Because alkanes have only single bonds, they contain the largest possible number of hydrogen atoms per carbon atom. As a result, they are called saturated hydrocarbons. Alkenes, alkynes, and aromatic hydrocarbons contain multiple bonds (double, triple, or delocalized π bonds). As a result, they contain less hydrogen than an alkane with the same number of carbon atoms. Collectively, they are called unsaturated hydrocarbons. On the whole, unsaturated molecules are more reactive than saturated ones.


How many isomers are there for propene, C3H6?

FIGURE 24.7 The alkene C4H8 has four structural isomers.


Alkenes are unsaturated hydrocarbons that contain at least one C═C bond. The simplest alkene is CH2═CH2, called ethene (IUPAC) or ethylene, which plays important roles as a plant hormone in seed germination and fruit ripening. The next member of the series is CH3—CH═CH2, called propene or propylene. Alkenes with four or more carbon atoms have several isomers. For example, the alkene C4H8 has the four structural isomers shown in FIGURE 24.7. Notice both their structures and their names.

The names of alkenes are based on the longest continuous chain of carbon atoms that contains the double bond. The chain is named by changing the ending of the name of the corresponding alkane from -am to -ene. The compound on the left in Figure 24.7, for example, has a double bond as part of a three-carbon chain; thus, the parent alkene is propene.

The location of the double bond along an alkene chain is indicated by a prefix number that designates the double-bond carbon atom that is nearest an end of the chain. The chain is always numbered from the end that brings us to the double bond sooner and hence gives the smallest-numbered prefix. In propene the only possible location for the double bond is between the first and second carbons; thus, a prefix indicating its location is unnecessary. For butene (Figure 24.7) there are two possible positions for the double bond, either after the first carbon (1-butene) or after the second carbon (2-butene).


How many distinct locations are there for a double bond in a five-carbon linear chain?

If a substance contains two or more double bonds, the location of each is indicated by a numerical prefix, and the ending of the name is altered to identify the number of double bonds: diene (two), triene (three), and so forth. For example, CH2═CH—CH2—CH═CH2 is 1,4-pentadiene.

The two isomers on the right in Figure 24.7 differ in the relative locations of their methyl groups. These two compounds are geometric isomers, compounds that have the same molecular formula and the same groups bonded to one another but differ in the spatial arrangement of these groups. (Section 23.4) In the cis isomer the two methyl groups are on the same side of the double bond, whereas in the trans isomer they are on opposite sides. Geometric isomers possess distinct physical properties and can differ significantly from each other in their chemical behavior.

FIGURE 24.8 Geometric isomers exist because rotation about a carbon-carbon double bond requires too much energy to occur at ordinary temperatures.

Geometric isomerism in alkenes arises because, unlike the C—C bond, the C═C bond resists twisting. Recall from Section 9.6 that the double bond between two carbon atoms consists of a σ and a π bond. FIGURE 24.8 shows a cis alkene. The carbon-carbon bond axis and the bonds to the hydrogen atoms and to the alkyl groups (designated R) are all in a plane, and the p orbitals that form the π bond are perpendicular to that plane. As Figure 24.8 shows, rotation around the carbon-carbon double bond requires the π bond to be broken, a process that requires considerable energy (about 250 kJ/mol). Because rotation doesn't occur easily around the carbon–carbon bond, the cis and trans isomers of an alkene cannot readily interconvert and, therefore, exist as distinct compounds.

SAMPLE EXERCISE 24.3 Drawing Isomers

Draw all the structural and geometric isomers of pentene, C5H10, that have an unbranched hydrocarbon chain.


Analyze We are asked to draw all the isomers (both structural and geometric) for an alkene with a five-carbon chain.

Plan Because the compound is named pentene and not pentadiene or pentatriene, we know that the five-carbon chain contains only one carbon–carbon double bond. Thus, we begin by placing the double bond in various locations along the chain, remembering that the chain can be numbered from either end. After finding the different unique locations for the double bond, we consider whether the molecule can have cis and trans isomers.

Solve There can be a double bond after either the first carbon (1-pentene) or second carbon (2-pentene). These are the only two possibilities because the chain can be numbered from either end. Thus, what we might erroneously call 3-pentene is actually 2-pentene, as seen by numbering the carbon chain from the other end:

Because the first C atom in 1-pentene is bonded to two H atoms, there are no cis-trans isomers. There are cis and trans isomers for 2-pentene, however. Thus, the three isomers for pentene are

(You should convince yourself that cis-3-pentene is identical to cis-2-pentene and trans-3-pentene is identical to trans-2-pentene. However, cis-2-pentene and trans-2-pentene are the correct names because they have smaller numbered prefixes.)


How many straight-chain isomers are there of hexene, C6H12?

Answer: five (1-hexene, cis-2-hexene, trans-2-hexene, cis-3-hexene, trans-3-hexene)


Alkynes are unsaturated hydrocarbons containing one or more C≡C bonds. The simplest alkyne is acetylene (C2H2), a highly reactive molecule. When acetylene is burned in a stream of oxygen in an oxyacetylene torch, the flame reaches about 3200 K. Because alkynes in general are highly reactive, they are not as widely distributed in nature as alkenes; alkynes, however, are important intermediates in many industrial processes.

Alkynes are named by identifying the longest continuous chain containing the triple bond and modifying the ending of the name of the corresponding alkane from -ane to -yne, as shown in Sample Exercise 24.4.

SAMPLE EXERCISE 24.4 Naming Unsaturated Hydrocarbons

Name the following compounds:


Analyze We are given the condensed structural formulas for an alkene and an alkyne and asked to name the compounds.

Plan In each case the name is based on the number of carbon atoms in the longest continuous carbon chain that contains the multiple bond. In the alkene, care must be taken to indicate whether cis-trans isomerism is possible and, if so, which isomer is given.


(a) The longest continuous chain of carbons that contains the double bond is seven carbons long, so the parent hydrocarbon is heptene. Because the double bond begins at carbon 2 (numbering from the end closer to the double bond), we have 2-heptene. With a methyl group at carbon atom 4, we have 4-methyl-2-heptene. The geometrical configuration at the double bond is cis (that is, the alkyl groups are bonded to the double bond on the same side). Thus, the full name is 4-methyl-cis-2-heptene.

(b) The longest continuous chain containing the triple bond has six carbons, so this compound is a derivative of hexyne. The triple bond comes after the first carbon (numbering from the right), making it 1-hexyne. The branch from the hexyne chain contains three carbon atoms, making it a propyl group. Because this substituent is located on C3 of the hexyne chain, the molecule is 3-propyl-1-hexyne.


Draw the condensed structural formula for 4-methyl-2-pentyne.


Addition Reactions of Alkenes and Alkynes

The presence of carbon–carbon double or triple bonds in hydrocarbons markedly increases their chemical reactivity. The most characteristic reactions of alkenes and alkynes are addition reactions, in which a reactant is added to the two atoms that form the multiple bond. A simple example is the addition of a halogen to ethylene:

The pair of electrons that forms the π bond in ethylene is uncoupled and is used to form two σ bonds to the two bromine atoms. The σ bond between the carbon atoms is retained.

Addition of H2 to an alkene converts it to an alkane:

The reaction between an alkene and H2, referred to as hydrogenation, does not occur readily at ordinary temperatures and pressures. One reason for the lack of reactivity of H2 toward alkenes is the stability of the H2 bond. To promote the reaction, a catalyst is used to assist in rupturing the H—H bond. The most widely used catalysts are finely divided metals on which H2 is adsorbed. (Section 14.7)

Hydrogen halides and water can also add to the double bond of alkenes, as in these reactions of ethylene:

The addition of water is catalyzed by a strong acid, such as H2SO4.

The addition reactions of alkynes resemble those of alkenes, as shown in these examples:

SAMPLE EXERCISE 24.5 Identifying the Product of a Hydrogenation Reaction

Write the condensed structural formula for the product of the hydrogenation of 3-methyl-1-pentene.


Analyze We are asked to predict the compound formed when a particular alkene undergoes hydrogenation (reaction with H2) and to write the condensed structural formula of the product.

Plan To determine the condensed structural formula of the product, we must first write the condensed structural formula or Lewis structure of the reactant. In the hydrogenation of the alkene, H2 adds to the double bond, producing an alkane.

Solve The name of the starting compound tells us that we have a chain of five C atoms with a double bond at one end (position 1) and a methyl group on C3:

Hydrogenation—the addition of two H atoms to the carbons of the double bond—leads to the following alkane:

Comment The longest chain in this alkane has five carbon atoms; the product is therefore 3-methylpentane.


Addition of HCl to an alkene forms 2-chloropropane. What is the alkene?

Answer: propene


As the understanding of chemistry has grown, chemists have advanced from simply cataloging reactions known to occur to explaining how they occur. An explanation of how a reaction occurs is called a mechanism. (Section 14.6)

The addition reaction between HBr and an alkene, for instance, is thought to proceed in two steps. In the first step, which is rate determining (Section 14.6), the HBr attacks the electron-rich double bond, transferring a proton to one of the double-bond carbons. In the reaction of 2-butene with HBr, for example, the first step is

The electron pair that formed the π bond is used to form the new C—H bond.

The second, faster step is addition of Br to the positively charged carbon. The bromide ion donates a pair of electrons to the carbon, forming the C—Br bond:

Because the rate-determining step involves both the alkene and the acid, the rate law for the reaction is second order, first order in the alkene and first order in HBr:

The energy profile for the reaction is shown in FIGURE 24.9. The first energy maximum represents the transition state in the first step, and the second maximum represents the transition state in the second step. The energy minimum represents the energies of the intermediate species, and Br.

To show electron movement in reactions like these, chemists often use curved arrows pointing in the direction of electron flow. For the addition of HBr to 2-butene, for example, the shifts in electron positions are shown as


What features of an energy profile allow you to distinguish between an intermediate and a transition state?

FIGURE 24.9 Energy profile for addition of HBr to 2-butene. The two maxima tell you that this is a two-step mechanism.

Aromatic Hydrocarbons

The simplest aromatic hydrocarbon, benzene (C6H6), is shown in FIGURE 24.10 along with some other aromatic hydrocarbons. Benzene is the most important aromatic hydrocarbon, and most of our discussion focuses on it.

FIGURE 24.10 Line formulas and common names of several aromatic compounds. The aromatic rings are represented by hexagons with a circle inscribed inside to denote delocalized π bonds. Each corner represents a carbon atom. Each carbon is bound to three other atoms—either three carbons or two carbons and a hydrogen––so that each carbon has the requisite four bonds.

Stabilization of π Electrons by Delocalization

The planar structure of benzene, with its 120° bond angles, suggests a high degree of unsaturation. You might therefore expect benzene to resemble the alkenes and to be highly reactive. Benzene and the other aromatic hydrocarbons, however, are much more stable than alkenes because the π electrons are delocalized in the π orbitals. (Section 9.6)

We can estimate the stabilization of the π electrons in benzene by comparing the energy required to form cyclohexane by adding hydrogen to benzene, to cyclohexene (one double bond) and to 1,4-cyclohexadiene (two double bonds):

From the second and third reactions, it appears that the energy required to hydrogenate each double bond is roughly 118 kJ/mol for each bond. Benzene contains the equivalent of three double bonds. We might expect, therefore, the energy required to hydrogenate benzene to be about 3 times –118, or –354 kJ/mol, if benzene behaved as though it were “cyclohexatriene,” that is, if it behaved as though it had three isolated double bonds in a ring. Instead, the energy released is 146 kJ less than this, indicating that benzene is more stable than would be expected for three double bonds. The difference of 146 kJ/mol between the “expected” heat (that is, enthalpy) of hydrogenation, –354 kJ/mol, and the observed heat of hydrogenation, –208 kJ/mol, is due to stabilization of the π electrons through delocalization in the π orbitals that extend around the ring.

Substitution Reactions

Although aromatic hydrocarbons are unsaturated, they do not readily undergo addition reactions. The delocalized π bonding causes aromatic compounds to behave quite differently from alkenes and alkynes. Benzene, for example, does not add Cl2 or Br2 to its double bonds under ordinary conditions. In contrast, aromatic hydrocarbons undergo substitution reactions relatively easily. In a substitution reaction one hydrogen atom of a molecule is removed and replaced (substituted) by another atom or group of atoms. When benzene is warmed in a mixture of nitric and sulfuric acids, for example, one of the benzene hydrogens is replaced by the nitro group, NO2:

More vigorous treatment results in substitution of a second nitro group into the molecule:

There are three isomers of benzene that contain two nitro groups—ortho-, meta-, and para-dinitrobenzene:

In the reaction of Equation 24.11, the principal product is the meta isomer.

Bromination of benzene, carried out with FeBr3 as a catalyst, is another substitution reaction:

In a similar reaction, called the Friedel-Crafts reaction, alkyl groups can be substituted onto an aromatic ring by reacting an alkyl halide with an aromatic compound in the presence of AlCl3 as a catalyst:


When the aromatic hydrocarbon naphthalene, shown in Figure 24.10, reacts with nitric and sulfuric acids, two compounds containing one nitro group are formed. Draw the structures of these two compounds.