyclonite (RDX) has the largest percentage of its molecular weight contributed by nitro groups.
Explosive
Mol Wt
NO2 groups
% NO2 groups
TNT
227.1
138
60.77
Nitroglycerine
227.1
138
60.77
Cyclonite
222.1
138
62.13
PETN
316.1
184
58.21
13.33 The functional groups most responsible for the water
solubility of these dyes are the two ionic iSO3 –Na1 groups.
13.35 Capsaicin is isolated from various species of peppers
(Capsicum and Solanaecae).
13.37 Following are the three possible resonance-contributing
structures for naphthalene.
Chapter 14 Alcohols, Ethers, and Thiols
14.1 (a) 2-Heptanol
(b) 2,2-Dimethyl-1-propanol
(c) cis-3-Isopropylcyclohexanol
14.2 (a) Primary (b) Secondary
(c) Primary (d) Tertiary
14.3 The structure of the major alkene product from each
reaction is enclosed in a box.
In each case, the major product contains the more substituted
double bond.
CH3
CH3
(a) CH3C " CHCH3 ⫹ CH2 " CCH2CH3
CH3
(b)
CH2
⫹
14.4
H2SO4
H2O
OH
2-Methylcyclohexanol
14.5
1-Methylcyclohexene (C)
1-Methylcyclohexanol (D)
Each secondary alcohol is oxidized to a ketone.
O
(a)
O
(b) CH3CCH2CH2CH3
14.6 (a) Ethyl isobutyl ether (b) Cyclopentyl methyl ether
14.7 (a) 3-Methyl-1-butanethiol (b) 3-Methyl-2-butanethiol
14.9 The difference is in the number of carbon atoms bonded
to the carbon bearing the OH group. For primary alcohols,
there is one; for a secondary alcohol, there are two and for a
tertiary alcohol, there are three.
14.11
OH
(a)
91123_00_Ans_pA8-A63 pp2.indd 31
OH
H2SO4
OH
(b)
OH
11/24/08 12:39:10 PM
Answers
OH
A32
■
14.31
(c)
(d) HO
OH
K2Cr2O7
OH
(a)
H2SO4
OH
OH (f)
(e)
OH
14.13 (a) Prednisone contains three ketones, one primary
alcohol, one tertiary alcohol, one disubstituted carbon–carbon
double bond, and one trisubstituted carbon–carbon double
bond.
(b) Estradiol contains one secondary alcohol and one disubstituted phenol.
14.15 Low-molecular-weight alcohols form hydrogen bonds
with water molecules through both the oxygen and hydrogen
atoms of their iOH groups. Low-molecular-weight ethers form
hydrogen bonds with water molecules only through the oxygen
atom of their iOi group. The greater extent of hydrogen
bonding between alcohol and water molecules makes the lowmolecular-weight alcohols more soluble in water than the lowmolecular-weight ethers.
14.17 Both types of hydrogen bonding are shown in the following illustration.
O 9 CH3
H
d⫺
CH3 9 O
H d⫹
H 9 O d⫺
d⫹
H
14.19
O
In order of increasing boiling point they are
CH3CH2CH3
CH3CH2OH
⫺42°C
78°C
CH3CH2CH2CH2OH HOCH2CH2OH
117°C
198°C
14.21 The thickness (viscosity) of these three liquids is
related to the degree of hydrogen bonding between their
molecules in the liquid state. Hydrogen bonding is strongest
between molecules of glycerol, weaker between molecules of
ethylene glycol, and weakest between molecules of ethanol.
14.23 In order of decreasing solubility in water, they are:
(a) ethanol > diethyl ether > butane
(b) 1,2-hexanediol > 1-hexanol > hexane
14.25 (a) T (b) T (c) F (d) F (e) F (f) F (g) T (h) F
(i) T (j) F
14.27 Phenols are weak acids, with pKa values approximately
equal to 10. Alcohols are considerably weaker acids and have
about the same acidity as water (pKa values around 16).
14.29
(a) CH3CH2CH2CH2OH
(b) CH3CH2CH2CH2OH
91123_00_Ans_pA8-A63 pp2.indd 32
H2SO4
heat
K2Cr2O7
H2SO4
CH3CH2CH " CH2 ⫹ H2O
(b)
HOCH2CH2CH2CH2OH
O
O
'
'
HOCCH2CH2COH
K2Cr2O7
H2SO4
14.33 (a) H2SO4, heat (b) H2O/H2SO4
(c) K2Cr2O7/H2SO4 (d) HBr (e) Br2
(f) H2/Pd (g) K2Cr2O7/H2SO4
(h) K2Cr2O7/H2SO4 (i) K2Cr2O7/H2SO4
14.35 2-Propanol (isopropyl alcohol) and glycerin (glycerol)
are derived from propene. 2-Propanol is the alcohol in rubbing
alcohol. Major uses of glycerol are in skin care products and
cosmetics. It is also the starting material for the synthesis of
nitroglycerin.
14.37 (a) Dicyclopentyl ether (b) Dipentyl ether
(c) Diisopropyl ether
14.39 (a) 2-Butanethiol (b) 1-Butanethiol
(c) Cyclohexanethiol
14.41 Because 1-butanol molecules associate by hydrogen
bonding in the liquid state, it has the higher boiling point
(117°C). There is little polarity to an S—H bond. The only
interactions among 1-butanethiol molecules in the liquid state
are the considerably weaker London dispersion forces. For this
reason, 1-butanethiol has the lower boiling point (98°).
14.43 (a) T (b) T (c) T (d) T (e) T (f) T
14.45 Nobel discovered that diatomaceous earth absorbs
nitroglycerin so that it will not explode without a fuse.
14.47 Dichromate ion is reddish-orange, chromium (III) ion is
green. When breath containing ethanol passes through a solution containing dichromate ion, ethanol is oxidized and dichromate ion is reduced to green chromium (III) ion.
14.49 Normal bond angles about tetrahedral carbon and
divalent oxygen are 109.5°. In ethylene oxide the CiCiO and
CiOiC bond angles are compressed to approximately 60°,
which results in angle strain within the molecule.
14.51 The molecular formula of each is C3H2ClF5O. They
have the same molecular formula but a different connectivity
of their atoms.
F
F
Cl
H!C!O!C!C!H
F
F
Enflurane
F
F
H
H
F!C!C!O!C!F
F
Cl
F
Isoflurane
14.53 CH 3CH 2OH 1 3O 2 h 2CO 2 1 3H 2O
14.55 The eight isomeric alcohols with the molecular formula
C5H12O are
CH3CH2CH2CO2H
11/24/08 12:39:58 PM
A33
■
Answers
OH
CH3CH2CH2CH2CH2OH
OH
CH3CH2CHCH2CH3
CH3CHCH2CH2CH3
1-Pentanol
2-Pentanol
3-Pentanol
OH
CH3
CH3CH2C ! OH
CH3
COOH
CH3CHCH2CH2OH
CH3
CH3
H
HO
CH3CH2CHCH2OH
CH3CCH2OH
CH3
CH3
2-Methyl-1-butanol
1-Pentanol
(2.3 g/100 mL)
1,4-Butanediol
(Infinitely soluble)
14.61 Each is prepared from 2-methyl-1-propanol (circled) as
shown in this flow chart.
CH3
H2O
H2SO4
CH3C " CH2
H2SO4
H2O
OH
CH3
CH3
CH3CHCH2OH
K2Cr2O7
H2SO4
C
CH3
15.2 The group of higher priority in each set is circled.
O
(a)
! CH2OH and ! CH2CH2COH
O
(b)
! CH2NH2 and ! CH2COH
15.3 The configuration is R and the compound is
(R)-glyceraldehyde.
15.4 (a) Structures 1 and 3 are one pair of enantiomers.
Structures 2 and 4 are a second pair of enantiomers.
(b) Compounds 1 and 2, 1 and 4, 2 and 3, and 3 and 4 are
diastereomers.
15.5 Four stereoisomers are possible for 3-methylcyclohexanol.
The cis isomer is one pair of enantiomers; the trans isomer is a
second pair of enantiomers.
15.6 Each stereocenter is marked by an asterisk and the
number of stereoisomers possible is shown under the structural formula.
NH2
HO
CH2CHCOOH
*
(a)
14.63 The three functional groups are a thiol, a primary
amine, and a carboxyl group.
Oxidation of the thiol gives a disulfide.
HO
21 2
O
HOCCHCH2S ! S ! CH2CHCOH
NH2
NH2
Cystine
Chapter 15 Chirality: The Handedness
of Molecules
15.1 The enantiomers of each part are drawn with two groups
in the plane of the paper, a third group toward you in front
91123_00_Ans_pA8-A63 pp2.indd 33
CHCH3
CH3
CH3CHCO2H
O
H3C
CH3
CH3CH2CH2CH2CH2OH
HOCH2CH2CH2CH2OH
CH3CCH3
CH3CH
OH
H
C
(b)
2,2-Dimethyl-1-propanol
14.57 Ethylene glycol has two iOH groups by which each
molecule participates in hydrogen bonding, whereas 1-propanol
has only one. The stronger intermolecular forces of attraction
between molecules of ethylene glycol give it the higher boiling
point.
14.59 Arranged in order of increasing boiling point, they are
CH3
C
H3C
3-Methyl-1-butanol
CH3
Hexane
(Insoluble)
H
CH3
3-Methyl-2-butanol
CH3CH2CH2CH2CH2CH3
HOOC
H
C
(a)
CH3CHCHCH3
2-Methyl-2-butanol
of the plane, and the fourth group away from you behind the
plane.
OH
(b)
CH2 " CHCHCH2CH3
*
21 2
*
OH
(c)
*
22
NH2
4
15.7 (a) T (b) T (c) T (d) F (e) T (f) T (g) T
(h) T (i) T
15.9 An achiral object has no handedness; it is an object
whose mirror image is superposable on the original. An example is methane, CH4.
15.11 Both constitutional isomers and stereoisomers have the
same molecular formula. Whereas stereoisomers have the same
11/22/08 1:18:24 AM
Answers
connectivity, constitutional isomers have a different connectivity of their atoms.
15.13 2-Pentanol has a stereocenter (carbon 2). 3-Pentanol
has no stereocenter.
15.15 The carbon of a carbonyl group has only three groups
bonded to it. To be a stereocenter, a carbon must have four different groups bonded to it.
15.17 Compounds (b), (c), and (d) contain stereocenters, here
marked by asterisks, and are chiral
(b)
*
(c)
Cl
*
*
OH
OH
(d)
OH
*
O
(22 4)
(one pair of
enantiomers)
(two pair of
enantiomers)
OH
*
*
HO
*
2-Pentanol
15.19
Following are the mirror images of each.
(a) H C
3
OH
OH
C
C
H
*
*
(d)
(21 2)
COOH
HOOC
CHO
A34
15.27 The specific rotation of its enantiomer is 141°.
15.29 (a) T (b) T (c) F (d) T
15.31 Of the eight alcohols with the molecular formula
C5H12O2 , only three are chiral.
CH3
OH
O
(c)
■
OH
*
2-Methyl-1-butanol
3-Methyl-2-butanol
15.33 Of the eight carboxylic acids with the molecular formula C6H12O2, only three are chiral.
O
CH3
H
*
O
*
OH
*
OH
CHO
15.35 Amoxicillin has four stereocenters.
(b)
H
C
OH
CH2OH
H
O
HOCH2
OH
O
C
HO
OH
*
CH ! C ! NH
HO
NH2
O
(c)
*
S
N
*
O
H
H
OH
HO
CH3
H3C
HO
*
HO
O
HO
H3C
HO
OH
(c)
CH3
15.37 Share your findings with others. You will find it interesting to compare them.
15.39 This molecule has eight stereocenters, here marked
with asterisks. There are 28 5 256 possible stereoisomers.
15.201 (a) T (b) F (c) T (d) F (e) T (f) T
15.23 Only parts (b) and(c) contain stereocenters.
OH
CH3
C
Amoxicillian
(d)
(b)
*
H3C
*
*
15.25 Stereocenters are marked with an asterisk. Under each
is the number of stereoisomers possible.
*
*
F
H
* *
*
*
O
O CH
3
O
CH3
H
O
Triamcinolone acetonide
OH
(a)
*
Chapter 16 Amines
*
(b)
OH
(22 4)
(Two pairs of
enantiomers)
2
(cis + trans)
16.1 Pyrrolidine has nine hydrogens; its molecular formula
is C4H9N. Purine has four hydrogens; its molecular formula is
C5H4N4.
16.2
(a)
91123_00_Ans_pA8-A63 pp2.indd 34
NH2
(b)
NH2
11/22/08 1:18:25 AM
A35
(c)
Answers
■
H2N
2° amines (3)
NH2
H
N
16.3
(a)
(c)
NH2
HO
N
H
(b)
N
H
N or
NH2
(b) CH3NH2 or
NH2
16.5 The product of each reaction is an ammonium salt.
(CH3CH2)3NHCl
Triethylammonium chloride
(b)
H
N
H
O
'
CH3CO
Piperidinium acetate
16.7 Each compound has a six-membered ring with three
double bonds.
16.9 Following is a structural formula for each amine.
NH2
(a)
CH3
Isopropylmethylamine
Diethylamine
CH3
(b)
CH3(CH2)6CH2NH2
(d)
H2N(CH2)5NH2
CH3
Ethyldimethylamine
16.13 (a) F (b) T (c) T
16.15 The association of 1-butanol molecules by hydrogen
bonding is stronger than the association of 1-butanamine molecules by hydrogen bonding because of the greater polarity of
an OiH bond compared with the polarity of an NiH bond.
16.17 Low-molecular-weight amines are polar molecules and
are soluble in water because they form relatively strong hydrogen bonds with water molecules. Hydrocarbons are nonpolar
molecules and do not interact with water molecules.
16.19 Amines are more basic than alcohols because nitrogen
is less electronegative than oxygen and, therefore, more willing
to donate its unshared pair of electrons to H1 in an acid–base
reaction to form a salt.
16.21 (a) Ethylammonium chloride
(b) Diethylammonium chloride
(c) Anilinium hydrogen sulfate
16.23 The form of amphetamine present at both pH 1.0 and
pH 7.4 is its conjugate acid.
H
H
N
NH2
(c)
H
N
3° amine (1)
N
(a)
CH3
Methylpropylamine
16.4 The stronger base is circled.
(a)
H
N
Conjugate acid
of amphetamine
NH2
(e)
(f)
(CH3CH2CH2CH2)3N
Br
16.11 For this molecular formula, there are four primary
amines, three secondary amines, and one tertiary amine. Only
2-butanamine is chiral.
CH3
16.25
(a)
O
'
CH3COH O
'
CH3CO
N
1° amines (4)
N
H
*
NH2
1-Butanamine
(Butylamine)
NH2
NH2
2-Butanamine
(sec-Butylamine)
(b)
H
N
(c)
NH2
HCl
CH3 H SO
2
4
H
NH2
2-Methyl-1propanamine
(Isobutylamine)
NH3Cl
2-Methyl-2propanamine
(tert-Butylamine)
91123_00_Ans_pA8-A63 pp2.indd 35
H
HSO4
N
CH3
11/22/08 1:18:28 AM
Answers
16.27 (a) Tamoxifen contains three aromatic (benzene) rings,
one carbon–carbon double bond, one ether, and one tertiary
amine.
(b) The amine is tertiary.
(c) Two stereoisomers are possible, one pair of cis-trans isomers.
(d) Insoluble in water and in blood.
16.29 Possible negative effects are long periods of sleeplessness, loss of weight, and paranoia.
16.31 Both coniine and nicotine have one stereocenter; two
stereoisomers (one pair of enantiomers) are possible for each.
The S enantiomer of each is shown below.
*
H
the 2° aliphatic amine is
the more basic nitrogen
Cl ⫺
H
N
*
N
*
*
*
N
CH2CH2CH3
CH3
H
N
(S)-Coniine
(S)-Nicotine
16.49 The structural formula of alanine is drawn on the left
showing a free amino group ( iNH2) and a free carboxyl group
( iCO2H). An acid–base reaction between these two groups
gives the internal salt shown at the right.
16.33 The four stereocenters of cocaine hydrochloride are
marked with asterisks. Following is the structural formula of
the salt formed by reaction of cocaine with HCl.
Cl ⫺
H3C ⫹ H
N
*
CH3CHCO2
NH2
NH3
an internal salt
17.1 (a) 3,3-dimethylbutanal (b) cyclopentanone
(c) 1-phenyl-1-propanone
17.2 Following are line-angle formulas for the eight aldehydes with the molecular formula C6H12O. In the three that
are chiral, the stereocenter is marked by an asterisk.
COCH3
*
H OCC H
6 5
*
CH3CHCO2H
Chapter 17 Aldehydes and Ketones
O
*
O
H
CHO
Cocaine•HCl
CHO
16.35 Neither Librium nor Valium is chiral. Each is achiral
(without handedness).
16.37 No. No unreacted HCl is present.
16.39 The amino group in each compound is secondary. Each
compound has one benzene ring, one phenolic iOH group, a
secondary alcohol on a carbon bonded to the benzene ring, and
the same configuration at its single stereocenter. The amine of
epinephrine is substituted with a methyl group, whereas that
on albuterol has a tertiary butyl group.
16.41 In order of decreasing ability to form intermolecular
hydrogen bonds, they are CH3OH . (CH3)2NH . CH3SH. An
OiH bond is more polar than an NiH bond, which is in turn,
more polar than an SiH bond.
16.43 Butane, the least polar molecule, has the lowest boiling point; 1-propanol, the most polar molecule, has the highest
boiling point.
CH3CH2CH2CH3
CH3CH2CH2NH2
CH3CH2CH2OH
⫺0.5°C
7.2°C
77.8°C
16.45 (a) Following is a structural formula for 1-phenyl-2amino-1-propanol.
Hexanal
*
*CHCHCH
3
NH2
1-Phenyl-2-amino-1-propanol
4-Methylpentanal
*
CHO
*
3-Methylpentanal
*
CHO
2-Methylpentanal
CHO
CHO
2,3-Dimethylbutanal
3,3-Dimethylbutanal
CHO
CHO
2,2-Dimethylbutanal
2-Ethylbutanal
17.3 (a) 2,3-dihydroxypropanal (b) 2-aminobenzaldehyde
(c) 5-amino-2-pentanone
17.4 Each aldehyde is oxidized to a carboxylic acid.
O
'
OH
91123_00_Ans_pA8-A63 pp2.indd 36
36A
(b) This molecule has two stereocenters. 22=4 stereoisomers
are possible.
16.47 (a) The secondary aliphatic amine is the more basic
nitrogen.
(b) The three stereocenters are marked by asterisks.
H
N
■
(a)
HO
'
O
OH
Hexanedioic acid
(Adipic acid)
11/22/08 1:18:34 AM
A37
■
Answers
O
'
*
OH
(b)
CHO
CHO
2,2-Dimethylpropanal
2-Methylbutanal
17.17
3-Phenylpropanoic acid
17.5 Each primary alcohol comes from the reduction of an
aldehyde. Each secondary alcohol comes from the reduction of
a ketone.
"O
(a)
O
O
(a) H
C
(b) CH3CH2CH
H
CHO
(c)
(d) CH3(CH2)8CHO
CHO
(b)
O
'
CH2CH
CH3O
(c)
17.6
O
'
HO
O
'
Shown first is the hemiacetal and then the acetal.
O
'
C!H
OH
s
C!OCH3
s
H
CH3OH
Benzaldehyde
CH3OH
A hemiacetal
OCH3
s
C!OCH3 H2O
s
H
An acetal
17.7 (a) A hemiacetal derived from 3-pentanone (a ketone)
and ethanol.
(b) Neither a hemiacetal nor an acetal. This compound is the
dimethyl ether of ethylene glycol.
(c) An acetal derived from 5-hydroxypentanal and methanol.
17.8 Following is the keto form of each enol.
O
O
'
C
(b)
s
OH
(c)
s
'
'
(a)
O
H
17.9 (a) T (b) T (c) T (d) F
17.11 In an aromatic aldehyde, the iCHO group is bonded
to an aromatic ring. In an aliphatic aldehyde, it is bonded to a
tetrahedral carbon atom.
17.13 Compounds (b), (c), (d), and (f) contain a carbonyl group.
17.15 Of the four aldehydes with the molecular formula
C5H10O, only one is chiral. The stereocenter in it is marked by
an asterisk.
CHO
CHO
Pentanal
91123_00_Ans_pA8-A63 pp2.indd 37
(f) HOCH2CHCHO
(e)
OH
17.19 (a) 4-heptanone (b) 2-methylcyclopentanone
(c) cis-2-methyl-2-butenone (d) 2-hydroxypropanal
(e) 1-phenyl 2-propanone (f) hexanedial
17.21 (a) T (b) T (c) T (d) T
17.23 The carbonyl group of acetone forms hydrogen bonds
with water. These hydrogen bonds are sufficient to make acetone soluble in water in all proportions. 4-Heptanone contains
a carbonyl which, through its hydrogen bonding with water
molecules, promotes water solubility. It also contains two threecarbon hydrocarbon groups bonded to the carbonyl carbon,
which inhibit water solubility. In 4-heptanone, the combined
hydrophobic effect of the two hydrocarbon groups is greater
than the hydrophilic effect of the single carbonyl group, making 4-heptanone insoluble in water.
17.25 Pentane is a nonpolar hydrocarbon and the only attractive forces between its molecules in the liquid state are the
very weak London dispersion forces. Pentane, therefore, has
the lowest boiling point. Pentanal and 1-butanol are both polar
molecules. Because 1-butanol has a polar OH group, its molecules can associate by hydrogen bonding. The intermolecular
attraction between molecules of 1-butanol are greater than
those between molecules of pentanal. 1-Butanol, therefore, has
a higher boiling point than pentanal.
17.27 Acetone molecules have no OiH or NiH group
through which to form intramolecular hydrogen bonds.
17.29 Only an aldehyde is oxidized by Tollens’ reagent. Under
the basic conditions of the reaction, the oxidation product is
a sodium salt of a carboxylic acid. Upon neutralization with
aqueous HCl, the oxidation product is isolated as a carboxylic
acid. Each oxidation product is shown as it would be before
treatment with HCl.
CO2Na
(a) CH3CH2CH2CO2 Na
(b)
(c) No reaction
(d) No reaction
17.31 Liquid aldehydes are very susceptible to oxidation by
atmospheric oxygen. To prevent this oxidation, they are generally stored under an atmosphere of nitrogen.
17.33 These experimental conditions reduce an aldehyde to a
primary alcohol and a ketone to a secondary alcohol. Products
(a) and (c) are chiral.
3-Methylbutanal
11/22/08 1:18:38 AM
Answers
OH
(a) CH3CHCH2CH3
*
(b) CH3(CH2)4CH2OH
OH
*
(c)
CH2OH
* CH3
(d)
OH
17.35
O
(a) HOCH2 ! C ! CH2OH (b) Soluble
O
O
(c) HOCH2CHCH2OH
(a)
(a)
OH
OH
CHCH3 (b)
CHCH3
(c) No reaction
(d) No reaction
17.39 Only compounds (a), (b), (d), and (f) will undergo ketoenol tautomerism because each has an H on an a-carbon.
17.41 Following are the keto forms of each enol.
A38
An example of hydrolysis is the acid-catalyzed reaction of an
acetal with a molecule of water to give an aldehyde or ketone
and two molecules of alcohol.
17.49 (a) To reduce the ketone to an alcohol, use either
NaBH4 followed by H2O or H2/M.
(b) To bring about dehydration of the alcohol to an alkene, use
H2SO4 and heat.
(c) To add HBr to the carbon–carbon double bond, use concentrated HBr.
(d) To reduce the carbon–carbon double bond, use H2/Pd.
(e) To add bromine to the carbon–carbon double bond, use a
solution of Br2 in CH2Cl2.
17.51 Compounds (a), (b), and (d) can be formed by the reduction of an aldehyde or a ketone.
OH
17.37
■
H
(b)
(d)
H
H
O
O
17.53
O
OH
(a) C6H5CCH2CH3
NaBH4
then H2O
H2SO4
C6H5CHCH2CH3
heat
C6H5CH " CHCH3
O
O
(a)
(b) CH3CCH2CH2CH2CH3
(b)
O
H2 / Pd
OH
O
CH2CCH3
(c)
17.43 Compounds (a), (c), (d), and (e) are acetals. Compound (b)
is a hemiacetal. Compound (f) is neither an acetal nor a hemiacetal.
17.45 Following are structural formulas for the products of
each hydrolysis.
O
(a) CH3CH2CCH2CH3 HOCH2CH2OH
O
CH
2CH3OH
(b)
OCH3
HCl
OH
OH
(a) HOCH2CH2CH2CH2CHCH3
(d)
OH
CHO CH3OH
17.47 Hydration refers to the addition of one or more molecules of water to a substance. An example of hydration is
the acid-catalyzed hydration of propene to give 2-propanol.
Hydrolysis refers to the reaction of a substance with water
with breaking (lysis) of one or more bonds in the substance.
91123_00_Ans_pA8-A63 pp2.indd 38
Cl
17.55 (a) Each compound is insoluble in water. Treat each
with dilute aqueous HCl. Aniline, an aromatic amine, reacts
with HCl to form a water-soluble salt. Cyclohexanone does not
react with this reagent and is insoluble in aqueous HCl.
(b) Treat each with a solution of Br2/CH2Cl2. Cyclohexene reacts to
discharge the red color of Br2 and to form 1,2-dibromocyclohexane,
a colorless compound. Cyclohexanol does not react with this
reagent.
(c) Treat each with a solution of Br2/CH2Cl2. Cinnamaldehyde,
which contains a carbon–carbon double bond, reacts to discharge
the red color of Br2 and to form 2,3-dibromo-3-phenylpropanal,
a colorless compound. Benzaldehyde does not react with this
reagent.
17.57 Each aldehyde or ketone in each will be reduced to an
alcohol.
O 2CH3OH
(c)
H2SO4
(b)
OH
OH
(d)
(c) HOCH2CHCH2OH
OH
CH3O
CHCH2CH3
(e)
CH2 OH
(f )
HO
CH3OH
11/22/08 1:18:42 AM
A39
■
Answers
17.59
(a)
(c)
O
'
CH2CCH3
s
Cl
(b)
O
OH
'
s
CH3CCH2CCH3
s
CH3
'
O
(e)
(f )
OH
s
17.67 Carbon 4 provides the iOH group and carbon 1
provides the iCHO group.
Following is a structural formula for the free aldehyde.
5
CH2OH
O
OH '
H C1
4 H
H
O
'
(a)
OH
K2Cr2O7
H2SO4
CHO
(b)
OH
K2Cr2O7
H2SO4
CO2 H
(c)
CH3
H
OH
s
O
*
CH3
'
O
'
CH3CCH3
O
'
K2Cr2O7
H2SO4
O
K2Cr2O7
H2SO4
18.1 (a) 2,3-dihydroxypropanoic acid
(b) 3-aminopropanoic acid
(c) 3,5-dihydroxy-3-methylpentanoic acid
18.2 Each acid is converted to its ammonium salt. Given
are both the IUPAC and common names of each acid and its
ammonium salt.
COONH4
COOH NH3
(a)
Butanoic acid
(Butyric acid)
OH
(b)
91123_00_Ans_pA8-A63 pp2.indd 39
OH
s
Chapter 18 Carboxylic Acids
O
'
CH3CH
(b)
K2Cr2O7
*
17.65 Given first is the alkene that undergoes
acid-catalyzed hydration to give the desired alcohol, then
the aldehyde or ketone that undergoes reduction to give the
desired alcohol.
O
O
'
H2SO4
(e)
The cyclic
hemiacetal
CH2 " CHCH3
OH
s
OH
*
(c)
H
17.69
(d)
Redraw to show
OH near CHO
O
' H
C
OH
CH2 " CH2
2
H
5-Hydroxyhexanal
(a)
3
OH OH
17.61 1-Propanol has the higher boiling point because of the
greater attraction between its molecules due to hydrogen bonding through its hydroxyl group.
17.63 (a) The hydroxyaldehyde is first redrawn to show the
OH group nearer the CHO group. Closing the ring in the hemiacetal formation gives the cyclic hemiacetal.
(b) 5-Hydroxyhexanal has one stereocenter, and two stereoisomers (one pair of enantiomers) are possible.
(c) The cyclic hemiacetal has two stereocenters, and four stereoisomers (two pairs of enantiomers) are possible.
O
'
*
CH3CHCH2CH2CH2CH
s
OH
! CH " CH2
(d)
CH3 O
s
'
! CCH2CH
s
CH3
(d)
'
O
OH
O
s
'
CH3CHCH2CH
O
'
! CCH3
Ammonium butanoate
(Ammonium butyrate)
OH
COOH NH3
COOHNH4
(S)-2-Hydroxypropanoic
acid
[(S)-Lactic acid]
Ammonium
(S)-2-hydroxypropanoate
[Ammonium (S)-lactate]
11/22/08 1:18:47 AM
Answers
18.3
A40
carboxyl groups is present as its carboxylic anion, giving a
charge of 21.
O
'
O O
' '
HOC ! CONa
H
OH HO !
(a)
■
O
'
H2O
O
Monopotassium oxalate
18.13 The dimer drawn here shows two hydrogen bonds.
Hydrogen bonds
(b)
O
'
HO
OH
O
H
" O H2O
H
O
O
18.15 The carboxyl group contributes to water solubility; the
hydrocarbon chain prevents water solubility.
18.17 In order of increasing boiling point, they are heptanal,
1-heptanol, and heptanoic acid.
18.7
COOH
O
'
CH3CH2CH2CH2CH2CH2CH
O2N !
(b)
H
O9H
18.5 (a) 3,4-dimethylpentanoic acid
(b) 2-aminobutanoic acid
(c) hexanoic acid
(a)
H9O
Heptanal
(bp 153°C)
H2NCH2CH2CH2COOH
COOH
(c)
CH3CH2CH2CH2CH2CH2CH2OH
1-Heptanol
(bp 176°C)
COOH
(d)
O
'
CH3CH2CH2CH2CH2CH2COH
COOH
18.9
Heptanoic acid
(bp 223°C)
O
'
C
ONa
(a)
(b)
O
'
CH3 ! C ! OLi
O
'
(d)
Na
O
(c)
O
'
CH3 ! C ! ONH4
18.19 In order of increasing solubility in water, they are decanoic acid, pentanoic acid, and acetic acid.
18.21 (a) T (b) T (c) F (d) F (e) T (f) F
(g) T (h) F (i) T
18.23 In order of increasing acidity, they are benzyl alcohol,
phenol, and benzoic acid.
18.25 Following are completed equations for these acid–base
reactions.
OH
O Na
'
O
NaOH
(a)
CH3
ONa
H2O
CH3
O
'
ONa
(e)
OH
(f )
(CH3CH2CH2COO)2Ca2
18.11 Oxalic acid (IUPAC name: ethanedioic acid) is a
dicarboxylic acid. In monopotassium oxalate, one of its
91123_00_Ans_pA8-A63 pp2.indd 40
COONa
HCl
(b)
OH
COOH
NaCl
OH
11/22/08 1:18:52 AM
A41
■
Answers
COOH
COOCH2CH3
⫹ H2NCH2CH2OH
(c)
(c)
COOCH2CH3
OCH3
⫺
COO
⫹
H3NCH2CH2OH
18.37 Following is a structural formula for methyl
2-hydoxybenzoate.
COOCH3
OCH3
OH
! COOH ⫹ NaHCO3
(d)
! COO⫺Na⫹ ⫹ CO2 ⫹ H2O
18.27 Dividing both sides of the Ka equation by [H3O1] gives
the desired relationship.
18.29 The pKa of lactic acid is 4.07. At this pH, lactic acid is
present as 50% CH3CH(OH)COOH and 50% CH3CH(OH)COO2.
At pH 7.45, which is more basic than pH 4.07, lactic acid is
present primarily as the anion, CH3CH(OH)COO2.
18.31 In part (a), the iCOOH group is a stronger acid than
the iNH31 group.
(a)
CH3CHCOOH ⫹ NaOH
s
NH3⫹
CH3CHCOO⫺Na⫹ ⫹ NaOH
s
NH3⫹
CH3CHCOO⫺Na⫹ ⫹ H2O
s
NH2
18.33 In part (a), the iNH2 group is a stronger base than the
iCOO2 group.
(a)
18.39
CO2⫺Na⫹
(a)
(a)
(b)
O
O
'
CH3CO !
91123_00_Ans_pA8-A63 pp2.indd 41
CO2CH3
(f )
(g) No reaction
18.41 Step 1: Treatment of benzoic acid with HNO3/H2SO4
brings about nitration of the aromatic ring.
Step 2: Treatment of 4-nitrobenzoic acid with H2/M brings
about catalytic reduction of the iNO2 group to an iNH2
group.
18.43 Each starting material is difunctional and can form
esters at both ends. The following equation shows reaction of
two molecules of adipic acid and two molecules of ethylene
glycol to form a triester. The product has a free carboxyl group
at one end and a free hydroxyl group at the other and ester
formation can continue at each end of the chain.
O
'
HO
'
O
O
'
OH OH
HO
'
O
OH OH
–3H2O
O
'
CH3CHCOOH ⫹ NaCl
s
NH3⫹
18.35 Following is a structural formula for the ester formed
in each reaction.
O
'
CH2CH2OH
(d)
CH3CHCOO⫺Na⫹ ⫹ HCl
s
NH2
CH3CHCOO⫺Na⫹ ⫹ HCl
s
NH3⫹
CO2⫺Na⫹
(b)
CO2⫺NH4⫹
(c)
CH3CHCOO⫺Na⫹ ⫹ NaCl
s
NH3⫹
(b)
Following are the expected organic products.
(e) No reaction
CH3CHCOO⫺Na⫹ ⫹ H2O
s
NH3⫹
(b)
Methyl 2-hydroxybenzoate
(Methyl salicylate)
HO
'
O
O
O
'
O
'
O
O
OH
Chapter 19 Carboxylic Anhydrides, Esters,
and Amides
19.1
(a)
O
'
CH3CNH !
(b)
O
'
! CNH2
19.2 Under basic conditions, as in part (a), each carboxyl
group is present as a carboxylate anion. Under acidic conditions, as in part (b), each carboxyl group is present in its
un-ionized form.
11/22/08 1:18:59 AM
Answers
O
'
COCH3
(b)
2NaOH
(a)
H2O
O
'
! CNH2 HCl
O
'
! COH NH4 Cl
O
'
CONa
19.9 The product is an amide.
2CH3OH
CO Na
'
O
CH3CH2O !
19.11 (a) Aspartame is chiral. It has two stereocenters. Four
stereoisomers are possible.
NH3
O
H2O
O
O
'
'
(b)
HCl
O
'
OH
O
'
! NH ! CCH3
*
O
'
O
O
H
N
'
O
'
A42
H2O
COCH3
'
O
O
'
■
OCH3
*
CH3CH2OH
Aspartame
H2O
heat
O
'
CH3CONa (CH3)2NH
NH
NaOH
H2O
heat
H2N
O
'
! CNH2 NaOH
H2 N
ONa
H2O
O
'
! CONa NH3
91123_00_Ans_pA8-A63 pp2.indd 42
'
O
O
O
O
'
19.5 (a) benzoic anhydride
(b) methyl decanoate
(c) N-methylhexanamide
(d) 4-aminobenzamide or p-aminobenzamide
(e) cyclopentyl ethanoate or cyclopentyl acetate
(f) ethyl 3-hydroxybutanoate
19.7 Each reaction brings about hydrolysis of the amide. Each
product is shown as it would exist under the specified reaction
conditions.
(a)
Hydrolysis in NaOH
*
O
O
'
(b)
NH2
Hydrolysis in HCl
HO
*
O
CH3OH
NH3
*
'
O
O
H3N
OH
*
'
O
'
O
'
(a)
O
'
CH3CN(CH3)2 NaOH
(b) Aspartame contains one carboxylate anion, one 1° ammonium ion, on amide group and one ester group.
(c) The net charge is zero.
(d) It is an internal salt. Therefore, expect it to be soluble in
water.
(e) Following are the products of hydrolysis of the ester and
amide bonds.
'
19.3 In aqueous NaOH, each carboxyl group is present as a
carboxylate anion, and each amine is present in its unprotonated form.
OH
CH3OH
19.13 Following are sections of two parallel nylon-66 chains,
with hydrogen bonds between NiH and C wO groups indicted
by dashed lines.
11/22/08 1:19:04 AM
Answers
H
s
N
O
H
Hydrogen
bonds
H
O
'
s
N
N
'
s
'
O
N
A sunscreen, absorbs UV radiation and then reirradiates the
energy as heat.
19.29 The portion derived from urea contains the atoms
iNHiCOiNHi.
19.31 Following is the structural formula of benzocaine.
N
H2N
O
H
19.15 In the anhydrides of carboxylic acids, the functional
group is two carbonyl (CwO) groups bonded to an oxygen
atom. In an anhydride of phosphoric acid, the functional group
is two phosphoryl (PwO) groups bonded to an oxygen atom.
19.17 The phosphate ester group is shown here in its doubly
ionized form and has a net charge of 22.
O
O
HOCH2CCH2OPO⫺
O⫺
Dihydroxyacetone phosophate
19.19 The box on the top encloses the COO of the ester group.
On the bottom is the structural formula of chrysanthemic acid.
Benzocaine
Ethyl 4-aminobenzote
19.33 The reaction involves hydrolysis of a phosphate ester
and a mixed anhydride of a carboxylic acid and a phosphate
acid. Two equivalents of water are required.
O
O
'
'
C ! O ! P ! O
s
O
O
HOCH
'
s
CH2 ! O ! P ! O
s
O
H
CH3'
2HPO4 2
HOCH
s
CH2 ! OH
O
O
Pyrethrin I
CH3
H
H
CH3'
OH
O
Chrysanthemic acid
19.21 (a) The cis/trans ratio refers to the cis-trans relationship between the ester group and the four carbon chain on the
cyclopropane ring. Specifically the repellant in the commercial
preparation consists of a minimum of 35% of the cis isomer and
a maximum of 65% of the trans isomer.
(b) Permethrin has two stereocenters, and four stereoisomers
(two pairs of enantiomers) are possible. The designation “(1/2)”
refers to the fact that the cis isomer is present as a racemic
mixture, as is the trans isomer.
19.23 The compound is salicin. Hydrolysis of the glucose unit
and oxidation of the primary alcohol to a carboxyl group gives
salicylic acid.
19.25 Both aspirin and ibuprofen contain a carboxylic acid
and a benzene ring. Naproxen also contains a carboxylic acid
and a benzene ring; in naproxen, the benzene ring is a part of a
naphthalene ring.
19.27 A sunblock prevents all ultraviolet radiation from
reaching protected skin by reflecting it away from the skin.
91123_00_Ans_pA8-A63 pp2.indd 43
2H2O
O
'
C ! O
O
'
CH3
H
O
'
COCH2CH3
s
'
H
N
s
O
H
O
'
'
■
s
A43
19.35 (a) Both lidocaine and carbocaine have an amide group
a tertiary aliphatic amine, and an aromatic ring.
(b) Both are derived from 2,6-dimethylaniline and a 2-alkylaminosubstituted carboxylic acid.
19.37 Hydrolysis gives hydrogen phosphate ion and the enol of
pyruvate. The enol then undergoes rapid keto-enol tautomerism
to give pyruvate ion. Recall from Chapter 17 that the keto form
generally predominates in cases of keto-enol tautomerism.
O
s
O! P " O
s
O
s
CH2 " C ! COO
H2O
hydrolysis
Phosphoenolpyruvate
OH
s
CH2 " C ! COO
HPO42
O O
s
s s
s
CH3 ! C ! C ! O
Pyruvate
11/22/08 1:19:07 AM
Answers
Chapter 20 Carbohydrates
20.1 Following are Fischer projections for the four
2-ketopentoses. They consist of two pairs of enantiomers.
One pair of enantiomers
D-Ribulose
CH2OH
HO
O
H
OH HO
H
HO
Unit of
b-D-glucopyranose
20.2 D-Mannose differs in configuration from D-glucose only
at carbon 2. One way to arrive at the structures of the a and b
forms of D-mannopyranose is to draw the corresponding a and b
forms of D-glucopyranose, and then invert the configuration in
each at carbon 2.
CH2OH
HO
OH HO
H
OH (b)
Anomeric
carbon
H
H
b-D-Mannopyranose
(b-D-Mannose)
HO
O
H
OH HO
H
H
Anomeric
carbon
OH (a)
H
a-D-Mannopyranose
(a-D-Mannose)
20.3 D-Mannose differs in configuration from D-glucose only
at carbon 2.
CH2OH
CH2OH
O
HO
HO
Anomeric
carbon
OH
2
b-D-Mannopyranose
(b-D-Mannose)
HO
91123_00_Ans_pA8-A63 pp2.indd 44
(b)
O
HO
HO
HO
CH2OH
O
HO
1
OH
CH2OH a-D-glucopyranose
O
O
3
OH
OH(a)
b-1,3-Glycosidic bond
20.7 The carbonyl group in an aldose is an aldehyde. In a
ketose, it is a ketone. An aldopentose is an aldose that contains
five carbon atoms. An aldoketose is a ketose that contains five
carbon atoms.
20.9 The three most abundant hexoses in the biological world
are D-glucose, D-galactose, and D-fructose. The first two are
aldohexoses. The third is a 2-ketohexose.
20.11 To say that they are enantiomers means that they are
nonsuperposable mirror images.
20.13 The D or L configuration in an aldopentose is determined by its configuration at carbon 4.
20.15 Compounds (a) and (c) are D-monosaccharides.
Compound (b) is an L-monosaccharide.
20.17 A 2-ketoheptose has four stereocenters and 16 possible
stereoisomers. Eight of these are D-2-ketoheptoses and eight
are L-2-ketoheptoses. Following is one of the eight possible
D-2-ketoheptoses.
CH2OH
s
C"O
s
*
HO !!
H
*s H
HO !!
s
*
H !!
OH
*s OH
H !!
s
CH2OH
CH2OH
H
OCH3 (a)
HO
L-Xylulose
O
HO
Unit of
CH2OH
CH2OH
s
s
C"O
C"O
s
s
HO !!
H
H
!!
s
s OH
H !!
OH
HO
!!
s
s H
CH2OH
CH2OH
H
OCH3 (a)
Anomeric
carbon
O
HO
HO
H
A second pair of enantiomers
H
CH2OH
H
20.5 The b-glycosidic bond is between carbon 1 of the left unit
and carbon 3 of the right unit.
L-Ribulose
D-Xylulose
A44
20.4 Following is a Haworth projection and a chair conformation for this glycoside.
H
CH2OH
CH2OH
s
s
C"O
C"O
s
s
H !!
s OH HO !!
s H
H !!
s OH HO !!
s H
CH2OH
CH2OH
■
2
OH (a)
a-D-Mannopyranose
(a-D-Mannose)
20.19 In an amino sugar, one or more iOH groups are
replaced by iNH2 groups. The three most abundant
amino sugars in the biological world are D-glucosamine,
D-galactosamine, and N-acetyl-D-glucosamine.
20.21 (a) A pyranose is a six-membered cyclic hemiacetal
form of a monosaccharide.
(b) A furanose is a five-membered cyclic hemiacetal form of a
monosaccharide.
20.23 Yes, they are anomers. No, they are not enantiomers;
that is, they are not mirror images. They differ in configuration
only at carbon 1 and, therefore, are diastereomers.
20.25 A Haworth projection shows the six-membered ring
as a planar hexagon. In reality, the ring is puckered and its
11/22/08 1:19:10 AM
A45
■
Answers
most stable conformation is a chair conformation with all bond
angles approximately 109.5°.
20.27 Compound (a) differs from D-glucose only in the
configuration at carbon 4. Compound (b) differs only at
carbon 3.
HO
HO
CH2OH
O
CH2OH
s
H!!
s OH
H!!
s OH
H!!
s OH
CH2OH
CH2OH
OH
(a)
HO
HO
OH
OH
OH
CHO
D-Galactose
CH2OH
O
HO
CH2OH
OH
(b)
HO
OH
OH
HO
OH
D-Allose
20.29 The specific rotation of a-L-glucose is 2112.2°.
20.31 A glycoside is a cyclic acetal of a monosaccharide. A glycosidic bond is the bond from the anomeric carbon to the iOR
group of the glycoside.
20.33 No, glycosides cannot undergo mutarotation because
the anomeric carbon is not free to interconvert between a and
b configurations via the open-chain aldehyde or ketone.
20.35 Following are Fischer projections of D-glucose and
D-sorbitol. The configurations at the four stereocenters of
D-glucose are not affected by this reduction.
D-Glucose
NaBH4
CH2OH
s
H!!
s OH
HO!!H
s
H!!
s OH
H!!
s OH
CH2OH
D-Sorbitol
20.37 Ribitol is the reduction product of D-ribose. b-D-ribose
1-phosphate is the phosphoric ester of the OH group on the
anomeric carbon of b-D-ribofuranose.
91123_00_Ans_pA8-A63 pp2.indd 45
HO
OH
b-D-Ribofuranose 1-phosphate
(b-D-Ribose 1-phosphate)
20.39 To say that it is a b-1,4-glycosidic bond means that the
configuration at the anomeric carbon (carbon 1 in this problem)
of the monosaccharide unit forming the glycosidic bond is beta
and that it is bonded to carbon 4 of the second monosaccharide
unit. To say that it is an a-1,6-glycosidic bond means that the
configuration at the anomeric carbon (carbon 1 in this problem) of the monosaccharide unit forming the glycosidic bond is
alpha and that it is bonded to carbon 6 of the second monosaccharide unit.
20.41 (a) Both monosaccharide units are D-glucose.
(b) They are joined by a b-1,4-glycosidic bond.
(c) It is a reducing sugar and
(d) it undergoes mutarotation.
CHO
CHO
s
H!!OH
s
H!!OH
s
H!!OH
s
H!!OH
s
CH2OH
CHO
s
H!!OH
s
HO!!H
s
H!!OH
s
H!!OH
s
CH2OH
H
H
Ribitol
CHO
s
H!!
s OH
HO!!
s H
HO!!
s H
H!!
s OH
CH2OH
HO
O
'
O ! P ! O
s
HOCH2
O
O
H
H
Units of
HO
CH2OH
O
D-glucose
O
A reducing sugar
because this carbon
is a hemiacetal and
in equilibium with
the open-chain aldehyde
CH2OH
O
HO
OH
HO
OH
b-1,4-Glycosidic bond
OH
20.43 An oligosaccharide contains approximately six to
ten monosaccharide units. A polysaccharide contains more—
generally many more—than ten monosaccharide units.
20.45 The difference lies in the degree of chain branching.
Amylose is composed of unbranched chains, whereas amylopectin is a branched network with the branches started by a-1,
6-glycosidic bonds.
20.47 Cellulose fibers are insoluble in water because the
strength of hydrogen bonding of a cellulose molecule in the
fiber with surface water molecules is not sufficient to overcome
the intermolecular forces that hold it in the fiber.
20.49 (a) In these structural formulas, the CH3CO (the acetyl
group) is abbreviated Ac.
CH2OH
H
HO
O
H
HO
CH2OH
O
OH (b)
OH(b)
HO
OH
H
H
NHAc
H
NHAc
11/22/08 1:19:11 AM
Answers
Following are Haworth and chair structures for this repeating
disaccharide.
CH2OH
CH2OH
H
HO
O
H
O
H
O
H
OH
H
H
NHAc
H
OH
OH
H
H
NHAc
H
HO
HO
CH2OH
O
4
1
O
CH2OH
O
Enzyme
catalysis
Dihydroxyacetone
phosphate
H ! C ! OH
'
s!OH
s
CH2OPO32
An enediol
intermediate
OH
NHAc HO
NHAc
20.51 Its lubricating power decreases.
20.53 With maturation, children develop an enzyme capable
of metabolizing galactose. Thus they are able to tolerate galactose as they mature. Until these children develop the ability to
metabolize galactose, substituting sucrose for lactose replaces
the galactose in lactose with fructose in sucrose.
20.55 L-Ascorbic acid is oxidized (there is loss of two hydrogen
atoms) when it is converted to L-dehydroascorbic acid. L-Ascorbic
acid is a biological reducing agent.
20.57 Types A, B, and O have in common D-galactose and
L-fucose. Only type A has N-acetyl-D-glucosamine.
20.59 Mixing types A and B blood will result in coagulation.
20.61 Consult Table 20.1 for the structural formula of D-altrose
and draw it. Then replace the iOH groups on carbons 2 and 6
with hydrogens.
CHO
s
HO!!
s H
H!!OH
s
H!!OH
s
H!!OH
s
CH2OH
Enzyme
catalysis
3-phosphate
20.75 (a) Coenzyme A is chiral. It has five stereocenters.
(b) The functional groups, starting from the left, are a thiol
(iSH), two amides, a secondary alcohol, a phosphate ester,
a phosphate anhydride, a phosphate ester, another phosphate
ester, a unit of 2-deoxyribose, and a (b-glycosidic bond to
adenine, a heterocyclic amine.
(c) Yes, it is soluble in water because of the presence of a number of polar CwO groups, one iOH group and three phosphate groups, all of which will interact with water molecules
by hydrogen bonding.
(d) Following are the products of hydrolysis of all amide, ester,
and glycosidic bonds.
O
'
HS !CH2CH2NH2 O ! C ! CH2CH2NH3 2-Aminoethanethiol
CHO
2s
H!!H
s
H!!OH
s
H!!OH
s
H!!OH
s
6
CH3
4-Aminobutanoic acid
(b-Alanine)
O
CH3
'
s
HO ! C ! CH ! C !OH
s
s
OH CH3
2,3-Dihydroxy-3-methylbutanoic acid
(Pantothenic acid)
2,6-Dideoxy-D-Altrose
(D-Digitoxose)
20.63 The monosaccharide unit of salicin is D-glucose.
20.65 Chitosan can be obtained from the shells of crustaceans
such as shrimp and lobsters.
20.67 The five-membered ring of fructose is nearly planar, so a Haworth projection is a good representation of its
structure.
CHO
s
H !!
s OH
CH2OPO32
D-Glyceraldehyde
D-Altrose
NH2
N
N
N
N
H
Adenine
91123_00_Ans_pA8-A63 pp2.indd 46
A46
20.69 In starch, a-glycosidic bonds join one glucose moiety to
another. Cellulose has b-glycosidic bonds. This difference means
that humans and other animals can digest starch, but not cellulose.
20.71 Amino sugars play an important structural role in polysaccharides, such as chitin, the hard shell of crabs, shrimp, and
lobsters. Amino sugars also play a role in the structures of the
blood group antigens.
20.73 The intermediate in this transformation is an enediol
formed by keto-enol tautomerism of dihydroxyacetone phosphate. Keto-enol tautomerism of this intermediate gives
D-glyceraldehyde-3-phosphate.
CH2OH
s
C"O
s
CH2OPO32
b-1,4-Glycosidic bond
■
HOCH2 O
OH
H
H
H
H
OH
H
b-2-Deoxy-D-ribose
O
'
3HO ! P ! OH
s
O
Dihydrogen phosphate
11/22/08 1:19:13 AM
A47
■
Answers
Chapter 21 Lipids
21.1 (a) It is an ester of glycerol and contains a phosphate
group; therefore it is a glycerophospholipid. Besides glycerol
and phosphate, it has a myristic acid and a linoleic acid component. The other alcohol is serine. Therefore, it belongs to the
subgroup of cephalins.
(b) The components present are glycerol, myristic acid, linoleic
acid, phosphate, and serine.
21.3 Hydrophobic means “water hating.” If the body did not
have such molecules, there could be no structure because the
water would dissolve everything.
21.5 The melting point would increase. The trans double
bonds would fit more in the packing of the long hydrophobic
tails, creating more order and therefore more interaction
between chains. This would require more energy to disrupt,
and hence a higher melting point.
21.7 The diglycerides with the highest melting points will be
the ones with two stearic acids (a saturated fatty acid). The
lowest melting points will be the ones with two oleic acids
(a monounsaturated fatty acid).
21.9 (b), because its molecular weight is higher.
21.11 lowest (c); then (b); highest (a)
21.13 The more long-chain groups, the lower the solubility;
lowest (a); then (b); highest (c).
21.15 glycerol, sodium palmitate, sodium stearate, sodium
linolenate
21.17 Complex lipids can be classified into two groups: phospholipids and glycolipids. Phospholipids contain an alcohol, two
fatty acids, and a phosphate group. There are two types: glycerophospholipids and sphingolipids. In glycerophospholipids,
the alcohol is glycerol. In sphingolipids, the alcohol is sphingosine. Glycolipids are complex lipids that contain carbohydrates.
21.19 The presence of cis double bonds in fatty acids produces
greater fluidity because they cannot pack together as closely as
saturated fatty acids.
21.21 Integral membrane proteins are embedded in the membrane. Peripheral membrane proteins are found on membrane
surfaces.
21.23 A phosphatidyl inositol containing oleic acid and arachidonic acid:
21.31 The carbon of the steroid D ring to which the
acetyl group is bonded in progesterone undergoes the most
substitution.
21.33 LDL from the bloodstream enters the cells by binding
to LDL receptor proteins on the surface. After binding, the LDL
is transported inside the cells, where cholesterol is released by
enzymatic degradation of the LDL.
21.35 Removing lipids from the triglyceride cores of VLDL
particles increases the density of the particles and converts
them from VLDL to LDL particles.
21.37 When serum cholesterol concentration is high, the synthesis of cholesterol in the liver is inhibited and the synthesis
of LDL receptors in the cell is increased. Serum cholesterol
levels control the formation of cholesterol in the liver by regulating enzymes that synthesize cholesterol.
21.39 Estradiol (E) is synthesized from progesterone (P)
through the intermediate testosterone (T). First the D-ring
acetyl group of P is converted to a hydroxyl group and T is
produced. The methyl group in T, at the junction of the rings
A and B, is removed and ring A becomes aromatic. The keto
group in P and T is converted to a hydroxyl group in E.
21.41 Steroid structures are shown in Section 21.10. The
major structural differences are at carbon 11. Progesterone
has no substituents except hydrogen, cortisol has a hydroxyl
group, cortisone has a keto group, and RU-486 has a large
p-aminophenyl group. The functional group at carbon 11
apparently has little importance in receptor binding.
21.43 They have a steroid ring structure, they have a
methyl group at carbon 13, they have a triply bonded group
at carbon 17, and all have some unsaturation in the A ring,
the B ring, or both.
21.45 Bile salts help solubilize fats. They are oxidation products of cholesterol themselves, and they bind to cholesterol,
forming complexes that are eliminated in the feces.
21.47 (a) Glycocholate:
H
OH OH
H
HO
Carboxylic acid
2° alcohol
(b) Cortisone:
H
Ketones
91123_00_Ans_pA8-A63 pp2.indd 47
NH
s
CH2
s
COO
OH
OH
'
O
H3C
CH2 ! OH
s
H3C C " O
s
! OH
1° alcohol
Ketone
3° alcohol
'
21.25 Complex lipids that contain ceramides include sphingomyelin, sphingolipids, and the cerebroside glycolipids.
21.27 The hydrophilic functional groups of (a) glucocerebroside: carbohydrate; hydroxyl and amide groups of the
cerebroside. (b) Sphingomyelin: phosphate group; choline;
hydroxyl and amide of ceramide.
21.29 Cholesterol crystals may be found in (1) gallstones,
which are sometimes pure cholesterol, and (2) joints of people
suffering from bursitis.
CH3
H3C
Arachidonate (20:4)
O
'
Oleate (18:1)
O CH2OC(CH2)2(CH2CH " CH)4(CH2)4CH3
' s
CH3(CH2)7CH"CH(CH2)7COCH O
s
'
Inositol
OH
CH2OPO H
s
O
OH H H
HO
2° alcohols
Amide
O
'
H3C
O
Carbon–carbon
double bond
11/22/08 1:19:15 AM
Answers
(c) PGE2:
Carbon–carbon
double bonds
Carboxylic acid
O
'
COOH
Ketone
HO
OH
2° alcohols
(d) Leukotriene B4:
Carbon–carbon
double bonds
2° alcohol
Carboxylic acid
OH
OH
COOH
2° alcohol
CH3
21.49 Aspirin slows the synthesis of thromboxanes by inhibiting the COX enzyme. Because thromboxanes enhance the blood
clotting process, the result is that strokes caused by blood clots
in the brain will occur less often.
21.51 Waxes consist primarily of esters of long-chain saturated acids and alcohols. Because of the saturated components,
wax molecules pack more tightly than those of triglycerides,
which frequently have unsaturated components.
21.53 The transporter is a helical transmembrane protein.
The hydrophobic groups on the helices are turned outward
and interact with the membrane. The hydrophilic groups of
the helices are on the inside and interact with the hydrated
chloride ions.
21.55 (a) Sphingomyelin acts as an insulator.
(b) The insulator is degraded, impairing nerve conduction.
21.57 a-D-galactose, b-D-glucose, b-D-glucose
21.59 They prevent ovulation.
21.61 It inhibits prostaglandin formation by preventing ring
closure.
21.63 NSAIDs inhibit cyclooxygenases (COX enzymes) that
are needed for ring closure. Leukotrienes have no ring in their
structure; therefore they are not affected by COX inhibitors.
21.65 (See Figure 21.2.) Polar molecules cannot penetrate the
bilayer. They are insoluble in lipids. Nonpolar molecules can
interact with the interior of the bilayer (“like dissolves like”).
21.67 Both groups are derived from a common precursor,
PGH2, in a process catalyzed by the COX enzymes.
21.69 Coated pits are concentrations of LDL receptors on the
surface of cells. They bind LDL and by endocytosis transfer it
inside the cell.
21.71 In facilitated transport, a membrane protein assists in
the movement of a molecule through the membrane with no
requirement for energy. In active transport, a membrane protein assists in the process, but energy is required. ATP hydrolysis usually supplies the needed energy.
21.73 Aldosterone has an aldehyde group at the junction of
the C and D rings. The other steroids have methyl groups.
21.75 The formula weight of the triglyceride is about
800 g/mol. This is 0.125 mol (100 g 4 800 g/mol 5 0.125 mol).
One mole of hydrogen is required for each mole of double
bonds in the triglyceride. There are three double bonds, so
91123_00_Ans_pA8-A63 pp2.indd 48
■
A48
the moles of hydrogen required for each 100 g 5 0.125 mol 3
3 5 0.375 mol of hydrogen gas. Converting to grams of
hydrogen, 0.375 3 2 g/mol 5 0.750 g hydrogen gas.
21.77 This lipid is a ceramide, a kind of sphingolipid.
21.79 Some proteins that are associated with membranes
associate exclusively with one side of the membrane rather
than the other.
21.81 Statements (c) and (d) are consistent with what is
known about membranes. Covalent bonding between lipids and
proteins [statement (e)] is not widespread. Proteins “float” in
the lipid bilayers rather than being sandwiched between them
[statement (a)]. Bulkier molecules tend to be found in the outer
lipid layer [statement (b)].
21.83 Statement (c) is correct. Transverse diffusion is only
rarely observed [statement (b)]. Proteins are bound to the
inside and outside of the membrane [statement (a)].
21.85 Both lipids and carbohydrates contain carbon, hydrogen, and oxygen. Carbohydrates have aldehyde and ketone
groups, as do some steroids. Carbohydrates have a number of
hydroxyl groups, which lipids do not have to a great extent.
Lipids have major components that are hydrocarbon in nature.
These structural features imply that carbohydrates tend to be
significantly more polar than lipids.
21.87 primarily lipid: olive oil and butter; primarily carbohydrate: cotton and cotton candy
21.89 The amounts are the key point here. Large amounts of
sugar can provide energy. Fat burning due to the presence of taurine plays a relatively minor role because of the small amount.
21.91 The other ends of the molecules involved in the ester
linkages in lipids, such as fatty acids, tend not to form long
chains of bonds with other molecules.
21.93 The bulkier molecules tend to be found on the exterior
of the cell because the curvature of the cell membrane provides
more room for them.
21.95 The charges tend to cluster on membrane surfaces.
Positive and negative charges attract each other. Two positive
or two negative charges repel each other, so unlike charges do
not have this repulsion.
Chapter 22 Proteins
22.1
O
O
H3N 9 CH 9 C 9 O H3N9CH9 C 9 O
CH9CH3
CH2
CH3
Phenylalanine
(Phe)
Valine
(Val)
O
O
H3N9 CH9 C 9 N9 CH 9 C 9 O H2O
H
CH9CH3
CH2
CH3
Valylphenylalanine
(Val-Phe)
11/22/08 1:19:16 AM
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Answers
22.2 a salt bridge
22.3 (a) storage (b) movement
22.5 protection
22.7 Tyrosine has an additional hydroxyl group on the phenyl
side chain.
22.9 arginine
22.11
H3N9 CH 9 C 9 N 9 CH 9 C 9 O
CH3
CH2
H
CH2
C"O
NH2
s
N
Alanylglutamine
(Ala-Gln)
H
Pyrrolidines
(heterocyclic
aliphatic amines)
22.13 They supply most of the amino acids we need in our
bodies.
22.15 These structures are similar except that one of the
hydrogens in the side chain of alanine has been replaced with a
phenyl group in phenylalanine.
22.17 Amino acids are zwitterions; therefore they all have
positive and negative charges. These molecules are very
strongly attracted to each other, so they are solids at low
temperatures.
22.19 All amino acids have a carboxyl group with a pKa
around 2 and an amino group with a pKa between 8 and 10.
One group is significantly more acidic and one is more basic.
To have an un-ionized amino acid, the hydrogen would have to
be on the carboxyl group and have vacated the amino group.
Given that the carboxyl group is the stronger acid, this would
never happen.
22.21
O
O
H3N 9 CH 9 C 9 N9 CH 9 C 9 O
H
CH2
CH3
CH2
C"O
NH2
Glutaminylalanine
(Gln-Ala)
22.37
O
O
NH39 CH9 C 9 N9 CH 9 C 9 N9 CH 9 COO
CH9OH
CH3
H
H3N! C! COO
s
CH2
s
COOH
H
CH2
H
CH2
CH2
CH2
CH2
S
NH
CH3
C "NH
NH2
22.23
H
H2N! C! COO
s
(CH2)4
s
NH3
22.25 the side-chain imidazole
22.27 The side chain of histidine is an imidazole with a nitrogen that reversibly binds to a hydrogen. When dissociated, it is
neutral; when associated, it is positive. Therefore, chemically it
is a base, even though it does have a pKa in the acidic range.
22.29 histidine, arginine, and lysine
22.31 Serine may be obtained by the hydroxylation of alanine.
Tyrosine is obtained by the hydroxylation of phenyl-alanine.
22.33 Thyroxine is a hormone that controls overall metabolic
rate. Both humans and animals sometimes suffer from low
levels of thyroxine, causing lack of energy and tiredness.
91123_00_Ans_pA8-A63 pp2.indd 49
O
O
COO
H
22.35
22.39 Only the peptide backbone contains polar units.
22.41
(a) NH3! CH ! C ! N! CH ! C ! N! CH! COOH
s
s
'
s
'
CH2 O H CH2 O H CH2
s
s
s
CH2
OH
SH
s
S!CH3
(b) pH 2 is shown above. At pH 7.0 would look like:
NH3! CH ! C ! N! CH ! C ! N! CH! COO
s
s
'
s
'
CH2 O H CH2 O H CH2
s
s
s
CH2
OH
SH
s
S!CH3
11/22/08 1:19:17 AM
Answers
(c) At pH 10:
NH2! CH ! C ! N ! CH! C ! N! CH! COO
s
s
'
s
'
CH2 O H CH2 O H CH2
s
s
s
CH2
OH
SH
s
S ! CH3
22.43 It would acquire a net positive charge and become more
water-soluble.
22.45 (a) 256 (b) 160,000
22.47 valine or isoleucine
22.49 (a) secondary (b) quaternary (c) quaternary
(d) primary
22.51 Above pH 6.0, the COOH groups are converted to COO2
groups. The negative charges repel each other, disrupting the
compact a-helix and converting it to a random coil.
22.53 (1) C-terminal end (2) N-terminal end
(3) pleated sheet (4) random coil
(5) hydrophobic interaction (6) disulfide bridge (7) a-helix
(8) salt bridge (9) hydrogen bonds
22.55 (a) Fetal hemoglobin has fewer salt bridges between
the chains.
(b) Fetal hemoglobin has a higher affinity for oxygen.
(c) Fetal hemoglobin has an oxygen saturation curve that is
in between myoglobin and maternal hemoglobin, so the graph
would look like the figure below:
100
Percent of saturation
Myoglobin
Fetal
hemoglobin
Maternal
hemoglobin
O2 pressure
22.57 The heme and the polypeptide chain form the quaternary structure of cytochrome c. This is a conjugated protein.
22.59 the intramolecular hydrogen bonds between the peptide backbone carbonyl group and the NiH group.
22.61 cysteine
22.63 Ions of heavy metals like silver denature bacterial proteins by reacting with cysteine iSH groups. The proteins, denatured by formation of silver salts, form insoluble precipitates.
22.65 (Chemical Connections 22A) Nutrasweet contains phenylalanine. People suffering from the genetic disease phenylketonuria must avoid phenylalanine as they cannot metabolize it,
and buildup in the body will have severe effects.
22.67 The symptoms of hunger, sweating, and poor coordination accompany diabetes when hypoglycemia is coming on.
22.69 The abnormal form has a higher percentage of
b-pleated sheet compared to the normal form.
22.71 The oxygen-binding behavior of myoglobin is hyperbolic, while that of hemoglobin is sigmoidal.
91123_00_Ans_pA8-A63 pp2.indd 50
A50
22.73 The two most common are prion diseases and Alzheimer’s
disease.
22.75 Even if it is feasible, it is not completely correct to call
the imaginary process that converts a-keratin to b-keratin
“denaturation.” Any process that changes a protein from a to b
requires at least two steps: (1) conversion from the a form to a
random coil, and (2) conversion from the random coil to the b
form. The term “denaturation” describes only the first half of
the process (Step 1). The second step would be called “renaturation.” The overall process is called denaturation followed by
renaturation. If we assume that the imaginary process actually
occurs without passing through a random coil, then the term
“denaturation” does not apply.
22.77 a quaternary structure because subunits are cross-linked
22.79 (a) hydrophobic (b) salt bridge (c) hydrogen bond
(d) hydrophobic
22.81 glycine
22.83 one positive charge on the amino group
22.85 These amino acids have side chains that can catalyze
organic reactions. They are polar or sometimes charged, and
the ability to make hydrogen bonds or salt bridges can help
catalyze the reaction.
22.87 Proteins can be denatured when the temperature is
only slightly higher than a particular optimum. For this reason, the health of a warm-blooded animal is dependent on the
body temperature. If the temperature is too high, proteins
could denature and lose function.
22.89 Even if you know all of the genes in an organism, not all the
genes code for proteins, nor are all genes expressed all the time.
22.91 A diet supplement full of collagen may help a person
lose weight, but it would be of little use for repairing muscle
tissue as collagen is not a good protein source. One third of its
amino acids are glycine and another third are proline. Muscle
repair requires high-quality protein to be effective.
Chapter 23 Enzymes
50
0
■
23.1 A catalyst is any substance that speeds up the rate of a
reaction and is not itself changed by the reaction. An enzyme
is a biological catalyst, which is either a protein or an RNA
molecule.
23.3 Yes. Lipases are not very specific
23.5 because enzymes are very specific and thousands of
reactions must be catalyzed in an organism
23.7 Lyases add water across a double bond or removes water
from a molecule, thereby generating a double bond. Hydrolases
use water to break an ester or amide bond, thereby generating
two molecules.
23.9 (a) isomerase (b) hydrolase (c) oxidoreductase
(d) lyase
23.11 Cofactor is more generic; it means a nonprotein part of
an enzyme. A coenzyme is an organic cofactor.
23.13 In reversible inhibition, the inhibitor can bind and then
be released. With noncompetitive inhibition, once the inhibitor is
bound, no catalysis can occur. With irreversible inhibition, once
the inhibitor is bound, the enzyme would be effectively dead, as
the inhibitor could not be removed and no catalysis could occur.
23.15 No, at high substrate concentration the enzyme surface
is saturated, and doubling of the substrate concentration will
produce only a slight increase in the rate of the reaction or no
increase at all.
11/22/08 1:19:18 AM
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Answers
23.17 (a) less active at normal body temperature
(b) The activity decreases.
23.19
(a)
5
•
4
Activity
3
•
2
•
1
0
•
•
pH 1
2
3
4
•
5
(b) 2
(c) zero activity
23.21 The active site of an enzyme is very specific for the size
and shape of the substrate molecules. Urea is a small molecule
and the urease active site is specific for it. Diethylurea has
the two ethyl groups attached. It is unlikely that diethylurea
would fit into an active site specific for urea.
23.23 The amino acid residues most often found at enzyme
active sites are His, Cys, Asp, Arg, and Glu.
23.25 The correct answer is (c). Initially the enzyme does not
have exactly the right shape for strongly binding a substrate,
but the shape of the active site changes to better accommodate
the substrate molecule.
23.27 Amino acid residues in addition to those at an enzyme
active site are present to help form a three-dimensional pocket
where the substrate binds. These amino acids act to make the
size, shape, and environment (polar or nonpolar) of the active
site just right for the substrate.
23.29 Caffeine is an allosteric regulator.
23.31 There is no difference. They are the same.
23.33
O
9 NH 9 CH 9 C 9
CH2
23.39 Just as with lactate dehydrogenase, there are five isozymes of PFK: M4, M3L, M2L2, ML3, and L4.
23.41 Two enzymes that increase in serum concentration following a heart attack are creatine phosphokinase and aspartate aminotransferase. Creatine phosphokinase peaks earlier
than aspartate aminotransferase and would be the best choice
in the first 24 hours.
23.43 Serum levels of the enzymes AST and ALT are monitored
for diagnosis of hepatitis and heart attack. Serum levels of AST
are increased after a heart attack, but ALT levels are normal. In
hepatitis, both enzymes are elevated. The diagnosis, until further
testing, would indicate the patient may have had a heart attack.
23.45 Chemicals present in organic vapors are detoxified in
the liver. The enzyme alkaline phosphatase is monitored to
diagnose liver problems.
23.47 It is not possible to administer chymotrypsin orally.
The stomach would treat it just as it does all dietary proteins:
degrade it by hydrolysis to free amino acids. Even if whole,
intact molecules of the enzyme were present in the stomach,
the low pH in the region would not allow activity for the
enzyme, which prefers an optimal pH of 7.8.
23.49 A transition-state analog is built to mimic the transition state of the reaction. It is not the same shape as the substrate or the product, but rather something in between. The
potency with which such analogs can act as inhibitors lends
credence to the theory of induced fit.
23.51 Succinylcholine has a chemical structure similar to that
of acetylcholine, so both can bind to the acetylcholine receptor
of the muscle end plate. The binding of either choline causes a
muscle contraction. However, the enzyme acetylcholinesterase
hydrolyzes succinylcholine only very slowly compared to acetylcholine. Muscle contraction does not occur as long as succinylcholine is still present and acts as a relaxant.
23.53 The most common reactions of kinases that we study in
this book are the ones that involve using ATP to phosphorylate
another molecule, be it an enzyme or a metabolite of a pathway.
One example would be glycogen phosphorylase kinase. This
enzyme catalyzes the following reaction as described in Chemical Connections 23E:
Phosphorylase 1 ATP h Phosphorylase-P 1 ADP
O
Another example is hexokinase from the pathway called glycolysis (chapter 28). Hexokinase catalyzes the following reaction:
O9P" O
Glucose 1 ATP h glucose 6-P 1 ADP
O
23.55 Many people have suffered psychological traumas that
haunt them for many years or even their entire lives. If longterm memories could be selectively blocked, it would offer relief
to patients suffering from something in their past.
23.57 In the enzyme pyruvate kinase, the wCH2 of the
substrate phosphoenolpyruvate sits in a hydrophobic pocket
formed by the amino acids Ala, Gly, and Thr. The methyl group
on the side chain of Thr, rather than the hydroxyl group, is in
the pocket. Hydrophobic interactions are at work here to hold
the substrate into the active site.
23.59 Researchers were trying to inhibit phosphodiesterases
because cGMP acts to relax constricted blood vessels. This
approach was hoped to help treat angina and high blood
pressure.
23.61 Phosphorylase exists in a phosphorylated form and an
unphosphorylated form, with the former being more active.
23.35
2ATP
ADP
kinase
Phosphorylase b
phosphatase
Phosphorylase a
2Pi
23.37 Glycogen phosphorylase is controlled by allosteric regulation and by phosphorylation. The allosteric controls are very
fast, so that when the level of ATP drops, for example, there is
an immediate response to the enzyme allowing more energy
to be produced. The covalent modification by phosphorylation
is triggered by hormone responses. They are a bit slower, but
more long lasting and ultimately more effective.
91123_00_Ans_pA8-A63 pp2.indd 51
11/22/08 1:19:19 AM
Answers
Phosphorylase is also controlled allosterically by several compounds, including AMP and glucose. While the two act semiindependently, they are related to some degree. The phosphorylated form has a higher tendency to assume the R state, which
is more active, and the unphosphorylated form has a higher
tendency to be in the less active T state.
23.63 In the processing of cocaine by specific esterase
enzymes, the cocaine molecule passes through an intermediate state. A molecule was designed that mimics this transition
state. This transition-state analog can be given to a host animal, which then produces antibodies to the analog. When these
antibodies are given to a person, they act like an enzyme and
degrade cocaine.
23.65 Cocaine blocks the reuptake of the neurotransmitter
dopamine, leading to overstimulation of the nervous system.
23.67 (a) Vegetables such as green beans, corn, and tomatoes
are heated to kill microorganisms before they are preserved by
canning. Milk is preserved by the heating process, pasteurization. (b) Pickles and sauerkraut are preserved by storage in
vinegar (acetic acid).
23.69 The amino acid residues (Lys and Arg) that are cleaved
by trypsin have basic side chains; thus they are positively
charged at physiological pH.
23.71 This enzyme works best at a pH of about 7.
23.73 a hydrolase
23.75 (a) The enzyme is called ethanol dehydrogenase or,
more generally, alcohol dehydrogenase. It could also be called
ethanol oxidoreductase.
(b) ethyl acetate esterase or ethyl acetate hydrolase
23.77 isozymes or isoenzymes
23.79 No, the direction a reaction goes is determined by the
thermodynamics of the reaction, including the concentration of
substrates and products. A reaction may only go in the forward
direction in a metabolic pathway due to an overwhelming concentration of the substrates along with immediate removal of
the products. However, the enzyme that catalyzes the reaction
would catalyze the reaction in either direction if it is thermodynamically possible.
23.81 The athlete may benefit from the stimulatory effect
of caffeine, but in a long race, the athlete would also become
dehydrated from the diuretic effect on the kidneys. One of the
most important factors to endurance performance is hydration,
so any substance that causes dehydration would be detrimental to performance in a long-distance event.
23.83 The structure of RNA makes it more likely to be able to
adopt a wider range of tertiary structures, so it can fold up to
form globular molecules similar to protein-based enzymes. It
also has an extra oxygen, which gives it an additional reactive
group to use in catalysis or an electronegative group, useful in
hydrogen bonding.
91123_00_Ans_pA8-A63 pp2.indd 52
A52
24.9 Upon binding of acetylcholine, the conformation of the
proteins in the receptor changes and the central core of the ion
channel opens.
24.11 The cobra toxin causes paralysis by acting as a nerve
system antagonist. It blocks the receptor and interrupts the
communication between neuron and muscle cell. The botulin
toxin prevents the release of acetylcholine from presynaptic
vesicles.
24.13 Taurine is a b -amino acid; its acidic group is iSO2OH
instead of iCOOH.
24.15 The amino group in GABA is in the gamma position;
proteins contain only alpha amino acids.
24.17 (a) norepinephrine and histamine
(b) They activate a secondary messenger, cAMP, inside the cell.
(c) amphetamines and histidine
24.19 It is phosphorylated by an ATP molecule.
24.21 Product from the MAO-catalyzed oxidation of
dopamine:
HO
Monoamine
oxidase (MAO)
NH3 H2O
HO
NAD
Dopamine
HO
NADH H
C
'
O
HO
H
NH4
24.23 (a) Amphetamines increase and (b) reserpine
decreases the concentration of the adrenergic neurotransmitter.
24.25 the corresponding aldehyde
24.27 (a) the ion-translocating protein itself
(b) It gets phosphorylated and changes its shape.
(c) It activates the protein kinase that does the phosphorylation of the ion-translocating protein.
24.29 They are pentapeptides.
24.31 The enzyme is a kinase. The reaction is the phosphorylation of inositol-1,4-diphosphate to inositol-1,4,5-triphosphate:
OP
OH
1
OH
ATP
OH
OH
5
4
OP
P !PO32
Chapter 24 Chemical Communications:
Neurotransmitters and Hormones
24.1 G-protein is an enzyme; it catalyzes the hydrolysis of
GTP to GDP. GTP, therefore, is a substrate.
24.3 A chemical messenger operates between cells; secondary
messengers signal inside a cell in the cytoplasm.
24.5 The concentration of Ca21 in neurons controls the
process. When it reaches 1024 M, the vesicles release the
neurotransmitters into the synapse.
24.7 anterior pituitary gland
■
OP
OH
1
OH
Kinase
ADP
OP
OH
5
4
OP
11/22/08 1:19:19 AM
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Answers
24.67 Aldosterone binds to a specific receptor in the nucleus.
24.33 Cyclic AMP
The aldosterone–receptor complex serves as a transcription
24.35 Protein Kinase
factor that regulates gene expression. Proteins for mineral
24.37 Glucagon initiates a series of reactions that eventually
metabolism are produced as a result.
activates protein kinase. The protein kinase phosphorylates
24.69 Large doses of acetylcholine will help. Decamethonium
two key enzymes in the liver, activating one and inhibiting
bromide is a competitive inhibitor of acetylcholine esterase.
the other. The combination of these effects lowers the level
The inhibitor can be removed by increasing substrate concenof fructose 2,6-bisphosphate, a key regulator of carbohydrate
tration.
metabolism. Fructose 2,6-bisphosphate stimulates glycolysis
24.71 Alanine is an a-amino acid in which the amino group is
and inhibits gluconeogenesis. Therefore, when fructose 2,
bonded to the same carbon as the carboxyl group. In b-alanine,
6-bisphosphate is decreased, gluconeogenesis is stimulated
the amino group is bonded to the carbon adjacent to the one to
and glycolysis is inhibited.
which the carboxyl group is bonded.
24.39 Insulin binds to insulin receptors on liver and muscle
cells. The receptor is an example of a protein called a tyrosine
a
b
kinase. A specific tyrosine residue becomes phosphorylated
CH3! CH! COO
CH2! CH2! COO
on the receptor, activating its kinase activity. The target pros
s
tein called IRS is then phosphorylated by the active tyrosine
NH3
NH3
kinase. The phosphorylated IRS acts as the second messenger.
Alanine
b-Alanine
It causes the phosphorylation of many target enzymes in the
(an a-amino acid)
(a b-amino acid)
cell. The effect is to reduce the level of glucose in the blood by
increasing the rate of pathways that use glucose and slowing
24.73 Effects of NO on smooth muscle are as follows: dilation
the rate of pathways that make glucose.
of blood vessels and increased blood flow; headaches caused by
24.41 Most receptors for steroid hormones are located in the
dilation of blood vessels in brain; increased blood flow in the
cell nucleus.
penis, leading to erections.
24.43 In the brain, steroid hormones can act as neurotrans24.75 Acetylcholine esterase catalyzes the hydrolysis of the
mitters.
neurotransmitter acetylcholine to produce acetate and choline.
24.45 Calmodulin, a calcium-ion-binding protein, activates
Acetylcholine transferase catalyzes the synthesis of acetylchoprotein kinase II, which catalyzes phosphorylation of other pro- line from acetyl—CoA and choline.
teins. This process transmits the calcium signal to the cell.
24.77 The reaction shown below is the hydrolysis of GTP:
24.47 Local injections of the toxin prevent release of acetylO
O
O
choline in that area.
Guanine
O
'
'
'
24.49 The neurofibrillar tangles found in the brains of
O ! P ! O ! P ! O ! P ! O
Alzheimer’s patients are composed of tau proteins. Mutated tau
s
s
s
proteins, which normally interact with the cytoskeleton, grow
HO
OH
O
O
O
into these tangles instead, thus altering normal cell structure.
24.51 Drugs that increase the concentration of the neuH2O
rotransmitter acetylcholine may be effective in the treatment of
Alzheimer’s disease. Acetylcholinesterase inhibitors, such as Aricept, inhibit the enzyme that decomposes the neurotransmitter.
O
O
O
Guanine
24.53 The neurotransmitter dopamine is deficient in ParO
'
'
'
kinson’s disease, but a dopamine pill would not be an effective
O ! P ! O ! P ! O O ! P ! OH
treatment. Dopamine cannot cross the blood-brain barrier.
s
s
s
24.55 Drugs like Cogentin that block cholinergic receptors
HO
OH
O
O
O
are often used to treat the symptoms of Parkinson’s disease.
These drugs lessen spastic motions and tremors.
24.79 Ritalin increases serotonin levels. Serotonin has a
24.57 Nitric oxide relaxes the smooth muscle that surrounds
calming effect on the brain. One advantage of this drug is that
blood vessels. This relaxation causes increased blood flow in
it does not increase levels of the stimulant dopamine.
the brain, which in turn causes headaches.
24.81 Proteins are capable of specific interactions at recogni24.59 Neurons adjacent to those damaged by the stroke begin
tion sites. This ability makes for useful selectivity in receptors.
to release glutamate and NO, which kills other cells in the
24.83 Adrenergic messengers, such as dopamine, are derivaregion.
tives of amino acids. For example, a biochemical pathway exists
24.61 Insulin-dependent (type 1) diabetes is caused by insufthat produces dopamine from the amino acid tyrosine.
ficient production of insulin by the pancreas. Administration of
24.85 Insulin is a small protein. It would go through protein
insulin relieves symptoms of this type of diabetes. Non-insulindigestion if taken orally and would not be taken up as the
dependent (type 2) diabetes is caused by a deficiency of insulin
whole protein.
receptors or by the presence of inactive insulin receptors. Other 24.87 Steroid hormones directly affect nucleic acid synthesis.
drugs are used to relieve symptoms.
24.89 Chemical messengers vary in their response times.
24.63 Monitoring glucose in the tears relieves the patient of
Those that operate over short distances, such as neurotranstaking many blood samples every day.
mitters, have short response times. Their mode of action fre24.65 Some possible dangers include enlargement of the prosquently consists of opening or closing channels in a membrane
tate, increases in chromosomal abnormalities, breast cancer
or binding to a membrane-bound receptor. Hormones must be
and early onset of puberty.
transmitted in the bloodstream, which requires a longer time
91123_00_Ans_pA8-A63 pp2.indd 53
11/22/08 1:19:21 AM
Answers
for them to take effect. Some hormones can and do affect protein synthesis, which makes the response time even longer.
24.91 Having two different enzymes for the synthesis and
breakdown of acetylcholine means that the rates of formation
and breakdown can be controlled independently.
Chapter 25 Nucleotides, Nucleic Acids,
and Heredity
25.1
O
H
N
O
O 9 P 9 O 9CH
2
O
O
H
O
N
H
OH
D-Ribose
2-Deoxy-D-ribose
25.15 The name “nucleic acid” derives from the fact that the
nucleosides are linked by phosphate groups, which are the
dissociated form of phosphoric acid.
25.17 anhydride bonds
25.19 In RNA, carbons 3 r and 5 r of the ribose are linked by
ester bonds to phosphates. Carbon 1 is linked to the nitrogen
base with an N-glycosidic bond.
25.21 (a)
O
N
OH
O
NH2
'
N
HO ! CH2 O
H
H
H
H
OH OH
Cytidine
NH2
'
N
O
N
HO !CH2 O
H
H
2
OH H
H
O
O
O 9 P 9 O 9 P 9 O 9 CH2
O
O
H
“Deoxy” because
there is no oxygen
at carbon 2
Deoxycytidine
25.13 D-Ribose and 2-deoxy-D-ribose have the same structure
except at carbon 2. D-Ribose has a hydroxyl group and hydrogen on carbon 2, whereas deoxyribose has two hydrogens.
H
H
OH
(b)
N
O
H
OH
NH2
N
N
O
N
N
O 9 P 9 O 9 CH2
O
O
H
H
H
H
N
O
A54
H
25.3 hemophilia, sickle cell anemia, etc.
25.5 (a) In eukaryotic cells, DNA is located in the cell nucleus
and in mitochondria.
(b) RNA is synthesized from DNA in the nucleus, but further use
of RNA (protein synthesis) occurs on ribosomes in the cytoplasm.
25.7 DNA has the sugar deoxyribose, while RNA has the
sugar ribose. Also, RNA has uracil, while DNA has thymine.
25.9 Thymine and uracil are both based on the pyrimidine
ring. However, thymine has a methyl substituent at carbon 5,
whereas uracil has a hydrogen. All of the other ring substituents are the same.
25.11
91123_00_Ans_pA8-A63 pp2.indd 54
CHO
s2
H!!H
s
H!!OH
s
H!!OH
s
CH2OH
H
H
H
CHO
s2
H!!OH
s
H!!OH
s
H!!OH
s
CH2OH
■
OH
H
25.23 (a) One end will have a free 5 r phosphate or hydroxyl
group that is not in phosphodiester linkage. That end is called
the 5 r end. The other end, the 3 r end, will have a 3 r free phosphate or hydroxyl group.
(b) By convention, the end drawn to the left is the 5 r end. A is
the 5 r end, and C is the 3 r end.
(c) The complementary strand would be GTATTGCCAT written from 5 r to 3 r.
25.25 two
25.27 electrostatic interactions
25.29 The superstructure of chromosomes consists of many
elements. DNA and histones combine to form nucleosomes
that are wound into chromatin fibers. These fibers are further
twisted into loops and minibands to form the chromosome
superstructure.
25.31 the double helix
25.33 DNA is wound around histones, collectively forming
nucleosomes that are further wound into solenoids, loops, and
bands.
25.35 rRNA
25.37 mRNA
25.39 Ribozymes, or catalytic forms of RNA, are involved in
post-transcriptional splicing reactions that cleave larger RNA
11/22/08 1:19:21 AM
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Answers
molecules into smaller, more active forms. For example, tRNA
molecules are formed in this way.
25.41 Small nuclear RNA is involved in splicing reactions of
other RNA molecules.
25.43 Micro RNAs are 22 bases long and prevent transcription
of certain genes. Small interfering RNAs vary from 22 to 30
bases and are involved in the degradation of specific mRNA
molecules.
25.45 Immediately after transcription, messenger RNA contains
both introns and exons. The introns are cleaved out by the action
of ribozymes that catalyze splicing reactions on the mRNA.
25.47 no
25.49 the specificity between the base pairs, A-T and G-C
25.51
Thymine
Adenine
AT pair
25.53 four
25.55 In semiconservative DNA replication, the new daughter
DNA helix is composed of one strand from the original (or parent) molecule and one new strand.
25.57
Histone9 (CH2)4 9 NH3 CH3 9 COO
acetylation
deacetylation
O
Histone 9 (CH2)4 9 N H 9 C9 CH3
25.59 Helicases are enzymes that break the hydrogen bonds
between the base pairs in double-helix DNA and thus help the
helix to unwind. This prepares the DNA for the replication
process.
25.61 pyrophosphate
25.63 The leading strand or continuous strand is synthesized
in the 5 r to 3 r direction.
25.65 DNA ligase
25.67 From the 5 r to the 3 r direction
25.69 One of the enzymes involved in the DNA base excision repair (BER) pathway is an endonuclease that catalyzes
the hydrolytic cleavage of the phosphodiester backbone. The
enzyme hydrolyzes on the 5 r side of the AP site.
25.71 a b-N-glycosidic bond between the damaged base and
the deoxyribose
25.73 Individuals with the inherited disease XP lack an
enzyme involved in the NER pathway. They are not able to
make repairs in DNA damaged by UV light.
91123_00_Ans_pA8-A63 pp2.indd 55
25.75 5 rATGGCAGTAGGC3 r
25.77 The anticancer drug fluorouracil causes the inhibition
of thymidine synthesis, thereby disrupting replication.
25.79 DNA polymerase, the enzyme that makes the phosphodiester bonds in DNA, is not able to work at the end of linear
DNA. This results in the shortening of the telomeres at each
replication. The telomere shortening acts as a timer for the cell,
allowing it to keep track of the number of divisions.
25.81 Because the genome is circular, even if the 5 r primers
are removed, there will always be DNA upstream that can act
as a primer for DNA polymerase to use as it synthesizes DNA.
25.83 A DNA fingerprint is made from the DNA of the child,
the mother, and any prospective fathers and used to eliminate
possible fathers.
25.85 Once a DNA fingerprint is made, each band in the
child’s DNA must come from one of the parents. Therefore,
if the child has a band, and the mother does not, then the
father must have that band. In this way, possible fathers
are eliminated.
25.87 One example might be that a life insurance company
could raise your rates or refuse to give you insurance if your
genetic profile had negative indicators. The same could happen with health insurance. Companies could begin to select
for people with certain positive traits thereby discriminating
against everyone else. Having that information could lead to
a new form of discrimination.
25.89 It detoxifies drugs and other synthetic chemicals by
adding a hydroxyl group to them.
25.91 A three-dimensional pocket of ribonucleotides where
substrate molecules are bound for catalytic reaction. Functional groups for catalysis include the phosphate backbone,
ribose hydroxyl groups, and the nitrogen bases.
25.93 (a) The structure of the nitrogen base uracil is shown
in Figure 25.1. It is a component of RNA. (b) Uracil with a
ribose attached by an N-glycosidic bond is called uridine.
25.95 native DNA
25.97 mol % A 5 29.3; mol % T 5 29.3; mol % G 5 20.7;
mol % C 5 20.7.
25.99 RNA synthesis is 5 r to 3 r
25.101 DNA replication requires a primer, which is RNA.
Because RNA synthesis does not require a primer, it makes
sense that RNA must have preceded DNA as a genetic material. This, added to the fact that RNA has been shown to be
able to catalyze reactions, means that RNA can be both an
enzyme and a heredity molecule.
25.103 The guanine-cytosine base pair has three hydrogen
bonds, while the adenine-thymine base pair has only two.
Therefore it takes more energy to separate DNA strands with
more G i C base pairs as it takes more energy to break their
three hydrogen bonds.
25.105 DNA is the blueprint for all of the components of
an organism. It is important that it have repair mechanisms
because if it is wrong, all of its products will always be wrong.
If correct DNA leads to incorrect RNA by some mutation,
then the products of the RNA may be wrong, but RNA is short
lived and the next time the RNA is produced, it will be correct. A good analogy is that of a cookbook. The words on the
page are the DNA. How you read them is the RNA. If you
misread the words, you may make the recipe wrong once. If
the book is printed wrong, however, you will always make the
recipe wrong.
11/22/08 1:19:23 AM
Answers
Chapter 26 Gene Expression
and Protein Synthesis
26.1 First, binding proteins must make the portion of the
chromosome where the gene is less condensed and more accessible. Second, the helicase enzyme must unwind the double
helix near the gene. Third, the polymerase must recognize the
initiation signal on the gene.
26.2 (a) CAU and CAC (b) GUA and GUG
26.3 valine 1 ATP 1 tRNAval
26.4 iCCT CGATTG i
iGGAGC TAAC i
26.5 (c); gene expression refers to both processes—transcription
and translation.
26.7 Protein translation occurs on the ribosomes.
26.9 Helicases are enzymes that catalyze the unwinding of
the DNA double helix prior to transcription. The helicases
break the hydrogen bonds between base pairs.
26.11 The termination signal is at the 5 r end of the template
strand that is being transcribed. It can also be said to be at the
3 r end of the coding strand.
26.13 The “guanine cap” methyl group is located on nitrogen
number 7 of guanine.
26.15 on the messenger RNA
26.17 The main subunits are the 60S and the 40S ribosomal
subunits, although these can be dissociated into even smaller
subunits.
26.19 326
26.21 Leucine, arginine, and serine have the most, with six
codons. Methionine and tryptophan have the fewest, with one
apiece.
26.23 The amino acid for protein translation is linked via
an ester bond to the 3 r end of the tRNA. The energy for producing the ester bond comes from breaking two energy-rich
phosphate anhydride bonds in ATP (producing AMP and two
phosphates).
26.25 (a) The 40S subunit in eukaryotes forms the preinitiation complex with the mRNA and the Met-tRNA that
will become the first amino acid in the protein. (b) The 60S
subunit binds to the pre-initiation complex and brings in the
next aminoacyl-tRNA. The 60S subunit contains the peptidyl
transferase enzyme.
26.27 Elongation factors are proteins that participate in the
process of tRNA binding and movement of the ribosome on the
mRNA during the elongation process in translation.
26.29 A special tRNA molecule is used for initiating protein
synthesis. In prokaryotes, it is tRNAfmet, which will carry a
formyl-methionine. In eukaryotes, there is a similar molecule,
but it carries methionine. However, this tRNA carrying methionine for the initiation of synthesis is different from the tRNA
carrying methionine for internal positions.
26.31 There are no amino acids in the vicinity of the nucleophilic attack that leads to peptide bond formation. Therefore,
the ribosome must be using its RNA portion to catalyze the
reaction, so it is a type of enzyme called a ribozyme.
26.33 Parts of the DNA involved are promoters, enhancers,
silencers, and response elements. Molecules that bind to DNA
include RNA polymerase, transcription factors, and other proteins
that may bind the RNA polymerase and a transcription factor.
26.35 The active site of aminoacyl-tRNA synthases (AARS)
contains the sieving portions that ensure that each amino acid
is linked to its correct tRNA. The two sieving steps work on the
basis of the size of the amino acid.
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A56
26.37 Both are DNA sequences that bind to transcription
factors. The difference is largely due to our own understanding of the big picture. A response element controls a set of
responses in a particular metabolic context. For example, a
response element may activate several genes when the organism is challenged metabolically by heavy metals, by heat, or by
a reduction in oxygen pressure.
26.39 Proteosomes play a role in post-translational degradation of damaged proteins. Proteins that are damaged by age or
proteins that have misfolded are degraded by the proteosomes.
26.41 (a) Silent mutation: Assume the DNA sequence is TAT
on the coding strand, which will lead to UAU on the mRNA.
Tyrosine is incorporated into the protein. Now assume a mutation in the DNA to TAC. This will lead to UAC in mRNA.
Again, the amino acid will be tyrosine. (b) Lethal mutation:
the original DNA sequence is GAA on the coding strand, which
transcribes into GAA on mRNA. This codes for the amino acid
glutamic acid. The DNA mutation TAA will lead to UAA, a stop
signal that incorporates no amino acid.
26.43 Yes, a harmful mutation may be carried as a recessive
gene from generation to generation, with no individual demonstrating symptoms of the disease. Only when both parents
carry recessive genes does an offspring have a 25% chance of
inheriting the disease.
26.45 Restriction endonucleases are enzymes that recognize
specific sequences on DNA and catalyze the hydrolysis of phosphodiester bonds in that region, thereby cleaving both strands
of the DNA.
26.47 Mutation by natural selection is an exceedingly long,
slow process that has occurred for centuries. Each natural
change in the gene has been ecologically tested and found
usually to have a positive effect or the organism is not viable.
Genetic engineering, where a DNA mutation is done very fast,
does not provide sufficient time to observe all of the possible
biological and ecological consequences of the change.
26.49 The discovery of restriction enzymes allowed scientists
to cut DNA at specific locations and link different pieces of
DNA together. This led to the ability to clone foreign DNA into
a host, leading to the ability to both amplify DNA of interest and also have it expressed. Without restriction enzymes
scientists would not be able to express a human protein in a
bacterial cell, for example, or to create the therapeutic gene
used in gene therapy.
26.51 The viral coat is a protective protein covering around
a virus particle. All of the components necessary to make the
coat—for example, amino acids and lipids—come from the host.
26.53 An invariant site is a location in a protein that has the
same amino acid in all species that have been studied. Studies
of invariant sites help establish genetic links and evolutionary
relationships.
26.55 A silent mutation is a change in the DNA that does not
lead to a change in the DNA product. This can happen when there
is a base change in the DNA but due to the redundancy of the
genetic code the change does not change the amino acid coded for.
26.57 A silent mutation may require a different tRNA molecule even though the same amino acid will be incorporated.
The pace of the ribosome movement during translation may be
different depending on the tRNA used, leading to the potential
for different folding patterns in the protein produced.
26.59 The protein p53 is a tumor suppressor. When its gene is
mutated, the protein no longer controls replication and the cell
begins to grow at an increased rate.
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26.61 The Duffy protein is found on the surface of human
red blood cells. It acts as a docking protein for malaria, so
mutations that lead to loss of the Duffy protein make the person resistant to malaria.
26.63 Several mutation types could affect production of protein Y. A mutation of gene X might change the protein sequence,
such as we see in Duffy or sickle cell anemia. These changes
can be minor or could lead to complete loss of function of the
protein. Another mutation in gene X might be a silent mutation, but as we saw in chemical connection 26D, even a silent
mutation could lead to a changed protein. Another possibility is
that a mutation affects not the gene X directly but rather the
promoter for gene X. If the promoter region is mutated, it might
lead to fewer RNA polymerase molecules binding and reduced
expression of the protein. Mutations could also affect enhancer
or silencer regions, altering the level of expression of protein Y.
26.65 (a) Transcription: The units include the DNA being
transcribed, the RNA polymerases, and a variety of transcription factors.
(b) Translation: mRNA, ribosomal subunits, aminoacyl-tRNA,
initiation factors, elongation factors.
26.67 Hereditary diseases cannot be prevented, but genetic
counseling can help people understand the risks involved in
passing a mutated gene to their offspring.
26.69 (a) Plasmid: a small, closed circular piece of DNA found
in bacteria. It is replicated in a process independent of the bacterial chromosome. (b) Gene: a section of chromosomal DNA
that codes for a particular protein molecule or RNA.
26.71 Each of the amino acids has four codons. All of the
codons start with G. The second base is different for each amino
acid. The third base may be any of the four possible bases. The
distinguishing feature for each amino acid is the second base.
26.73 The hexapeptide is Ala-Glu-Val-Glu-Val-Trp.
Chapter 27 Bioenergetics: How the Body
Converts Food to Energy
27.1 ATP
27.3 (a) 2 (b) the outer membrane
27.5 Cristae are folded membranes originating from the inner
membrane. They are connected to the inner membrane by
tubular channels.
27.7 There are two phosphate anhydride bonds:
Phosphate
anhydride bonds
Phosphate
ester bond
NH2
N
O
O
O
'
'
'
HO ! P ! O ! P ! O ! P ! O ! CH2 O
N
s
s
s
H
H
O
O
O
H
N
N
H
OH OH
27.9 Neither; they yield the same energy.
27.11 It is a phosphate ester bond.
27.13 The two nitrogen atoms that are part of CwN bonds
are reduced to form FADH2.
27.15 (a) ATP (b) NAD+ and FAD (c) acetyl groups
91123_00_Ans_pA8-A63 pp2.indd 57
27.17 An amide bond is formed between the amine portion of
mercaptoethanolamine and the carboxyl group of pantothenic
acid (see Figure 27.7).
27.19 No. The pantothenic acid portion is not the active
part. The active part is the i SH group at the end of the
molecule.
27.21 Both fats and carbohydrates are degraded to acetyl
coenzyme A.
27.23 a-ketoglutarate
27.25 Succinate is oxidized by FAD, and the oxidation product
is fumarate.
27.27 Fumarase is a lyase (it adds water across a double
bond).
27.29 No, but GTP is produced in Step 햶.
27.31 It allows the energy to be released in small packets.
27.33 carbon–carbon double bonds occur in cis-aconitate and
fumarate.
27.35 a-Ketoglutarate transfers its electrons to NAD1, which
becomes NADH 1 H1.
27.37 Mobile electron carriers of the electron transport chain:
cytochrome c and CoQ
27.39 When H+ passes through the ion channel, the proteins
of the channel rotate. The kinetic energy of this rotatory
motion is converted to and stored as the chemical energy
in ATP.
27.41 This process takes place in the inner membranes of the
mitochondria.
27.43 (a) 0.5 (b) 12
27.45 Ions reenter the mitochondrial matrix through the
proton-translocating ATPase.
27.47 The F1 portion of ATPase catalyzes the conversion of
ADP to ATP.
27.49 The molecular weight of acetate 5 59 g/mol, so
1 g acetate 5 1 4 59 5 0.017 mol acetate. Each mole of
acetate produces 12 moles of ATP [see Problem 27.43(b)],
so 0.017 mol 3 12 5 0.204 mol ATP. This gives 0.204 mol
ATP 3 7.3 kcal/mol 5 1.5 kcal.
27.51 (a) Muscles contract by sliding the thick filaments
(myosin) and the thin filaments (actin) past each other.
(b) The energy comes from the hydrolysis of ATP.
27.53 ATP transfers a phosphate group to the serine residue
at the active site of glycogen phosphorylase, thereby activating
the enzyme.
27.55 No. It would harm humans because they would not
synthesize enough ATP molecules.
27.57 This amount of energy (87.6 kcal) is obtained from
12 mol of ATP (87.6 kcal 4 7.3 kcal/mol ATP 5 12 mol ATP).
Oxidation of 1 mol of acetate yields 12 mol ATP. The molecular
weight of CH3COOH is 60 g/mol, so the answer is 60 g or 1 mol
CH3COOH.
27.59 The energy of motion appears first in the ion channel,
where the passage of H+ causes the proteins lining the channel
to rotate.
27.61 They are both hydroxy acids.
27.63 Myosin, the thick filament in muscle, is an enzyme that
acts as an ATPase.
27.65 Isocitrate has two stereocenters.
27.67 The ion channel is the F0 portion of the ATPase;
it is made of 12 subunits.
27.69 No, it largely comes from the chemical energy as a
result of the breaking of bonds in the O2 molecule.
11/22/08 1:19:25 AM
Answers
27.71 It removes two hydrogens from succinate to produce
fumarate.
27.73 The carbon dioxide that we exhale is released by the
two oxidative decarboxylation steps in the citric acid cycle.
27.75 Because of the central role of citric acid in metabolism,
it can be considered a good nutrient.
27.77 Complex II does not generate enough energy to produce
ATP. The rest do.
27.79 Citrate isomerizes to isocitrate to convert a tertiary
alcohol to a secondary alcohol. Tertiary alcohols cannot be
oxidized, but secondary alcohols can be oxidized to produce a
keto group.
27.81 Iron is found in iron-sulfur clusters in proteins and is
also part of the heme group of cytochromes.
27.83 Mobile electron carriers transfer electrons on their path
from one large, less mobile protein complex to another.
27.85 ATP and reducing agents such as NADH and FADH2,
which are generated by the citric acid cycle, are needed for
biosynthetic pathways.
27.87 Biosynthetic pathways are likely to feature reduction
reactions because their net effect is to reverse catabolism,
which is oxidative.
27.89 ATP is not stored in the body. It is hydrolyzed to provide energy for many different kinds of processes and thus
turns over rapidly.
27.91 The citric acid cycle generates NADH and FADH2,
which are linked to oxygen by the electron transport chain.
Chapter 28 Specific Catabolic Pathways:
Carbohydrate, Lipid, and Protein Metabolism
28.1 According to Table 28.2, the ATP yield from stearic acid
is 146 ATP. This makes 146/18 5 8.1 ATP/carbon atom. For
lauric acid (C12):
Step 햲 Activation
Step 햳 Dehydrogenation five times
Step 햴 Dehydrogenation five times
Six C2 fragments in common pathway
–2 ATP
10 ATP
15 ATP
72 ATP
Total
95 ATP
95/12 5 7.9 ATP per carbon atom for lauric acid. Thus stearic
acid yields more ATP/C atom.
28.3 They serve as building blocks for the synthesis of proteins.
28.5 The two C3 fragments are in equilibrium. As the glyceraldehyde phosphate is used up, the equilibrium shifts and
converts the other C3 fragment (dihydroxyacetone phosphate)
to glyceraldehyde phosphate.
28.7 (a) Steps 햲 and 햴 (b) Steps 햷 and 햺
28.9 ATP inhibition takes place at Step 햺. It inhibits the
pyruvate kinase by feedback regulation.
28.11 NADPH is the compound in question.
28.13 Each mole of glucose produces two moles of lactate, so
three moles of glucose give rise to six moles of lactate.
28.15 According to Table 28.1, two moles of ATP are produced
directly in the cytoplasm.
28.17 Two net ATP molecules are produced in both cases.
28.19 Enzymes that catalyze the phosphorylation of substrates using ATP are called kinases. Therefore, the enzyme
that transforms glycerol to glycerol 1-phosphate is called
glycerol kinase.
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28.21 (a) The two enzymes are thiokinase and thiolase.
(b) “Thio” refers to the presence of a sulfur-containing group,
such as i SH. (c) Both enzymes insert a CoAi SH into a
compound.
28.23 Each turn of fatty acid b-oxidation yields one acetyl CoA, one FADH2, and one NADH. After three turns,
CH3(CH2)4COi CoA remains from the original lauric acid;
three acetyl CoA, three FADH2, and three NADH 1 H+ are
produced.
28.25 Using data from Table 28.2, we obtain a figure of
112 moles of ATP for each mole of myristic acid.
28.27 The body preferentially uses carbohydrates as an
energy source.
28.29 (a) The transformation of acetoacetate to b-hydroxybutyrate is a reduction reaction. (b) Acetone is produced by
decarboxylation of acetoacetate.
28.31 It enters the citric acid cycle.
28.33 Oxidative deamination of alanine to pyruvate:
CH3 9 CH 9 COO NAD H2O
NH3
CH3 9 C 9 COO NADH H NH4
O
28.35 One of the nitrogens comes from ammonium ion
through the intermediate carbamoyl phosphate. The other
nitrogen comes from aspartate.
28.37 (a) The toxic product is ammonium ion. (b) The body
gets rid of it by converting it to urea.
28.39 Tyrosine is considered a glucogenic amino acid because
pyruvate can be converted to glucose when the body needs it.
28.41 It is stored in ferritin and reused.
28.43 Muscle cramps come from lactic acid accumulation.
28.45 The bicarbonate/carbonic acid buffer counteracts the
acidic effects of ketone bodies.
28.47 The reaction is a transamination:
COO
C"O
CH2 9 CH 9 COO CH2
NH3
CH2
COO
Phenylalanine
a-Ketoglutarate
COO
CH 9 NH3
CH2 9 C 9 COO CH2
O
CH2
COO
Phenylpyruvate
Glutamate
28.49 Black and blue are due to the hemoglobin in congealed
blood, green to biliverdin, and yellow to bilirubin.
28.51 Production of ethanol in yeast takes place as a result
of glycolysis, giving a net yield of two ATP molecules for each
mole of glucose metabolized.
11/22/08 1:19:26 AM
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28.53 Glucose can be converted to ribose by the pentose
phosphate pathway.
28.55 The step in glycolysis in which a phosphate group is
transferred from phosphoenolpyruvate (PEP) to ADP to produce ATP indicates that the energy of the phosphate group in
PEP is higher than that in ATP.
28.57 Carbamoyl phosphate has an amide group and a
phosphate group.
28.59 Pyruvate can be converted to oxaloacetate.
28.61 Table 28.1 takes into account the fact that glucose can
be metabolized further by the citric acid cycle, which produces
NADH and FADH2. These coenzymes pass electrons to oxygen,
giving rise to ATP in the process.
28.63 Lactate plays a key role in regenerating NAD+.
28.65 Amino acids can be catabolized to yield energy, but
usually only under starvation conditions.
28.67 Catabolism, oxidative, energy-yielding; anabolism,
reductive, energy-requiring.
28.69 If you look at the balanced chemical equations for the
two processes, they are the exact opposite of each other. They
differ in that photosynthesis requires energy from the sun and
occurs only in some organisms such as plants, whereas aerobic
catabolism of glucose releases energy and occurs in organisms
of all sorts.
28.71 Sugars are already partially oxidized, so their pathway of complete oxidation is further advanced, producing less
energy.
28.73 The reactions of glycolysis take place in the cytosol.
Because of their charge the compounds that form a part of the
pathway are not as prone to crossing the cell membrane to the
exterior as they would be if they were uncharged. The reactions
of the citric acid cycle take place in mitochondria, which have
double membrane. The intermediates of the citric acid cycle
tend to stay within mitochondria even without a charge.
28.75 ATP production takes place in connection with the
re-oxidation of the NADH and FADH2 produced in the citric
acid cycle.
29.11 Maltose is a disaccharide that is composed of two glucose units linked by an a-1,4-glycosidic bond.
UDP-glucose 1 glucose h maltose 1 UDP
29.13 UTP consists of uracil, ribose, and three phosphates.
29.15 (a) Fatty acid biosynthesis occurs primarily in the
cytoplasm. (b) No, fatty acid degradation occurs in the
mitochondrial matrix.
29.17 In fatty acid biosynthesis, a three-carbon compound,
malonyl ACP, is repeatedly added to the synthase.
29.19 Carbon dioxide is released from malonyl ACP, leading
to the addition of two carbons to the growing fatty acid chain.
29.21 It is an oxidation step because the substrate is oxidized
with concomitant removal of hydrogen. The oxidizing agent
is O2. NADPH is also oxidized during this step.
29.23 NADPH is bulkier than NADH because of its extra
phosphate group; it also has two more negative charges.
29.25 No, the body makes other unsaturated fatty acids, such
as oleic acid and arachidonic acid.
29.27 The activated components needed are sphingosine, acyl
CoA, and UDP-glucose.
29.29 All of the carbons in cholesterol originate in acetyl CoA.
A C5 fragment called isopentenyl pyrophosphate is an important intermediate in steroid biosynthesis.
3 Acetyl-CoA h mevalonate
C6
C2
h isopentenyl pyrophosphate 1 CO2
C5
29.31 The amino acid product is aspartic acid.
29.33 The products of the transamination reaction shown are
valine and a-ketoglutarate.
O
'
(CH3)2CH ! C ! COO OOC ! CH2 ! CH2 ! CH! COO
s
NH3
The keto form
of valine
Glutamate
O
'
(CH3)2CH ! CH ! COO OOC ! CH2 ! CH2 ! C ! COO
s
NH3
Chapter 29 Biosynthetic Pathways
29.1 Different pathways allow for flexibility and overcome
unfavorable equilibria. Separate control of anabolism and
catabolism becomes possible.
29.3 The main biosynthesis of glycogen does not use inorganic
phosphate because the presence of a large inorganic phosphate
pool would shift the reaction to the degradation process such
that no substantial amount of glycogen would be synthesized.
29.5 Photosynthesis is the reverse of respiration:
6CO 2 1 6H 2O h C 6H 12O 6 16O 2
C 6H 12O 6 1 6O 2 h 6CO 2 1 6H 2O
Photosynthesis
Respiration
29.7 A compound that can be used for gluconeogenesis:
(a) from glycolysis: pyruvate
(b) from the citric acid cycle: oxaloacetate
(c) from amino acid oxidation: alanine
29.9 Glucose needs for the brain are met by gluconeogenesis,
because the other pathways metabolize glucose, and only gluconeogenesis manufactures it.
91123_00_Ans_pA8-A63 pp2.indd 59
Valine
a-Ketoglutarate
29.35 NADPH is the reducing agent in the process of carbon
dioxide being incorporated into carbohydrates.
29.37 Acetyl-CoA carboxylase (ACC) is a key enzyme in fatty
acid biosynthesis. It exists in two forms, one found in liver and
one in muscle tissue. The one found in muscle affects weight
loss and may become a target for anti-obesity drugs.
29.39 The bonds that connect the nitrogen bases to the
ribose units are b-N-glycosidic bonds just like those found in
nucleotides.
29.41 The amino acid produced by this transamination is
phenylalanine.
29.43 The structure of a lecithin (phosphatidyl choline) is
shown in Section 21.6. Synthesis of a molecule of this sort
requires activated glycerol, two activated fatty acids, and activated choline. Each activation requires one ATP molecule, for a
total number of four ATP molecules.
11/22/08 1:19:27 AM
Answers
29.45 The compound that reacts with glutamate in a transamination reaction to form serine is 3-hydroxypyruvate. The
reverse of the reaction is shown below:
COO
COO
s
s
CH ! NH3 C " O
s
s
CH2OH
CH2
s
CH2
s
COO
Serine
a-Ketoglutarate
COO
COO
s
s
C " O CH ! NH3
s
s
CH2OH
CH2
s
CH2
s
COO
3-Hydroxypyruvate
Glutamate
29.47 HMG-CoA is hydroxymethylglutaryl CoA. Its structure
is shown in Section 29.4. Carbon 1 is the carbonyl group linked
to the thio group of CoA.
29.49 Heme is a porphyrin ring with iron at the center. Chlorophyll is a porphyrin ring with magnesium at the center.
29.51 Fatty acid biosynthesis takes place in the cytoplasm,
requires NADPH, and uses malonyl CoA. Fatty acid catabolism
takes place in the mitochondrial matrix, produces NADH and
FADH2, and has no requirement for malonyl CoA.
29.53 Photosynthesis has high requirements for light energy
from the Sun.
29.55 Lack of essential amino acids would hinder the synthesis of the protein part. Gluconeogenesis can produce sugars
even under starvation conditions.
29.57 Separation of catabolic and anabolic pathways allows
for greater efficiency, especially in control of the pathways.
29.59 If laboratory rats are fed all the amino acids but minus
one of the essential ones, they will be unable to synthesize
protein. Administering the essential amino acid later will not
be useful because the other amino acids have already been
metabolized.
Chapter 30 Nutrition
30.1 No, nutrient requirements vary from person to person.
30.3 Sodium benzoate is not catabolized by the body; therefore, it does not comply with the definition of a nutrient—
components of food that provide growth, replacement, and
energy. Calcium propionate enters mainstream metabolism by
conversion to succinyl-CoA and catabolism by the citric acid
cycle and thus is a nutrient.
30.5 The Nutrition Facts label found on all foods must list the
percentage of Daily Values for four important nutrients: vitamins A and C, calcium, and iron.
30.7 Chemically, fiber is cellulose, a polysaccharide that
cannot be degraded by humans. It is important for proper
operation of dietary processes, especially in the colon.
30.9 The basal caloric requirement is calculated assuming the
body is completely at rest. Because most of us perform some
activity, we need more calories than this basic minimum.
30.11 1833 Cal
30.13 No. Using diuretics would be a temporary fix at best.
30.15 The product would be different-sized oligosaccharide
fragments much smaller than the original amylose molecules.
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30.17 No. Dietary maltose, the disaccharide composed of
glucose units linked by an a-1,4-glycosidic bond, is rapidly
hydrolyzed in the stomach and small intestines. By the time it
reaches the blood, it is the monosaccharide glucose.
30.19 linoleic acid
30.21 No. Lipases degrade neither; they degrade triacylglycerols.
30.23 Yes, it is possible for a vegetarian to obtain a sufficient
supply of adequate proteins; however, the person must be very
knowledgeable about the amino acid content of vegetables, so
as to allow for protein complementation.
30.25 Dietary proteins begin degradation in the stomach,
which contains HCl in a concentration of about 0.5%. Trypsin
is a protease present in the small intestines that continues
protein digestion after the stomach. Stomach HCl denatures
dietary protein and causes somewhat random hydrolysis of
the amide bonds in the protein. Fragments of the protein are
produced. Trypsin catalyzes hydrolysis of peptide bonds only on
the carboxyl side of the amino acids Arg and Lys.
30.27 It is expected that many of the prisoners will develop
deficiency diseases in the near future.
30.29 Limes provided sailors with a supply of vitamin C to
prevent scurvy.
30.31 Vitamin K is essential for proper blood clotting.
30.33 The only disease that has been proven scientifically to
be prevented by vitamin C is scurvy.
30.35 Vitamins E and C and the carotenoids may have significant effects on respiratory health. This may be due to their
activity as antioxidants.
30.37 There is a sulfur atom in biotin and in vitamin B1 (also
called thiamine).
30.39 The original Food Guide Pyramid did not consider the
differences between types of nutrients. It assumed that all fats
were to be limited and that all carbohydrates were healthy. The
new guidelines recognizes that polyunsaturated fats are necessary and that carbohydrates from whole grains are better for
you than those from refined sources. The new pyramid also recognizes the importance of exercise, which the original did not.
30.41 All proteins, carbohydrates, and fats in excess have
metabolic pathways that lead to increased levels of fatty acids.
However, there is no pathway that allows fats to generate a
net surplus of carbohydrates. Thus fat stores cannot be used to
make carbohydrates when a person’s blood glucose is low.
30.43 All effective weight loss is based on increasing activity
while limiting caloric intake. However, it is more effective to
concentrate on increasing activity than on limiting intake.
30.45 Theoretically speaking, if humans had the glyoxylate
pathway, dieting would be easier. By eliminating the two decarboxylation steps of the citric acid cycle, there is no loss of carbon from the acetyl-CoA. Therefore, carbon compounds could be
removed from the pathway to form glucose. A person could diet
and use fat stores to power the body’s systems and maintain
blood glucose levels.
30.47 (a) Most studies show that the artificial sweeteners
Sucralose and acesulfame-K are not metabolized in any measurable amounts. (b) Digestion of aspartame can lead to high
levels of phenylalanine.
30.49 Iron is an important cofactor in many biological compounds. The most obvious is the part iron plays in hemoglobin.
It is the iron that directly binds the oxygen that is the source
of respiration for our metabolism. Iron must be consumed in
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the diet to maintain iron levels for hemoglobin and many other
compounds.
30.51 Factors that affect absorption include the solubility of
the compound of iron, the presence of antacids in the digestive
tract, and the source of the iron.
30.53 arginine
30.55 Carbohydrate loading before the event and consuming
carbohydrates during the event
30.57 Caffeine acts as a central nervous system stimulant,
which provides a feeling of energy that athletes often enjoy. In
addition, caffeine reduces insulin levels and stimulates oxidation of fatty acids, which would be beneficial to endurance
athletes. However, caffeine is also a diuretic and can lead to
dehydration in long-distance events.
30.59 Cost is the most significant downside to organic food,
as organic food can be up to 100% more expensive than nonorganic. The type of food is also an important consideration as
pesticides or other chemicals are transferred from the food to
the consumer while others are not. For example, if a pesticide
is concentrated in a banana peel, that is not as big a problem
as if it were concentrated in the banana itself. Pesticides or
other chemicals are more hazardous to children and pregnant
women than to others.
30.61 The vitamin pantothenic acid is part of CoA.
(a) Glycolysis: Pyruvate dehydrogenase uses CoA as a coenzyme.
(b) Fatty acid synthesis: The first step involves the enzyme
fatty acid synthase.
30.63 Proteins that are ingested in the diet are degraded to
free amino acids, which are then used to build proteins that
carry out specific functions. Two very important functions
include structural integrity and biological catalysis. Our proteins are constantly being turned over—that is, continuously
being degraded and rebuilt using free amino acids.
30.65 The very tip of the Food Guide Pyramid displays fats,
oils, and sweets, with the cautionary statement, “Use sparingly.” We can omit sweets completely from the diet; however,
complete omission of fats and oils is dangerous. We must have
dietary fats and oils that contain the two essential fatty acids.
The essential fatty acids may be present as components in
other food groups—that is, the meat, poultry, and fish group.
30.67 Walnuts are not just a tasty snack—they are a healthy
one. Walnuts have protein. In fact, nuts are included in a group
of the U.S. Department of Agriculture’s Food Guide Pyramid.
Walnuts are also a good source of vitamins and minerals,
including vitamins E and B, biotin, potassium, magnesium,
phosphorus, zinc, and manganese.
30.69 No, the lecithin is degraded in the stomach and intestines long before it could get into the blood. The phosphoglyceride is degraded to fatty acids, glycerol, and choline, which are
absorbed through the intestinal walls.
30.71 Patients who have undergone ulcer surgery are administered digestive enzymes that may have been lost during the
procedure. The enzyme supplement should contain proteases
to help break down proteins as well as lipases to assist in fat
digestion.
Chapter 31 Immunochemistry
31.1 Examples of external innate immunity include action by
the skin, tears, and mucus.
31.3 The skin fights infection by providing a barrier against
penetration of pathogens. The skin also secretes lactic acid and
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fatty acids, both of which create a low pH, thereby inhibiting
bacterial growth.
31.5 Innate immunity processes have little ability to change
in response to immune dangers. The key features of adaptive
(acquired) immunity are specificity and memory. The acquired
immune system uses antibody molecules designed for each type of
invader. In a second encounter with the same danger, the response
is more rapid and more prolonged than the first response.
31.7 T cells originate in the bone marrow, but grow and
develop in the thymus gland. B cells originate and grow in the
bone marrow.
31.9 Macrophages are the first cells in the blood that encounter potential threats to the system. They attack virtually
anything that is not recognized as part of the body, including
pathogens, cancer cells, and damaged tissue. Macrophages
engulf an invading bacterium or virus and kill it with (NO),
nitric oxide, and then digest it.
31.11 protein-based antigens
31.13 Class II MHC molecules pick up damaged antigens.
A targeted antigen is first processed in lysosomes, where it is
degraded by proteolytic enzymes. An enzyme, GILT, reduces
the disulfide bridges of the antigen. The reduced peptide antigens unfold and are further degraded by proteases. The peptide
fragments remaining serve as epitopes that are recognized by
class II MHC molecules.
31.15 MHC molecules are transmembrane proteins that
belong to the immunoglobulin superfamily. They are originally
present inside cells until they become associated with antigens
and move to the surface membrane.
31.17 If we assume that the rabbit has never been exposed to
the antigen, the response will occur 1–2 weeks after the injection of antigen.
31.19 (a) IgE molecules have a carbohydrate content of
10–12%, which is equal to that of IgM molecules. IgE molecules
have the lowest concentration in the blood. The blood concentration of IgE is about 0.01–0.1 mg/100 mL of blood.
(b) IgE molecules are involved in the effects of hay fever and
other allergies. They also offer protection against parasites.
31.21 The two Fab fragments would be able to bind to
an antigen. These fragments contain the variable protein
sequence regions and hence are able to change during
synthesis against a specific antigen.
31.23 Immunoglobulin superfamily refers to all of the
proteins that have the standard structure of a heavy chain and
a light chain.
31.25 Antibodies and antigens are held together by weak,
noncovalent interactions: hydrogen bonds, electrostatic interactions (dipole–dipole), and hydrophobic interactions.
31.27 The DNA for the immunoglobulin superfamily has
multiple ways of recombining during cell development. The
diversity is a reflection of the number of permutations and
ways of combining various constant regions, variable regions,
joining regions, and diversity regions.
31.29 T cells carry on their surfaces unique receptor proteins
that are specific for antigens. These receptors (TcR), which are
members of the immunoglobulin superfamily, have constant
and variable regions. They are anchored in the T-cell membrane by hydrophobic interactions. They are not able to bind
antigens alone, but rather need additional protein molecules
called cluster determinants that act as coreceptors. When TcR
molecules combine with cluster determinant proteins, they
form T-cell-receptor complexes (TcR complexes).
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31.31 The components of the TcR complexes are (1) accessory
protein molecules called cluster determinants and (2) the T-cell
receptor.
31.33 CD4
31.35 They are adhesion molecules that help dock antigenpresenting cells to T cells. They also act as signal transducers.
31.37 Cytokines are glycoproteins that interact with cytokine
receptors on macrophages, B cells, and T cells. They do not
recognize and bind antigens.
31.39 Chemokines are a class of cytokines that send messages between cells. They attract leukocytes to the site of
injury and bind to specific receptors on the leukocytes.
31.41 All chemokines are low-molecular-weight proteins that
have four cysteine residues that are linked in very specific
disulfide bonds: Cys1iCys3 and Cys2iCys4.
31.43 The T cells mature in the thymus gland. During
maturation, those cells that fail to interact with MHC and
thus cannot respond to foreign antigens are eliminated by a
special selection process. T cells that express receptors that
may interact with normal self antigens are eliminated by the
same selection process.
31.45 A signaling pathway that controls the maturation of
B cells is the phosphorylation pathway activated by tyrosine
kinase and deactivated by phosphatase.
31.47 the cytokines and chemokines
31.49 helper T cells
31.51 It is hard to find because the virus mutates quickly.
Also, one of its docking proteins changes conformation when it
docks, so that antibodies elicited against undocked proteins are
ineffective. It binds to several proteins that inhibit antiviral
factors, and cloaks its outer membrane with sugars that are
very similar to the natural sugars found on host cells.
31.53 Vaccines rely on the immune system’s ability to recognize a foreign molecule and make specific antibodies to it.
HIV hides from the immune system in a variety of ways, and
it changes often. The body makes antibodies, but they are not
very effective at finding or neutralizing the virus.
31.55 It had been known since the 1800s that the mayapple
had anticancer properties. It was later found that a chemical
found in the mayapple, picropodophyllin, inhibits spindle formation during mitosis in dividing cells. As all chemotherapy
agents do, they hinder rapidly dividing cells, like cancer cells,
more then regular cells.
31.57 Most cancer cells have specific proteins on their surface
that help allow their identification as cancerous. Monoclonal
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antibodies are very specific for the molecules they will bind to,
making them an excellent choice for a weapon against cancer.
The antibodies will attack the cancer cell and only the cancer
cell if the monoclonal antibody is specific enough.
31.59 Fluorescence labeling studies show that breast cancer
cells have elevated levels of the HER2 protein. In addition,
drugs designed to attack HER2 are very successful at targeting
breast cancer cells.
31.61 Many cancers are linked to dimerization of specific
cell receptors. Tyrosine kinase is a type of cell receptor that
functions via dimerization. Specific monoclonal antibodies
are being designed to block the dimerization of these tyrosine
kinases.
31.63 Jenner noticed that milkmaids, who would often be
exposed to cowpox, rarely if ever got smallpox.
31.65 Allergies to antibiotics can be very potent. A person
may show no symptoms with the first exposure, but a second or
third may produce severe reactions or even be fatal.
31.67 Sex workers in some countries use constant low doses
of antibiotics in an attempt to avoid sexually transmitted
diseases. The unfortunate side effect of this practice has been
to allow the evolution of strains of gonorrhea that are antibiotic resistant.
31.69 One of the molecules on the streptococcus bacteria
resembles a protein found in the valves of the heart. The body’s
attempt to fight strep throat can lead to antibodies that attack
not only the bacteria but also the person’s own heart valves.
This is the danger in rheumatic fever.
31.71 Stem cells can be transformed into other cell types.
Scientists are working to find ways to use stem cells to repair
damaged nerve tissue or brain tissue. In some animal models,
brain cell function has been restored after a stroke by adding
stem cells to the brain in the area of the damage.
31.73 IgA molecules are the first line of defense since they
are found in tears and mucous secretions. They can intercept
invaders before they get into the bloodstream.
31.75 Chemokines (or, more generally, cytokines) help
leukocytes migrate out of a blood vessel to the site of injury.
Cytokines help the proliferation of leukocytes.
31.77 A compound called 12:13 dEpoB, a derivative of
epothilon B, is being studied as an anticancer vaccine.
31.79 Tumor necrosis factor receptors are located on the surfaces of several cell types, but especially on tumor cells.
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