Jones M., Fleming S.A. Organic Chemistry

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3.3 Derivatives and Isomers of Alkenes 109

to show that the two hydrogens on the end (called terminal) methylene group are

not the same. One, H

, is on the same side of the double bond as the methyl group,

whereas the other, H

a′

, is on the opposite side from the methyl. Replacing hydro-

gens H

and H

a′

with X gives two different compounds.

The two molecules on the right of Figure 3.20 are interconvertible only by rota-

tion about the double bond (Fig. 3.21). We know the rotation shown in Figure 3.21

usammen ( Z ):

Hydrogens

on same side

Entgegen (E ):

Hydrogens

on opposite sides

rotate

FIGURE 3.21 The (hypothetical)

interconversion of two isomers of

by rotation

about the carbon–carbon

double bond.

requires about 66 kcal/mol, too high an amount of energy for interconversion to be

common.The molecule with the two hydrogens on the same side of the double bond

is designated as cis or (Z) (for zusammen, German for “together”). The compound

with the hydrogens on the opposite sides is called trans or (E) (for entgegen, German

for “opposite”). Recall the use of cis and trans in our discussion of ring compounds

(p. 84).

If we let X  methyl we get the butenes (C

).The four isomeric butenes are

1-butene,cis-2-butene,trans-2-butene, and 2-methylpropene (or isobutene, but also

sometimes called isobutylene).These are shown in Figure 3.22.The numbers in the

replace H

with CH

replace H

with CH

replace H

with CH

replace H

with CH

trans-2-Butene

(E )-2-butene

cis-2-Butene

( Z )-2-butene

2-Methylpropene

isobutene

1-Butene

WEB 3D

FIGURE 3.22 Replacement of the

four different hydrogens in propene

by a methyl group leads to the four

butene isomers.

names 1-butene, cis-2-butene, and trans-2-butene are used to locate the position of

the double bond.We will shortly discuss the naming convention for alkenes, but you

might try to figure it out here from the names and structures of the butenes.

Not every alkene is capable of cis/trans (Z/E) isomerism! Ethylene and propene

are not, and of the butenes only 2-butene has such isomers. Dealing with cis and

trans isomers is an essential skill that can be “cemented” forever right now. Future

difficulties can be avoided if you try Problem 3.4. It is simple but worthwhile.

Once we get to alkenes containing more than four carbons, the systematic naming

protocol takes over. Five-carbon compounds (C

) are called pentenes, six-carbon

compounds (C

) are called hexenes,and so on. Be alert for isomerism of the cis/trans

(Z/E) kind. It takes some practice to find it.

CONVENTION ALERT

110 CHAPTER 3 Alkenes and Alkynes

PROBLEM SOLVING

Alkenes are flat! In the minimum energy form of an alkene, the two carbon

atoms making up the double bond and the four atoms directly attached to it

(atoms 1, 2, 3, and 4) are coplanar. It is very easy to forget that fact, or ignore it.

Don’t, because it leads immediately to the idea that alkenes have discernable

sides.Thus, cis and trans isomers become possible for many alkenes.

PROBLEM 3.4 Which of the following alkenes is capable of cis/trans (Z/E)

isomerism?

PROBLEM 3.5 Write all the isomers of the pentenes (C

) and hexenes

). Be alert for isomerism of the cis/trans (Z/E) kind in these molecules.

3.4 Nomenclature

Most of the rules for naming alkenes are similar to those used for alkanes, with a

number attached to indicate the position of the double bond. Chains are numbered

so as to give the double bond the smallest possible number. One important rule that

is different in the alkenes is that the name is based on the longest chain containing

a double bond whether or not it is also the longest chain in the molecule. When

molecules contain more than one double bond, they are called polyenes, and are

named as dienes, trienes, and so on. In these cases, the longest straight chain is

defined as the one with the greatest number of double bonds. Cyclic molecules are

named as cycloalkenes.These rules are illustrated with the molecules in Figure 3.23.

1-Butene

not 3-butene

1,3-Butadiene

3-Methyl-1,3-heptadiene

not 2-vinyl-2-hexene

3-Propyl-1-nonene

not 4-vinyldecane

1-Methylcyclohexene 3-Methylcyclohexene

not 6-methylcyclohexene

2-Methyl-1,3-cyclohexadiene

not 3-methyl-1,3-cyclohexadiene

FIGURE 3.23 Examples of the naming protocols for alkene nomenclature.

3.4 Nomenclature 111

PROBLEM 3.6 Name the pentenes and hexenes of Problem 3.5.

PROBLEM 3.7 Recall p. 88 in Chapter 2 where the

C NMR spectrometer was

described.That machine will find a signal for each different carbon in a molecule.

How many signals will each of the following molecules show in its

C NMR

spectrum?

Finally, there are examples in which the numbering of a substituent becomes

important. Usually, it is the double bond numbering that must be considered

first. The double bond is usually given the lowest possible number (Fig. 3.24a).

An important exception is the hydroxyl (OH) group, which has priority

over the double bond, and is given the lower number (Fig. 3.24b). Only if

there are two possible names in which the double bond has the same number

do you have to consider the position of the substituent. In such a case,

give preference to the name with the lower number for the substituent

(Fig. 3.24c).

Usually the cis/trans naming system is adequate to distinguish pairs of iso-

mers, but not always. Problem 3.10 and Figure 3.25 give examples in which

cis/trans naming is inadequate. In the molecule 1-bromo-1-chloropropene there

are clearly two isomers, but it is not obvious which is cis and which is trans. It

was to solve such problems that the old cis/trans system was elaborated into the

(Z/E) nomenclature by three European chemists, R. S. Cahn (1899–1981), C. K.

Ingold (1893–1970), and V. Prelog (1906–1998). The Cahn–Ingold–Prelog

priority system ranks the groups attached to the double bonds, with the higher

priority being 1 and the next 2. The compound with the higher priority groups

on the same side of the double bond is (Z), and the other is (E). This priority

system has other, very important uses to be encountered in Chapter 4, and so we

will spend some time here elaborating it, even though we will see it again in

another context.

(E)-4-Chloro-2-pentene

trans-4-Chloro-2-pentene

ot trans-2-chloro-3-pentene

135

(E)-2-Buten-1-ol

trans-2-Buten-1-ol

not trans-2-buten-4-ol

12 56

65 21

(Z)-2-Chloro-3-hexene

cis-2-Chloro-3-hexene

not cis-5-chloro-3-hexene

H H

(a)

(b)

(c)

FIGURE 3.24 The double bond takes

precedence over most substituents,

but when two names put the double

bond at the same numbered position,

we use the name in which the

substituent has the lower number.

PROBLEM 3.9 Draw the following molecules: cis-2-pentene, 2-chloro-1-pentene,

trans-3-penten-2-ol, 4-bromocyclohexene, 1,3,6-cyclooctatriene.

CBr

CC C

FIGURE 3.25 Sometimes the cis/trans

convention is inadequate to

distinguish two isomers. Which of

these two isomers of 1-bromo-

1-chloropropene would you call cis

and which trans?

(a) ethylene

(b) propene

(d) cis-2-butene

(e) trans-2-butene

(f) isobutene (2-methylpropene)

PROBLEM 3.8 How many signals will each of the molecules in Problem 3.4 show

in its

C NMR spectrum?

112 CHAPTER 3 Alkenes and Alkynes

WORKED PROBLEM 3.10 In Figure 3.23, the structure of 3-methyl-1,3-heptadiene

is not completely specified by that name. Even if the cis/trans nomenclature is

applied, problems remain. What’s the difficulty? Can you devise a system for

resolving the ambiguity?

ANSWER In describing fully the structure of 3-methyl-1,3-heptadiene, we

need to specify how groups are oriented in space (the stereochemistry), in

order to distinguish the two possible forms. In one of the structures, the two

saturated alkyl groups (methyl and propyl) are cis while in the other they

are trans.

Often, cis/trans will do the job, but here these terms fail.There is but one H, so

you cannot find two hydrogens “on the same side” (cis), and two hydrogens “on

opposite sides” (trans). That is the problem; it is up to you to devise a protocol

for resolving it. For more about the device used by organic chemists, the Z/E

system, read on in the text.

cis CH

and CH

trans CH

and CH

H H

3.5 The Cahn–Ingold–Prelog Priority System

1. The first step in using the priority system is to distinguish atoms on the basis of

atomic number. The atom of higher atomic number has the higher priority

(Fig. 3.26). Thus, a methyl group, attached to the double bond through a carbon

atom (atomic number 6), has a higher priority than the hydrogen attached to the

same carbon (atomic number 1). Similarly, in the second compound, oxygen (atomic

number 8) has a higher priority than the boron attached to the same carbon

(atomic number 5).

Higher priority

(atomic number = 8)

Lower priority

(atomic number = 5)

Higher priority

(atomic number = 6)

Lower priority

(atomic number = 1)

C C

FIGURE 3.26 The substituent with

the higher atomic number gets the

higher priority.

How is this first rule used in practice? Consider the two isomers in Figure 3.25.

Bromine has a higher atomic number than chlorine, so the isomer in which the

CONVENTION ALERT

3.5 The Cahn–Ingold–Prelog Priority System 113

higher-priority methyl group and the higher-priority bromine are on the same side

is (Z), and the isomer in which the lower-priority hydrogen is on the same side as

the higher-priority bromine is (E) (Fig. 3.27).

( Z )-Isomer: The high-priority

groups are on the same side

( E )-Isomer: The high-priority

groups are on opposite sides

CBr

CC C

Higher priority

(right C)

Higher priority

(left C)

Higher priority

(left C)

Lower priority

(right C)

Lower priority

(right C)

Lower priority

(left C)

Lower priority

(left C)

Higher priority

(right C)

FIGURE 3.27 Use of the

Cahn–Ingold–Prelog priority system

in naming alkenes.

2. For isotopes, atomic mass is used to break the tie in atomic number. Thus deuteri-

um (atomic number 1, atomic mass 2) has a higher priority than hydrogen (atomic

number 1, atomic mass 1) (Fig. 3.28).

Higher priority

(atomic number = 1,

atomic mass = 2)

Lower priority

(atomic number = 1,

atomic mass = 1)

FIGURE 3.28 When the atomic numbers are the same, the

heavier isotope gets the higher priority.

3. Nonisotopic ties are broken by looking at the groups attached to the tied atoms. For

example, 1-chloro-2-methyl-1-butene has a double bond to which a methyl and an

ethyl group are attached. Both these groups are connected to the double bond

through carbons, so Rule 1 does not differentiate them (Fig. 3.29).To break the tie,

we look at the groups attached to these carbons. The carbon of the methyl group is

attached to three hydrogens, whereas the “first”carbon of the ethyl group is attached

to two hydrogens and a carbon. Therefore the ethyl group gets the higher priority

(Fig. 3.30). If the groups are still tied after this procedure, one simply looks farther

out along the chain to break the tie, as in Problem 3.11.

Same priority

(atomic number = 6)

Same priority

(atomic number = 6)

1-Chloro-2-methyl-1-butene

CCH

FIGURE 3.29 In this molecule,

the two C atoms attached to

the double bond of course have the

same atomic number and, therefore,

the same priority according to

Cahn–Ingold–Prelog Rule 1.This

isomer is Z.

This C is attached to

two H atoms and

one C = higher priority

This C is attached to three

H atoms = lower priority

FIGURE 3.30 The red methyl C is attached to three H atoms.The red ethyl C is attached

to C, H, and H. The ethyl C gets the higher priority. Because the two higher priority

groups, ethyl and Cl, are on the same side of the double bond, this isomer is Z.

114 CHAPTER 3 Alkenes and Alkynes

PROBLEM 3.11 Determine which of the following molecules is (Z) and which is (E).

PROBLEM 3.12 Make drawings of the following molecules:

(a) (E)-3-fluoro-3-hexene

(b) (E)-4-ethyl-3-heptene

4. Multiple bonds attached to alkenes are treated as multiplied single bonds. A dou-

ble bond to carbon is considered to be two single bonds to the carbons in the double

bond, as shown in Figure 3.31. This convention results in an isopropenyl group

being of higher priority than a tert-butyl group, for example (Fig. 3.32). As we will

shortly see, it is also possible to connect two carbon atoms through a

triple bond

(see Section 3.9). The priority system treats triple bonds in a similar fashion.

C C

FIGURE 3.31 The priority of doubly

bonded carbons is determined by

adding two single carbon bonds as

shown. A triple bond is treated

similarly.

tert-Butyl

group

Isopropenyl

group

Higher priority

Lower priority

FIGURE 3.32 An isopropenyl group

has a higher priority than a tert-butyl

group.

PROBLEM 3.13 In each of the following pairs of isomers, which is (Z), and which

is (E)?

(c)

(b)

(a)

(d)

3.6 Relative Stability of Alkenes: Heats of Formation 115

Summary

The technique of assigning (Z) and (E) is to determine the priority (high or low)

at each carbon of the double bond.The isomer with the two high-priority groups

on the same side of the double bond is (Z).The isomer with the two high-priority

groups on opposite sides is (E). Two examples are given in Figure 3.33.

FIGURE 3.33 Two examples of the

Cahn–Ingold–Prelog priority system

at work.

3.6 Relative Stability of Alkenes: Heats of Formation

The heat of formation of a compound is the enthalpy of formation from its

constituent elements in their standard states.The standard state of an element is the

most stable form of the element at 25 °C and 1 atm pressure. For an element in its

standard state, is taken as zero.For carbon,the standard state is graphite.Thus,

for graphite, as well as simple gases (e.g., H

, and N

), is 0 kcal/mol.The

more negative—or less positive—a compound’s is, the more stable the com-

pound is. A negative for a compound means that its formation from its con-

stituent elements is exothermic—heat is liberated in the reaction.In contrast,a positive

means that the constituent elements are more stable than the compound and

its formation is endothermic—energy must be applied.

Remember: Bonding is an energy-releasing process. We expect the formation of

a molecule from its constituent atoms to be an exothermic process.For example, con-

sider the simple formation of methane from carbon and hydrogen. Carbon in its

standard state (graphite) and gaseous H

have , whereas for

methane . The formation of methane from graphite and

hydrogen releases 17.8 kcal/mol. In other words, this reaction is exothermic by

17.8 kcal/mol (Fig. 3.34).

=-17.8 kcal>mol¢H °

= 0 kcal>mol¢H °

¢H °

(≤H °

)

Energy

C (Graphite) + H

(Gas)

ΔH

⬚ = 0 kcal/mol ΔH

⬚ = –17.8 kcal/mol

(Gas)

17.8 kcal/mol

C + 2 H

FIGURE 3.34 The formation of

methane from graphite and gaseous

hydrogen is exothermic.

The two high-priority groups are

on the same side

—this molecule

is (Z)-2-methoxy-2-pentene

Here the situation is different

—

the two high-priority groups are

on opposite sides; this molecule

is (E )-2-methoxy-2-pentene

Higher priority

(right C)

Higher priority

(left C)

Higher priority

(left C)

Lower priority

(right C)

Lower priority

(right C)

Lower priority

(left C)

Lower priority

(left C)

Higher priority

(right C)

116 CHAPTER 3 Alkenes and Alkynes

Thus, the double bond in 1-hexene is monosubstituted and the double bond in

2-hexene or 3-hexene is disubstituted.The trans isomer is the more stable compound

(more negative ). Figure 3.35 shows the reason; in the cis isomer there is a cis

ethyl–ethyl eclipsing that is absent in the trans compound. In the trans isomer, both

alkyl groups are eclipsed by hydrogen, and the destabilizing ethyl–ethyl repulsion

is missing.

¢H °

Table 3.1 gives the heats of formation of a number of hexenes. First look at the

pair of disubstituted alkenes, the isomers cis- and trans-3-hexene. “Substituted”

simply means that an atom other than hydrogen is attached to the double bond.

TABLE 3.1 Heats of Formation for Some Hexenes

Isomer

10.0 Least stable

cis 11.2

trans 12.1

16.0

16.6 Most stable(CH

)

C(CH

)

(CH

)

CHCH

(kcal/mol)≤H

Z———————U

cis-3-Hexene

trans-3-Hexene

Here is the high-energy

destabilizing interaction

in the cis compound

—

the eclipsed ethyl–ethyl

“bumping”

Newman projections

FIGURE 3.35 Newman projections show the unfavorable ethyl–ethyl eclipsing interaction

in cis-3-hexene.

Both disubstituted isomers cis- and trans-3-hexene are more stable than the

monosubstituted 1-hexene, and the trisubstituted isomer is more stable than

either disubstituted molecule. Tetrasubstituted 2,3-dimethyl-2-butene has the

lowest energy of all (lowest heat of formation, most stable). From these obser-

vations, it would appear that the degree of substitution (the number of alkyl

groups attached to the double bond) is important in stabilizing the molecule.

In general, the more substituted the double bond, the more stable it is

(Table 3.1).

The reason for this has been the subject of some controversy. One good way to

look at the question focuses on the different kinds of carbon–carbon bonds present

in isomeric molecules of different substitution patterns.In all the molecules in Table

3.1, three kinds of σ carbon–carbon linkages are present: sp

/sp

, sp

/sp

, and sp

/sp

3.6 Relative Stability of Alkenes: Heats of Formation 117

bonds (Fig. 3.36). An electron in a 2s orbital is at lower energy than an electron

in a 2p orbital. The more s character in an orbital, the more an electron in it is

stabilized. In these isomeric hexenes, the more substituted the carbon–carbon

double bond is, the more relatively low-energy (strong) sp

/sp

bonds are present.

Monosubstituted

σ Bond sp

/sp

Disubstituted

σ Bond sp

/sp

Trisubstituted Tetrasubstituted

/sp

σ Bond sp

/sp

σ Bond sp

/sp

CC CC

FIGURE 3.36 Three kinds of

carbon–carbon σ bonds in hexene

isomers: sp

/sp

overlap, sp

/sp

overlap, and sp

/sp

overlap.

By contrast, less substituted alkenes have more relatively high-energy (weaker)

/sp

bonds. The more strong bonds present, the more stable is the molecule.

Figure 3.37 shows the number of sp

/sp

, sp

/sp

, and sp

/sp

bonds in the isomer-

ic hexenes of Table 3.1. 2,3-Dimethyl-2-butene, the molecule with the strongest

bonds, is the most stable isomer of the set, and 1-hexene, with the weakest set of

bonds, is the least stable.

This analysis is somewhat primitive because it ignores the carbon–hydrogen bonds, which are also different

in differently substituted alkenes. However, the changes in carbon–carbon bonds dominate. You might try to

do a full comparison using detailed bond energies.

C C

σ Bonds

/sp

Energy

Most

stable

Least

stable

FIGURE 3.37 The isomeric hexenes

have different numbers of three types

of carbon–carbon σ bonds.

118 CHAPTER 3 Alkenes and Alkynes

PROBLEM 3.14 Carry out an analysis similar to that of Figure 3.37 for the series

of molecules: 1-pentene, 2-methyl-2-butene, (Z)-2-pentene.

3.7 Double Bonds in Rings

Figure 3.38 shows a number of cycloalkenes, ring compounds containing one or more

double bonds. Note that even small rings can contain double bonds.

HC CH

WEB 3D

FIGURE 3.38 Some cyclic alkenes

(cycloalkenes).

1,3,5-Cyclo-

heptatriene

1,3-Cyclohexadiene 1,4-Cyclohexadiene

Cyclopentadiene

Cyclobutadiene

1,3,5,7-Cyclo-

octatetraene

WEB 3DWEB 3D WEB 3D

FIGURE 3.39 Some cyclic polyenes, ring compounds containing more than one double bond.

PROBLEM 3.15 Name all the compounds in Figure 3.38.

Once the ring becomes larger than cyclopropene, it becomes possible for it to

contain more than one double bond. Several such molecules are shown in

Figure 3.39.

PROBLEM 3.16 How many signals will appear in the

C NMR spectrum (p. 88)

of the compounds in Figure 3.39?

PROBLEM 3.17 Write the structures for 1,3,5-cyclohexatriene, 1,3,5,7-

cyclooctatetraene, 3-methyl-1,4-cyclohexadiene, 2-fluoro-1,3-cyclohexadiene,

and 2-bromo-1,4-cycloheptadiene.