Chemistry, Student Solutions Manual
Chapter 10
Chapter 10 Organic Chemistry
Solutions to Problems in Chapter 10
10.1
sp2
sp2
All other carbon atoms sp3
10.3 (a) 2-Methylpentane
(b) 2,3-Dimethylbutane
(c) 2-Methylpentane
(d) 2,4-Dimethylpentane
(e) 3-Methylpentane
10.5
a)
b)
c)
d)
e)
10.7 (a) o-Bromonitrobenzene
(b) p-Dicyclohexylbenzene
(c) m-Dinitrobenzene
10.9
a)
CH3
b)
c)
CH3
d)
CH3
Cl
e)
CH3
O2N
NO2
CH3
NO2
Cl
Cl
Cl
Cl
Cl
Cl
10.11
(a) 1-Chloropropane (or n-propyl chloride)
(b) 2-Chlorobutane
(c) 2-Chloro-3,3-dimethylbutane
(d) 2-Chloro-3-methylbutane
(e) 3-Chloro-3-methylheptane
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10.13 (a) Secondary 2°
(b) Primary 1°
(c) Secondary 2°
(d) Tertiary 3°
(e) Secondary 2°
10.15
OH
a)
OH
OH
b)
OH
c)
d)
OH
e)
OH
10.17 (a) Tertiary 3°
(b) Primary 1°
(c) Primary 1°
(d) Secondary 2°
(e) Tertiary 3°
10.19 (a) Cyclohexanone
(b) Butanal
(c) Ethyl butanoate
(d) Butanoic acid
(e) Butanamide
10.21 (a) No E/Z isomerization
(b) No E/Z isomerization
(c) E-2-pentene Z-2-pentene
(d) No E/Z isomerization
(e) No E/Z isomerization
10.23 (a) Achiral
(b) Achiral
(c) Chiral
(d) Chiral
(e) Chiral
10.25
(R)
(S)
NH2
H
N
O
O
(R)
H
S
N
(R)
CO2H
Amoxicillin
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10.27 Good leaving groups are conjugate bases of strong acids (like Cl– and Br–); that is,
weak bases. Poor leaving groups are conjugate bases of weak acids (like OH–); that
is, strong bases.
(a) Br– = good leaving group
(b) CH3–O– = poor leaving group
(c) NH2– = poor leaving group
(d) Cl– = good leaving group
(e) CH3COO– (acetate anion) = good leaving group
10.29
OH + HBr
Br
(a)
NH 2
Br + NH 3
(b)
Cl
OH
+ H2O
(c)
OCH3
I
+ OCH3
(d)
Cl
CN
+ NaCN
(e)
10.31
CH3
CH3
CH3
CH3
Cl
+ Cl2
AlCl3
Cl
Cl
There are three unique mono-chlorination products for toluene
10.33
a)
b)
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c)
d)
e)
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Chapter 10
10.35
H 2O
Br
H
C
H2
CH3
H
H
10.37
(a)
(b)
+ H2(g)
Pt
+ H2(g)
Pt
+ H2(g)
Pt
+ H2(g)
Pt
+ H2(g)
Pt
(c)
(d)
CH3
H
CH3
H
(e)
10.39
Br
H Br
H
H
Br
10.41
H
H
O
(from HCl)
H
H
H
O
H
O
H
H
O
H
H
H
H
O
H
10.43
HBr
(a)
Br
(or Z-isomer)
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H
H
O
H
Chemistry, Student Solutions Manual
Chapter 10
Br
HBr
(b)
Br
HBr
(c)
Br
HBr
(d)
HBr
Br
(e)
10.45 Markovnikov’s rule states that in the addition of HX to an alkene, the hydrogen
atom adds to the carbon atom of the double bond that already has the greater
number of hydrogen atoms. An example is
H
H Br
H
+
(major)
(minor)
H
H
Br
Br
10.47
Cl
Cl
I Cl
I
I
10.49
OH
H2
cat. HCl
H2O
Pt
(a)
HBr
Br
NaOCH3
OCH3
(b)
HBr
(c)
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t-BuOK
Br
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Chapter 10
OH
CN
HCl
NaCN
(d)
Br
NaOEt
HBr
Br
(e)
10.51
Cl
Cl2, AlCl3
(a)
CH3
CH3
HNO3, H2SO4
NO2
(b)
NO2
NO2
CH3–Cl, AlCl3
CH3
(c)
tert-Bu–Cl
FeCl3
(d)
CH3
CH3
H2SO4
(e)
SO3H
10.53 The Pauling electronegativity number for carbon is 2.5 and for hydrogen is 2.1.
This small difference (0.4) means that the two electrons in the C–H bond are
essentially equally shared between the two atoms. The result is a covalent bond that
is essentially non-polar.
10.55 Hexane is an essentially non-polar molecule with relatively weak dispersion forces
as the main intermolecular forces. Water is a highly polar molecule with the
comparatively stronger hydrogen bonding as the main intermolecular force. Nonpolar hexane has no groups that can act as hydrogen-bond donors or acceptors. As a
result, the intermolecular forces holding water molecules together (hydrogen bonds)
are much stronger than any forces between water molecules and hexane molecules,
thus the liquids are immiscible.
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10.57 In CCl4, even thought there is an electronegative chlorine (χ = 3.0) covalently
bonded to a less electronegative carbon (χ = 2.5) the four dipoles cancel each other
out and the net result is a non-polar molecule. In CH2Cl2, there are two non-polar
bonds (C–H) and two polar (C–Cl) bonds. These dipoles do not cancel each other
out and the result is a polar molecule.
10.59 The carbonyl group of acetone would have a dipole, as shown below. The + end
of the dipole is centred on the carbon atom, whereas the – end is centred on the
oxygen atom. Attack by a nucleophile (such as the OH– ion) would occur at the
positive end of the dipole, as shown below.
–
O
H3C
+
CH3
O
H3C
O
CH3
OH–
H3C
CH3
OH
10.61 There are two factors that make acetic acid more acidic than ethanol. First is the
presence of a second electronegative oxygen atom. This helps make the negative
charge that exists on the acetate anion more favourable overall. Second is the fact
that the negative charge on the acetate anion can be delocalized over two oxygen
atoms through a process known as resonance, which helps to stabilize the negative
charge. There is neither an extra electronegative atom nor resonance stabilization to
help make the ethoxide anion more stable.
10.63 The configuration changes only because the atomic number for sulphur is higher
than the atomic number for oxygen. This makes the CH2–SH substituent a higher
priority than the COOH substituent on cystine. On serine, the COOH substituent
has a higher priority (because of two oxygen atoms) than the CH2–OH group. This
change in priority results in a change in the configuration of the two different
chirality centres.
10.65 The two stereoisomers (2R,3R)-2,3-dibromopentane and (2S,3S)-2,3dibromopentane are enantiomers of each other. The two stereoisomers (2R,3S)-2,3dibromopentane and (2S,3R)-2,3-dibromopentane are also enantiomers of each
other. The (2S,3S)-2,3-dibromopentane and the (2R,3R)-2,3-dibromopentane
stereoisomers are diastereomers of the (2R,3S)-2,3-dibromopentane and the
(2S,3R)-2,3-dibromopentane. This is illustrated below.
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Br
Br
Br
(2R,3R)-2,3-dibromopentane
Br
(2R,3S)-2,3-dibromopentane
Diastereomers
Enantiomers
Br
Enantiomers
Br
Br
Br
(2S,3S)-2,3-dibromopentane
(2S,3R)-2,3-dibromopentane
10.67 In the absence of acid, the leaving group on tert-butanol would be a hydroxide ion,
which is the conjugate base of a weak acid (H2O). Hydroxide ion is, therefore, a
very poor leaving group. By adding acid, the first step in the mechanism is to
protonate the oxygen atom of the alcohol. In this case then, the leaving group is a
molecule of water, the conjugate base of a strong acid (H3O+). This makes water a
very good leaving group.
10.69 In the addition of HCl to a double bond, the first step is protonation of the double
bond and the formation of a carbocation on one of the carbon atoms. The presence
of the highly electronegative fluorine atoms strongly disfavours formation of this
carbocation.
10.71 The critical step in the SN1 reaction is the formation of the carbocation
intermediate. This carbocation is formally an sp2-hybridized carbon atom that is
planar, with 120° bond angles and an empty p orbital. The incoming nucleophile
has no preference as to which side of the carbocation to attack to form the new
covalent bond. In this example, the result is an equimolar mixture of the (S)-1bromo-1-phenylethane, the result of attack from the “top” of the carbocation, and
(R)-1-bromo-1-phenylethane from attack at the “bottom” of the carbocation.
10.73 The reactions show that with a small base (i.e., NaOCH3), substitution is the result
via an SN2 mechanism. The larger base (NaO-tert—Bu) results in only the
elimination product being formed. The conclusion is that elimination is favoured by
larger bases.
10.75 The tert-butyl chloride reacts faster, because under SN1 conditions the ratedetermining step is formation of the carbocation. Tertiary carbocations are more
stable than primary carbocations and this is demonstrated by this example. It is
much easier to form a tertiary than a primary carbocation. If you picture each C–H
bond on the adjacent methyl groups as a “cloud” of electrons (and this is a pretty
accurate picture), then each bond is able to donate a little of this cloud into the
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empty p orbital on the carbocation through a process called hyperconjugation.
Branched alkyl groups have a stronger electron donation effect than straight-chain
alkyl groups. It is this slight donation of each of the CH bonds in the alkyl branches
that accounts for the extra stability of the tertiary carbocation.
10.77 The fact that more energy is released when Z-2-butene is hydrogenated must mean
that there was more energy stored in this double bond than in E-2-butene. This leads
to the conclusion that E-alkenes are more stable (i.e., lower in energy) than Zalkenes. The reason for this is that in Z-2-butene, the two methyl groups that are on
the same side of the double bond can interact with each other, raising the energy of
the molecule. In E-2-butene, this unfavourable interaction is removed as the two
methyl groups are as far apart as possible, lowering the overall energy of the
system.
10.79 The cycloheptatrienyl cation is an aromatic system, the same as benzene. This is
demonstrated by the fact that all of the bond lengths are of equal length and the
molecule is planar owing to the presence of the sp2-hybridized positively charged
carbon and the sp2-hybridized carbons of the double bonds.
10.81 The molecular orbital for the π bond would be formed by combination of the two
atomic p orbitals. The “in-phase” overlap would then result in a region of greatest
probability for finding the two electrons in the bonding molecular orbital between
the two carbon atoms. This overlap would exist in a region above and below the
plane of the carbon–carbon sigma bond, preventing the rotation about this bond.
Any rotation out of the plane would result in loss of the favourable atomic orbital
overlap that generates the bonding molecular orbital.
10.83 The enantiomer of D-glucose is L-glucose and is the complete mirror image of the
molecule, as shown below. All of the chiral centres in D-glucose have the (S)
configuration, whereas all of the chiral centres in L-glucose have the (R)
configuration.
H OH
HO H
HO
HO
HO
H
H
OH
1 H
OH
OH
D-Glucose
H 1
HO
HO
OH
OH
H
H
L-Glucose
10.85 In general, an E2 reaction can take place on any substrate with a hydrogen atom on
a carbon atom adjacent to the leaving group. For the E2 reaction, it makes little
difference if the leaving group is on a primary, secondary, or tertiary carbon atom.
However, the SN2 mechanism is favoured by primary and some secondary but never
by a tertiary substrate. Moreover, the size of the base has an effect. Large, strong,
and bulky bases, such as tert-butoxide, favour the elimination pathway. Smaller
bases that are derived from weaker acids favour the substitution pathway.
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10.87 The electronegativity numbers for hydrogen (2.1) and boron (2.0) mean that the
hydrogen atom is more electronegative than the boron atom and that the electrons in
the B–H bond are polarized toward the hydrogen atom. This means that in an
electrophilic addition, the mechanism occurs as shown below, where the boron
actually adds first to the double bond. The second step is the attack of an H– (a
hydride ion) on the carbocation. This mechanism agrees with the definition of
Markovnikov’s rule in Problem 10.86, since the most stable carbocation is still the
intermediate in the reaction.
H
BH2
H
BH2
H
BH2
10.89 In this reaction the relative energy difference between starting material and
product is the same as in Problem 10.88, and there will be a net release of 63
kJ/mol. The major and significant difference is that this reaction proceeds with the
initial formation of a carbocation intermediate. This intermediate will be higher in
energy than either the starting material or the product but will have a measurable
existence (unlike a transition state). Therefore, the diagram has to take this into
account as shown below.
H3C CH3
C
CH3
Relative Energy
63 kJ/mole
C(CH3)3–Br
C(CH3)3–Cl
Reaction Co-ordinate
10.91 This trend is explainable based on the electronegativity of chlorine (3.0) compared
with hydrogen (2.1). The electronegative chlorine atom will help to stabilize the
negative charge that forms on the conjugate base of each of these acids. Each
chlorine induces a dipole in the C–Cl bond that is then transferred through the other
C–C bonds to the oxygen atom that carries the negative charge. The more chlorine
atoms on the acid, the stronger this inductive effect and the more stable the anion,
which results in a stronger base. This is shown below.
O
Cl
Cl
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O
Cl
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10.93 Note: There in fact may be more than one possible synthesis of each product from
the given starting material; the solutions suggested below may not be the only
possible legitimate pathway.
O
H2O (cat. H2SO4)
+
OH
H2SO4 (cat.)
HO
CH3
O
CH3
O
HBr
+
NaOCH2CH3
Br
HBr
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Br
NH3
NH2
191