Download 4 ORGANIC CHEMISTRY: STRUCTURE AND NOMENCLATURE

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M O D U L E
4
ORGANIC CHEMISTRY:
STRUCTURE AND
NOMENCLATURE OF
HYDROCARBONS
O1.1
What Is an Organic Compound?
O1.2
The Saturated Hydrocarbons, or Alkanes
O1.3
The Cycloalkanes
O1.4
Rotation around C—C Bonds
O1.5
The Nomenclature of Alkanes
O1.6
The Unsaturated Hydrocarbons: Alkenes and Alkynes
O1.7
Alkene Stereoisomers
O1.8
The Reactions of Alkanes, Alkenes, and Alkynes
O1.9
Naturally Occurring Hydrocarbons and Their Derivatives
O1.10 Aromatic Hydrocarbons and Their Derivatives
O1.11 The Chemistry of Petroleum Products
O1.12 The Chemistry of Coal
O1.13 Chiral Stereoisomers
O1.14 Optical Activity
O1.1 WHAT IS AN ORGANIC COMPOUND?
When you drive up to the pump at some gas stations you are faced with a variety of choices.
As you filled the tank, you might wonder, “What is ‘unleaded’ gas, and why would anyone
want to add ‘lead’ to gas?” Or, “What would I get for my money if I bought premium gas,
with a higher octane number?” Or, “What would happen if I filled the tank at the pump
marked ‘diesel’?”
You then stop to buy drugs for a sore back that has been bothering you since you helped
a friend move into a new apartment. Once again, you are faced with choices (see Figure O1.1). You could buy aspirin, which has been used for more than a hundred years. Or
Tylenol, which contains acetaminophen. Or a more modern pain-killer, such as ibuprofen.
While you are deciding which drug to buy, you might wonder, “What is the difference
between these drugs?,” and even, “How do they work?”
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OH
O
B
CH3 H EC H
OH
CH
OP
EOH
C
O
B
OOC O CH3
Aspirin
(acetylsalicylic acid)
CH3 H E NH
C
B
Tylenol
O (acetaminophen)
CH3 H ECH2
CH
CH3
Advil
(ibuprofen)
FIGURE O1.1 The structures of the active ingredients in three common painkillers.
You then drive to campus, where you sit in a “plastic” chair to eat a sandwich that has
been wrapped in “plastic,” without worrying about why one of the plastics is flexibile while
the other is rigid. While you’re eating, a friend stops by and starts to tease you about the
effect of your diet on the level of cholesterol in your blood, which brings up the questions,
“What is cholesterol?” and “Why do so many people worry about it?”
Answers to each of these questions fall within the realm of a field known as organic
chemistry. For more than 200 years, chemists have divided materials into two categories.
Those isolated from plants and animals were classified as organic, while those that trace
back to minerals were inorganic. At one time, chemists believed that organic compounds
were fundamentally different from those that were inorganic because organic compounds
contained a vital force that was only found in living systems.
The first step in the decline of the vital force theory occurred in 1828, when Friederich
Wöhler synthesized urea from inorganic starting materials. Wöhler was trying to make ammonium cyanate (NH4OCN) from silver cyanate (AgOCN) and ammonium chloride
(NH4Cl). What he expected is described by the following equation.
AgOCN(aq) NH4Cl(aq) 88n AgCl(s) NH4OCN(aq)
The product he isolated from the reaction, however, had none of the properties of cyanate
compounds. It was a white, crystalline material that was identical to urea, H2NCONH2,
which could be isolated from urine.
Expected Product
H
A
HONOH
A
H
Ammonium cyanate
OOCqN
Observed Product
O
H
H
B
G
D
NOCON
D
G
H
H
Urea
Neither Wöhler nor his contemporaries claimed that his results disproved the vital force
theory. But his results set in motion a series of experiments that led to the synthesis of a
variety of organic compounds from inorganic starting materials. This inevitably led to the
disappearance of “vital force” from the list of theories that had any relevance to chemistry,
although it did not lead to the death of the theory, which still had proponents more than
90 years later.
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If the difference between organic and inorganic compounds isn’t the presence of some
mysterious vital force required for their synthesis, what is the basis for distinguishing between these classes of compounds? Most compounds extracted from living organisms contain carbon. It is therefore tempting to identify organic chemistry as the chemistry of carbon. But that definition would include compounds such as calcium carbonate (CaCO3) as
well as the elemental forms of carbon—diamond and graphite—that are clearly inorganic.
We will therefore define organic chemistry as the chemistry of compounds that contain both
carbon and hydrogen.
Even though organic chemistry focuses on compounds that contain carbon and hydrogen, more than 95% of the compounds that have been isolated from natural sources or
synthesized in the laboratory are organic. The special role of carbon in the chemistry of
the elements is the result of a combination of factors, including the number of valence electrons on a neutral carbon atom, the electronegativity of carbon, and the atomic radius of
carbon atoms (see Table O1.1).
TABLE O1.1 Physical Properties
of Carbon
Electronic configuration
Electronegativity
Covalent radius
1s2 2s2 2p2
2.54
0.077 nm
Carbon has four valence electrons (2s2 2p2), and it must either gain four electrons or
lose four electrons to reach a rare gas configuration. The electronegativity of carbon is too
small for carbon to gain electrons from most elements to form C4 ions, and too large for
carbon to lose electrons to form C4 ions. Carbon therefore forms covalent bonds with a
large number of other elements, including the hydrogen, nitrogen, oxygen, phosphorus,
and sulfur found in living systems.
Because they are relatively small, carbon atoms can come close enough together to
form strong CPC double bonds or even CqC triple bonds. Carbon also forms strong double and triple bonds to nitrogen and oxygen. It can even form double bonds to elements
such as phosphorus and sulfur that do not form double bonds to themselves.
When the unmanned Viking spacecraft carried out experiments designed to search for
evidence of life on Mars, the experiments were based on the assumption that living systems
contain carbon, and the absence of any evidence for carbon-based life on that planet was
presumed to mean that no life existed. Several factors make carbon essential to life.
•
•
•
Carbon atoms form strong bonds to other carbon atoms.
Carbon forms strong bonds to other nonmetals, such as N, O, P, and S.
Carbon forms multiple bonds to other nonmetals, including C, N, O, P, and S atoms.
These factors provide an almost infinite variety of potential structures for organic compounds, such as vitamin C shown in Figure O1.2. No other element can provide the variety of combinations and permutations necessary for life to exist.
O1.2 THE SATURATED HYDROCARBONS, OR ALKANES
Compounds that contain only carbon and hydrogen are known as hydrocarbons. Those
that contain as many hydrogen atoms as possible are said to be saturated. The saturated
hydrocarbons are also known as alkanes.
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O
HO
C
C
HO
C
O
CH
CH O CH2OH
OH
FIGURE O1.2 Structure of vitamin C.
The simplest alkane is methane: CH4. The Lewis structure of methane can be generated by combining the four electrons in the valence shell of a neutral carbon atom with
four hydrogen atoms to form a compound in which the carbon atom shares a total of eight
valence electrons with the four hydrogen atoms.
HT
H
HT
TP
CT
R
TH
H C
H
H
P
H
Methane is an example of a general rule that carbon is tetravalent; it forms a total of four
bonds in almost all of its compounds. To minimize the repulsion between pairs of electrons
in the four COH bonds, the geometry around the carbon atom is tetrahedral, as shown in
Figure O1.3.
FIGURE O1.3 Ball-and-stick model of methane.
Exercise O1.1
Use the fact that carbon is usually tetravalent to predict the formula of ethane, the alkane
that contains two carbon atoms.
Solution
As a rule, compounds that contain more than one carbon atom are held together by COC
bonds. If we assume that carbon is tetravalent, the formula of the compound must be C2H6.
H H
A A
HOCOCOH
A A
H H
Ethane
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The alkane that contains three carbon atoms is known as propane, which has the formula C3H8 and the following skeleton structure.
H H H
A A A
HOCOCOCOH
A A A
H H H
Propane
The four-carbon alkane is butane, with the formula C4H10.
H H H H
A A A A
HOCOCOCOCOH
A A A A
H H H H
Butane
The names, formulas, and physical properties for a variety of alkanes with the generic formula CnH2n2 are given in Table O1.2. The boiling points of the alkanes gradually increase
with the molecular weight of the compounds. At room temperature, the lighter alkanes are
gases; the midweight alkanes are liquids; and the heavier alkanes are solids, or tars.
TABLE O1.2
The Saturated Hydrocarbons, or Alkanes
Name
Molecular
Formula
Melting
Point (oC)
Boiling
Point (oC)
State
at 25oC
Methane
Ethane
Propane
Butane
Pentane
Hexane
Heptane
Octane
Nonane
Decane
Undecane
Dodecane
Eicosane
Triacontane
CH4
C2H6
C3H8
C4H10
C5H12
C6H14
C7H16
C8H18
C9H20
C10H22
C11H24
C12H26
C20H42
C30H62
182.5
183.3
189.7
138.4
129.7
95
90.6
56.8
51
29.7
24.6
9.6
36.8
65.8
164
88.6
42.1
0.5
36.1
68.9
98.4
124.7
150.8
174.1
195.9
216.3
343
449.7
Gas
Gas
Gas
Gas
Liquid
Liquid
Liquid
Liquid
Liquid
Liquid
Liquid
Liquid
Solid
Solid
The alkanes in Table O1.2 are all examples of straight-chain hydrocarbons, in which
the carbon atoms form a chain that runs from one end of the molecule to the other. The
generic formula for the compounds can be understood by assuming that they contain chains
of CH2 groups with an additional hydrogen atom capping either end of the chain. Thus,
for every n carbon atoms there must be 2n 2 hydrogen atoms: CnH2n2.
Because two points define a line, the carbon skeleton of the ethane molecule is linear,
as shown in Figure O1.4.
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FIGURE O1.4 A molecule of ethane.
Because the bond angle in a tetrahedron is 109.5°, alkane molecules that contain three or
carbon atoms can no longer be thought of as “linear,” as shown in Figure O1.5.
Propane
Butane
FIGURE O1.5 Ball-and-stick models of propane and butane.
In addition to the straight-chain examples considered so far, alkanes also form branched
structures. The smallest hydrocarbon in which a branch can occur has four carbon atoms.
The compound has the same formula as butane (C4H10), but a different structure. Compounds with the same formula and different structures are known as isomers (from the
Greek words isos, “equal,” and meros, “parts”). When it was first discovered, the branched
isomer with the formula C4H10 was therefore given the name isobutane.
CH3
A
CH3OCHOCH3
Isobutane
The best way to understand the difference between the structures of butane and isobutane
is to compare the ball-and-stick models of the compounds shown in Figure O1.6.
Butane
Isobutane
FIGURE O1.6 The two isomers with the formula C4H10.
Butane and isobutane are called constitutional isomers because they literally differ in their
constitution. One contains two CH3 groups and two CH2 groups; the other contains three
CH3 groups and one CH group.
There are three constitutional isomers of pentane, C5H12. The first is “normal” pentane, or n-pentane.
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CH3OCH2OCH2OCH2OCH3
7
n-Pentane
A branched isomer is also possible, which was originally named isopentane. When a more
highly branched isomer was discovered, it was named neopentane (the new isomer of pentane).
CH3
A
CH3OCHOCH2OCH3
CH3
A
CH3OC OCH3
A
CH3
Isopentane
Neopentane
Ball-and-stick models of the three isomers with the formula C5H12 are shown in Figure O1.7.
n-Pentane
Isopentane
Neopentane
FIGURE O1.7 The three isomers with the formula C5H12.
Exercise O1.2
The following structures all have the same molecular formula: C6H14. Which of the structures represent the same molecule?
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CH3
CH3
G
CHOCH2
D
G
CH3
A CH2OCH3
CH3
G D
CH
A
CH2
G
CH2
B
A
CH3
CH3
A
CH3OCOCH3
A
C CH2
G
CH3
Solution
There is no difference between compounds A and B; they both contain a five-carbon chain
with a branch on the second carbon. Compound C, on the other hand, contains a fourcarbon chain with two branches on the second carbon atom.
Exercise O1.3
Determine the number of constitutional isomers of hexane, C6H14.
Solution
There are five constitutional isomers of hexane. There is a straight-chain, or normal, isomer.
CH3OCH2OCH2OCH2OCH2OCH3
There are two isomers with a single carbon branch.
CH3
A
CH3OCHOCH2OCH2OCH3
CH3
A
CH3OCH2OCHOCH2OCH3
And there are two isomers with two branches.
CH3 CH3
A
A
CH3OCHOCHOCH3
CH3
A
CH3OCOCH2OCH3
A
CH3
Checkpoint
Use a set of molecular models to confirm that the following compounds are isomers,
CH3 CH3
A
A
CH3OCHOCHOCH3
CH3
A
CH3OCOCH2OCH3
A
CH3
whereas the following are different ways of writing the structure of the same compound.
CH3 CH3
A
A
CH3OCHOCHOCH3
CH3
A
CH3OCHOCHOCH3
A
CH3
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As we have seen, there are two constitutional isomers with the formula C4H10, three
isomers of C5H12, and five isomers of C6H14. The number of isomers of a compound increases rapidly with additional carbon atoms. There are over 4 billion isomers for C30H62,
for example.
O1.3 THE CYCLOALKANES
If the carbon chain that forms the backbone of a straight-chain hydrocarbon is long enough,
we can envision the two ends coming together to form a cycloalkane. One hydrogen atom
has to be removed from each end of the hydrocarbon chain to form the COC bond that
closes the ring. Cycloalkanes therefore have two less hydrogen atoms than the parent alkane
and a generic formula of CnH2n.
The smallest alkane that can form a ring is cyclopropane, C3H6, in which the three carbon atoms lie in the same plane. The angle between adjacent COC bonds is only 60°, which
is very much smaller than the 109.5° angle in a tetrahedron, as shown in Figure O1.8. Cyclopropane is therefore susceptible to chemical reactions that can open up the threemembered ring.
FIGURE O1.8 Ball-and-stick model of cyclopropane.
Any attempt to force the four carbons that form a cyclobutane ring into a plane of
atoms would produce the structure shown in Figure O1.9, in which the angle between adjacent COC bonds would be 90°. One of the four carbon atoms in the cyclobutane ring is
therefore displaced from the plane of the other three to form a “puckered” structure that
is vaguely reminiscent of the wings of a butterfly.
FIGURE O1.9 Cyclobutane.
The angle between adjacent COC bonds in a planar cyclopentane molecule would be
108°, which is close to the ideal angle around a tetrahedral carbon atom. Cyclopentane is
not a planar molecule, as shown in Figure O1.10, because displacing two of the carbon
atoms from the plane of the other three produces a puckered structure that relieves some
of the repulsion between the hydrogen atoms on adjacent carbon atoms in the ring.
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FIGURE O1.10 Structure of cyclopentane.
By the time we get to the six-membered ring in cyclohexane, a puckered structure can
be formed by displacing a pair of carbon atoms at either end of the ring from the plane of
the other four members of the ring. One of the carbon atoms is tilted up, out of the ring,
whereas the other is tilted down to form the “chair” structure shown in Figure O1.11.
FIGURE O1.11 Chair form of cyclohexane.
O1.4 ROTATION AROUND COC BONDS
It is easy to fall into the trap of thinking about the ethane molecule as if it were static.
Nothing could be further from the truth. At room temperature, the average velocity of an
ethane molecule is about 500 m/s—more than twice the speed of a Boeing 747. While it
moves through space, the molecule is tumbling around its center of gravity like an airplane
out of control. At the same time, the COH and COC bonds are vibrating like springs at
rates as fast as 9 1013 s1.
There is another way in which the ethane molecule can move. The CH3 groups at either end of the molecule can rotate with respect to each other around the COC bond.
When this happens, the molecule passes through an infinite number of conformations that
have slightly different energies. The highest energy conformation corresponds to a structure in which the hydrogen atoms are “eclipsed.” If we view the molecule along the COC
bond, the hydrogen atoms on one CH3 group would obscure those on the other, as shown
in Figure O1.12.
FIGURE O1.12 The eclipsed conformation of ethane.
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The lowest energy conformation is a structure in which the hydrogen atoms are “staggered,” as shown in Figure O1.13.
FIGURE O1.13 The staggered conformation of ethane.
The difference between the eclipsed and staggered conformations of ethane are best illustrated by viewing these molecules along the COC bond, as shown in Figure O1.14.
Staggered
Eclipsed
FIGURE O1.14 The eclipsed and staggered conformations of ethane viewed along the COC bond.
The difference between the energies of these conformations is relatively small, only about
12 kJ/mol. But it is large enough that rotation around the COC bond isn’t smooth. Although the frequency of the rotation is on the order of 1010 revolutions per second, the
ethane molecule spends a slightly larger percentage of the time in the staggered conformation.
The different conformations of a molecule are often described in terms of Newman
projections. These line drawings show the six substituents on the COC bond as if the structure of the molecule were projected onto a piece of paper by shining a bright light along
the COC bond in a ball-and-stick model of the molecule. Newman projections for the different staggered conformations of butane are shown in Figure O1.15.
CH3
H
H
CH3
CH3
H
H
H
H
H
CH3
CH3
H
H
H3C
H
H
H
FIGURE O1.15 Newman projections for the staggered conformations of butane.
Because of the ease of rotation around COC bonds, there are several conformations
of some of the cycloalkanes described in the previous section. Cyclohexane, for example,
forms both the “chair” and “boat” conformations shown in Figure O1.16. The difference
between the energies of the chair conformation, in which the hydrogen atoms are staggered, and the boat conformation, in which they are eclipsed, is about 30 kJ/mol. As a result, even though the rate at which the two conformations interchange is about 1 105 s1,
we can assume that most cyclohexane molecules at any moment in time are in the chair
conformation.
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Boat
Chair
FIGURE O1.16 The chair and boat conformations of cyclohexane.
O1.5 THE NOMENCLATURE OF ALKANES
Common names such as pentane, isopentane, and neopentane are sufficient to differentiate between the three isomers with the formula C5H12. They become less useful, however,
as the size of the hydrocarbon chain increases.
The International Union of Pure and Applied Chemistry (IUPAC) has developed a
systematic approach to naming alkanes and cycloalkanes based on the following steps.
•
Find the longest continuous chain of carbon atoms in the skeleton structure. Name
the compound as a derivative of the alkane with that number of carbon atoms. The
following compound, for example, is a derivative of pentane because the longest
chain contains five carbon atoms.
H
A
HOCOH
A
H A H H H
A A A A A
HOCOCOCOCOCOH
A A A A A
H H H H H
•
Name the substituents on the chain. Substituents derived from alkanes are named
by replacing the -ane ending with -yl. This compound contains a methyl (OCH3)
substituent.
H
A
HOCOH
A
H A H H H
A A A A A
HOCOCOCOCOCOH
A A A A A
H H H H H
•
Number the chain starting at the end nearest the first substituent and specify the
carbon atoms on which the substituents are located. Use the lowest possible numbers. This compound, for example, is 2-methylpentane, not 4-methylpentane.
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H
A
HOCOH
A
H A H H H
A
A
A
A A
HOC1OC2OC3OC4OC5OH
A
A
A
A A
H H H H H
•
•
Use the prefixes di-, tri-, and tetra- to describe substituents that are found two, three,
or four times on the same chain of carbon atoms.
Arrange the names of the substituents in alphabetical order.
Exercise O1.4
Use the IUPAC system to name the following compound.
CH3
CH3
A
A
CH3OCOCH2OCHOCH3
A
CH3
Solution
The compound is a derivative of pentane because the longest chain contains five carbon
atoms. There are three identical CH3 substituents on the backbone. Two of the methyl
groups are on the second carbon, and one is on the fourth. The compound is therefore
2,2,4-trimethylpentane. Since it contains a total of eight carbon atoms, it is also known by
the common name isooctane. This compound is the “octane” used as a standard against
which to measure octane numbers.
Exercise O1.5
Use the IUPAC system to name the following compound.
CH3
A
CH3OCHOCH2OCHOCH2OCH3
A
CH2
A
CH2
A
CH3
Solution
The longest continuous chain in the skeleton structure of the compound contains seven
carbon atoms. It is therefore named as a derivative of heptane.
1
CH3
A
4
CH3OCHOCH2OCHOCH2OCH3
2
3
A
5
CH2
A
6
CH2
A
7
CH3
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The heptane chain contains two substituents: a methyl group on the second carbon atom
and an ethyl group on the fourth carbon atom.
CH3
A
CH3OCHOCH2OCHOCH2OCH3
A
Methyl
Ethyl
CH2
A
CH2
A
CH3
Because the substituents are listed in alphabetical order, the systematic name for the compound is 4-ethyl-2-methylheptane.
O1.6 THE UNSATURATED HYDROCARBONS: ALKENES AND
ALKYNES
Carbon not only forms the strong COC single bonds found in alkanes, it also forms strong
CPC double bonds. Compounds that contain CPC double bonds were once known as
olefins (literally, “to make an oil”) because they were hard to crystallize. (They tend to remain oily liquids when cooled.) These compounds are now called alkenes. The simplest
alkene has the formula C2H4 and the following Lewis structure.
HH
EH
CPC
E
H
H
H
The relationship between alkanes and alkenes can be understood by thinking about
the following hypothetical reaction. We start by breaking the bond in an H2 molecule so
that one of the electrons ends up on each of hydrogen atoms. We do the same thing to one
of the bonds between the carbon atoms in an alkene. We then allow the unpaired electron
on each hydrogen atom to interact with the unpaired electron on a carbon atom to form
a new COH bond.
HOH
HH
EH
CPC
E
H
H
H
HT HT
POC
POH
HOC
A A
H H
H H
A A
HOCOCOH
A A
H H
Thus, in theory, we can transform an alkene into the parent alkane by adding an H2 molecule across a CPC double bond. In practice, the reaction occurs only at high pressures
in the presence of a suitable catalyst, such as piece of nickel metal.
H
H
CPC
H
H
H2
Ni
H H
A A
HOC OCOH
A A
H H
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Because an alkene can be thought of as a derivative of an alkane from which an H2 molecule has been removed, the generic formula for an alkene with one CPC double bond is
CnH2n.
Alkenes are examples of unsaturated hydrocarbons because they have fewer hydrogen
atoms than the corresponding alkanes. They were once named by adding the suffix -ene to
the name of the substituent that carried the same number of carbon atoms.
CH2PCH2
CH2PCH2OCH3
Ethylene
Propylene
The IUPAC nomenclature for alkenes names the compounds as derivatives of the parent
alkanes. The presence of the CPC double bond is indicated by changing the -ane ending
on the name of the parent alkane to -ene.
CH3OCH3
CH2PCH2
Ethane
Ethene
CH3OCH2OCH3
CH3OCHPCH2
Propane
Propene
The location of the CPC double bond in the skeleton structure of the compound is indicated by specifying the number of the carbon atom at which the CPC bond starts.
CH2PCHOCH2OCH3
CH3OCHPCHOCH3
1-Butene
2-Butene
The names of substituents are then added as prefixes to the name of the alkene.
Exercise O1.6
Use the IUPAC system to name the following compound.
CH3
A
CH3OCHPCOCH2OCHOCH3
A
CH3
Solution
The compound is a derivative of hexane because the longest carbon chain contains six carbon atoms. It contains a CPC double bond, which means it is a hexene. Because the double bond links the second and third carbon atoms, it is a 2-hexene. Because the CH3 substituents are on the third and fifth carbon atoms, the compound is 3,5-dimethyl-2-hexene.
Compounds that contain CqC triple bonds are called alkynes. These compounds have
four less hydrogen atoms than the parent alkanes, so the generic formula for an alkyne
with a single CqC triple bond is CnH2n2. The simplest alkyne has the formula C2H2 and
is known by the common name acetylene.
HOCqCOH
Acetylene
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The IUPAC nomenclature for alkynes names the compounds as derivatives of the parent
alkane, with the ending -yne replacing -ane.
CCH2CH3
CCH2CH3
CH3
A
CCH2CHCH2CH3
CH3C
HC
CH3C
2-Pentyne
1-Butyne
5-Methyl-2-heptyne
In addition to compounds that contain one double bond (alkenes) or one triple bond
(alkynes), we can also envision compounds with two double bonds (dienes), three double
bonds (trienes), or a combination of double and triple bonds.
CH3CHPCHCH2C CH
CH2 PCHCH PCH2
4-Hexen-1-yne
1,3-Butadiene
O1.7 ALKENE STEREOISOMERS
The geometry around the CPC double bond in an alkene plays an important role in the
chemistry of the compounds. To understand why, let’s return to the hypothetical intermediate from the previous section in which we have a C2H4 molecule with an unpaired electron on each of the carbon atoms.
The sigma () bond skeleton in the molecule is formed by the overlap of sp2 hybridized
orbitals on each carbon atom with either a 1s orbital on a hydrogen atom or the sp2 hybridized orbital on the other carbon atom. This leaves one unpaired electron in an empty
2p orbital on each carbon atom. The orbitals that hold these electrons interact to form a
pi () bond.1
pi bond
C
C
The geometry around a CPC double bond is therefore different from the geometry around
a COC single bond. Because of the double bond, the six atoms in a C2H4 molecule must
all lie in the same plane, as shown in Figure O1.17. The presence of the bond restricts
rotation around a CPC double bond. There is no way to rotate one end of the bond relative to the other without breaking the bond. Because the bond is relatively strong
(~270 kJ/mol), rotation around the CPC double bond cannot occur at room temperature.
Alkenes therefore form stereoisomers that differ in the way substituents are arranged
around the CPC double bond. The isomer with similar substituents on the same side of
1
Students sometimes ask, “Why are there three lines between the carbon atoms in this drawing if there are
only two bonds?” The answer is: Both of the curved lines are needed to represent a single bond.
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17
H
C
H
H
C
H
FIGURE O1.17 The planar C2H4 molecule.
the double bond is called the cis isomer, from a Latin stem meaning “on this side.” The isomer in which similar substituents are across from each other, is called the trans isomer, from
a Latin stem meaning “across.” Consider, for example, the terms “transcontinental,” or “transAtlantic.” The cis isomer of 2-butene, for example, has both CH3 groups on the same side of
the double bond. In the trans isomer the CH3 groups are on opposite sides of the double bond.
CH3
CH3
D
G
CPC
D
G
H
H
cis-2-Butene
H
D
G
CPC
D
G
H
CH3
CH3
trans-2-Butene
Cis/trans isomers have similar chemical properties but different physical properties. cis-2Butene, for example, freezes at 138.9°C, whereas trans-2-butene freezes at 105.6°C.
Exercise O1.7
Name the straight-chain constitutional and stereoisomers of pentene (C5H10).
Solution
There are two constitutional isomers of pentene in which all the carbon atoms lie in the
same chain.
CH2 PCHCH2CH2CH3
CH3CH PCHCH2CH3
1-Pentene
2-Pentene
There are no cis/trans isomers of 1-pentene because there is only one way of arranging the
substituents around the double bond.
H
CH CH2CH3
D 2
G
CPC
D
G
H
H
Cis and trans isomers are possible, however, for 2-pentene.
CH2CH3
D
G
CPC
D
G
H
H
CH3
cis-2-Pentene
H
D
G
CPC
D
G
H
CH2CH3
CH3
trans-2-Pentene
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O1.8 THE REACTIONS OF ALKANES, ALKENES, AND ALKYNES
Alkanes
In the absence of a spark or a high-intensity light source, alkanes are generally inert to
chemical reactions. However, anyone who has used a match to light a gas burner, or dropped
a match onto charcoal coated with lighter fluid, should recognize that alkanes burst into
flame in the presence of a spark. It doesn’t matter whether the starting material is the
methane found in natural gas,
CH4(g) 2 O2(g) 88n CO2(g) 2 H2O(g)
the mixture of butane and isobutane used in disposable cigarette lighters,
2 C4H10(g) 13 O2(g) 88n 8 CO2(g) 10 H2O(g)
the mixture of C5 to C6 hydrocarbons in charcoal lighter fluid,
C5H12(g) 8 O2(g) 88n 5 CO2(g) 6 H2O(g)
or the complex mixture of C6 to C8 hydrocarbons in gasoline.
2 C8H18(l) 25 O2(g) 88n 16 CO2(g) 18 H2O(g)
Once the reaction is ignited by a spark, the hydrocarbons burn to form CO2 and H2O and
give off between 45 and 50 kJ of energy per gram of fuel consumed.
In the presence of light, or at high temperatures, alkanes react with halogens to form
alkyl halides. Reaction with chlorine gives an alkyl chloride.
light
CH4(g) Cl2(g) 888n
CH3Cl(g) HCl(g)
Reaction with bromine gives an alkyl bromide.
light
CH4(g) Br2(l) 888n
CH3Br(g) HBr(g)
Alkenes and Alkynes
Unsaturated hydrocarbons such as alkenes and alkynes are much more reactive than the
parent alkanes. They react rapidly with bromine, for example, to add a Br2 molecule across
the CPC double bond.
CH3CH P CHCH3 Br2
Br
A
CH3CHCHCH3
A
Br
2,3-Dibromobutane
This reaction provides a way to test for alkenes or alkynes. Solutions of bromine in CCl4
have an intense red-orange color. When Br2 in CCl4 is mixed with a sample of an alkane,
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19
no change is initially observed. When it is mixed with an alkene or alkyne, the color of Br2
rapidly disappears.
The reaction between 2-butene and bromine to form 2,3-dibromobutane is just one example of the addition reactions of alkenes and alkynes. Hydrogen bromide (HBr) adds
across a CPC double bond to form the corresponding alkyl bromide, in which the hydrogen ends up on the carbon atom that had more hydrogen atoms to begin with. Addition
of HBr to 2-butene, for example, gives 2-bromobutane.
Br
A
CH3CHCH2CH3
CH3CH P CHCH3 HBr
2-Bromobutane
As noted in Section O1.6, H2 adds across double (or triple) bonds in the presence of a suitable catalyst to convert an alkene (or alkyne) to the corresponding alkane.
CH3CH P CHCH3 H2
Pt
CH3CH2CH2CH3
In the presence of an acid catalyst, it is even possible to add a molecule of water across a
CPC double bond.
CH3CH P CHCH3 H2O
OH
A
CH3CHCH2CH3
H2SO4
Addition reactions provide a way to add new substituents to a hydrocarbon chain and
thereby produce new derivatives of the parent alkanes.
In theory, two products can form when an unsymmetric reagent such as HBr is added
to an unsymmetric CPC double bond. In practice, only one product is obtained. When
HBr is added to 2-methylpropene, for example, the product is 2-bromo-2-methylpropane,
not 1-bromo-2-methylpropane.
CH3
G
C PCH2 HBr
D
CH3
CH3
A
CH3 O C O CH3
A
Br
CH3
A
(not CH3 O CH O CH2Br)
In 1870, after careful study of many examples of addition reactions, the Russian chemist
Vladimir Markovnikov formulated a rule for predicting the product of the reactions.
Markovnikov’s rule states that the hydrogen atom adds to the carbon atom that already has
the larger number of hydrogen atoms when HX adds to an alkene. Thus, water (H—OH)
adds to propene to form the product in which the OH group is on the middle carbon
atom.
OH
CH3CH PCH2 HOH
H
CH3CHCH2
H
Carbon in the
double bond with
the most H atoms
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O1.9 NATURALLY OCCURRING HYDROCARBONS AND THEIR
DERIVATIVES
Complex hydrocarbons and their derivatives are found throughout nature. Natural rubber,
for example, is a hydrocarbon that contains long chains of alternating CPC double bonds
and COC single bonds.
CH3
CH3
CH3
A
A
A
CH2H
CN ECH2...
CN ECH2H
C
E
E
E
E N
CH
CH
C
C
C
2
2
...CH2
A
A
A
H
H
H
Writing the structure of complex hydrocarbons can be simplified by using a line notation
in which a carbon atom is assumed to be present wherever a pair of lines intersect and
enough hydrogen atoms are present to satisfy the tendency of carbon to form a total of
four bonds.
...
...
There are a variety of techniques for isolating both pleasant- and foul-smelling compounds known as essential oils from natural sources, particularly from plants. These compounds are not “essential,” in the sense of being vital to life. They were given that name
because they give off a distinct “essence,” or smell.
The essential oils are used in perfumes and medicines. Some of the compounds can be
isolated by gently heating, or steam distilling, the crushed flowers of plants. Others can be
extracted into nonpolar solvents or absorbed onto grease-coated cloths in which the plants
are wrapped. Many of the essential oils belong to classes of compounds known as terpenes
and terpenoids. The terpenes are hydrocarbons that usually contain one or more CPC
double bonds. The terpenoids are oxygen-containing analogs of the terpenes.
Examples of terpenes include -pinene and -pinene, the primary components of turpentine that give rise to its characteristic odor.
CH3
CH3
CH2
B
CH3
CH3
-Pinene
CH3
-Pinene
Camphor and menthol are examples of terpenoids.
CH3
CH3
OH
CH3
Camphor
O
Menthol
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21
Both of the compounds have a fragrant, penetrating odor and taste cool. Camphor is
used as a moth repellent. Menthol is a mild anesthetic that is added to some brands of
cigarettes.
Although the terpenes and terpenoids discussed so far have very different structures,
they have one important property in common: They all contain 10 carbon atoms, neither
more nor less. Each of the compounds can be traced back to a reaction in which a pair of
five-carbon molecules are fused. Thus, it isn’t surprising that we can also find sesquiterpenes (15 carbon atoms), diterpenes (20 carbons), triterpenes (30 carbons), and so on. Important examples of these compounds include vitamin A and the -carotene that gives carrots their characteristic color.
CH3
CH3
CH3
OH
CH3
CH3
Vitamin A
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
CH3
-Carotene
Steroids aren’t terpenes or terpenoids in the literal sense because they don’t contain
the characteristic number of carbon atoms. Consider cholesterol, for example, which is one
of the most important steroids.
CH3
CH3
Cholesterol
HO
Analysis of the structure suggests the formula C27H46O, which doesn’t fit the pattern expected of a terpenoid. The biosynthetic precursor of the molecule, however, is a 30-carbon
triterpene that is converted into cholesterol by a series of enzyme-catalyzed reactions.
By definition, the steroids are compounds that have the basic structure formed by fusing three six-membered rings and a five-membered ring. The most important property of
the molecule is the fact that, with the exception of the OOH group on the lower left-hand
corner of the molecule, there is nothing about the structure of the compound that would
make it soluble in water.
In this day of cholesterol-free products, it is useful to question the label on certain products such as peanut butter that are advertised as cholesterol-free. That is like saying that
the Sahara desert is rain-free. Peanut butter is made from peanuts and cholesterol isn’t a
characteristic ingredient in plants; it is synthesized by animals, particularly mammals. It is
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also useful to note that placing someone on a cholesterol-free diet won’t reduce their cholesterol level to zero. Even on a low-cholesterol diet, the individual will synthesize about 0.80
gram of cholesterol per day. The key question is, Is there an excess of cholesterol in the
bloodstream? If there is, a diet that reduces the intake of cholesterol might be important.
O1.10 AROMATIC HYDROCARBONS AND THEIR DERIVATIVES
At the turn of the nineteenth century, one of the signs of living the good life was having
gas lines connected to your house, so that you could use gas lanterns to light the house after dark. The gas burned in the lanterns was called coal gas because it was produced by
heating coal in the absence of air. The principal component of coal gas was methane, CH4.
In 1825, Michael Faraday was asked to analyze an oily liquid with a distinct odor that
collected in tanks used to store coal gas at high pressures. Faraday found that the compound had the empirical formula CH. Ten years later, Eilhardt Mitscherlich produced the
same material by heating benzoic acid with lime. Mitscherlich named the substance benzin, which became benzene when translated into English. He also determined that the molecular formula of the compound was C6H6.
Benzene must be an unsaturated hydrocarbon because it has far less hydrogen than the
equivalent saturated hydrocarbon: C6H14. But benzene is too stable to be an alkene
or alkyne. Alkenes and alkynes rapidly add Br2 to their CPC and CqC bonds, whereas
benzene reacts with bromine only in the presence of a catalyst, FeBr3. Furthermore, when
benzene reacts with Br2 in the presence of FeBr3, the product of the reaction is a compound in which a bromine atom has been substituted for a hydrogen atom, not added to
the compound the way an alkene adds bromine.
FeBr3
C6H6 Br2 8888n C6H5Br HBr
Other compounds were eventually isolated from coal that had similar properties. Their formulas suggested the presence of multiple CPC bonds, but the compounds were not reactive enough to be alkenes. Because they often had a distinct odor, or aroma, they became
known as aromatic compounds.
The structure of benzene was a recurring problem throughout most of the nineteenth
century. The first step toward solving the problem was taken by Friedrich August Kekulé
in 1865. (Kekulé’s interest in the structure of organic compounds may have resulted from
the fact that he first enrolled at the University of Giessen as a student of architecture.)
One day, while dozing before a fire, Kekulé dreamed of long rows of atoms twisting in a
snakelike motion until one of the snakes seized hold of its own tail. This dream led Kekulé
to propose that benzene consists of a ring of six carbon atoms with alternating COC single bonds and CPC double bonds. Because there are two ways in which the bonds can alternate, Kekulé proposed that benzene was a mixture of two compounds in equilibrium.
H
H
H
H
H
H
H
H
H
H
H
H
Kekulé’s explanation of
the structure of benzene
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23
Kekulé’s structure explained the formula of benzene, but it did not explain why benzene failed to behave like an alkene. The unusual stability of benzene wasn’t understood
until the development of the theory of resonance. This theory states that molecules for
which two or more satisfactory Lewis structures can be drawn are an average, or hybrid,
of the structures. Benzene, for example, is a resonance hybrid of the two Kekulé structures.
H
H
H
H
H
H
H
H
H
H
H
H
The resonance structures
for benzene
The difference between the equilibrium and resonance descriptions of benzene is subtle,
but important. In the equilibrium approach, a pair of arrows is used to describe a reversible
reaction, in which the molecule on the left is converted into the one on the right, and vice
versa. In the resonance approach, a double-headed arrow is used to suggest that a benzene
molecule does not shift back and forth between two different structures; it is a hybrid mixture of the structures.
One way to probe the difference between Kekulé’s idea of an equilibrium between two
structures and the resonance theory in which benzene is a hybrid mixture of the structures
would be to study the lengths of the carbon–carbon bonds in benzene. If Kekulé’s idea
was correct, we would expect to find a molecule in which the bonds alternate between
relatively long COC single bonds (0.154 nm) and significantly shorter CPC double
bonds (0.133 nm). When benzene is cooled until it crystallizes, and the structure of the
molecule is studied by X-ray diffraction, we find that the six carbon–carbon bonds in
the molecule are the same length (0.1395 nm). The crystal structure of benzene is therefore more consistent with the resonance model of bonding in benzene than the original
Kekulé structures.
The resonance theory does more than explain the structure of benzene—it also explains why benzene is less reactive than an alkene. The resonance theory assumes that molecules that are hybrids of two or more Lewis structures are more stable than those that
aren’t. It is the extra stability that makes benzene and other aromatic derivatives less reactive than normal alkenes. To emphasize the difference between benzene and a simple alkene,
many chemists replace the Kekulé structures for benzene and its derivatives with an aromatic ring in which the circle in the center of the ring indicates that the electrons in the
ring are delocalized; they are free to move around the ring.
H
H
H
H
H
H
It is the delocalization of electrons around the aromatic ring that makes benzene less reactive than a simple alkene.
Aromatic compounds were being extracted from coal tar as early as the 1830s. As a
result, many of the compounds were given common names that are still in use today. A
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few of these compounds are shown below.
CH3
OH
OCH3
Phenol
Toluene
CO2H
NH2
Anisole
Aniline
Benzoic acid
There are three ways in which a pair of substituents can be placed on an aromatic ring.
In the ortho (o) isomer, the substituents are in adjacent positions on the ring. In the meta
(m) isomer, they are separated by one carbon atom. In the para (p) isomer, they are on
opposite ends of the ring. The three isomers of dimethylbenzene, or xylene, are shown
below.
CH3
CH3
CH3
CH3
CH3
Ortho
Meta
CH3
Para
Exercise O1.8
Predict the structure of para-dichlorobenzene, one of the active ingredients in mothballs.
Solution
Para isomers of benzene contain two substituents at opposite positions in the sixmembered ring. para-Dichlorobenzene therefore has the following structure.
Cl
Cl
p-Dichlorobenzene
Aromatic compounds can contain more than one six-membered ring. Naphthalene, anthracene, and phenanthrene are examples of aromatic compounds that contain two or more
fused benzene rings.
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H
Naphthalene (C10H8)
H
H
H
H
H
Anthracene (C14H10)
H
H
H
H
H
H
Phenanthrene (C14H10)
A ball-and-stick model of anthracene is shown in Figure O1.18.
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25
FIGURE O1.18 Anthracene.
O1.11 THE CHEMISTRY OF PETROLEUM PRODUCTS
The term petroleum comes from the Latin stems petra, “rock,” and oleum, “oil.” It is used
to describe a broad range of hydrocarbons that are found as gases, liquids, or solids beneath the surface of the earth. The two most common forms are natural gas and crude oil.
Natural gas is a mixture of lightweight alkanes. A typical sample of natural gas when
it is collected at its source contains 80% methane (CH4), 7% ethane (C2H6), 6% propane
(C3H8), 4% butane and isobutane (C4H10), and 3% pentanes (C5H12). The C3, C4, and C5
hydrocarbons are removed before the gas is sold. The commercial natural gas delivered to
the customer is therefore primarily a mixture of methane and ethane. The propane and
butanes removed from natural gas are usually liquefied under pressure and sold as liquefied petroleum gases (LPG).
Natural gas was known in England as early as 1659. But it didn’t replace coal gas as an
important source of energy in the United States until after World War II, when a network
of gas pipelines was constructed. By 1980, annual consumption of natural gas had grown
to more than 55,000 billion cubic feet, which represented almost 30% of total U.S. energy
consumption.
The first oil well was drilled by Edwin Drake in 1859, in Titusville, PA. It produced
up to 800 gallons per day, which far exceeded the demand for the material. By 1980, consumption of oil had reached 2.5 billion gallons per day. About 225 billion barrels of oil
were produced by the petroleum industry between 1859 and 1970. Another 200 billion barrels were produced between 1970 and 1980. The total proven world reserves of crude oil
in 1970 were estimated at 546 billion barrels, with perhaps another 800 to 900 billion barrels of oil that remained to be found. It took 500 million years for the petroleum beneath
the earth’s crust to accumulate. At the present rate of consumption, we might exhaust the
world’s supply of petroleum by the 200th anniversary of the first oil well.
Crude oil is a complex mixture that is between 50 and 95% hydrocarbon by weight.
The first step in refining crude oil involves separating the oil into different hydrocarbon
fractions by distillation. A typical set of petroleum fractions is given in Table O1.3. Since
there are a number of factors that influence the boiling point of a hydrocarbon, these
petroleum fractions are complex mixtures. More than 500 different hydrocarbons have been
identified in the gasoline fraction, for example.
About 10% of the product of the distillation of crude oil is a fraction known as straightrun gasoline, which served as a satisfactory fuel during the early days of the internal combustion engine. As the automobile engine developed, it was made more powerful by increasing the compression ratio.2 Straight-run gasoline burns unevenly in high-compression
engines, producing a shock wave that causes the engine to “knock,” or “ping.” As the petroleum industry matured, it faced two problems: increasing the yield of gasoline from each
barrel of crude oil and decreasing the tendency of gasoline to knock when it burned.
2
Modern cars run at compression ratios of about 91, which means the gasoline–air mixture in the cylinder is
compressed by a factor of nine before it is ignited.
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TABLE O1.3
Petroleum Fractions
Fraction
Boiling Range ( oC)
Number of Carbon Atoms
Natural gas
Petroleum ether
Gasoline
Kerosene
Fuel oils
Lubricants
Asphalt or coke
20
20–60
40–200
150–260
260
400
Residue
C1 to C4
C5 to C6
C5 to C12, but mostly C6 to C8
Mostly C12 to C13
C14 and higher
C20 and above
Polycyclic
The relationship between knocking and the structure of the hydrocarbons in gasoline
is summarized in the following general rules.
•
•
•
•
Branched alkanes and cycloalkanes burn more evenly than straight-chain alkanes.
Short alkanes (C4H10) burn more evenly than long alkanes (C7H16).
Alkenes burn more evenly than alkanes.
Aromatic hydrocarbons burn more evenly than cycloalkanes.
The most commonly used measure of a gasoline’s ability to burn without knocking is its
octane number. Octane numbers compare a gasoline’s tendency to knock against the tendency of a blend of two hydrocarbons—heptane and 2,2,4-trimethylpentane, or isooctane—
to knock. Heptane (C7H16) is a long, straight-chain alkane, which burns unevenly and produces a great deal of knocking. Highly branched alkanes such as 2,2,4-trimethylpentane
are more resistant to knocking. Gasolines that match a blend of 87% isooctane and 13%
heptane are given an octane number of 87.
There are three ways of reporting octane numbers. Measurements made at high speed
and high temperature are reported as motor octane numbers. Measurements taken under
relatively mild engine conditions are known as research octane numbers. The road index
octane numbers reported on gasoline pumps are an average of the two. Road index octane
numbers for a few pure hydrocarbons are given in Table O1.4.
TABLE O1.4
Hydrocarbon Octane Numbers
Hydrocarbon
Road Index Octane Number
Heptane
2-Methylheptane
Hexane
2-Methylhexane
1-Heptene
Pentane
1-Pentene
Butane
Cyclohexane
2,2,4-Trimethylpentane (isooctane)
Benzene
Toluene
0
23
25
44
60
62
84
91
97
100
101
112
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By 1922 a number of compounds had been discovered that could increase the octane
number of gasoline. Adding as little as 6 mL of tetraethyllead (shown in Figure O1.19) to
a gallon of gasoline, for example, can increase the octane number by 15 to 20 units. This
discovery gave rise to the first “ethyl” gasoline, and it enabled the petroleum industry to
produce aviation gasolines with octane numbers greater than 100.
CH3
G
CH2
CH Pb CH2
D 2
CH3
CH2
G
CH3
CH3
D
FIGURE O1.19 Structure of the tetraethyllead used to make “leaded”
gasoline.
Another way to increase the octane number is thermal reforming. At high temperatures (500–600°C) and high pressures (25–50 atm), straight-chain alkanes isomerize to form
branched alkanes and cycloalkanes, thereby increasing the octane number of the gasoline.
Running the reaction in the presence of hydrogen and a catalyst such as a mixture of silica (SiO2) and alumina (Al2O3) results in catalytic reforming, which can produce a gasoline with even higher octane numbers. Thermal or catalytic reforming and gasoline additives such as tetraethyllead increase the octane number of the straight-run gasoline obtained
from the distillation of crude oil, but neither process increases the yield of gasoline from
a barrel of oil.
The data in Table O1.3 suggest that we could increase the yield of gasoline by “cracking” the hydrocarbons that end up in the kerosene or fuel oil fractions into smaller pieces.
Thermal cracking was discovered as early as the 1860s. At high temperatures (500°C) and
high pressures (25 atm), long-chain hydrocarbons break into smaller pieces. A saturated
C12 hydrocarbon in kerosene, for example, might break into two C6 fragments. Because
the total number of carbon and hydrogen atoms remains constant, one of the products of
the reaction must contain a CPC double bond.
CH3(CH2)10CH3 888n CH3(CH2)4CH3 CH2PCH(CH2)3CH3
The presence of alkenes in thermally cracked gasolines increases the octane number (70)
relative to that of straight-run gasoline (60), but it also makes thermally cracked gasoline
less stable for long-term storage. Thermal cracking has therefore been replaced by catalytic
cracking, which uses catalysts instead of high temperatures and pressures to crack longchain hydrocarbons into smaller fragments for use in gasoline.
About 87% of the crude oil refined in 1980 went into the production of fuels such as
gasoline, kerosene, and fuel oil. The remainder went for nonfuel uses, such as petroleum
solvents, industrial greases and waxes, or starting materials for the synthesis of petrochemicals. Petroleum products are used to produce synthetic fibers such as nylon, Orlon,
and Dacron, and other polymers such as polystyrene, polyethylene, and synthetic rubber.
They also serve as raw materials in the production of refrigerants, aerosols, antifreeze, detergents, dyes, adhesives, alcohols, explosives, weed killers, insecticides, and insect repellents. The H2 given off when alkanes are converted to alkenes or when cycloalkanes are
converted to aromatic hydrocarbons can be used to produce a number of inorganic petrochemicals, such as ammonia, ammonium nitrate, and nitric acid. As a result, most fertilizers as well as other agricultural chemicals are also petrochemicals.
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O1.12 THE CHEMISTRY OF COAL
Coal can be defined as a sedimentary rock that burns. It was formed by the decomposition of plant matter, and it is a complex substance that can be found in many forms. Coal
is divided into four classes: anthracite, bituminous, subbituminous, and lignite. Elemental
analysis gives empirical formulas such as C137H97O9NS for bituminous coal and C240H90O4NS
for high-grade anthracite. A model for one portion of the extended structure for coal is
shown in Figure O1.20.
H
H
O
S
H
H
H
H
O
H
NH2
H
H
H
HOCOH H
H H
H2
H
H
HH
S
HOCOH
H H
H
H
S
O
N
H
H
C
f
H
H
H2
fi
HO H
H
O
H
H
O
H
i
C
C
f i
H H
H
O
H
S
H
N
i
H H
H
H
N
H
H
H
H
H2
O
H
H O
f
O
HOCOH
H2
H2
H
H
H2
H
H
H
N
H
O
S
H
CH3
H
HOCOH
H
S
O SO
H2
H
H
HOCOH
S
H
COC
H
O
i
H
H
HOCOH
H
H
O
H
H
H
HOCOH
H
H HH
C PO
MO
H2
H
i
C
H H
COC
S
H2
H
HH
O
H
O
C
H
CH2CH3
O
H
H
H
C
f
H
H
HH
f
H H
O
H
H
H H
H
H2
FIGURE O1.20 Model for a portion of the extended structure of coal.
Anthracite coal is a dense, hard rock with a jet-black color and a metallic luster. It contains between 86% and 98% carbon by weight, and it burns slowly, with a pale blue flame
and very little smoke. Bituminous coal, or soft coal, contains between 69% and 86% carbon by weight and is the most abundant form of coal. Subbituminous coal contains less
carbon and more water, and it is therefore a less efficient source of heat. Lignite coal, or
brown coal, is a very soft coal that contains up to 70% water by weight.
The total energy consumption in the United States for 1990 was 86 1015 kJ. Of that
total, 41% came from oil, 24% from natural gas, and 23% from coal. Coal is unique as a
source of energy in the United States, however, because none of the 2118 billion pounds
used in 1990 was imported. Furthermore, the proven reserves are so large we can continue
using coal at that level of consumption for at least 2000 years.
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At the time this text was written, coal was the most cost-efficient fuel for heating. The
cost of coal delivered to the Purdue University physical plant was $1.41 per million kilojoules of heating energy. The equivalent cost for natural gas would have been $5.22, and
#2 fuel oil would have cost $7.34. Although coal is less expensive than natural gas and oil,
it is more difficult to handle. As a result, there has been a long history of efforts to turn
coal into either a gaseous or a liquid fuel.
Coal Gasification
As early as 1800, coal gas was made by heating coal in the absence of air. Coal gas is rich
in CH4 and gives off up to 20.5 kJ per liter of gas burned. Coal gas—or town gas, as it was
also known—became so popular that most major cities and many small towns had a local
gas house in which it was generated, and gas burners were adjusted to burn a fuel that produced 20.5 kJ/L. Gas lanterns, of course, were eventually replaced by electric lights. But
coal gas was still used for cooking and heating until the more efficient natural gas (38.3
kJ/L) became readily available.
A slightly less efficient fuel known as water gas can be made by reacting the carbon in
coal with steam.
C(s) H2O(g) 88n CO(g) H2(g)
Ho 131.3 kJ/molrxn
Water gas burns to give CO2 and H2O, releasing roughly 11.2 kJ per liter of gas consumed.
Note that the enthalpy of reaction for the preparation of water gas is positive, which means
that the reaction is endothermic. As a result, the preparation of water gas typically involves
alternating blasts of steam and either air or oxygen through a bed of white-hot coal. The
exothermic reactions between coal and oxygen to produce CO and CO2 provide enough
energy to drive the reaction between steam and coal.
Water gas formed by the reaction of coal with oxygen and steam is a mixture of CO,
CO2, and H2. The ratio of H2 to CO can be increased by adding water to the mixture, to
take advantage of a reaction known as the water-gas shift reaction.
CO(g) H2O(g) 88n CO2(g) H2(g)
Ho 41.2 kJ/molrxn
The concentration of CO2 can be decreased by reacting the CO2 with coal at high temperatures to form CO.
C(s) CO2(g) 88n 2 CO(g)
Ho 172.5 kJ/molrxn
Water gas from which the CO2 has been removed is called synthesis gas because it can
be used as a starting material for a variety of organic and inorganic compounds. It can be
used as the source of H2 for the synthesis of ammonia, for example.
N2(g) 3 H2(g) 88n 2 NH3(g)
It can also be used to make methyl alcohol, or methanol.
CO(g) 2 H2(g) 88n CH3OH(l)
Methanol can then be used as a starting material for the synthesis of alkenes, aromatic
compounds, acetic acid, formaldehyde, and ethyl alcohol (ethanol). Synthesis gas can also
be used to produce methane, or synthetic natural gas (SNG).
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CO(g) 3 H2(g) 88n CH4(g) H2O(g)
2 CO(g) 2 H2(g) 88n CH4(g) CO2(g)
Coal Liquefaction
The first step toward making liquid fuels from coal involves the manufacture of synthesis
gas (CO and H2) from coal. In 1925, Franz Fischer and Hans Tropsch developed a catalyst that converted CO and H2 at 1 atm and 250–300°C into liquid hydrocarbons. By 1941,
Fischer–Tropsch plants produced 740,000 tons of petroleum products per year in Germany.
Fischer–Tropsch technology is based on a complex series of reactions that use H2 to
reduce CO to CH2 groups linked to form long-chain hydrocarbons.
CO(g) 2 H2(g) 88n (CH2)n(l) H2O(g)
Ho 165 kJ/molrxn
The water produced in the reaction combines with CO in the water-gas shift reaction to
form H2 and CO2.
CO(g) H2O(g) 88n CO2(g) H2(g)
Ho 41.2 kJ/molrxn
The overall Fischer–Tropsch reaction is therefore described by the following equation.
2 CO(g) H2(g) 88n (CH2)n(l) CO2(g)
Ho 206 kJ/molrxn
At the end of World War II, Fischer–Tropsch technology was under study in most industrial nations. The low cost and high availability of crude oil, however, led to a decline in
interest in liquid fuels made from coal. The only commercial plants using this technology
today are in the Sasol complex in South Africa, which uses 30.3 million tons of coal per
year.
Another approach to liquid fuels is based on the reaction between CO and H2 to form
methanol, CH3OH.
CO(g) 2 H2(g) 88n CH3OH(l)
Methanol can be used directly as a fuel, or it can be converted into gasoline with catalysts
such as the ZSM-5 zeolite catalyst developed by Mobil Oil Company.
As the supply of petroleum becomes smaller and its cost continues to rise, a gradual
shift may be observed toward liquid fuels made from coal. Whether this takes the form of
a return to a modified Fischer–Tropsch technology, the conversion of methanol to gasoline, or other alternatives, only time will tell.
O1.13 CHIRAL STEREOISOMERS
The cis/trans isomers formed by alkenes aren’t the only example of stereoisomers. To understand the second example of stereoisomers, it might be useful to start by considering a
pair of hands. For all practical purposes, they contain the same “substituents”—four fingers and one thumb on each hand. If you clap them together, you will find even more similarities between the two hands. The thumbs are attached at about the same point on the
hand, significantly below the point where the fingers start. The second fingers on both
hands are usually the longest, then the third fingers, then the first fingers, and finally the
“little” fingers.
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In spite of the many similarities, there is a fundamental difference between a pair of
hands that can be observed by trying to place your right hand into a left-hand glove. Your
hands have two important properties: (1) each hand is the mirror image of the other, and
(2) the mirror images are not superimposable. The mirror image of the left hand looks like
the right hand, and vice versa, as shown in Figure O1.21.
FIGURE O1.21 Each hand is the mirror image of the other, but
neither hand is superimposable on the other.
Objects that possess a similar handedness are said to be chiral (literally, “handed”). Those
that do not are said to be achiral. Gloves are chiral. (It is difficult, if not impossible, to
place a right-hand glove on your left hand or a left-hand glove on your right hand.) Mittens, however, are often achiral. (Either mitten can fit on either hand.) Feet and shoes are
both chiral, but socks are not.
In 1874 Jacobus van’t Hoff and Joseph Le Bel recognized that a compound that contains a single tetrahedral carbon atom with four different substituents could exist in two
forms that were mirror images of each other. Consider the CHFClBr molecule, for example, which contains four different substituents on a tetrahedral carbon atom. Figure O1.22
shows one possible arrangement of the substituents and the mirror image of the structure.
By convention, solid lines are used to represent bonds that lie in the plane of the paper.
Wedges are used for bonds that come out of the plane of the paper toward the viewer;
dashed lines describe bonds that go behind the paper.
≥
H
A
C
Br ( E
F
Cl
≥
H
A
C
E ( Br
F
Cl
FIGURE O1.22 One form of the CHFClBr molecule and its mirror
image.
If we rotate the molecule on the right by 180° around the COH bond we get the structures shown in Figure O1.23. Chiral molecules are said to be optically active because they
are capable of rotating a plane of polarized light.
H
A
C
E ( Cl
F
Br
≥
≥
H
A
C
E ( Br
F
Cl
FIGURE O1.23 To predict whether a molecule is optically active, we have
to decide whether the molecule and its mirror image can be superimposed. We can start by rotating the mirror image by 180° around the
C—H bond.
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These structures are different because they cannot be superimposed on one another, as
shown in Figure O1.24.
≥ ≥
H
HA
AC
EC
Cl
FE
Br
F
Br
Cl
FIGURE O1.24 The compound CHFClBr is chiral because the structures are not superimposable.
CHFClBr is therefore a chiral molecule that exists in the form of a pair of stereoisomers
that are mirror images of each other. As a rule, any tetrahedral atom that carries four different substituents is a stereocenter, or a stereogenic atom. Compounds that contain a single stereocenter are always chiral. Some compounds that contain two or more stereocenters are achiral because of the symmetry of the relationship between the stereocenters.
The prefix “en-” often means “to make, or cause to be,” as in “endanger.” It is also
used to strengthen a term, to make it even more forceful, as in “enliven.” Thus, it isn’t
surprising that a pair of stereoisomers that are mirror images of each other are called enantiomers. They are literally compounds that contain parts that are forced to be across from
each other. Stereoisomers that aren’t mirror images of each other are called diastereomers.3
The cis/trans isomers of 2-butene, for example, are stereoisomers, but they are not mirror
images of each other. As a result, they are diastereomers.
Exercise O1.9
Which of the following compounds would form enantiomers because the molecule is
chiral?
CH3
A
CH3O COCH2CH3
A
Br
CH3
A
BrCH2OC OCH2CH3
A
H
2-Bromo-2-methylbutane
1-Bromo-2-methylbutane
Solution
The second carbon atom in 2-bromo-2-methylbutane contains two identical CH3 substituents. As a result, the compound is achiral and does not form enantiomers.
CH2CH3
A
C
E ( ¨ CH3
Br
CH3
3
2-Bromo-2-methylbutane
The prefix “dia-” is often used to indicate “opposite directions,” or “across,” as in “diagonal.”
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33
The second carbon atom in 1-bromo-2-methylbutane carries four different substituents: H,
BrCH2, CH3, and CH2CH3. As a result, the molecule is chiral and it forms enantiomers.
CH2CH3
A
EC ( ¨ CH3
BrCH2
H
and
CH2CH3
A
EC ( ¨ H
BrCH2
CH3
Enantiomers of 1-bromo-2-methylbutane
Note: Every object—with the possible exception of the vampires found on late-night TV—
has a mirror image. The relevant question is whether the mirror image can be superimposed on the source of the image. If it can, the object is not chiral (achiral). If it cannot,
the object is chiral and it can exist as a pair of stereoisomers.
O1.14 OPTICAL ACTIVITY
The Difference between Enantiomers on the Macroscopic Scale
If you could analyze the light that travels toward you from a lamp, you would find the electric and magnetic components of the radiation oscillating in all of the planes parallel to the
path of the light. However, if you analyzed light that has passed through a polarizer, such
as a Nicol prism or the lens of polarized sunglasses, you would find that the oscillations
were now confined to a single plane.
In 1813 Jean Baptiste Biot noticed that plane-polarized light was rotated either to the
right or the left when it passed through single crystals of quartz or aqueous solutions of
tartaric acid or sugar. Because they interact with light, substances that can rotate planepolarized light are said to be optically active. Those that rotate the plane clockwise (to the
right) are said to be dextrorotatory (from the Latin word dexter, “right”).4 Those that rotate the plane counterclockwise (to the left) are called levorotatory (from the Latin word
laevus, “left”).
In 1848 Louis Pasteur noted that sodium ammonium tartrate forms two different kinds
of crystals that are mirror images of each other, much as the right hand is a mirror image
of the left hand. By separating one type of crystal from the other with a pair of tweezers
he was able to prepare two samples of the compound. One was dextrorotatory when dissolved in aqueous solution; the other was levorotatory. Since the optical activity remained
after the compound had been dissolved in water, it could not be the result of macroscopic
properties of the crystals. Pasteur therefore concluded that there must be some asymmetry in the structure of the compound that allowed it to exist in two forms.
Once techniques were developed to determine the three-dimensional structure of a molecule, the source of the optical activity of a substance was recognized: Compounds that are
optically active contain molecules that are chiral. Chirality is a property of a molecule that
results from its structure. Optical activity is a macroscopic property of a collection of the
molecules that arises from the way they interact with light. Compounds, such as CHFClBr,
that contain a single stereocenter are the simplest to understand. One enantiomer of the chiral compound is dextrorotatory; the other is levorotatory. To decide whether a compound
should be optically active, we look for evidence that the molecules are chiral.
4
You might remember that “dextro” means right by noting that the predominantly right-handed world in
which we live uses words such as dextrous to mean unusually skilled at the use of one’s hands—in particular,
the “right” hand.
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The instrument with which optically active compounds are studied is a polarimeter,
shown in Figure O1.25. Imagine a horizontal line that passes through the zero of a coordinate system. By convention, negative numbers are placed on the left and positive numbers on the right of zero. Thus levorotatory compounds are indicated with a negative sign
() and dextrorotatory compounds with a positive sign ().
Emergent light
with rotated
plane of
polarization
Nicol
prism
polarizer
Unpolarized
light
α
Nicol
prism
analyzer
Incident
plane-polarized
light
Polarimeter tube containing
solution of an optical isomer
FIGURE O1.25 A polarimeter for studying the rotation of plane-polarized light.
The magnitude of the angle through which an enantiomer rotates plane-polarized light
depends on four quantities: (1) the wavelength of the light, (2) the length of the cell through
which the light passes, (3) the concentration of the optically active compound in the solution through which the light passes, and (4) the specific rotation of the compound, which
reflects the relative ability of the compound to rotate plane-polarized light. The specific
rotation of the dextrorotatory isomer of glucose is written as follows:
[] 20
D 3.12
When the spectrum of sunlight was first analyzed by Joseph von Fraunhofer in 1814, he
observed a limited number of dark bands in the spectrum, which he labeled A–H. We now
know that the D band in the spectrum is the result of the absorption by sodium atoms of
light that has a wavelength of 589.6 nm. The “D” in the symbol for specific rotation indicates that it is light of this wavelength that was studied. The “20” indicates that the experiment was done at 20°C. The “” sign indicates that the compound is dextrorotatory;
it rotates light clockwise. Finally, the magnitude of the measurement indicates that when
a solution of the compound with a concentration of 1.00 g/mL was studied in a 10-cm cell,
it rotated the light by 3.12°.
The magnitude of the rotations observed for a pair of enantiomers is always the same.
The only difference between the compounds is the direction in which they rotate planepolarized light. The specific rotation of the levorotatory isomer of the compound would
therefore be 3.12°.
The Difference between Enantiomers on the Molecular Scale
An unambiguous system for differentiating between enantiomers is based on the Latin
terms for left (sinister) and right (rectus).
•
Arrange the four substituents in order of decreasing atomic number of the atoms
attached to the stereocenter. (The substituent with the highest atomic number gets
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35
the highest priority.) The substituents in 2-bromobutane, for example, would be
listed in the order: Br CH3 CH2CH3 H.
1st priority
Br
A
CH3OCOCH2CH3
A
H
2nd priority
2nd priority
3rd priority
•
When two or more substituents have the same priority—such as the CH3 and
CH2CH3 groups in 2-bromobutane—continue working down the substituent chain
until you find a difference. In 2-bromobutane, we would give the CH2CH3 group a
higher priority than the CH3 group because the next point down the chain is a CH3
group in the CH2CH3 substituent and an H atom in the CH3 group. Thus, the four
substituents on 2-bromobutane would be listed in the order: Br CH2CH3 CH3 H.
1st priority
Br
A
CH3OCO CH2CH3
A
H
3rd priority
2nd priority
4th priority
•
View the enantiomer from the direction that places the substituent with the lowest
priority as far from the eye as possible. In the following example, this involves rotating the molecule counterclockwise around the C—CH2CH3 bond and tilting it
slightly around an axis that lies in the plane of the paper. When this is done, the
substituent that has the lowest priority is hidden from the eye.
CH2CH3
A
C
E ( ¨ H
CH3
Br
•
CH2CH3
%
C
; '
CH3
Br
Trace a path that links the substituents in decreasing order of priority. If the path
curves to the right—clockwise—the molecule is the rectus or R enantiomer. If it
curves to the left—counterclockwise—it is the sinister or S enantiomer.
2nd priority
CH2CH3
%
C
; '
CH3
Br
3rd priority
1st priority
(S)-2-Bromobutane
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In this example, the path curves to the left, so the enantiomer is the (S)-2-bromobutane
stereoisomer.
It is important to recognize that the (R)/(S) system is based on the structure of an individual molecule and the ()/() system is based on the macroscopic behavior of a large
collection of molecules. The most complete description of an enantiomer combines aspects
of both systems. The enantiomer analyzed in this section is best described as (S)-()-2bromobutane. It is the (S) enantiomer because of its structure and the () enantiomer because samples of the enantiomer with this structure are levorotatory; they rotate planepolarized light clockwise.
KEY TERMS
Achiral
Addition reaction
Alkane
Alkene
Alkyl bromide
Alkyl chloride
Alkyl halide
Alkyne
Branched hydrocarbon
Catalytic cracking
Catalytic reforming
Chiral
Cis/trans isomer
Coal
Coal gas
Conformation
Constitutional isomer
Crude oil
Cycloalkane
Dextrorotatory
Diastereomer
Enantiomer
Essential oil
Hydrocarbon
Isomer
Levorotatory
Markovnikov’s rule
Natural gas
Newman projection
Octane number
Optical activity
Organic chemistry
Specific rotation
Stereocenter
Stereogenic atom
Steroid
Stereoisomer
Straight-chain
hydrocarbon
Straight-run gasoline
Synthesis gas
Terpene
Terpenoid
Tetravalent
Thermal cracking
Thermal reforming
Town gas
Unsaturated
hydrocarbon
Vital force
Water gas
Water-gas shift reaction
PROBLEMS
The Special Role of Carbon
1. Explain why carbon forms covalent bonds, not ionic bonds, with so many other elements.
2. Carbon forms relatively strong double bonds, not only with itself, but with other nonmetals such as nitrogen, oxygen, phosphorus, and sulfur. Why is this important in terms
of the great diversity of carbon compounds?
3. In The Life Puzzle, Graham Cairns-Smith proposed three rules for speculating on the
origin of life: (1) take a good look at atoms and see what they can do, (2) take a good
look at organisms and see how they work, and then (3) put an organism together in
the easiest way you can think of. He reports two hypotheses to explain the striking
fundamental biochemical similarity between life in all its forms. That “the first organisms on Earth had a similar composition to modern forms.” Or that “modern biochemical
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37
uniformity is a product of evolution—a tribute to the effectiveness of natural
selection—rather than an indication that life depends uniquely on amino acids, purines
and so on.” Speculate on a few of the relative merits of these hypotheses, or propose
an alternative hypothesis.
The Saturated Hydrocarbons: Alkanes and Cycloalkanes
4. Use examples to explain the difference between saturated and unsaturated hydrocarbons and between straight-chain and branched hydrocarbons.
5. Explain why it is better to describe butane as a “straight-chain hydrocarbon” than it
is to describe it as a “linear hydrocarbon.”
6. Use the fact that straight-chain alkanes have a CH3 group at either end and a chain of
CH2 groups down the middle to explain why alkanes have the generic formula CnH2n2.
Write the generic formulas for cycloalkanes, alkenes, and alkynes.
7. Describe the difference between n-pentane, isopentane, and neopentane. Classify the
compounds as either stereoisomers or constitutional isomers.
8. Predict the number of constitutional isomers of heptane, C7H16.
9. Write the molecular formula for the saturated hydrocarbon that has the following carbon skeleton and name the compound.
C OCOCOC OC
A A
C C
A
C
10. Write the molecular formula for the saturated hydrocarbon that has the following carbon skeleton and name the compound.
C OC C OC OC
A A
CO C O C O C O C
A
C OC
11. Explain why it is possible to isolate different constitutional isomers of butane, but not
different conformations of butane.
12. Two of the conformations of butane can be described by the following diagrams, known
as Newman projections, which look down the C2—C3 bond in butane. Which of these
conformations is more favorable?
H
H
CH3
CH3
H
H
H
H
A
CH3
H
H
CH3
B
13. Provide the systematic (IUPAC approved) name for the following compound.
CH3
A
CH3CH2CHCHCH2CHCH3
A
A
CH3
CH2CH3
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14. Write the formula for the compound known as 1-methyl-1-chlorocyclopentane.
15. Draw the structure of 2,3,4-trimethyl-4-ethyloctane.
16. Provide the systematic for the compound in the following Newman projection.
CH2CH3
Br
Br
H
H
CH3
17. One way to decide whether a pair of structures represent different compounds is to assign a systematic name to each structure. Use this approach to decide whether the following are isomers or different descriptions of the same compound.
CH3
G
CHOCH2
D
i
CH3
CHOCH3
f
CH3
CH3
CH3
A
A
CH3CHCH2CHCH3
The Unsaturated Hydrocarbons: Alkenes and Alkynes
18. Draw the structures of all the alkenes that have the formula C6H12 and name the
compounds.
19. Draw the structures of all the alkynes that have the formula C5H8 and name the
compounds.
20. Explain why it is a mistake to name a compound 3-pentene. What would be the correct name of the compound?
21. Use the VSED theory to predict the shape of the tetrafluoroethylene (C2F4) molecule
that is the starting material used to make Teflon.
22. Explain why alkenes can form both constitutional isomers and stereoisomers. What
characteristic feature of a pair of alkenes can be used to decide whether they are constitutional isomers or stereoisomers?
23. Explain why alkynes can have constitutional isomers but not stereoisomers.
24. Describe the hybridization of each of the carbon atoms in the following compound.
CH3CHPCHCH2CqCH
25. Which of the following compounds does not have the same molecular formula as the
others?
(a) cyclopentane (b) methylcyclobutane (c) 1-pentene (d) pentane
(e) 1,1-dimethylcyclopropane
26. Which of the following compounds have cis/trans isomers?
(a) CHCl3 (b) F2CPCF2 (c) Cl2CPCHCH3 (d) FClCPCFCl (e) H2CPCHF
27. Which of the following compounds have cis/trans isomers?
(a) 1-pentene (b) 2-pentene (c) 2-methyl-2-butene
(d) 1-chloro-2-butene (e) 2-pentyne
28. Which of the following pairs of compounds would be constitutional isomers?
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39
(a) CH3CH2OCH2CH3 and CH3CH2CH2CH2OH
(b) (CH3)2CHCH3 and CH3CH2CH2CH3
CH3
A
(c) CH3CH2CH2CH3 and CH2CH2
A
CH3
Cl
Cl
Cl
H
(d)
and
H
H
H
Cl
29. Determine whether the following compounds are isomers. Explain why or why not.
30.
31.
32.
33.
34.
Draw the structure for cis-2,3-dichloro-2-hexene.
Draw the structure for 2,6-dimethyl-3-heptyne.
Draw the structure for trans-6-methyl-2-heptene.
Draw the structure for cyclohexene.
Draw the structure for 2,3-dimethyl-2-pentene. Explain why the compound can’t exist
as a pair of cis/trans isomers.
The Reactions of Alkanes, Alkenes, and Alkynes
35. Write a balanced equation for the reaction in which isooctane, C8H18, burns in oxygen
to form CO2 and H2O vapor.
36. Define the term addition reaction and give an example of an addition reaction of an
alkene.
37. Predict the product of the addition reaction between Br2 and 2-pentene.
38. Predict the product of the addition reaction between water and 1-pentene in the presence of sulfuric acid.
39. Predict the product of the addition reaction between excess H2 and 2-pentyne in the
presence of a platinum metal catalyst.
40. Which of the following is a product of the reaction between chlorine and ethylene?
(a) CH3CHCl2 (b) CH3CH2Cl (c) ClCH2CH2Cl (d) Cl2CHCHCl2
41. Which of the following does not rapidly react with Br2 dissolved in CCl4?
(a) pentane (b) 1-pentene (c) 2-pentene (d) 1-pentyne (e) 2-pentyne
42. Assume that HI adds to a CPC double bond according to Markovnikov’s rule. Draw
the structure and name the product of the reaction between hydrogen iodide and
2-methyl-2-pentene.
43. Initial attempts to confirm Markovnikov’s rule for the addition of HBr to an alkene
were complicated by the fact that the reactions did not always give the same product.
We now know that HBr adds to a CPC double according to Markovnikov’s rule
unless the solvent is contaminated with a source of free radicals. Draw the structures
and name the products of both the Markovnikov addition of HBr to 1-pentene and
the anti-Markovnikov addition of HBr that occurs in the presence of a free radical
contaminant.
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THE STRUCTURE OF HYDROCARBONS
The Aromatic Hydrocarbons and Their Derivatives
44. Draw the structures of the following aromatic compounds: aniline, anisole, benzene,
and benzoic acid.
45. Draw the structures of ortho-, meta-, and para-bromotoluene.
46. TNT is an abbreviation for 2,4,6-trinitrotoluene. Toluene is a derivative of benzene in
which a methyl (CH3) group is substituted for one of the hydrogen atoms. Trinitrotoluene is a derivative of toluene in which NO2 groups have replaced three more hydrogen atoms on the benzene ring. Draw the structures of all the possible isomers of
trinitrotoluene. Label the isomer that is 2,4,6-trinitrotoluene.
47. On an exam, a student described benzene as a mixture of two structures that are rapidly
being converted from one to the other. What is wrong with that answer?
48. Explain why the 12 atoms in a benzene molecule all lie in the same plane.
The Chemistry of Petroleum Products and Coal
49. Explain why a mixture of CO and H2 can be used as a fuel. What are the products of
the combustion of the mixture, which was once known as “water gas.”
50. Natural gas, petroleum ether, gasoline, kerosene, and asphalt are all different forms of
hydrocarbons that give off energy when burned. Describe how the substances differ.
What happens to the boiling points of the mixtures as the average length of the hydrocarbon chain increases?
51. Which of the following compounds has the largest octane number?
(a) n-butane (b) n-pentane (c) n-hexane (d) n-octane
52. Which of the following won’t increase the octane number of gasoline?
(a) increasing the concentration of branched-chain alkanes (b) increasing the concentration of cycloalkanes (c) increasing the concentration of aromatic hydrocarbons (d) increasing the average length of the hydrocarbon chains.
53. Describe how thermal cracking and catalytic cracking increase the amount of highoctane gasoline that can be obtained from a barrel of oil. Explain why neither thermal
reforming nor catalytic reforming can achieve this.
54. Coal gas can be obtained when coal is heated in the absence of air. Water gas can be
obtained when coal reacts with steam. Describe the difference between the gases and
explain why much more water gas can be extracted from a ton of coal.
Optical Activity
55. Describe the difference between constitutional isomers and stereoisomers. What characteristic feature of a molecule could be used to distinguish between the two forms of
isomers?
56. Objects that cannot be superimposed on their mirror images are said to be chiral, and
chiral molecules are optically active. Which of the following molecules are optically
active?
(a) CH4 (b) CH3Cl (c) CHCl3 (d) CHFCl2 (e) CHFClBr
57. Which of the following compounds are optically active?
CH3
A
(a) C2H4 (b) C6H6 (c) C6H4Cl2 (d) CH3CH2CHCH3
58. Coniine is the active ingredient in the poison known as hemlock that was given to
EQA
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THE STRUCTURE OF HYDROCARBONS
41
Socrates in 399 B.C. Use the following diagram of the structure of coniine to predict
whether the compound is optically active.
N
A
H
59. Use the structure of caffeine shown below to predict whether the compound is optically active.
CH3
O
CH3
N
O
N
N
N
CH3
60. Determine the number of centers of chirality in the structure of vitamin C shown in
Figure O1.2.
61. Predict whether the following compound is chiral.
CH2CH2CH2CH3
A
CH3ONOCH2CH3
A
CH(CH3)2
62. Which of the following compounds, which play an important role in the chemistry of
biological systems, is chiral?
CO2H
A
CH2
A
HOOCOCO2H
A
CH2
A
CO2H
Citric acid
CO2H
A
CH2
A
HOCOCO2H
A
HOOCH
A
CO2H
CO2H
A
CH2
A
CH2
A
HOOCOCO2H
A
CH2
A
CO2H
Isocitric acid
Homocitric acid
63. Which (if any) of the carbon atoms in 2-bromo-3-methylbutane are stereocenters?
64. Determine whether the following isomers are enantiomers or diastereomers. What
characteristic feature of the molecule can be used to make the decision?
CH2CH3
A
C
D ( CH3
H
Br
CH2CH3
A
C
D ( Br
H
CH3
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THE STRUCTURE OF HYDROCARBONS
65. Determine whether the following isomers are enantiomers or diastereomers. What
characteristic feature of the molecule can be used to make the decision?
66. Determine whether the following compound is the R or S enantiomer of 2-bromobutane.
CH2CH3
A
C
D ( Br
CH3 H