Brown 5e ISV SM 4-1

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-1-butene 3,3-Dimethyl-1-butyne

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

(a) (b)

trans-2,2-Dimethyl-3-hexenecis-4-Methyl-2-pentene

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

(a)Cl

Cl

Br

(b) (c)

(E)-1-Chloro-2,3-dimethyl-2-pentene

(Z)-1-Bromo-1-chloro-1-propene

(E)-2,3,4-Trimethyl-3-heptene

4.4 Write the IUPAC name for each cycloalkene:

(a) (b) (c)

1-Isopropyl-4-methylcyclohexene Cyclooctene

4-tert-Butylcyclohexene

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?

4-2 Brown 5e ISV SM

CH3C=CHCH2CH2C=CHCH2CH2C=CHCH2OH

CH3 CH3 CH3

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

identical groups (CH3). The number of possible cis-trans isomers is therefore 22 = 4.

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. The four trans double bonds in the side chain of retinal are labeled a-d. Double bond c

(between carbons 11 and 12) is isomerized to its cis isomer by an enzyme in the body.

Which of the other three double bonds in the side chain of retinal would yield the least

stable isomer of cis-retinal if it were to be isomerized? (Hint: Think steric strain.)

O

H11

12

a b c d

Consider the isomerization of a, b, and d to give the following cis-retinal isomers:

OH

a

OH

b

OH

d

Brown 5e ISV SM 4-3

The isomerization of double bond a places two methyl groups in close proximity, as

indicated in the box above. This isomer would have the highest energy (lowest stability) of

all the possible cis-retinal isomers.

4C. 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.

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.

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.

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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.

End-of-Chapter Problems

Structure of Alkenes and Alkynes

4.7 Describe what will happen when trans-3-heptene is added to the following compounds:

(a) Cyclohexane. Both cyclohexane and trans-3-heptene are nonpolar compounds. The

alkene will dissolve in cyclohexane, forming a homogenous solution.

(b) Ammonia (l). NH3 is a polar solvent (recall its trigonal pyramidal shape). When

nonpolar trans-3-heptene is added to ammonia, an immiscible mixture is formed.

4.8 Each carbon atom in ethane and in ethylene is surrounded by eight valence electrons and

has four bonds to it. Explain how the VSEPR model (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

Brown 5e ISV SM 4-5

double bond is counted as a single region of electron density), resulting in a trigonal

planar arrangement (120°).

4.9 Explain the difference between saturated and unsaturated.

A compound that is saturated does not contain any carbon-carbon bonds. Compounds

that are unsaturated contain one or more carbon-carbon bonds.

4.10 Use valence-shell electron-pair repulsion (VSEPR) to predict all bond angles about each

of the highlighted carbon atoms.

To make these predictions, determine the number of regions of electron density

associated with each carbon of interest (don’t forget the implicit hydrogens). If there are

two regions of electron density, the bond angle is 180°; three regions, 120°; and four

regions, 109.5°. Note that these respectively correlate to sp, sp2, and sp

3 hybridization.

(a)109.5°

120°

(b) CH2OH

120°(c) H C C CH CH2

120°180°(d)

120°

4.11 For each highlighted carbon atom in Problem 4.10, identify which orbitals are used to

form each sigma bond and which are used to form each pi bond.

It is useful to first identify the hybridization of carbon. Hybrid orbitals are used to form

bonds, while any remaining p orbitals can be used to form bonds. Recall that a double

bond consists of one bond and bond, and that a triple bond consists of one bond

and two bonds.

(a) (b) CH2OH

H H

H

(c) H C C CH C (d)

H

H

bonds from sp2

bond from p

bonds from sp3

bonds from sp2

bond from p

bonds from sp2

bond from p bonds from sp2

bond from p

bonds from sp

bonds from p

4-6 Brown 5e ISV SM

4.12 Predict all bond angles about each highlighted carbon atom:

OHBr

Br

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

109.5°

120°

120°

109.5°

109.5°

180°109.5°

4.13 For each highlighted carbon atom in Problem 4.12, identify which orbitals are used to

form each sigma bond and which are used to form each pi bond.

As with Problem 4.11, hybrid orbitals are used to form bonds, while remaining p

orbitals are used to form bonds.

OH

Br

Br

(a) (b)

(c) (d)

bonds from sp3

bonds from sp2

bond from p bonds from sp2

bond from p

bonds from sp3

bonds from sp3

bonds from sp bonds from p

bonds from sp3

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.

C1

C2

C3

H

HH

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.

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(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 bond with carbon 1 while the other p orbital forms the

bond with carbon 2. Accordingly, these two 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.

Nomenclature of Alkenes and Alkynes

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

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

C

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

cyclobutene 1-methylcyclopropene 2-methylcyclopropene

4-8 Brown 5e ISV SM

4.19 Write the IUPAC name for each compound:

(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. Write the correct name for the intended compound.

(a) 1,1-dibromo-2-cyclopentene

(b) 3,4-diethyl-3-pentene

(c) 4-ethyl-3-hexene

(d) 5-chloro-5-methylcyclohexene

(e) 1-tert-butyl-1-butene

(f) 3-bromo-2-chlorocyclohexene

(a) 1,1-dibromo-2-cyclopentene

For cycloalkenes the carbons of the double bond must be carbons 1 and 2. Therefore the

bromines are at position 3.

Answer: 3,3-dibromocyclopentene

a) b) c) d) e) f)

Brown 5e ISV SM 4-9

(b) 3,4-diethyl-3-pentene

The longest chain containing the double bond has 6 carbons. Also a 3-pentene is not

possible because that would have a double bond one position from the end making it a 2-

pentene.

Answer: 3-ethyl-4-methyl-3-hexene.

The chain is numbered to give the lower value for the ethyl group because it is

alphabetically before methyl.

(c) 4-ethyl-3-hexene

The correct numbering puts the ethyl group at the lower 3-position.

Answer: 3-ethyl-3-hexene

(d) 5-chloro-5-methylcyclohexene

The correct numbering puts the substituents at the lower 4-position

Answer: 4-chloro-4-methylcyclohexene

4-10 Brown 5e ISV SM

(e) 1-tert-butyl-1-butene

There is a longer chain of six carbons.

Answer: 2,2-dimethyl-3-hexene

(f) 3-bromo-2-chlorocyclohexene

The correct numbering puts the first encountered substituent (chlorine) at a lower

numbered position (1).

Answer: 6-bromo-1-chlorocyclohexene

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

Brown 5e ISV SM 4-11

4.22 Indicate which of the following alkenes show E/Z isomerism. For each that does, draw

both isomers.

(a) 3-hexene

(b) 3-ethyl-3-hexene

(c) 3-methyl-3-hexene

(d) 3-methyl-2-hexene

(e) 3,4-dimethyl-3-hexene

(f) 3-ethyl-4-methyl-3-hexene

(a) 3-hexene

Yes

(b) 3-ethyl-3-hexene

No, there are two ethyl groups on the same carbon of the double bond.

(c) 3-methyl-3-hexene

Yes

(d) 3-methyl-2-hexene

Yes

4-12 Brown 5e ISV SM

(e) 3,4-dimethyl-3-hexene

Yes

(f) 3-ethyl-4-methyl-3-hexene

No, there are two ethyl groups on the same carbon of the double bond.

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.

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.

d)b) BrBr Br

Br

ZE

BrF Br

F

ZE

Brown 5e ISV SM 4-13

C C

BrBr

H H

C C

HBr

H Br

C C

BrH

H Br

no net dipole moment

net

net

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.

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) –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) (E)-5-bromo-2-pentene

(b) (Z)-3-ethyl-2-hexene

(c) (Z)-1-fluoro-2,4,4-trimethyl-2-

pentene

Cl Cl& are E/Z isomers

don't have E/Z isomers&ClCl

Br

4-14 Brown 5e ISV SM

4.29 Draw all possible bromoalkenes with the formula C4H7Br. Indicate which exhibit E/Z

isomerism and which do not.

A C=C double bond exhibits E/Z isomerism when each carbon of the C=C bond contains

two different groups.

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

(a) Z

(b) E

(c) Z

(d) Z

(e) E

(f) E

Brown 5e ISV SM 4-15

(g) E

(h) Z

(i) Z

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

4.32 Draw all possible isomers (including cis-trans and E/Z isomers) with the molecular

formula C3H4Br2 .

Cl

Cl

4-16 Brown 5e ISV SM

4.33 Draw structural formulas for all alkenes with the molecular formula C6H12. Indicate

which alkenes exhibit E/Z isomerism and which do not.

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4.34 -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

-ocimene.

12

3

4

5

6

7

8

(3Z)-3,7-Dimethyl-1,3,6-octatriene

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 srain.

4.36 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.

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

(c) and

OHH

OH

H

same compounds but

different conformations

(d) and

same compounds but

different conformations

4.37 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. How many of the carbon-

4-18 Brown 5e ISV SM

carbon double bonds in lycopene have the possibility for cis-trans isomerism? Use the

E,Z system to assign the configuration of all applicable double bonds.

cis-trans Isomerism is possible for eleven of the carbon-carbon double bonds, which are

indicated (*) in the structure below. All of these double bonds have the E configuration.

* * * * * * * *

* * *

4.38 As you might suspect, -carotene, a precursor of vitamin A, was first isolated from

carrots. Dilute solutions of -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 -carotene and lycopene. What are the similarities?

What are the differences?

(b) Use the E,Z system to assign the configuration of all applicable double bonds.

E E E E E E E E EZ

Z

The main structural difference between -carotene and lycopene is that -carotene

has six-membered rings at the two ends of the structure. These rings are formed by

the cyclization of two sets of carbon atoms (new bonds indicated in bold).

4.39 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

O

HO

H

OHZoapatanol

*

*

(a) Specify the configuration about the carbon-carbon double bond to the seven-

membered ring, according to the E,Z system.

Brown 5e ISV SM 4-19

The double bond has the E configuration.

(b) 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.

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.

4.40 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

O

CH3

O

H

O

CH3

CH3

H

HCH

C

CH3OC

CH3

O

Pyrethrin II

*

*

O

CH3 OCCH3

O

CH2

O

H

H

CH3

OH

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 sp

3 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.

4-20 Brown 5e ISV SM

Looking Ahead

4.41 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.

1,3-Butadiene 1-Butene

1.47 Å 1.51 Å

The indicated bond in 1,3-butadiene is formed by the overlap of two sp2-hybridized

carbons, while the indicated bond in 1-butene is formed by the overlap of one sp3-

hybridized and one sp2-hybridized carbon. sp

2-Hybridized orbitals are smaller because

they have greater s character (an s orbital holds its electrons closer to the nucleus of an

atom), so the bond made from two sp2 hybrids is shorter.

4.42 What effect might the ring size in the following cycloalkenes have on the reactivity of the

C=C double bond in each?

120°111°108°90°

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).

4.43 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) OCH3

(b) CN (c) 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.

Brown 5e ISV SM 4-21

4.44 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.

OH

O

OH

O

OH

O

Oleic acid

Linolenic acid

Linoleic acid

4.45 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.

4-22 Brown 5e ISV SM

H

H

COOH

HOOC

COOH

H

HOOC

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).

Group Learning Activities

4.46 Take turns coming up with structures that fit the following criteria. For each structure you

come up with, explain to the group why your answer is correct.

(a) An alkene of formula C6H12 that cannot be named using cis-trans or E,Z.

If an alkene cannot be named using cis-trans or E,Z, then at least one of the two

alkene carbons must have two identical substituents. For example:

(b) A compound of formula C7H12 that does not contain a pi bond.

In order for a compound to have that formula and not contain a pi bond, the

compound must contain a ring. Your answer must have a ring, but don’t forget that

you can have three-, four-, five-, six-, and seven-membered rings.

(c) A compound of formula C6H10 that does not contain a methylene group.

A methylene group is an alkyl –CH2− group. An example of a compound that does

not contain the methylene group is:

(d) An alkene that uses “vinyl” in its IUPAC name.

A vinyl group is the –CH=CH2 group. An example of such a compound is:

Brown 5e ISV SM 4-23

Vinylcyclohexane

(e) A compound that can be named with the E,Z system but not with the cis/trans

system.

This would be an alkene where the use of the cis-trans system would lead to

ambiguity. An example would be:

OH

OH

(f) A compound that can be named with the cis-trans system but not with the E,Z

system.

Recall that the cis-trans system is also used to name the relative spatial positions of

substituents bonded to sp3-hybridized ring carbons. The E,Z system is used only to

name alkenes. Examples would be:

(g) A trans cycloalkene that has no ring or angle strain. (Hint: you may need to use a

model kit to explain.)

With small cycloalkenes, because of geometric constraints, the alkene must be cis

with respect to the carbon atoms of the ring. However, with large cycloalkenes, it is

geometrically possible to have the alkene in a trans configuration.

cis trans with small ring(geometrically impossible)

trans with large ring