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Alkenes
Structure, Nomenclature, Stability,
and an Introduction to Reactivity
E isomer
of 2-butene
I
n Chapter 3, we saw that alkanes are hydrocarbons that
contain only carbon–carbon single bonds. Hydrocarbons
that contain a carbon–carbon double bond are called alkenes.
Alkenes play many important roles in biology. Ethene, for example, is a
plant hormone—a compound that controls the plant’s growth and other
changes in its tissues. Ethene affects seed germination, flower maturation,
and fruit ripening. Many of the flavors and fragrances produced by certain
plants also belong to the alkene family.
Z isomer
of 2-butene
OH
citronellol
in rose and
geranium oils
limonene
in lemon and
orange oils
b-phellandrene
oil of eucalyptus
We will start our study of alkenes by looking at how they are named, their structures, and their relative stabilities. Then we will take a look at a reaction of an alkene,
paying close attention to why alkenes react the way they do. We will also look at how
chemists portray the step-by-step process by which a reaction occurs and the energy
changes that occur during a reaction. You will see that some of this material is based
on information with which you are familiar while some is new. Both will add to your
foundation of knowledge that will be built on in subsequent chapters.
Ethene is the hormone that causes
tomatoes to ripen.
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PHEROMONES
Insects communicate by releasing pheromones—chemical substances that other insects of the same species detect with
their antennae. There are sex, alarm, and trail pheromones, and many of these are alkenes. Interfering with an insect’s
ability to send or receive chemical signals is an environmentally safe way to control insect populations. For example, traps with synthetic sex attractants have been used to capture such crop-destroying insects as the gypsy moth and the boll weevil.
muscalure
sex attractant of the housefly
3-D Molecules:
Limonene; b -Phellandrene;
Multifidene
The general molecular formula for
a hydrocarbon is CnH2n 2 , minus two
hydrogens for every P bond or ring
present in the molecule.
4.1
multifidene
sex attractant of
brown algae
Molecular Formulas
We have seen that the general molecular formula for a noncyclic alkane is CnH 2n + 2
(Section 3.0). You also learned that the general molecular formula for a cyclic alkane
is CnH 2n because the cyclic structure reduces the number of hydrogens by two
(Section 3.3).
The general molecular formula for a noncyclic alkene is also CnH 2n because, as a
result of the double bond, an alkene has two fewer hydrogens than an alkane with the
same number of carbons. Thus, the general molecular formula for a cyclic alkene must
be CnH 2n - 2 . We can, therefore, make the following statement: The general molecular
formula for a hydrocarbon is CnH 2n + 2 , minus two hydrogens for every bond and/or
ring in the molecule.
CH3CH2CH2CH2CH3
CH3CH2CH2CH
an alkane
C5H12
CnH2n+2
an alkene
C5H10
CnH2n
CH2
a cyclic alkane
C5H10
CnH2n
a cyclic alkene
C5H8
CnH2n−2
Because alkanes contain the maximum number of C ¬ H bonds possible—that is,
they are saturated with hydrogen—they are called saturated hydrocarbons. In contrast, alkenes are called unsaturated hydrocarbons because they have fewer than the
maximum number of hydrogens.
CH3CH2CH2CH3
a saturated hydrocarbon
PROBLEM 1 ◆
CH3CH
CHCH3
an unsaturated hydrocarbon
SOLVED
Determine the molecular formula for each of the following:
a. a 5-carbon hydrocarbon with one p bond and one ring
b. a 4-carbon hydrocarbon with two p bonds and no rings
c. a 10-carbon hydrocarbon with one p bond and two rings
For a 5-carbon hydrocarbon with no p bonds and no rings,
C nH 2n + 2 = C 5H 12 . A 5-carbon hydrocarbon with one p bond and one ring has four
fewer hydrogens, because two hydrogens are subtracted for every p bond or ring present in
the molecule. Its molecular formula, therefore, is C 5H 8 .
SOLUTION TO 1a
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Section 4.2
PROBLEM 2 ◆
Nomenclature of Alkenes
83
SOLVED
Determine the total number of double bonds and/or rings for the hydrocarbons with the
following molecular formulas:
a. C 10H 16
c. C 8H 16
b. C 20H 34
For a 10-carbon hydrocarbon with no p bonds and no rings,
C nH 2n + 2 = C 10H 22 . Thus, a 10-carbon compound with molecular formula C 10H 16 has
six fewer hydrogens. Therefore, the total number of its double bonds and/or rings is three.
SOLUTION TO 2a
4.2
Nomenclature of Alkenes
The functional group is the center of reactivity in a molecule. In an alkene, the double bond is the functional group. The IUPAC system uses a suffix to denote certain
functional groups. The systematic name of an alkene, for example, is obtained by replacing the ane at the end of the parent hydrocarbon’s name with the suffix ene. Thus,
a two-carbon alkene is called ethene and a three-carbon alkene is called propene.
Ethene also is frequently called by its common name: ethylene.
H2C
CH2
CH3CH
systematic name: ethene
common name:
ethylene
CH2
propene
propylene
cyclopentene
cyclohexene
The following rules are used to name a compound that has a functional group suffix:
1. The longest continuous chain containing the functional group (in this case, the
carbon–carbon double bond) is numbered in a direction that gives the functional
group suffix the lowest possible number. The position of the double bond is indicated by the number immediately preceding the name of the alkene. For example, 1-butene signifies that the double bond is between the first and second
carbons of butene; 2-hexene signifies that the double bond is between the second and third carbons of hexene. (The four alkene names shown above do not
need a number, because there is no ambiguity.)
4
3
2
CH3CH2CH
1
1
CH2
CH3CH
2
1-butene
3
1
4
2
3
CH3CH
CHCH3
4
5
6
CHCH2CH2CH3
2-hexene
2-butene
Notice that 1-butene does not have a common name. You might be tempted to
call it “butylene,” which is analogous to “propylene” for propene, but butylene is
not an appropriate name. A name must be unambiguous, and “butylene” could
signify either 1-butene or 2-butene.
2. The name of a substituent is cited before the name of the longest continuous
chain that contains the functional group, together with a number to designate
the carbon to which the substituent is attached. Notice that if there is a functional group suffix and a substituent, the functional group suffix gets the lowest
possible number.
CH3
1
2
CH3CH
3
4
5
CHCHCH3
4-methyl-2-pentene
Number the longest continuous chain
containing the functional group in the
direction that gives the functional group
suffix the lowest possible number.
2
1
CH2CH3
3
CH3C
4
5
6
7
CHCH2CH2CH3
3-methyl-3-heptene
When there is only a substituent, the
substituent gets the lowest possible
number.
When there is only a functional group
suffix, the functional group suffix gets
the lowest possible number.
When there is both a functional group
suffix and a substituent, the functional
group suffix gets the lowest possible
number.
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Alkenes
3. If a chain has more than one double bond, we first identify the chain that contains all the double bonds by its alkane name, replacing the “ne” ending with the
appropriate suffix: diene, triene, etc. The chain is numbered in the direction that
yields the lowest number in the name of the compound.
1
2
3
4
5
1
CH2
CH
CH2
CH
CH2
CH3CH
2
1,4-pentadiene
Substituents are cited in alphabetical
order.
3
4
5
CH
CH
CHCH2CH3
6
7
5
2,4-heptadiene
3
2
1
CH
CH
CH2
1,3-pentadiene
4. If a chain has more than one substituent, the substituents are cited in alphabetical
order, using the same rules for alphabetizing that you learned in Section 3.2. The
appropriate number is then assigned to each substituent.
2
1
4
3
6
5
Br Cl
8
7
CH3CH2CHCHCH2CH
7
3,6-dimethyl-3-octene
A substituent receives the lowest
possible number only if there is no
functional group suffix or if the same
number for the functional group suffix
is obtained in both directions.
4
CH3CH
6
5
4
3
2
CH2
1
5-bromo-4-chloro-1-heptene
5. If the same number for the alkene functional group suffix is obtained in both
directions, the correct name is the name that contains the lowest substituent number.
CH3CH2CH2C
CH3
CHCH2CHCH3
CH3CHCH
CH3
Br
2,5-dimethyl-4-octene
not
4,7-dimethyl-4-octene
because 2 < 4
CCH2CH3
CH3
2-bromo-4-methyl-3-hexene
not
5-bromo-3-methyl-3-hexene
because 2 < 3
6. A number is not needed to denote the position of the double bond in a cyclic
alkene because the ring is always numbered so that the double bond is between
carbons 1 and 2. In determining a substituent number, you need to move around
the ring in the direction (clockwise or counterclockwise) that puts the lowest
number into the name.
Tutorial:
Alkene nomenclature
2
3
1
CH2CH3
5
1
4
3-ethylcyclopentene
CH3
CH3
CH2CH3
4
2
5
CH3
3
4,5-dimethylcyclohexene
4-ethyl-3-methylcyclohexene
7. Numbers are needed if the ring has more than one double bond.
CH3
1,3-cyclohexadiene
1,4-cyclohexadiene
2-methyl-1,3-cyclopentadiene
Remember that the name of a substituent is stated before the name of the parent hydrocarbon, and the functional group suffix is stated after the name of the parent hydrocarbon.
[substituent] [parent hydrocarbon] [functional group suffix]
The sp 2 carbons of an alkene are called vinylic carbons. An sp 3 carbon that is
adjacent to a vinylic carbon is called an allylic carbon.
vinylic carbons
RCH2
CH
CH
allylic carbons
CH2R
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Section 4.3
The Structure of Alkenes
Two groups containing a carbon–carbon double bond are used in common names—
the vinyl group and the allyl group. The vinyl group is the smallest possible group
that contains a vinylic carbon; the allyl group is the smallest possible group that
contains an allylic carbon. When “allyl” is used in nomenclature, the substituent must
be attached to the allylic carbon.
H2C
CH
H2C
the vinyl group
H2C
CHCl
H2C
systematic name: chloroethene
common name:
vinyl chloride
CHCH2
the allyl group
CHCH2Br
3-bromopropene
allyl bromide
PROBLEM 3
Draw the structure for each of the following compounds:
a. 3,3-dimethylcyclopentene
b. 6-bromo-2,3-dimethyl-2-hexene
c. ethyl vinyl ether
d. allyl alcohol
PROBLEM 4 ◆
Give the systematic name for each of the following compounds:
a. CH3CHCH
c. BrCH2CH2CH
CHCH3
CCH3
CH2CH3
CH3
CH3
b. CH3CH2C
CCHCH3
d.
H3C
CH3 Cl
4.3
CH3
The Structure of Alkenes
The structure of the smallest alkene (ethene) was described in Section 1.8. Other
alkenes have similar structures. Each double-bonded carbon of an alkene has three sp 2
orbitals that lie in a plane with angles of 120°. Each of these sp 2 orbitals overlaps an
orbital of another atom to form a s bond. Thus, one of the carbon–carbon bonds in a
double bond is a s bond, formed by the overlap of an sp 2 orbital of one carbon with an
sp 2 orbital of the other carbon. The second carbon–carbon bond in the double bond
(the p bond) is formed from side-to-side overlap of the remaining p orbitals of the sp 2
carbons. Because three points determine a plane, each sp 2 carbon and the two atoms
singly bonded to it lie in a plane. In order to achieve maximum orbital–orbital overlap,
the two p orbitals must be parallel to each other. Therefore, all six atoms of the doublebond system are in the same plane.
H3C
CH3
C
H3C
C
CH3
the six carbon atoms
are in the same plane
Tutorial:
Common names of
alkyl groups
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It is important to remember that the p bond represents the cloud of electrons that
is above and below the plane defined by the two sp 2 carbons and the four atoms
bonded to them.
p orbitals overlap to form
a p bond
H3C
3-D Molecule:
2,3-Dimethyl-2-butene
CH3
C
C
CH3
H3C
PROBLEM 5 ◆
SOLVED
For each of the following compounds, tell how many of its carbon atoms lie in the
same plane:
CH3
CH3
a.
b.
c.
CH3
d.
CH3
H3C
The two sp 2 carbons (blue dots) and the carbons bonded to each of
the sp carbons (red dots) lie in the same plane. Therefore, five carbons lie in the same
plane.
SOLUTION TO 5a
2
CH3
4.4
Cis–Trans Isomerism
We have just seen that the two p orbitals that form the p bond must be parallel to
achieve maximum overlap. Therefore, rotation about a double bond does not readily
occur. If rotation were to occur, the two p orbitals would cease to overlap—in other
words, the p bond would break (Figure 4.1).
Because rotation about a double bond does not readily occur, an alkene such as
2-butene can exist in two distinct forms: The hydrogens bonded to the sp 2 carbons can
be on the same side of the double bond or on opposite sides of the double bond. The
p bond is broken
H3C
CH3
C
H
H3C
C
H
C
H
cis isomer
the hydrogens are on the
same side of the double bond
H
H
H3C
C
C
CH3
H
C
CH3
trans isomer
the hydrogens are on
opposite side of the double bond
▲ Figure 4.1
Rotation about the carbon–carbon double bond would break the p bond.
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Section 4.4
Cis–Trans Isomerism
87
isomer with the hydrogens on the same side of the double bond is called the cis isomer,
and the isomer with the hydrogens on opposite sides of the double bond is called the
trans isomer. A pair of isomers such as cis-2-butene and trans-2-butene is called
cis–trans isomers or geometric isomers. This should remind you of the cis–trans isomers of disubstituted cyclohexanes you encountered in Section 3.12— the cis isomer
had its substituents on the same side of the ring, and the trans isomer had its substituents on opposite sides of the ring.
H3C
CH3
C
C
H
H
H
H3C
C
C
CH3
H
cis-2-butene
3-D Molecules:
cis-2-Butene; trans-2-Butene
trans-2-butene
If one of the sp 2 carbons of the double bond is attached to two identical substituents, there is only one possible structure for the alkene. In other words, cis and
trans isomers are not possible for an alkene that has identical substituents attached to
one of the double-bonded carbons.
cis and trans isomers are not possible for these compounds because
two substituents on an sp2 carbon are the same
CH3
H
C
CH3CH2
C
C
Cl
H
CH3
3-D Molecule:
2-Methyl-2-pentene
C
CH3
H
Cis and trans isomers can be interconverted only when the molecule absorbs sufficient heat or light energy to cause the p bond to break, because once the p bond is
broken, rotation can occur easily about the remaining s bond (Section 3.8).
CH2CH3
H3C
C
H
C
H
H
H3C
> 180 ºC
or
h
C
H
C
CH2CH3
trans-2-pentene
cis-2-pentene
PROBLEM 6 ◆
a. Which of the following compounds can exist as cis–trans isomers?
b. For those compounds, draw and label the cis and trans isomers.
1. CH3CH
CHCH2CH3
3. CH3CH
2. CH3CH
CCH3
4. CH3CH2CH
CH3
CHCH3
CH2
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CIS–TRANS INTERCONVERSION IN VISION
When rhodopsin absorbs light, a double bond interconverts between the cis and trans forms. This process plays an
important role in vision.
trans double bond
cis double bond
light
N
N
opsin
opsin
rhodopsin
OPSIN
OPSIN
cis form
3-D Molecules:
cis-Retinal; trans-Retinal
trans form
4.5
The E,Z System of Nomenclature
As long as each of the sp 2 carbons of an alkene is bonded to only one substituent, we
can use the terms cis and trans to designate the structure of the alkene. If the hydrogens
are on the same side of the double bond, it is the cis isomer; if they are on opposite
sides of the double bond, it is the trans isomer. But how can we designate the isomers
of a compound such as 1-bromo-2-chloropropene?
Cl
Br
C
H
CH3
Br
C
C
CH3
H
C
Cl
Which isomer is cis and which is trans?
The Z isomer has the high-priority
groups on the same side.
The cis–trans system of nomenclature cannot be used for such a compound because
there are four different groups on the two sp 2 carbons. The E,Z system of nomenclature was devised for these kinds of compounds.
To name an isomer by the E,Z system, we first determine the relative priorities of
the two groups bonded to one of the sp 2 carbons and then the relative priorities of the
two groups bonded to the other sp 2 carbon. (The rules for assigning relative priorities
will soon be explained.) If the high-priority groups are on the same side of the double
bond, the isomer has the Z configuration (Z is for zusammen, German for “together”).
If the high-priority groups are on opposite sides of the double bond, the isomer has the
E configuration (E is for entgegen, German for “opposite”).
low priority
low priority
C
high priority
high priority
low priority
C
C
high priority
the Z isomer
high priority
C
low priority
the E isomer
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Section 4.5
The E,Z System of Nomenclature
89
The relative priorities of the two groups bonded to an sp 2 carbon are determined
using the following rules:
• Rule 1. The relative priorities of the two groups depend on the atomic numbers
of the atoms that are bonded directly to the sp 2 carbon. The greater the atomic
number, the higher is the priority of the group.
For example, in the following compounds, one of the sp 2 carbons is bonded
to a Br and to an H:
high priority
high priority
Cl
Br
C
CH3
Br
C
H
The greater the atomic number of the
atom bonded to an sp2 carbon, the
higher is the priority of the substituent.
C
C
H
CH3
the Z isomer
Cl
the E isomer
Br has a greater atomic number than H, so Br has a higher priority than H.
The other sp 2 carbon is bonded to a Cl and to a C. Cl has the greater atomic number, so Cl has a higher priority than C. (Notice that you use the atomic number
of C, not the mass of the CH 3 group, because the priorities are based on the
atomic numbers of atoms, not on the masses of groups.) The isomer on the left
has the high-priority groups (Br and Cl) on the same side of the double bond,
so it is the Z isomer. (Zee groups are on Zee Zame Zide.) The isomer on the
right has the high-priority groups on opposite sides of the double bond, so it is
the E isomer.
• Rule 2. If the two substituents attached to the sp 2 carbon start with the same
atom (there is a tie), you must move outward from the point of attachment and
consider the atomic numbers of the atoms that are attached to the “tied” atoms.
In the following compounds, one of the sp 2 carbons is bonded both to a Cl
and to a C:
CH3
CHCH3
ClCH2
C
Cl
C
CH3
CHCH3
Cl
C
CH2OH
the Z isomer
ClCH2
C
CH2OH
the E isomer
Cl has a greater atomic number than C, so Cl has the higher priority. Both
atoms bonded to the other sp 2 carbon are C’s (from a CH 2OH group and a
CH(CH 3)2 group), so there is a tie at this point. The C of the CH 2OH group is
bonded to O, H, and H, and the C of the CH(CH 3)2 group is bonded to C, C,
and H. Of these six atoms, O has the greatest atomic number, so CH 2OH has
a higher priority than CH(CH 3)2 . (Note that you do not add the atomic numbers; you take the single atom with the greatest atomic number.) The E and Z
isomers are as shown.
• Rule 3. If an atom is doubly bonded to another atom, the priority system treats
it as if it were singly bonded to two of those atoms. If an atom is triply bonded to
another atom, the priority system treats it as if it were singly bonded to three of
those atoms.
If the atoms attached to sp2 carbons
are the same, the atoms attached to
the “tied” atoms are compared; the
one with the greatest atomic number
belongs to the group with the
higher priority.
Tutorial:
E and Z Nomenclature
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For example, one of the sp 2 carbons in the following pair of isomers is bonded to a
CH 2CH 3 group and to a CH “ CH 2 group:
CH
HOCH2
C
HC
C
CH2
C
C
CH2CH3
HC
the Z isomer
If an atom is doubly bonded to another
atom, treat it as if it were singly bonded
to two of those atoms.
If an atom is triply bonded to another
atom, treat it as if it were singly bonded
to three of those atoms.
Cancel atoms that are identical in the
two groups; use the remaining atoms to
determine the group with the higher
priority.
CH2CH3
HOCH2
C
C
CH
CH2
the E isomer
Because the atoms immediately bonded to the sp 2 carbon are both C’s, there is a tie.
The triple-bonded C is considered to be bonded to C, C, and C; the other C is bonded
to O, H, and H. Of the six atoms, O has the greatest atomic number, so CH 2OH has a
higher priority than C ‚ CH. Both atoms that are bonded to the other sp2 carbon are
C’s, so there is a tie there as well. The first carbon of the CH 2CH 3 group is bonded
to C, H, and H. The first carbon of the CH “ CH 2 group is bonded to an H and
doubly bonded to a C. Therefore, it is considered to be bonded to H, C, and C. One C
cancels in each of the two groups, leaving H and H in the CH 2CH 3 group and C and H
in the CH “ CH 2 group. C has a greater atomic number than H, so CH “ CH 2 has a
higher priority than CH 2CH 3 .
PROBLEM 7 ◆
Assign relative priorities to each set of substituents:
a. ¬ Br, ¬ I, ¬ OH, ¬ CH 3
b. ¬ CH 2CH 2OH, ¬ OH, ¬ CH 2Cl, ¬ CH “ CH 2
PROBLEM 8
Draw and label the E and Z isomers for each of the following compounds:
a. CH3CH2CH
CHCH3
b. CH3CH2CH2CH2
CH3CH2C
CCH2Cl
CH3CHCH3
PROBLEM 9
Indicate whether each of the following is an E isomer or a Z isomer:
a.
b.
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Section 4.6
The Relative Stabilities of Alkenes
PROBLEM-SOLVING STRATEGY
Draw the structure of (E)-1-bromo-2-methyl-2-butene.
First draw the compound without specifying the isomer; this allows you to identify the
substituents bonded to the sp 2 carbons. Then determine the relative priorities of the two
substituents on each of the sp 2 carbons.
CH3
BrCH2C
CHCH3
One sp 2 carbon is attached to a CH 3 and an H; CH 3 has the higher priority. The other sp 2
carbon is attached to a CH 3 and a CH 2Br; CH 2Br has the higher priority. To get the E isomer, draw the compound with the two high-priority substituents on opposite sides of the
double bond.
BrCH2
C
H
C
H3C
CH3
Now continue on to Problem 10.
PROBLEM 10 ◆
Draw (Z)-3-isopropyl-2-heptene.
4.6
The Relative Stabilities of Alkenes
Alkyl substituents that are bonded to the sp 2 carbons of an alkene have a stabilizing
effect on the alkene.
relative stabilities of alkyl-substituted alkenes
most stable
R
R
C
R
R
>
C
R
R
C
R
R
>
C
H
H
C
R
R
>
C
H
H
C
H
C
least stable
H
We can, therefore, make the following statement: The more alkyl substituents bonded
to the sp 2 carbons of an alkene, the greater is its stability. (Some students find it easier to look at the number of hydrogens bonded to the sp 2 carbons. In terms of
hydrogens, we can use the following rule: The fewer hydrogens bonded to the sp 2
carbons of an alkene, the greater is its stability.)
PROBLEM 11
a. Which of the following compounds is the most stable?
CH2CH3
CH2CH3
CH2CH3
CH2CH3
CH2CH3
CH2CH3
b. Which is the least stable?
Both trans-2-butene and cis-2-butene have two alkyl groups bonded to their sp 2
carbons. The trans isomer, in which the large substituents are farther apart, is more stable than the cis isomer, in which the large substituents are closer together.
The fewer hydrogens bonded to the
sp2 carbons of an alkene, the more
stable it is.
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the cis isomer has steric strain
H H
H
the trans isomer does not have steric strain
H
H
C
C
H
C
C
H
H
H
H
C
H
C
H
H
C
C
H
H
cis-2-butene
H
trans-2-butene
When the large substituents are on the same side of the molecule, their electron clouds
can interfere with each other, causing strain in the molecule and making it less stable.
This kind of strain is called steric strain (Section 3.8). When the large substituents are
on opposite sides of the molecule, their electron clouds cannot interact; thus, the molecule has less steric strain and is more stable.
PROBLEM 12 ◆
Rank the following compounds in order of decreasing stability:
trans-3-hexene, cis-3-hexene, cis-2,5-dimethyl-3-hexene
4.7
How Alkenes React • Curved Arrows
There are many millions of organic compounds. If you had to memorize how each of
them reacts, studying organic chemistry would be a horrendous experience. Fortunately, organic compounds can be divided into families, and all the members of a family
react in similar ways. What determines the family to which an organic compound
belongs is its functional group. The functional group is the center of reactivity of a
molecule. You will find a table of common functional groups inside the back cover of
this book. You are already familiar with the functional group of an alkene: the
carbon–carbon double bond. All compounds with a carbon–carbon double bond react
in similar ways, whether the compound is a small molecule like ethene or a large
molecule like cholesterol.
H
H
C
+ HBr
C
H
H
H
H
H
C
C
H
Br H
ethene
+ HBr
HO
HO
Br
H
cholesterol
To further reduce the need to memorize, one needs to understand why a functional
group reacts the way it does. It is not sufficient to know that a compound with a carbon–
carbon double bond reacts with HBr to form a product in which the H and Br atoms
have taken the place of the p bond; we need to understand why the compound reacts
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How Alkenes React • Curved Arrows
with HBr. In each chapter that discusses the reactivity of a particular functional group,
we will see how the nature of the functional group allows us to predict the kind of reactions it will undergo. Then, when you are confronted with a reaction you have never
seen before, knowledge of how the structure of the molecule affects its reactivity will
help you predict the products of the reaction.
In essence, organic chemistry is about the interaction between electron-rich atoms
or molecules and electron-deficient atoms or molecules. It is these forces of attraction
that make chemical reactions happen. From this follows a very important rule that
determines the reactivity of organic compounds: Electron-rich atoms or molecules are
attracted to electron-deficient atoms or molecules. Each time you study a new functional group, remember that the reactions it undergoes can be explained by this very
simple rule.
Therefore, to understand how a functional group reacts, you must first learn to recognize electron-deficient and electron-rich atoms and molecules. An electron-deficient
atom or molecule is called an electrophile. An electrophile has an atom that can
accept a pair of electrons. Thus, an electrophile looks for electrons. Literally, “electrophile” means “electron loving” (phile is the Greek suffix for “loving”).
Electron-rich atoms or molecules are
attracted to electron-deficient atoms
or molecules.
+
H+
CH3CH2
these are electrophiles because they
can accept a pair of electrons
An electron-rich atom or molecule is called a nucleophile. A nucleophile has a pair
of electrons it can share. Some nucleophiles are neutral and some are negatively
charged. Because a nucleophile has electrons to share and an electrophile is seeking
electrons, it should not be surprising that they attract each other. Thus, the preceding
rule can be restated as a nucleophile reacts with an electrophile.
HO
−
Cl
−
CH3NH2
H2O
these are nucleophiles because they
have a pair of electrons to share
We have seen that a p bond is weaker than a s bond (Section 1.14). The p bond,
therefore, is the bond that is most easily broken when an alkene undergoes a reaction.
We also have seen that the p bond of an alkene consists of a cloud of electrons above
and below the s bond. This cloud of electrons causes an alkene to be an electron-rich
molecule—it is a nucleophile. (Notice the relatively electron-rich pale orange area in
the electrostatic potential maps for cis- and trans-2-butene in Section 4.3.) We can,
therefore, predict that an alkene will react with an electrophile and, in the process, the
p bond will break. So if a reagent such as hydrogen bromide is added to an alkene, the
alkene (a nucleophile) will react with the partially positively charged hydrogen (an
electrophile) of hydrogen bromide and a carbocation will be formed. In the second
step of the reaction, the positively charged carbocation (an electrophile) will react with
the negatively charged bromide ion (a nucleophile) to form an alkyl halide.
CH3CH
d+
CHCH3 + H
d−
Br
CH3CH
+
CHCH3 + Br−
H
a carbocation
CH3CH
Br
CHCH3
H
2-bromobutane
an alkyl halide
The description of the step-by-step process by which reactants (e.g., alkene + HBr)
are changed into products (e.g., alkyl halide) is called the mechanism of the reaction. To
help us understand a mechanism, curved arrows are drawn to show how the electrons
move as new covalent bonds are formed and existing covalent bonds are broken. In other
A nucleophile reacts with an
electrophile.
93
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Curved arrows show the flow
of electrons; they are drawn from an
electron-rich center to an electrondeficient center.
words, the curved arrows show which bonds are formed and which are broken. Because
the curved arrows show how the electrons flow, they are drawn from an electron-rich
center (at the tail of the arrow) to an electron-deficient center (at the point of the arrow).
Notice that these arrows represent the simultaneous movement of two electrons (an electron pair). These are called “curved” arrows to distinguish them from the “straight”
arrows used to link reactants with products in chemical reactions.
p bond has broken
d+
CH3CH
CHCH3 + H
d−
Br
CH3CH
+
CHCH3 +
H
Br
−
new s bond
has formed
For the reaction of 2-butene with HBr, an arrow is drawn to show that the two electrons of the p bond of the alkene are attracted to the partially positively charged hydrogen of HBr. The hydrogen, however, is not free to accept this pair of electrons
because it is already bonded to a bromine, and hydrogen can be bonded to only one
atom at a time (Section 1.4). Therefore, as the p electrons of the alkene move toward
the hydrogen, the H ¬ Br bond breaks, with bromine keeping the bonding electrons.
Notice that the p electrons are pulled away from one carbon, but remain attached to
the other. Thus, the two electrons that formerly formed the p bond now form a s bond
between carbon and the hydrogen from HBr. The product of this first step in the
reaction is a carbocation because the sp 2 carbon that did not form the new bond with
hydrogen has lost a share in a pair of electrons (the electrons of the p bond). It is,
therefore, positively charged.
In the second step of the reaction, a lone pair on the negatively charged bromide ion
forms a bond with the positively charged carbon of the carbocation. Notice that both
steps of the reaction involve the reaction of an electrophile with a nucleophile.
CH3CH
+
CHCH3 +
H
It will be helpful to do the exercise on
drawing curved arrows in the Study
Guide/Solution Manual (Special Topic II).
Br
−
new s bond
CH3CH
Br
CHCH3
H
Solely from the knowledge that an electrophile reacts with a nucleophile and a p
bond is the weakest bond in an alkene, we have been able to predict that the product
of the reaction of 2-butene and HBr is 2-bromobutane. Overall, the reaction involves
the addition of 1 mole of HBr to 1 mole of the alkene. The reaction, therefore, is
called an addition reaction. Because the first step of the reaction involves the addition of an electrophile (H +) to the alkene, the reaction is more precisely called an
electrophilic addition reaction. Electrophilic addition reactions are the characteristic
reactions of alkenes.
At this point, you may think that it would be easier just to memorize the fact that 2bromobutane is the product of the reaction, without trying to understand the mechanism that explains why 2-bromobutane is the product. Keep in mind, however, that the
number of reactions you encounter is going to increase substantially, and it will be impossible to memorize them all. If you strive to understand the mechanism of each reaction, the unifying principles of organic chemistry will become apparent, making
mastery of the material much easier and a lot more fun.
PROBLEM 13
Which of the following are electrophiles, and which are nucleophiles?
H−
CH3O−
CH3C
CH
+
CH3CHCH3
NH3
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How Alkenes React • Curved Arrows
95
A FEW WORDS ABOUT CURVED ARROWS
1. Make certain that the curved arrows are drawn in the direction of the electron flow and never against the flow. This
means that an arrow will always be drawn away from a negative charge and/or toward a positive charge.
incorrect
correct
O
CH3
−
O
Br
C
+
−
Br
CH3
CH3
O
O
Br
C
CH3
CH3
CH3
H
O
+
C
CH3
CH3
CH3
O
−
H + H+
+
CH3
H
Br
−
CH3
H
O
+
C
CH3
O
H + H+
H
correct
incorrect
2. Curved arrows are drawn to indicate the movement of electrons. Never use a curved arrow to indicate the movement of an
atom. For example, you can’t use an arrow as a lasso to remove the proton, as shown here:
incorrect
correct
+
H
O
+
O
O
+
CH3CCH3 + H
CH3CCH3
H
CH3CCH3
O
CH3CCH3 + H+
3. The arrow starts at the electron source. It does not start at an atom. In the following example, the arrow starts at the electrons
of the p bond, not at a carbon atom:
CH3CH
CHCH3 + H
Br
+
CH3CH
CHCH3 +
Br
−
H
correct
CH3CH
CHCH3 + H
Br
+
CH3CH
CHCH3 +
H
incorrect
PROBLEM 14 ◆
Use curved arrows to show the movement of electrons in each of the following reaction steps:
O
O
a. CH3C
O
H + HO
−
CH3C
O− + H2O
Br
+
+ Br
b.
+
+
OH
O
c. CH3COH + H
+
O
H
H
CH3
d. C
CH3
CH3
Cl
C+ + Cl−
CH3
CH3COH + H2O
Br
−
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4.8
Using a Reaction Coordinate Diagram
to Describe a Reaction
We have just seen that the addition of HBr to 2-butene is a two-step reaction (Section
4.7). In each step of the reaction, the reactants pass through a transition state as they
are converted into products. The structure of the transition state lies somewhere between the structure of the reactants and the structure of the products. The structure of
the transition state for each of the steps is shown below in brackets. Bonds that break
and bonds that form, as reactants are converted to products, are partially broken and
partially formed in the transition state. Dashed lines are used to show partially broken
or partially formed bonds. Similarly, atoms that either become charged or lose their
charge during the course of the reaction are partially charged in the transition state.
Transition states are always shown in brackets with a double-dagger superscript.
‡
CH3CH
CHCH3 + HBr
d+
CH3CH
CH3CHCH2CH3 + Br−
+
CHCH3
H
Mechanistic Tutorial:
Addition of HBr to an alkene
d− Br
transition state
CH3CHCH2CH3
+
+
−
Br
‡
d+
CH3CHCH2CH3
CH3CHCH2CH3
d−Br
Br
transition state
Figure 4.2 N
a
Free energy
Reaction coordinate diagrams for
the two steps in the addition of HBr
to 2-butene: (a) the first step;
(b) the second step.
b
transition state
∆G‡
carbocation
Free energy
The more stable the species, the lower is
its energy.
The energy changes that take place in each step of the reaction can be described by
a reaction coordinate diagram (Figure 4.2). In a reaction coordinate diagram, the
total energy of all species is plotted against the progress of the reaction. A reaction
progresses from left to right as written in the chemical equation, so the energy of the
reactants is plotted on the left-hand side of the x-axis and the energy of the products is
plotted on the right-hand side.
Figure 4.2a shows that, in the first step of the reaction, the alkene is converted into
a carbocation that is less stable than the reactants. Remember that the more stable the
species, the lower is its energy. Because the product of the first step is less stable than
the reactants, we know that this step consumes energy. We see that as the carbocation
is formed, the reaction passes through the transition state. Notice that the transition
state is a maximum energy state on the reaction coordinate diagram.
The carbocation reacts in the second step with the bromide ion to form the final
product (Figure 4.2b). Because the product is more stable than the reactants, we know
that this step releases energy.
transition state
∆G‡
product
carbocation
reactants
Progress of the reaction
Progress of the reaction
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Section 4.8
Using a Reaction Coordinate Diagram to Describe a Reaction
A chemical species that is the product of one step of a reaction and is the reactant
for the next step is called an intermediate. Thus, the carbocation is an intermediate.
Although the carbocation is more stable than either of the transition states, it is still
too unstable to be isolated. Do not confuse transition states with intermediates:
Transition states have partially formed bonds, whereas intermediates have fully
formed bonds.
Because the product of the first step is the reactant in the second step, we can hook
the two reaction coordinate diagrams together to obtain the reaction coordinate diagram for the overall reaction (Figure 4.3).
Transition states have partially formed
bonds. Intermediates have fully
formed bonds.
> Figure 4.3
Reaction coordinate diagram for
the addition of HBr to 2-butene.
intermediate
Free energy
CH3CHCH2CH3
+
Br−
CH3CH
CHCH3
HBr
CH3CHCH2CH3
Br
Progress of the reaction
A reaction is over when the system reaches equilibrium. The relative concentrations
of products and reactants at equilibrium depend on their relative stabilities: The more
stable the compound, the greater is its concentration at equilibrium. Thus, if the products are more stable (have a lower free energy) than the reactants, there will be a higher concentration of products than reactants at equilibrium. On the other hand, if the
reactants are more stable than the products, there will be a higher concentration of reactants than products at equilibrium. We see that the free energy of the final products
in Figure 4.3 is lower than the free energy of the initial reactants. Therefore, we know
that there will be more products than reactants when the reaction has reached equilibrium. A reaction that leads to a higher concentration of products compared with the
concentration of reactants is called a favorable reaction.
How fast a reaction occurs is indicated by the energy “hill” that must be climbed for
the reactants to be converted into products. The higher the energy barrier, the slower is
the reaction. The energy barrier is called the free energy of activation. The free energy of activation for each step is indicated by ¢G ‡ in Figure 4.2. It is the difference between the free energy of the transition state and the free energy of the reactants:
The more stable the compound,
the greater is its concentration at
equilibrium.
≤G‡ (free energy of the transition state) (free energy of the reactants)
We can see from the reaction coordinate diagram that the free energy of activation
for the first step of the reaction is greater than the free energy of activation for the second step. In other words, the first step of the reaction is slower than the second step.
This is what you would expect because the molecules in the first step of this reaction
must collide with sufficient energy to break covalent bonds, whereas no bonds are broken in the second step.
97
The higher the energy barrier,
the slower is the reaction.
98
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Alkenes
The reaction step that has its transition state at the highest point on the reaction
coordinate is called the rate-determining step. The rate-determining step controls the
overall rate of the reaction because the overall rate cannot exceed the rate of
the rate-determining step. In Figure 4.3, the rate-determining step is the first step—the
addition of the electrophile (the proton) to the alkene.
What determines how fast a reaction occurs? The rate of a reaction depends on the
following factors:
1. The number of collisions that take place between the reacting molecules in a given
period of time. The greater the number of collisions, the faster is the reaction.
2. The fraction of the collisions that occur with sufficient energy to get the reacting
molecules over the energy barrier. If the free energy of activation is small, more
collisions will lead to reaction than if the free energy of activation is large.
3. The fraction of the collisions that occur with the proper orientation. For example,
2-butene and HBr will react only if the molecules collide with the hydrogen of
HBr approaching the p bond of 2-butene. If collision occurs with the hydrogen
approaching a methyl group of 2-butene, no reaction will take place, regardless of
the energy of the collision.
rate of a reaction =
number of collisions
per unit of time
×
fraction with
sufficient energy
×
fraction with
proper orientation
Increasing the concentration of the reactants increases the rate of a reaction because
it increases the number of collisions that occur in a given period of time. Increasing the
temperature at which the reaction is carried out also increases the rate of a reaction because it increases both the frequency of collisions (molecules that are moving faster
collide more frequently) and the number of collisions that have sufficient energy to get
the reacting molecules over the energy barrier (molecules that are moving faster collide with greater energy).
PROBLEM 15 ◆
Draw a reaction coordinate diagram for
a. a fast reaction with products that are more stable than the reactants.
b. a slow reaction with products that are more stable than the reactants.
c. a slow reaction with products that are less stable than the reactants.
PROBLEM 16
Given the following reaction coordinate diagram for the reaction of A to give D, answer the
following questions:
Free energy
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B
A
D
Progress of the reaction
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Summary
a.
b.
c.
d.
e.
f.
g.
h.
99
How many intermediates are there?
Which intermediate is the most stable?
How many transition states are there?
Which transition state is the most stable?
Which is more stable, reactants or products?
What is the fastest step in the reaction?
What is the reactant of the rate-determining step?
Is the overall reaction favorable?
PESTICIDES: NATURAL
AND SYNTHETIC
Long before chemists learned how to create
compounds that would protect plants from predators, plants
were doing the job themselves. Plants had every incentive to
synthesize pesticides—when you can’t run, you need to find
another way to protect yourself. But this poses a question:
Which pesticides are more harmful—those synthesized by
chemists or those synthesized by plants? Unfortunately,
we don’t know the answer because federal laws require that
all human-made pesticides be tested for any cancer-causing
effects, but do not require plant-made pesticides to be
tested. In addition, risk evaluations of chemicals are usually
done on rats, and a chemical that is carcinogenic in a rat may
not be carcinogenic in a human. Some chemicals are harmful
only at high doses, and rats are exposed to much greater concentrations of the chemical during testing than humans are
exposed to during normal activities. For example, we all need
sodium chloride for survival, but high concentrations are poisonous. We associate alfalfa sprouts with healthy eating, but
monkeys fed very large amounts of alfalfa sprouts develop an
immune system disorder.
Summary
Alkenes are hydrocarbons that contain a double bond. The
double bond is the functional group or center of reactivity
of the alkene. The functional group suffix of an alkene is
ene. The general molecular formula for a hydrocarbon is
CnH 2n + 2 , minus two hydrogens for every p bond or ring in
the molecule. Because alkenes contain fewer than the maximum number of hydrogens, they are called unsaturated
hydrocarbons.
Because of restricted rotation about the double bond, an
alkene can exist as cis–trans isomers. The cis isomer has
its hydrogens on the same side of the double bond; the
trans isomer has its hydrogens on opposite sides of the
double bond. The Z isomer has the high-priority groups on
the same side of the double bond; the E isomer has the
high-priority groups on opposite sides of the double bond.
The relative priorities depend on the atomic numbers of the
atoms bonded directly to the sp 2 carbons. The more alkyl
substituents bonded to the sp 2 carbons of an alkene, the
greater is its stability. Trans alkenes are more stable than
cis alkenes because of steric strain.
All compounds with a particular functional group react
similarly. Due to the cloud of electrons above and below its
p bond, an alkene is an electron-rich molecule (a nucleophile). Nucleophiles are attracted to electron-deficient
atoms or molecules, called electrophiles. Alkenes undergo
electrophilic addition reactions. The description of the
step-by-step process by which reactants are changed into
products is called the mechanism of the reaction. Curved
arrows show which bonds are formed and which are broken and the direction of the electron flow that accompany
these changes.
A reaction coordinate diagram shows the energy
changes that take place in a reaction. The more stable the
species, the lower is its energy. As reactants are converted
into products, a reaction passes through a maximum energy
transition state. An intermediate is the product of one
step of a reaction and the reactant for the next step. Transition states have partially formed bonds; intermediates have
fully formed bonds. The rate-determining step has its
transition state at the highest point on the reaction
coordinate.
The relative concentrations of products and reactants at
equilibrium depend on their relative stabilities. The more
stable the compound, the greater is its concentration at
equilibrium. The free energy of activation, ¢G ‡, is the energy barrier of a reaction. It is the difference between the
free energy of the reactants and the free energy of the transition state. The smaller the ¢G ‡, the faster is the reaction.
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Problems
17. Give the systematic name for each of the following compounds:
a.
CH2CH3
b.
c.
CH3
CH3
18. Squalene, a hydrocarbon with molecular formula C 30H 50 , is obtained from shark liver. (Squalus is Latin for “shark.”) If squalene
is a noncyclic compound, how many p bonds does it have?
19. Draw and label the E and Z isomers for each of the following compounds:
a. CH3CH2CH2CH2
CH3CH2C
b. HOCH2CH2C
CCH2Cl
O
CC
CH
CH C(CH3)3
CH(CH3)2
20. For each of the following pairs, indicate which member is more stable:
CH3
a. CH3C
CH3
CH3
CHCH2CH3 or CH3CH
CH3
CHCHCH3
b.
or
21. a. Give the structures and the systematic names for all alkenes with molecular formula C 4H 8 , ignoring cis–trans isomers.
(Hint: There are three.)
b. Which of the compounds have E and Z isomers?
22. Draw the structure for each of the following:
a. (Z)-1,3,5-tribromo-2-pentene
b. (Z)-3-methyl-2-heptene
c. (E)-1,2-dibromo-3-isopropyl-2-hexene
d. vinyl bromide
e. 1,2-dimethylcyclopentene
f. diallylamine
23. Determine the total number of double bonds and/or rings for the hydrocarbons with the following molecular formulas:
a. C 12H 20
b. C 40H 56
24. Name the following compounds:
a.
c.
b.
d.
Br
e.
f.
25. Draw curved arrows to show the flow of electrons responsible for the conversion of the reactants into the products:
H
−
O
H
+ H C
H
H
C H
Br
H
H
H2O +
C
+ Br−
C
H
26. Draw three alkenes with molecular formula C 5H 10 that do not have cis–trans isomers.
H
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Problems
101
27. Tell whether each of the following compounds has the E or the Z configuration:
CH2CH3
H3C
a.
C
c.
C
CH3CH2
CH2Br
H3C
C
C
Br
CH2CH2Cl
CH2CH2CH2CH3
O
CH(CH3)2
H3C
b.
C
HC
d.
C
CH2CH
C
CH2Br
CH3C
CH2
C
C
CH2CH2Cl
HOCH2
28. Which of the following compounds is the most stable? Which is the least stable?
3,4-dimethyl-2-hexene; 2,3-dimethyl-2-hexene; 4,5-dimethyl-2-hexene
29. Assign relative priorities to each set of substituents:
¬ CH(CH 3)2 ,
¬ CH “ CH 2 ,
a. ¬ CH 2CH 2CH 3 ,
¬ CH 2OH
¬ NH 2 , ¬ OH,
b. ¬ CH 2NH 2 ,
¬ CH “ CH 2 ,
¬ Cl,
c. ¬ COCH 3 ,
¬ CH 3
¬C‚N
30. Determine the molecular formula for each of the following:
a. a 5-carbon hydrocarbon with two p bonds and no rings
b. an 8-carbon hydrocarbon with three p bonds and one ring
31. Give the systematic name for each of the following compounds:
CH3
a. CH3CH2CHCH
CHCH2CH2CHCH3
Br
CH3
Br
CH2CH3
H3C
b.
c.
C
CH3CH2
CH2CH3
H3C
d.
C
CH2CH2CHCH3
C
H3C
C
CH2CH2CH2CH3
CH3
32. Draw a reaction coordinate diagram for a two-step reaction in which the products of the first step are less stable than the reactants, the
reactants of the second step are less stable than the products of the second step, the final products are less stable than the initial
reactants, and the second step is a the rate-determining step. Label the reactants, products, intermediates, and transition states.
33. Molly Kule was a lab technician who was asked by her supervisor to add names to the labels on a collection of alkenes that showed
only structures on the labels. How many did Molly get right? Correct the incorrect names.
a. 3-pentene
f. 5-chloro-3-hexene
b. 2-octene
g. 5-bromo-2-pentene
c. 2-vinylpentane
h. (E)-2-methyl-1-hexene
d. 1-ethyl-1-pentene
i. 2-methylcyclopentene
e. 5-ethylcyclohexene
j. 2-ethyl-2-butene
34. Determine the number of double bonds and/or p bonds and then draw possible structures, for compounds with the following
molecular formulas:
a. C 3H 6
b. C 3H 4
c. C 4H 6
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35. Draw a reaction coordinate diagram for the following reaction in which C is the most stable and B is the least stable of the three
species and the transition state going from A to B is more stable than the transition state going from B to C:
k1
k2
k -1
k -2
A ERF B ERF C
a.
b.
c.
d.
e.
f.
g.
How many intermediates are there?
How many transition states are there?
Which step has the greater rate constant in the forward direction?
Which step has the greater rate constant in the reverse direction?
Of the four steps, which has the greatest rate constant?
Which is the rate-determining step in the forward direction?
Which is the rate-determining step in the reverse direction?
36. The rate constant for a reaction can be increased by ______ the stability of the reactant or by ______ the stability of the transition
state.
37. a-Farnesene is a compound found in the waxy coating of apple skins. To complete its systematic name, include the E or Z
designation after the number indicating the location of each double bond.
a-farnesene
3,7,11-trimethyl-(1,3,6,10)-dodcatetraene
38. Give the structures and the systematic names for all alkenes with molecular formula C 6H 12 , ignoring cis–trans isomers.
(Hint: There are 13.)
a. Which of the compounds have E and Z isomers?
b. Which of the compounds is the most stable?