Chapter 4: Alkenes and Alkynes 99

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Chapter 4: Alkenes and Alkynes

Problems

4.1 Write the IUPAC name of each unsaturated hydrocarbon:

(a) (b) (c)

3,3-dimethyl-1-pentene 2,3-dimethyl-2-butene 3,3-dimethyl-1-butyne

4.2 Name each alkene, and, using the cis-trans system, specify its configuration:

(a) (b)

cis-4-methyl-2-pentene trans-2,2-dimethyl-3-hexene

4.3 Name each alkene and specify its configuration by the E,Z system:

Cl (a) (b) (c) Cl Br (E)-1-chloro-2,3- (Z)-1-bromo-1- (E)-2,3,4- dimethyl-2-pentene chloro-1-propene trimethyl-3-heptene

4.4 Write the IUPAC name for each cycloalkene:

(a) (b) (c)

1-isopropyl-4- 4-tert-butylcyclohexene cyclooctene methylcyclohexene

100 Chapter 4: Alkenes and Alkynes

4.5 Draw structural formulas for the other two cis-trans isomers of 2,4-heptadiene:

cis,trans-2,4-heptadiene cis,cis-2,4-heptadiene

4.6 How many cis-trans isomers are possible for the following unsaturated alcohol?

CH3 CH3 CH3

CH3C=CHCH2CH2C=CHCH2CH2C=CHCH2OH

There are three carbon-carbon double bonds, but cis-trans isomerism is only possible for the two that are indicated by arrows. cis-trans Isomerism is not possible for the alkene on the far left of the molecule because one of the two alkene carbons is bonded to two 2 identical groups (CH3). The number of possible cis-trans isomers is therefore 2 = 4.

Quick Quiz

1. Ethylene and acetylene are constitutional isomers. False. Constitutional isomers must have the same molecular formula but different atom connectivity (bonding sequence). Ethylene and acetylene are respectively C2H4 and C2H2.

2. Alkanes that are liquid at room temperature are insoluble in water and when added to water will float on water. True. Alkanes are nonpolar and therefore water-insoluble, and they have a lower density than water.

3. The bulk of the ethylene used by the chemical industry worldwide is obtained from nonrenewable resources. True. Ethylene is derived from the cracking of hydrocarbons.

4. Alkenes and alkynes are nonpolar molecules. True. Alkenes and alkynes do not contain any polar bonds, so the molecules do not have a net dipole moment.

5. The IUPAC name of CH3CH=CHCH3 is 1,2-dimethylethylene. False. The longest carbon chain is four carbons long, which leads to the name 2-butene.

6. Cyclohexane and 1-hexene are constitutional isomers. True. They have the same molecular formula but a different connectivity of atoms.

7. The IUPAC name of an alkene is derived from the name of the longest chain of carbon atoms that contains the double bond. True. The chain is also numbered such that the double bond has the lowest possible number. Chapter 4: Alkenes and Alkynes 101

8. There are two classes of unsaturated hydrocarbons, alkenes and alkynes. False. Arenes, which are compounds based on benzene, are also unsaturated hydrocarbons.

9. Both geraniol and menthol (Figure 4.2) show cis-trans isomerism. True. Geraniol can exhibit cis-trans isomerism at one of the double bonds. Menthol can show cis-trans isomerism with respect to the substituents bonded to the cyclohexane ring.

10. Terpenes are identified by their carbon skeletons, namely, one that can be divided into five-carbon isoprene units. True. All terpenes consist of isoprene building blocks.

11. 1,2-Dimethylcyclohexene shows cis-trans isomerism. False. The double bond is a part of a six-membered ring and must be cis with respect to the ring. There is no cis-trans isomerism with respect to the methyl groups on the ring because they are both bonded to sp2 carbons and are planar.

12. 2-Methyl 2-butene shows cis-trans isomerism. False. Carbon 2 is bonded to two identical substituents, two methyl groups.

13. Both ethylene and acetylene are planar molecules. True. The carbon atoms in ethylene are sp2-hybridized, and all the carbon and hydrogen atoms lie in the same plane. Acetylene has sp-hybridized carbons, and all the atoms of acetylene lie in the same plane.

14. The physical properties of alkenes are similar to those of alkanes with the same carbon skeletons. True. These properties include density, melting point, and boiling point.

15. Isoprene is the common name for 2-methyl-1,3-butadiene. True. Realize that once this compound is incorporated into a terpene, it need not bear the same order of single and double bonds. Other chemical modifications can also be performed on the terpene. However, the five-carbon skeleton of isoprene is always retained in a terpene.

Chemical Connections

4A. Explain the basis for the saying “A rotten apple can spoil the barrel.”

Ethylene is a natural ripening agent for apples. A rotten apple, or one that is overripe, releases ethylene and causes the other apples in the barrel to ripen.

4B. Based on the information in this chemical connection, what can you deduce about the physical properties of leaf cell membranes?

Isoprene is dissolved in the leaf cell membranes before it is released into the atmosphere. Because isoprene is a nonpolar hydrocarbon, the membranes must also be relatively nonpolar. At higher temperatures, isoprene is more volatile and more easily released. 102 Chapter 4: Alkenes and Alkynes

End-of-Chapter Problems

Structure of Alkenes and Alkynes

4.7 Classify each of the following as either an alkane, alkene, alkyne, or arene.

(a) (b) (c) (d) (e)

(a) alkene (b) alkane (c) arene (d) arene (e) alkene

4.8 Give the hybridization of each of the highlighted carbons in problem 4.7.

(a) sp2 (b) sp3 (c) sp2 (d) sp (e) sp2

4.9 Each of the highlighted carbons in problem 4.7 is attached to two other carbons. Give the approximate C—C—C bond angle (with the highlighted carbon being the central carbon) for each of these compounds.

(a) 120° (b) 109.5° (c) 120° (d) 180° (e) 120°

4.10 How many hydrogens are attached to of each of the highlighted carbons in problem 4.7?

(a) 1 (b) 2 (c) 1 (d) 0 (e) 1

4.11 Cyclohexene (see below) is stable, but cyclohexyne is not. Offer an explanation why.

cyclohexene cyclohexyne

There is too much “angle strain” in cyclohexyne. The alkyne carbons want to have linear geometry (bond angle = 180°), but the geometry of the ring is forcing the angle to be closer to 120°,

Chapter 4: Alkenes and Alkynes 103

4.12 Each carbon atom in ethane and in ethylene is surrounded by eight valence electrons and has four bonds to it. Explain how VSEPR (Section 1.3) predicts a bond angle of 109.5° about each carbon in ethane, but an angle of 120° about each carbon in ethylene.

Although each carbon atom in ethane and ethylene is surrounded by eight valence electrons and has four bonds to it, the shape and geometry about each carbon is based on the number of regions of electron density associated with it. In ethane, each carbon has four regions of electron density and a tetrahedral shape, which has bond angles of 109.5°. Whereas, each carbon in ethylene has three regions of electron density (remember that a double bond is counted as a single region of electron density), resulting in a trigonal planar arrangement (120°).

4.13 Explain the difference between saturated and unsaturated.

A compound that is saturated does not contain any carbon-carbon π bonds. Compounds that are unsaturated contain one or more carbon-carbon π bonds.

4.14 Following is the structure of 1,2-propadiene (allene). In it, the plane created by H−C−H of carbon 1 is perpendicular to that created by H−C−H of carbon 3.

H H 1 2 3 C C C H H 1,2-Propadiene (Allene)

(a) State the orbital hybridization of each carbon in allene.

Carbons 1 and 3 are sp2-hybridized, while carbon 2 is sp-hybridized.

(b) Account for the molecular geometry of allene in terms of the orbital overlap model. Specifically, explain why all four hydrogen atoms are not in the same plane.

Carbon 2, which is the carbon that bears two double bonds, is sp-hybridized. In sp hybridization, the two remaining p orbitals are perpendicular to each other. One of the two p orbitals forms the π bond with carbon 1 while the other p orbital forms the π bond with carbon 2. Accordingly, these two π bonds are perpendicular to each other. As a result, the trigonal plane formed by the CH2 on the left is perpendicular to the one formed by the CH2 group on the right.

104 Chapter 4: Alkenes and Alkynes

Nomenclature of Alkenes and Alkynes

4.15 Nine different compounds have the molecular formula C4H6. Draw all nine.

C 1-butyne 2-butyne 1,3-butadiene 1,2-butadiene

cyclobutene 1-methylcyclopropene 2-methylcyclopropene

4.16 Give IUPAC names for seven of the compounds you drew in question 4.15 (there are two you won’t be able to name based on what you’ve learned so far).

(See names under structures above.)

4.17 Would any of the compounds you drew in question 4.15 exist as cis-trans isomers?

No.

4.18 Draw a structural formula for each compound:

(a) 4-ethyl-3-methyl-1-heptene (b) 2,3-dimethylcyclopentene (c) 2-methyl-3-hexyne (d) 4-methyl-1,3-pentadiene (e) cyclopentadiene (f) trans-3,6-dimethylcyclohexene

(a) (b) (c) (d)

(e) (f)

Chapter 4: Alkenes and Alkynes 105

4.19 Write the IUPAC name for each compound:

(a) (b) (c)

(d) (e) (f)

(a) 2-methyl-2-pentene (b) 3-methylcyclohexene (c) 3-methyl-1-pentyne (d) 6-methyl-1,5-heptadiene (e) 1,3,5,7-cyclooctatetraene (f) 2-ethyl-1-pentene

4.20 Explain why each name is incorrect, and then write a correct name for the intended compound:

(a) 2-Ethyl-1-propene

4 3 The parent chain is four carbons long. 2 Correct name: 2-methyl-1-butene 1

(b) 5-Isopropylcyclohexene 2 3 4 The numbering of the ring is incorrect. 1 Correct name: 4-isopropylcyclohexene

(c) 4-Methyl-4-hexene The numbering of the chain is incorrect, and it is also necessary to indicate the configuration (E or Z). 2 4 6 Correct name: 3-Methyl-2-hexene (E isomer shown) 1 3 5

106 Chapter 4: Alkenes and Alkynes

(f) 2-Ethyl-2-hexene The parent chain is five carbons long, and it is also necessary to indicate the configuration (E or Z). Correct name: 3-methyl-3-heptene (E isomer shown)

Cis-Trans (E/Z) Isomerization in Alkenes and Cycloalkenes

4.21 Which of these alkenes show cis–trans isomerism? For each that does, draw structural formulas for both isomers. (a) 1-butene (b) 2-butene (c) 2-methyl-1-butene (d) 2-methylpropene (e) 3-methyl-1-butene (f) 2-methyl-2-butene Only 2-butene shows cis-trans isomerism.

cis trans

4.22 Which of these alkenes show cis-trans isomerism? For each that does, draw structural formulas for both isomers.

(a) 1-Pentene (b) 2-Pentene (c) 3-Ethyl-2-pentene (d) 2,3-Dimethyl-2-pentene (e) 2-Methyl-2-pentene (f) 2,4-Dimethyl-2-pentene

Chapter 4: Alkenes and Alkynes 107

Only 2-pentene can show cis-trans isomerism, because it is the only alkene where each of the two alkene carbons is bonded to two different substituents.

cis-2-pentene trans-2-pentene

4.23 Which alkenes can exist as pairs of E/Z isomers? For each alkene that does, draw both isomers.

(a) CH2= CFBr (b) BrCH=CHBr (c) CH3CH=CBr2 (d) BrCH=CHF

Only (b) and (d) can exist as E/Z isomers.

Br F Br F (b) Br Br (d) Br Br E Z E Z

4.24 There are three compounds with the molecular formula C2H2Br2. Two of these compounds have a dipole greater than zero, and one has no dipole. Draw structural formulas for the three compounds, and explain why two have dipole moments but the third one has none.

The three compounds are 1,1-dibromoethene, cis-1,2-dibromoethene, and trans-1,2- dibromoethene. The dipole moments of the bonds in trans-1,2-dibromoethene cancel out.

net H Br Br H CC Br Br CC net H Br CC H Br H H no net dipole moment

108 Chapter 4: Alkenes and Alkynes

4.25 Draw two compounds of formula C3H5Cl that don’t have E/Z isomers and two compounds of formula C3H5Cl that are a pair of E/Z isomers.

Cl & Cl don’t have E/Z isomers

& Cl Cl are E/Z isomers

4.26 Draw a molecule of formula C4H7Br that would be classified as trans by the cis-trans system, but “Z” by the E/Z system.

4.27 Arrange the groups in each set in order of increasing priority:

(a) −CH3, −CH2CH3,−CH2Cl (b) −F, −OH, −NH2 (c) −CH2Br, −Cl, −CH2CH2l (d) −COOH, −CH=CH2,−C(CH3)3

(a) –CH3, –CH2CH3, –CH2Cl (b) –NH2, –OH ,–F (c) –CH2CH2I, –CH2Br, –Cl (d) –CH=CH2, –C(CH3)3, –COOH

4.28 Name each alkene and specify its configuration using the E,Z system.

(a) Br

(b)

(a) (E)-5-bromo-2-pentene (b) (Z)-3-ethyl-2-hexene (c) (Z)-1-fluoro-2,4,4-trimethyl-2-pentene Chapter 4: Alkenes and Alkynes 109

4.29 Draw the structural formula for at least one bromoalkene with molecular formula C5H9Br that: (a) Shows E,Z isomerism For an alkene to show E,Z isomerism, each of the two alkene carbons must be bonded to two different substituents. Thus, draw any structural isomer where each alkene carbon is bonded to two different groups. Two examples include:

Br (E)-2-bromo-2-pentene (Z)-1-bromo-1-pentene

(b) Does not show E,Z isomerism Draw any structural isomer where at least one of the alkene carbons is bonded to two identical substituents. Two examples include:

Br Br 5-bromo-1-pentene 1-bromo-3-methyl-2-butene

4.30 Classify each of the following compounds as either E or Z.

Cl OCH3 (a) (b)

IBr CH3O

OCH3 O (c) (d) Cl

Br Cl OH (e) F (f)

F3COCH3 Cl O F (g) (h)

F3C OH (i) HS “O

(a) Z (b) E (c) Z (d) Z (e) E (f) E (g) E (h) Z (i) Z 110 Chapter 4: Alkenes and Alkynes

4.31 Draw a molecule of formula C4H4Cl2 that would contain both an E and a Z double bond.

Cl Cl

4.32 Explain why each name is incorrect or incomplete, and then write a correct name: In all four molecules, cis-trans isomerism (and therefore E,Z isomerism) is not possible because at least one of the two alkene carbons is bonded to two identical substituents. Correct names are indicated below each structure.

(a) (Z)-2-Methyl-1-pentene (b) (E)-3,4-Diethyl-3-hexene

2-methyl-1-pentene 3,4-diethyl-3-hexene

2,3-dimethyl-2-hexene 2,3-dimethyl-1,3-butadiene

4.33 Draw structural formulas for all compounds with the molecular formula C5H10 that are

(a) Alkenes that do not show cis-trans isomerism.

1-pentene 2-methyl-2-butene 3-methyl-1-butene 2-methyl-1-butene

(b) Alkenes that do show cis-trans isomerism.

trans-2-pentene cis-2-pentene

Chapter 4: Alkenes and Alkynes 111

cyclopentane methylcyclobutane ethylcyclopropane 1,1-dimethylcyclopropane (d) Cycloalkanes that do show cis-trans isomerism.

cis-1,2-dimethylcyclopropane trans-1,2-dimethylcyclopropane

4.34 β-Ocimene, a triene found in the fragrance of cotton blossoms and several essential oils, has the IUPAC name (3Z)-3,7-dimethyl-1,3,6-octatriene. Draw a structural formula for β-ocimene.

1 2 3 5 7 (3Z)-3,7-dimethyl-1,3,6-octatriene 4 6 8

4.35 Which would you expect to be more stable: cis- or trans-2,5-dimethyl-3-hexene? Briefly explain why.

The trans isomer would be more stable due to a lesser amount of steric strain.

4.36 Which would you expect to be more stable: cis- or trans-cyclohexene? Briefly explain why.

The cis isomer would be more stable, as trans-cyclohexene would have a large amount of angle strain.

112 Chapter 4: Alkenes and Alkynes

4.37 Determine whether the structures in each set represent the same molecule, cis-trans isomers, or constitutional isomers. If they are the same molecule, determine whether they are in the same or different conformations as a result from rotation about a carbon–carbon single bond.

(a) and

(b) and

OH (c) and H HOH

(d) and

The two molecules in (a) are cis-trans isomers. The two structures in each of (b), (c), and (d) are identical, and the molecules in each of these pairs are conformations caused by rotation about the indicated carbon-carbon single bond(s).

(a) and cis-trans isomers

(b) and same compounds but different conformations

(d) and same compounds but different conformations

Chapter 4: Alkenes and Alkynes 113

Terpenes

4.38 Following is the structural formula of lycopene, a deep-red compound that is partially responsible for the red color of ripe fruits, especially tomatoes. Approximately 20 mg of lycopene can be isolated from 1 kg of fresh, ripe tomatoes.

(a) Show that lycopene is a terpene; that is, show that its carbon skeleton can be divided into two sets of four isoprene units with the units in each set joined head-to-tail.

***** * * * ** *

three head-to-tail linkages three head-to-tail linkages

tail-to-tail linkage joining the two sets of four isoprene units

(b) How many of the carbon-carbon double bonds in lycopene have the possibility for cis-trans isomerism? Lycopene is the all-trans isomer.

cis-trans Isomerism is possible for eleven of the carbon-carbon double bonds, which are indicated (*) in the structure above.

4.39 As you might suspect, β-carotene, a precursor of vitamin A, was first isolated from carrots. Dilute solutions of β-carotene are yellow – hence its use as a food coloring. In plants, it is almost always present in combination with chlorophyll to assist in the harvesting of the energy of sunlight. As tree leaves die in the fall, the green of their chlorophyll molecules is replaced by the yellows and reds of carotene and carotene- related molecules.

(a) Compare the carbon skeletons of β-carotene and lycopene. What are the similarities? What are the differences?

cross-linked tail-to-tail linkage

cross-linked

The main structural difference between β-carotene and lycopene is that β-carotene has six-membered rings at the two ends of the structure. To form each ring, two isoprene units are cross-linked. 114 Chapter 4: Alkenes and Alkynes

Both β-carotene and lycopene contain two sets of four isoprene units that are joined in the middle via a tail-to-tail linkage. In both molecules, all of the double bonds that can exhibit cis-trans isomerism are in the trans configuration (with respect to the main chain).

(b) Show that β-carotene is a terpene. β-Carotene has eight isoprene units, which are highlighted in the structure.

4.40 α-Santonin, isolated from the flower heads of certain species of artemisia, is an anthelmintic – that is, a drug used to rid the body of worms (helminths). It has been estimated that over one-third of the world's population is infested with these parasites. Farnesol is an alcohol with a florid odor:

7 5 1 8 3 OH 6 4 2 O 9 10 O 11 O 12 Santonin Farnesol

Locate the three isoprene units in santonin, and show how the carbon skeleton of farnesol might be coiled and then cross–linked (a carbon–carbon bond formed between two carbons) to give santonin. Two different coiling patterns of the carbon skeleton of farnesol can lead to santonin. Try to find them both.

6 8 4 2 1 5 9 5 7 3 4 10 6 10 12 12 O 2 O 8 3 1 7 9 11 11 O O O O

4.41 Lindestrene is one of the principle components of the fragrance of myrrh. Determine whether or not lindestrene is a terpene by locating the isoprene units present in the structure.

It is a terpene.

Chapter 4: Alkenes and Alkynes 115

4.42 Caryophyllene is one of the principle components of the fragrance of cloves.

(a) Determine whether or not caryophyllene is a terpene by locating the isoprene units present in the structure. (b) Assign the configuration (E or Z) to the double bond in caryophyllene for which this is possible.

(a) It is a terpene. (b) E

4.43 In many parts of South America, extracts of the leaves and twigs of Montanoa tomentosa are used as a contraceptive, to stimulate menstruation, to facilitate labor, and as an abortifacient. The compound responsible for these effects is zoapatanol:

O HO

Zoapatanol OH O H (a) Show that the carbon skeleton of zoapatanol can be divided into four isoprene units . three head-to-tail linkages

O HO

* * OH O H (b) Specify the configuration about the carbon-carbon double bond to the seven- membered ring, according to the E,Z system.

The double bond has the E configuration.

(c) How many cis-trans isomers are possible for zoapatanol? Consider the possibilities for cis-trans isomerism in cyclic compounds and about carbon-carbon double bonds.

116 Chapter 4: Alkenes and Alkynes

Possibilities are indicated (*). The two ring substituents (the −OH group and the chain containing a ketone) can be cis or trans. The double bond on the right also exhibits cis-trans isomerism. Thus, there are a total of 22 = 4 possibilities.

4.44 Pyrethrin II and pyrethrosin are natural products isolated from plants of the chrysanthemum family. Pyrethrin II is a natural insecticide and is marketed as such.

(a) Label all carbon-carbon double bonds in each about which cis-trans isomerism is possible.

Alkenes where cis-trans isomerism is possible are indicated (*).

CH2 H H O * H CH3 O H O CH CH 3 3 O CH O H * 3 H C H * C CH3 CH2 CH3 O CCH CH3OC 3 O O

Pyrethrin II Pyrethrosin

(b) Why are cis-trans isomers possible about the three-membered ring in pyrethrin II, but not about its five-membered ring?

With respect to the five-membered ring, the only substituent bonded to a tetrahedral carbon is the ester group. Recall that cis-trans isomerism in cycloalkanes results from the different spatial arrangement of two substituents bonded to two different tetrahedral (sp3) carbons. There are no other substituents bonded to an sp3 ring carbon, so it is not possible to make a same-side or opposite-side relative comparison between the ester group and another substituent on the ring. Note that the two substituents bonded to the cycloalkene ring have a planar geometry.

(c) Show that the ring system of pyrethrosin is composed of O three isoprene units. O O Isoprene units are highlighted in the structure on the right. (For clarity, the stereochemistry and implicit hydrogens have been omitted.) O CCH3 O Chapter 4: Alkenes and Alkynes 117

4.45 Lanosterol is a key steroid from which most others are made biologically. Determine whether or not lanosterol is a terpene by locating the isoprene units present in the structure.

Although lanosterol is derived from a terpene precursor, it has undergone a skeletal rearrangement that no longer permits one to account for all the carbons as isoprene units (see below).

Looking Ahead

4.46 Explain why the central carbon–carbon single bond in 1,3-butadiene is slightly shorter than the central carbon–carbon single bond in 1-butene.

1.47 Å 1.51 Å

1,3-butadiene 1-butene

The two double bonds in 1,3-butadiene are conjugated, which allows contributing structures to be drawn, as shown in the following diagram. In the contributing structure on the right, the central carbon–carbon bond is a double bond, which is shorter than a single bond. Recall that the actual structure is best represented as an “average” of the two contributing structures, and in the average structure, the central carbon–carbon bond has partial double bond character.

4.47 What effect might the ring size in the following cycloalkenes have on the reactivity of the C=C double bond in each?

90° 108° 111° 120°

118 Chapter 4: Alkenes and Alkynes

The ideal bond angle for an alkene is 120°, because the alkene carbon is sp2-hybridized. As the ring size becomes smaller, the bond angle deviates further from the ideal angle, which increases bond strain. Smaller cycloalkenes are therefore more reactive towards both ring opening (to yield an acyclic molecule) and addition reactions (Chapter 5).

4.48 What effect might each substituent have on the electron density surrounding the alkene C=C bond; that is, how does each substituent affect whether each carbon of the C−C bond is partially positive or partially negative?

(a) (b) (c) OCH3 CN Si(CH3)3

The electron density surrounding each alkene is affected by the electronegativity of the substituent near it. In molecules (a) and (b), the presence of oxygen and nitrogen, both of which are more electronegative than carbon, the alkenes have a reduced electron density, and the alkene carbon closest to the oxygen or nitrogen atom has a partial positive charge. Whereas, the silicon atom in (c) is lower in electronegativity than carbon, so the alkene has a higher electron density.

4.49 In Section 21.1 on the biochemistry of fatty acids, we will study the following three long- chain unsaturated carboxylic acids. Each has 18 carbons and is a component of animal fats, vegetable oils, and biological membranes. Because of their presence in animal fats, they are called fatty acids.

Oleic acid CH3(CH2)7CH=CH(CH2)7COOH

Linoleic acid CH3(CH2)4CH=CHCH2CH=CH(CH2)7COOH

Linolenic acid CH3CH2CH=CHCH2CH=CHCH2CH=CH(CH2)7COOH

(a) How many cis-trans isomers are possible for each fatty acid?

The number of possible cis-trans isomers corresponds to 2n, where n is the number of carbon-carbon double bonds that can exhibit cis-trans isomerism. Thus, oleic, linoleic, and linolenic acid respectively have 2, 4, and 8 cis-trans isomers.

(b) These three fatty acids occur in biological membranes almost exclusively in the cis configuration. Draw line-angle formulas for each fatty acid, showing the cis configuration about each carbon–carbon double bond. Chapter 4: Alkenes and Alkynes 119

OH O

oleic acid OH

linoleic acid O

linolenic acid

4.50 Assign an E or Z configuration and a cis or trans configuration to these carboxylic acids, each of which is an intermediate in the citric acid cycle. Under each is given its common name.

H COOH HOOC COOH

HOOC H H CH2COOH Fumaric acid Aconitic acid

Fumaric acid has the E configuration, and it can be designated as a trans alkene. Aconitic acid has the Z configuration, and it can be designated as a trans alkene (note that while the two COOH groups are cis to each other, the C5 parent chain is trans).