CHAPTER 15
Practice exercises
15.1
15.3
(a)
(b)
(c)
C9H18: isobutylcyclopentane
C11H22: sec-butylcycloheptane
C6H2: 1-ethyl-1-methylcyclopropane
15.5
(a)
(b)
(c)
3,3-dimethyl-1-pentene
2,3-dimethyl-2-butene
3,3-dimethyl-1-butyne
15.7
(a)
(b)
(c)
(E)-1-chloro-2,3-dimethyl-2-pentene
(Z)-1-bromo-1-chloropropene
(E)-2,3,4-trimethyl-3-heptene
15.9
cis,trans-2,4-heptadiene
15.11
cis,cis-2,4-heptadiene
(a)
2-iodopropane
(b)
1-iodo-1-methylcyclohexane
15.13
CH3
+
CH3
HI
I
Step 1: Protonation of the alkene to give the most stable 3˚ carbocation intermediate:
slow step
CH3
H
+
I
rate-determining
step
CH3
I
Step 2: Nucleophilic attack of the iodide anion on the 3˚ carbocation intermediate to give
the product:
+
CH3
I
CH3
I
15.15
CH3
H2O
CH3
H3O+
OH
Step 1: Protonation of the alkene to give the most stable 3˚ carbocation intermediate:
CH3
H
O+ H
H
slow step
rate-determining
step
+
H
CH3
Step 2: Nucleophilic attack of the water on the 3˚ carbocation intermediate to give the
protonated alcohol:
O
H
H
+
CH3
+
H
O H
H
CH3
O
Step 3: The protonated alcohol loses a proton to form the product:
H
O
+
CH3
15.17
(a)
(b)
(c)
H
H
CH3
O
H
+ H 3O +
OH
2-phenyl-2-propanol
(E)-3,4-diphenyl-3-hexene
3-methylbenzoic acid or m-methylbenzoic acid
Review questions
15.1
A hydrocarbon is a compound composed only of hydrogen and carbon atoms.
15.3
In saturated hydrocarbons, each carbon is bonded to four other atoms, either hydrogen or
carbon atoms. Unsaturated hydrocarbons have carbon atoms that have a double or triple
bond to another carbon atom.
15.5
(a)
(b)
(c)
(d)
(e)
(f)
15.7
(CH2)2CH3
(a) CH3(CH2)4CH(CH3)2
(b) CH3(CH2)2CH(CH2)2CH3
(CH2)2CH3
(c)
CH3(CH2)2C(CH2)4CH3
CH3
15.9
(a)
(b)
(c)
(d)
(e)
(f)
different compounds
constitutional isomers
constitutional isomers
different compounds
constitutional isomers
constitutional isomers
15.11
(a)
(b)
(c)
(d)
(e)
(f)
2-methylpentane
2,5-dimethylhexane
3-ethyloctane
2,2,3-trimethylbutane
isobutylcyclopentane
1-tert-butyl-2,4-dimethyl-cyclohexane
15.13
(a)
1,3-dimethylbutane
The longest chain is pentane.
The IUPAC name is 2-methylpentane.
(b)
4-methylpentane
The pentane chain is numbered incorrectly.
The IUPAC name is 2-methylpentane.
(c)
2,2-diethylbutane
The longest chain in pentane.
The IUPAC name is 3-ethyl-3-methylpentane.
(d)
2-ethyl-3-methylpentane
The longest chain is hexane.
The IUPAC name is 3,4-dimethylhexane.
(e)
2-propylpentane
The longest chain is heptane.
The IUPAC name is 4-methylheptane.
(f)
2,2-diethylheptane
The longest chain is octane.
The IUPAC name is 3-ethyl-3-methyloctane.
(g)
2,2-dimethylcyclopropane
(h)
1-ethyl-5-methylcyclohexane
The ring is numbered incorrectly.
The IUPAC name is 1,1-dimethylcyclopropane.
The ring is numbered incorrectly.
The IUPAC name is
1-ethyl-3-methylcyclohexane.
15.15
No, because alkanes do not have rings or C—C double bonds and so all conformations
are usually interconvertable by rotation about a C—C single bond. However, there are
some extremely crowded molecules which are locked into specific conformers. These are
called ‘conformational isomers’ or ‘locked conformational isomers’ and are not correctly
defined by cis-trans nomenclature.
15.17
There are two enantiomers of the trans-1,2-dimethylcyclopropane..
CH3
CH2CH3
cyclopentane
methylcyclobutane
ethylcyclopropane
CH3
CH3
CH3
CH3
H
CH3
H
H
H
CH3
1,1-dimethylcyclopropane
15.19
cis-1,2-dimethylcyclopropane
(a) trans-2-methylhex-3-ene
trans-1,2-dimethylcyclopropane
(b) 2-methyl-3-hex-3-yne
(d) 3-ethyl-3-methylpent-1-yne
(e) 2,3-dimethylbut-2-ene
(c) 2-methylbut-1-ene
(f) cis-pent-2-ene
15.21
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
2-isobutylhept-1-ene
1,4,4-trimethylcyclopentene
1,3-cyclopentadiene
3,3-dimethylbut-1-yne
2,4-dimethylpent-2-ene
oct-1-yne
2,2,5-dimethylhex-3-yne
3-methylpent-1-yne
15.23
(a)
(b)
(c)
(c)
2,2-dimethylhex-3-yne
oct-2,5-diyne
3,6-dimethylhept-2-ene-4-yne
hept-1,4-diyne
15.25
(a)
Correct name: but-2-ene.You must select the longest carbon chain containing the
‘ene’ functional group.
3
1
4
2
but-2-ene
(b)
Not:
2
H3C
3
1
1-methylpropene
Correct name: pent-2-ene. You must number the chain to give the first carbon of
the double bond the lowest possible number.
5
Not:
2
3
5
pent-3-ene
Correct name: 1-methylcyclohexene. By default, the first carbon of the double
bond in a ring is given the number 1. In this, case this can also be the ring carbon
with the methyl group attached.
1
Not:
2
2
1
2-methylcyclohexene
1-methylcyclohexene
(d)
This name is not incorrect even though there is no indication of where the double
bond is positioned. The indication of the two methyls on the third carbon means
that the double bond is attached to the first carbon. Technically, the correct name
is 3,3-dimethylpent-1-ene.
(e)
Correct name: 4-hexyne. As with (b) the chain carbons are numbered to give the
first carbon of the double bond the lowest possible number.
7
6
5
4
CH3CH2CH2C
(f)
4
3
(a)
(b)
(c)
(d)
(e)
(f)
(b)
(c)
3 2
1
CCH2CH3
Correct name: 2-isopropyl-2-butene. The longest chain containing the function
group is a pentene.
5
15.27
1
3
1
pent-2-ene
(c)
4
2
4
No
Yes
Yes
No
Yes
No
1
2
(e)
15.29
For alkenes to exist as a pair of cis-trans isomers, both carbons of the double bond must
have two different substituents. Thus, only (b) and (d) can exist as a cis or trans isomer.
(b)
H3C
H
Br
H3C
H
cis-bromoprop-1-ene
H
H
Br
trans-bromoprop-1-ene
(d)
15.31
Molecules (a) and (c) do not show cis-trans isomerism.
(b)
15.33
(a)
(d)
Alkenes that do not show cis-trans isomerism are:
pent-1-ene
(b)
2-methylbut-2-ene
3-methylbut-1-ene
Alkenes that do show cis-trans isomerism are:
trans-pent-2-ene
cis-pent-2-ene
2-methylbut-1-ene
(c)
Cycloalkanes that do not show cis-trans isomerism are:
CH2CH3
CH3
methylcyclobutane
cyclopentane
(d)
ethylcyclopropane
H3 C
CH3
H
H
+
CH 3 C HCH 3 is more stable.
CH3
(b)
CH 3 C CH 2 CH 3 is more stable.
+
15.37
CH3
CH3C=CH2
+
6O2
4CO2 +
15.39
H2
(a)
H2C CH2
CH3CH3
Ni
H2 O
(b)
1,1-dimethyl
cyclopropane
H2C CH2
CH3CH2OH
H2SO4
(c)
H2C=CH2
(d)
H2C CH2
(e)
H2C CH 2
CH3
H
H
trans-1,2-dimethylcyclopropane
cis-1,2-dimethylcyclopropane
(a)
CH3
Cycloalkanes that do show cis-trans isomerism are:
H 3C
15.35
H3C
CH3CH2Br
Br2
BrCH2CH2Br
HCl
CH3CH2Cl
4H2O
15.41
H2 / Pd/C
H2
H2
Lindlar
catalyst
Na/NH3
Br2
Br
Br
15.43
15.45
(a)
1-bromo-2-chloro-4-ethylbenzene
(b)
4-iodo-1,2-dimethylbenzene
Br
Cl
I
(c)
2,4,6-trinitrotoluene
O2N
(d)
4-phenylpentan-2-ol
NO2
OH
NO2
(e)
p-cresol
(f)
2,4-dichlorophenol
OH
OH
Cl
Cl
(g)
1-phenylcyclopropanol
(h)
styrene (phenylethene)
(j)
2,4-dibromoaniline
OH
(i)
m-bromophenol
NH2
Br
OH
Br
Br
(k)
isobutylbenzene
(l)
m-xylene
15.47
isopropylbenzene
Cl
OH
(1)
/AlCl3
(2)
/H2SO4
(3)
/H2SO4
Review problems
15.49
(1) Alkanes are less dense than water.
(2) As alkane molar mass increases, density increases.
(3) Constitutional isomers have similar densities.
15.51
Boiling points of unbranched alkanes are related to their surface area: the larger the
surface area, the greater the strength of the dispersion forces and the higher the boiling
point. The relative increase in molecular size per CH2 group is greatest between CH4 and
CH3—CH3 and becomes progressively smaller as molecular mass increases. Therefore,
the increase in boiling point per CH2 group is greatest between CH4 and CH3—CH3, and
becomes progressively smaller for the higher alkanes.
15.53
(a)
(b)
(c)
(d)
(e)
15.55
The three structures with molecular formula C2H2Br are 1,1-dibromoethane, cis-1,2dibromoethane and trans-1,2-dibromoethane. The dipole moment of the C—Br bond is
shown in the structures below by a solid arrow. trans-1,2-dibromoethane has no dipole
moment because the two C—Br dipoles point in opposite directions and therefore cancel
each other out. However, 1,1-dibromoethane and cis-1,2-dibromoethane both have a net
dipole moment (in the direction shown by the dotted arrow) since the two C—Br dipoles
do not point in opposite directions.
+
No
Yes
Yes
It is a liquid.
It is less dense than water.
Br
Br
Br
+
C C
H
C C
H
1,1-dibromoethane
15.57
(a)
H
Br
Br
H
C C
H
cis-1,2-dibromoethane
H
Br
No dipole moment
trans-1,2-dibromoethane
CH3
CH3 CH
CH3
C C
H
H
cis-4-methylpent-2-ene
(b)
(c)
15.59
15.61
(a)
2,2-dichlorobutane
CH3
CH3 CH
H
C C
H
CH3
trans-4-methylpent-2-ene
CH3
CH3CHCH2CH CH2
4-methylpent-1-ene
15.63
(b)
butan-2-ol
(c)
2,2,3,3-tetrabromobutane
A hydrocarbon of formula C5H8 must have a triple bond, or two double bonds, or a
combination of a double bond and a ring. Since it only reacts with one mole of H2
indicates that the compound is an alkyne. Lindlar’s catalyst gives a cis alkene, which in
turn is brominated to give a dibromoalkane which has two chiral carbons. Only the final
possibility shown below meets this last criterion.
H
H
Br
H
Br
H
H
OR
H
H
Br
H
Br
H
H
OR
H
Br
Br
H
H
H
15.65
(a)
The formation of the more stable tertiary carbocation intermediate proceeds at a
faster rate, therefore 2-methylbut-2-ene reacts faster than trans-but-2-ene.
I
+
trans-but-2-ene
secondary
carbocation
2-iodobutane
I
+
2-methylbut-2-ene
(b)
tertiary
carbocation
2-iodo-2-methyl
butane
(major product)
The formation of the more stable tertiary carbocation intermediate proceeds at a
faster rate, therefore 1-methylcyclohexene reacts faster than cyclohexene.
I
+
secondary
carbocation
cyclohexene
CH3
+
iodocyclohexane
CH3
CH3
I
1-methylcyclohexene
15.67
tertiary
carbocation
1-iodo-1-methyl
cyclohexane
(major product)
The first step in the reaction is the protonation of the alkene to generate a carbocation.
The reaction path that produces the more stable carbocation occurs at a faster rate, thus
producing the observed regioselectivity.
CH3
HI
(a) CH3C CHCH2CH3
CH3
CH3CCH2CH2CH3
I
CH3
H 2O
(b) CH3C CHCH2CH3
CH3
CH3CCH2CH2CH3
H2SO4
OH
15.69
(a)
(b)
(c)
15.71
Chlorine undergoes an anti-addition to alkenes.
Cl
Cl2
(a)
Cl2
(b)
Cl
CH3
(c)
15.73
H3 C
Cl
Cl
Cl
Cl2
Cl
CH2
(d)
Cl
Cl2
Cl
Both cis- and trans-hex-3-ene yield hexan-3-ol upon acid-catalysed hydration because
they both form the same carbocation intermediate that leads to hexan-3-ol according to
the partial mechanism:
H
H
+
O H
OH
+
H
H
+
O H
15.75
(a)
H 2O
H 3O +
OH
CH3
CH3
H2O
OH
(b)
H3
CH3
(c)
CH3
O+
H2O
OH
H3O+
CH3
OH
H2O
(d) H3C
15.77
CH3
H3O+
H3C
Step 1: Protonation of the alkene gives a stable 3˚ carbocation intermediate:
CH3
CH3C
CH2
H
+
O CH3
H
CH3
H3 C C +
CH3
CH3OH
Step 2: Nucleophilic attack of methanol on the carbocation intermediate yields a
protonated ether intermediate.
CH3
H 3C C +
CH3
CH3 CH
+
3
H 3C C O
CH3 H
CH3
O
H
Step 3: Proton loss by the protonated ether intermediate yields the ether and regenerates
the acid catalyst.
CH3
+
CH2
H3 C C O
CH3
H
15.79
CH3
O
H
CH3
H3C C O CH3
CH3
H
H
+
O CH3
The skeleton and location of the double bonds in hydrocarbon A are identified from the
structure of the brominated product:
Br
2 mol Br2
Br
Br
Br
1,2,3,4-tetrabromo-2-methylbutane
Hydrocarbon A (C5H8)
2-methylbuta-1,3-diene
15.81
Br
Br2
(a)
Br
OH
H2O
(b)
H2SO4
Br
HBr
(c)
(d)
H2
Pt
15.83
Br
HBr
(a)
(b)
H2O
or
H2SO4
Br2
Br
(c)
Br
OH
15.85
Resonance accounts for the three equivalent structures of naphthalene:
15.87
(a)
H H
H H
(b)
NH
(c)
O
- 8 π electrons
- Non-aromatic
- 14 π electrons
- Not planar
- Non-aromatic
- 10 π electrons
- Planar
- Aromatic
O
O
H
B
(d)
- 6 π electrons
- Planar
- Aromatic
(e)
(f)
- 6 π electrons
- Not a 2p orbital
on every atom
- Non-aromatic
- 8 π electrons
- Non-aromatic
Although compound (b) contains 14 π electrons (a Hückel number), a closer inspection of
the molecule reveals that the protons inside the ring are held too close together and force
the ring to be non-planar. Molecular modelling of structure (b) confirms its non-planarity
and it therefore non-aromatic.
In compound (c), only one of the electron pairs on oxygen is in a 2p orbital, the other
electron pair is in an sp2 hybridised orbital, perpendicular to the π bonds. The nitrogen
lone pair in compound (c) is also in a 2p orbital for a total of 10 π electrons in a planar
ring making it aromatic.
In compound (d), the boron atom contributes an empty 2p orbital, giving an aromatic
compound with six π electrons in a seven-membered ring with seven 2p orbitals.
Compound (f), is similar to compound (c) where the oxygen has one electron pair in a 2p
orbital and the other in an sp2 hybridised orbital with a total of eight π electrons, giving a
non-aromatic compound.
15.89
Two monochloronaphthalenes are possible when naphthalene is treated with Cl2/AlCl3:
Cl
Cl2
AlCl3
Cl
15.91
2
+
AlCl3
CH2Cl2
CH2
+ HCl
Step 1: Formation of Lewis acid-Lewis base complex:
ClCH2
Cl
Al
Cl
Cl
Cl
ClCH2
Cl
+
Cl
Al
Cl
Cl
Step 2: Nucleophilic attack of benzene on the electrophilic Lewis acid-Lewis base
complex and the formation of a resonance-stabilised carbocation (two more resonance
structures exist):
ClCH2
Cl
+
Cl
Al
CH2
Cl
+ AlCl4-
Cl
Cl
H
+
Step 3: Deprotonation of the carbocation to give benzyl chloride, HCl, and AlCl3:
CH2
Cl
Cl
+
H
Cl
Al
Cl
Cl
CH2
Cl
+ AlCl3 +
HCl
Step 4: Formation of Lewis acid-Lewis base complex between benzyl chloride and
AlCl3:
CH2
Cl
Al
Cl
Cl
CH2
Cl
Cl
Al
Cl
Cl
Cl
Step 5: Dissociation of the complex to give a resonance-stabilised benzyl cation (only
one out of five contributing structures is shown) and AlCl4–:
Cl
CH2
Cl
Al
Cl
+
Cl
CH2
Cl
Cl
Al
Cl
Cl
Step 6: Nucleophilic attack of the second molecule of benzene on the benzylic cation to
form another resonance-stabilised carbocation (only one structure is shown):
+
CH2
CH2
H +
Step 7: Deprotonation of the carbocation intermediate to regenerate the aromatic ring
(diphenylmethane), HCl, and AlCl3:
Cl
Cl Al
Cl
CH2
+ AlCl3 + HCl
CH2
Cl
H
+
Additional exercises
15.93
A molecular model of cyclohexane shows hindered rotation about the C—C single bonds,
allowing the possibility of cis-trans isomers in disubstituted six-membered rings. The
much larger cyclododecane ring is flexible enough to allow unrestricted rotation about
the C—C single bonds, thus no cis-trans isomerisation exists.
CH3
CH3
CH3
CH3
CH3
cis-1,2-dimethylcyclohexane
trans-1,2-dimethylcyclohexane
CH3
1,2-dimethylcyclododecane
15.95
109.5o
(a)
(b)
120o
120o
CH2OH
180o H
H
(c) H C C C C
120o H
(d)
120o
15.97
109.5o
OH
120o
(a)
(b)
(a)
(b)
(c)
180o
(d)
Br
120o
109.5o
15.99
109.5o
Br
109.5o
Carbons 1 and 3 are sp2 hybridised. Carbon 2 is sp hybridised.
π 2pz-2pz
y-axis
H
H
H
H
π 2py-2py
z-axis
Each of the terminal carbons is sp2 hybridised and thus has three sp2 orbitals oriented
120˚ apart. Two of these sp2 hybrid orbitals form sigma bonds with hydrogen atoms and
the third forms a sigma bond with the central carbon. Each terminal carbon has one 2p
orbital (C1 has a 2pz orbital and C3 has a 2py orbital) that each form a π bond with the 2pz
and 2py orbitals on the central carbon atom. Because these two 2p orbitals on the centre
carbon atom are at right angles to each other, the 2p orbitals on the terminal carbons must
be at right angles to each other to permit full orbital overlap.
15.101 Molecules of the trans isomer can ‘pack together’ more tightly than those of the cis
isomers, which means that the dispersion forces controlling intermolecular bonding are
greater for the trans than for the less symmetrical cis isomer.
COOH
COOH
COOH
COOH
COOH
COOH
The trans isomer of octadec-9-enoic acid can pack together closely and so it has a higher
melting point (44–45° C) than the cis isomer.
COOH
COOH
COOH
COOH
15.103 (a)
OH
+ H2O2
OH
(b)
Yes
enzyme
catalyst
O
+ 2H2O + heat
O