Volume 4

Ethylene, the Simplest Alkene • How Can We Explain That CH2=CH2 Is Planar? • Alkenes with Three or More Carbon Atoms. Double Bond Position Isomerism and Cis-Trans Isomerism. Nomenclature of Alkenes • Chemical Properties of Alkenes: Addition Reactions of the C=C Double Bond. Markovnikov's Rule • Mechanism of Addition Reactions of Alkenes • Hydrogenation of Alkenes • Radical Polymerization of Ethylene • Natural Occurrence, Production, and Applications of Alkenes • Exercises

4.3.1. Ethylene, the Simplest Alkene. Alkenes, also known as olefins, are hydrocarbons containing one carbon-carbon double bond, C=C. Hydrocarbons featuring at least one C=C double bond and/or C≡C triple bond (section 4.4) on the molecule are referred to as unsaturated hydrocarbons. In contrast, alkanes and cycloalkanes lacking such multiple carbon-carbon bonds are called saturated hydrocarbons because every carbon atom in their molecules is "saturated" with hydrogen atoms.

The simplest alkene is ethylene, C2H4, or H2C=CH2. Ethylene is a nontoxic, colorless gas with boiling and melting points of -104 and -169 oC, respectively. Like all other hydrocarbons, ethylene is flammable and, like all other nontoxic gases except oxygen, C2H4 is an asphyxiant. The molecular weight of ethylene is 28, meaning that it is as dense as nitrogen and minutely less dense than air, whose average molecular mass is 29.
Digression. In contrast with odorless methane and ethane, ethylene has a faint smell that has been described as "pleasant sweet". By my perception, the odor of ethylene is neither pleasant nor sweet, yet it is not obnoxious, either. To me, the smell of ethylene is similar to that of pure dmiethylsufloxide (DMSO), (CH3)2SO, an important and broadly used solvent. Some may not agree with me since we all have somewhat different odor perception. If you happen to come by a sample of DMSO and are curious enough to smell it, you may. DMSO is not toxic (less toxic than ethyl alcohol!) and, in fact, widely used in medicine. However, keep in mind that old samples of DMSO are often contaminated with other sulfur compounds, mainly dimethylsulfide, (CH3)2S, which gives it a strong unpleasant rotten egg - garlic odor even if present in minute quantities. Pure DMSO is devoid of this odor and has a very faint smell. If you accidentally spill nontoxic DMSO on your skin, do not panic. Just rinse the area with water. But do not be surprised if just minutes after the exposure you get bad breath and/or start smelling of garlic. DMSO penetrates the skin easily (which is why it is used in medicine) and metabolizes very quickly to give rise to malodorous dimethylsulfide.
Since the 1950s, ethylene has been the most important industrial organic chemical. The current annual world consumption of ethylene is approximately 200 million tons and growing. The world as we know it would not exist without ethylene. We will consider some important applications of ethylene later in this section. Meanwhile, I would like to touch on two interesting facts about this molecule. One is that ethylene is an important plant hormone that accelerates ripening and regulates many other processes in plants, including the shedding of leaves and the opening of flowers. Second, the first organometallic compound ever reported (back in 1831) was Zeise's salt, a platinum complex (compound) of ethylene.

4.3.2. How Can We Explain That CH2=CH2 Is Planar? It has been established experimentally that the molecule of ethylene is planar. All six atoms of the H2C=CH2 molecule lie in one plane, with all H-C-H and C-C-H angles being approximately 120o (Figure 4-21).
Figure 4-21. The molecule of ethylene (C2H4) is planar (right image: modified from source).

The planar geometry of ethylene is in contrast with the already familiar to us tetrahedral geometry of methane, ethane, and other alkanes. To rationalize this difference and the entire structure of ethylene, we will use the hybridization concept again. In the current case, however, only two of the three 2p orbitals on the carbon atom hybridize with the 2s orbital. The third 2p orbital is not uninvolved, remaining "as is" (Figure 4-22). To indicate the number of s and p orbitals mixing together in the case of ethylene, this type of hybridization is referred to as sp2.
Figure 4-22. Formation of three identical sp2 orbitals from one 2s orbital and two 2p orbitals (sp2 hybridization). One of the three 2p orbitals remains non-hybridized.

Now we have to position the three sp2 hybrids and one original non-hybrid p orbital in space such that (a) all three sp2 orbitals are equivalent and (b) repulsions between the three sp2 orbitals and one 2p orbital carrying one electron each is minimized. There is only one way to fulfill both requirements: the three sp2 orbitals lie in one plane to form an equilateral triangle and the non-hybrid p orbital is perpendicular to that plane (Figure 4-23).
Figure 4-23. Orientations of orbitals of an sp2-hybridized carbon atom.

Let us now place one more sp2-hybridized atom in the same plane next to the first one. One sp2 orbital of one of the two C atoms overlaps with an sp2 orbital of the other C atom (Figure 4-24). This end-to-end overlap results in the formation of a C-C bond. Simultaneously, the non-hybridized p orbitals of the two carbon atoms also overlap to form one more C-C bond. Note that the p non-hybrids overlap in a different manner, side-to side.
Figure 4-24. Formation of the double C=C bond of ethane by end-to-end overlap of the sp2 orbitals (red) and side-to-side overlap of the non-hybridized p orbitals (blue).

The end-to-end type of orbital overlap is not new to us. For example, this is the way the Cl-Cl bond in Cl2 and the C-C bonds of alkanes are formed. The side-to-side overlap, however, is something we have not encountered before. To distinguish between these two different types of covalent bonds, those resulting from an end-to-end overlap of orbitals are called σ-bonds (sigma-bonds) and those from a side-to-side orbital overlap π-bonds (pi-bonds).

The remaining four single-electron-occupied sp2 orbitals then overlap with 1s orbitals of four hydrogen atoms to give a molecule of ethylene (Figure 4-25).
Figure 4-25. Ethylene model derived from two sp2-hybridized C atoms and four H atoms.

So, we have just built a model of ethylene using the sp2 hybridization concept and simple logic. This model accounts for the following experimentally established facts.

- The molecule of ethylene is planar;

- The H-C-H and C-C-H angles in the molecule of ethylene are approximately 120o; and

- The C=C double bond of ethylene is considerably stronger and shorter (bond length 1.33 Å) than the C-C single bond of ethane (bond length 1.54 Å).

4.3.3. Alkenes with Three or more Carbon Atoms. Double Bond Position Isomerism and Cis-Trans Isomerism. Nomenclature of Alkenes. By replacing one or more H atoms of ethylene with alkyl groups we derive alkenes containing more than two carbon atoms. Due to the presence of a C=C double bond, alkenes have two hydrogen atoms less than alkanes composed of the same number of carbon atoms. The general formula of alkenes is, consequently, CnH2n.

Naming alkenes is easy, as long as you remember the names of linear alkanes. Just change the suffix ane in the name of an alkane to ene (Table 3).

Table 3. Nomenclature of linear alkenes CnH2n vs. linear alkanes CnH2n+2.
The closest homologue of ethylene is propene, CH3-CH=CH2, which, like propane, exists as only one isomer. Let us now consider the next homologue, butene. First off, skeletal isomerism that is characteristic of alkanes is as characteristic of alkenes (Figure 4-26).
Figure 4-26. Skeletal isomers of butane and butene.

There is, however, another type of isomerism that exists for alkenes but not for alkanes. This isomerism is new to us. For the unbranched chain of four carbon atoms (butene), one can easily envision two isomers, one with the C=C bond at the end of the chain, CH2=CH-CH2-CH3, and one in the middle, CH3-CH=CH-CH3 (Figure 4-27). This type of isomerism is called double bond position isomerism because the isomers differ in the place of the double bond within the same carbon skeleton.
Figure 4-27. Double bond position isomers of butene.

Let us take a closer look at the butene molecule featuring the double bond in the center. As shown in Figure 4-28, this alkene can exist as two isomers. One of the two has both CH3 groups on the same side of the molecule. This isomer is called cis. In the other isomer, called trans, the CH3 groups are on opposite sides of the molecule.
Figure 4-28. Butene: cis and trans isomers.

Cis-trans isomerism is a type of geometric isomerism, the occurrence of compounds that have the same composition and connectivity yet differ in spatial arrangements of the atoms. Why is it that geometric isomerism is characteristic of alkenes but not for alkanes? As mentioned above, there is free rotation about the C-C single bond of alkanes. In contrast, rotation about the C=C bond is not possible without breaking its π-component (Figures 4-24 and 4-25), as clearly shown in this animated cartoon. It is entirely due to the rigidity of the C=C double bond that cis and trans isomers of alkenes exist.

Cis and trans isomers of the same alkene have different properties, including physical, chemical, and even biological. For example, the boiling point of cis-butene is ~4 oC, whereas trans-butene boils at ~1 oC. A remarkable example of how important the cis-trans geometry around a C=C bond can be, is muscalure, the sex pheromone of the common housefly. Housefly females produce muscalure to attract males for mating. Muscalure is a cis linear alkene containing 23 carbon atoms with the double bond in the 9th position (Figure 4-29). While muscalur is sensed by housefly males even if present in miniscule concentrations, the trans isomer of muscalure is virtually inactive!
Figure 4-29. Structures of muscalure and its trans isomer.

So, there are three types of isomerism that are encountered in alkenes: (1) skeletal; (2) double bond position; and (3) cis-trans. The nomenclature of alkenes takes into account all three.

For a simple straight chain alkene devoid of any branches, we have to use the names presented in Table 3. In front of the name we place a number to indicate the position of the double bond. The numbering of the carbon chain begins at the end nearest the double bond. Some examples illustrating this rule are presented in Figure 4-30.
Figure 4-30. Examples of straight chain (unbranched) alkenes with names.

To name a branched alkene, we first find the longest carbon chain that includes both carbon atoms of the C=C double bond. The numbering of the chain begins at the end nearest the double bond. Note the important difference in the numbering rules for alkanes and alkenes. For alkanes, we have to find the longest chain, whereas for alkenes it should be the longest chain involving the C=C link. This point is illustrated by the two structures of an alkane and an alkene featuring the same carbon skeleton (Figure 4-31).
Figure 4-31. Difference in the numbering of C atoms of an alkane and an alkene with the same carbon skeleton. For alkanes, the longest chain should be found. For alkenes, the longest chain containing both carbon atoms of the C=C bond is selected.

Next we have to name the parent alkene (Table 3). For the structure in Figure 4-31 (right), it is hexene. The next step is to indicate the position of the double bond in the chain. In our case, it is 1-hexene. Now we have to specify all of the substituents and their positions. There is an ethyl group in the 2nd position and a methyl group in the 3rd position. Hence the full name is: 2-ethyl-3-methyl-1-hexene. This name bears little resemblance to that of the alkane with the same carbon skeleton (Figure 31, left), 3,4-dimethylheptane.

Finally, we have to specify if our alkene is cis or trans. Obviously, 1-alkenes, the ones with the C=C bond at the end of the main chain (also known as terminal alkenes), cannot be cis or trans because the terminal C atom of the double bond bears two identical substituents, H atoms. For alkenes containing an internal C=C bond, the cis or trans geometry is specified by placing cis- or trans- (usually italicized) at the very beginning of the name, as exemplified in Figure 4-32.
Figure 4-32. Examples of cis-trans isomers with names.

4.3.4. Chemical Properties of Alkenes: Addition Reactions of the C=C Double Bond. Markovnikov's Rule. Like alkanes and other hydrocarbons, alkenes burn in oxygen or air, for example:

C2H4 + 3 O2 = 2 CO2 + 2 H2O

In contrast with chemically inert alkanes, however, alkenes readily react with a broad variety of substances. The high reactivity of alkenes stems from the propensity of the π-bond to "open up", thereby creating two valences for bonding to a variety of atoms or groups of atoms. For example, ethylene quickly reacts with bromine or chlorine even in dilute aqueous solutions. In these reactions, a molecule of Br2 or Cl2 adds across the double bond to give 1,2-dibromoethane or 1,2-dichloroethane, respectively (Figure 4-33).
Figure 4-33. Addition of Br2 or Cl2 to ethylene.

Higher alkenes such as propylene react with halogens similarly (Figure 4-34).
Figure 4-34. Addition of halogens to propylene.

There are two key differences between reactions of halogens with alkanes and with alkenes.

1. Reactions of alkanes with halogens are substitution (or displacement) reactions; a hydrogen atom of an alkane is substituted with a halogen atom. In contrast, the reactions of alkenes with halogens are addition reactions, meaning that both atoms of a halogen molecule add to the carbon atoms of the C=C double bond.

2. To make alkanes react with Cl2 or Br2, initiation by light is required. These reactions are governed by a radical chain mechanism (see previous subsection). In contrast, alkenes react with halogens even in the dark because no initial light-induced homolytic halogen-halogen bond cleavage is needed for the transformation to occur. These reactions do not involve radicals of any sort.

Alkenes also add hydrogen halides, such as HBr and HCl (Figure 4-35).
Figure 4-35. Addition of HBr to ethylene.

The important question arises as to which of the two possible products would be formed in the reaction of HBr with propylene? The two carbon atoms connected by the double bond are equivalent in ethylene, but nonequivalent in propylene. So, which of the two carbons is going to get the H atom and which the Br atom of the HBr molecule? Figure 4-36 answers this question.
Figure 4-36. Addition of HBr to propylene.

The reaction of propylene with HBr is governed by Markovnikov's rule, which states that in the addition of hydrogen halides HX to the C=C bond of an asymmetric alkene, the H atom of HX bonds to the carbon atom bearing more hydrogen substituents. This rule is named after Vladimir Markovnikov (1838-1904), the Russian chemist who established the positional selectivity of such addition reactions in the 1860s.

It might help to memorize Markovnikov's rule by recalling Percy Bysshe Shelley's famous aphorism "the rich get richer" and applying it to the addition of HX to alkenes. If one of the two carbons of the C=C bond is more rich in hydrogen (bonded to more H atoms) than the other, then it is this already richer in hydrogen carbon that will become even richer in hydrogen after the addition of HX.

A few more examples illustrating Markovnikov's rule are presented in Figure 4-37. Let us take a closer look at the bottom equation in this figure, showing that water also adds across the C=C bond. This reaction of hydration of propylene obeys Markovnikov's rule, too, and requires an acid catalyst to occur. With phosphoric acid as the catalyst, the hydration of propylene is run on a large industrial scale to manufacture isopropanol (rubbing alcohol). Likewise, ethanol is made from ethylene and water on an even larger scale (Figure 4-38).
Figure 4-37. Addition of HX to three different alkenes.
Figure 4-38. Addition of water to ethylene.

4.3.5. Mechanism of Addition Reactions of Alkenes. The mechanism of addition of halogens, hydrogen halides, and water to alkenes is known in considerable detail. We will discuss a simplified version of this mechanism in order to adapt it to our introductory course.

The key to the reactivity of alkenes is the π-bond, the one formed by the side-to-side overlap of the non-hybridized p orbitals on the carbon atoms (Figures 4-24 and 4-25). The π-bond is polarizable, which means that this nonpolar covalent bond can become polar covalent in the electric field of ions or dipoles. This key characteristic feature of the π-bond is illustrated in Figure 4-39.
Figure 4-39. Polarization of the π-bond of ethylene in the presence of H+ of dissociated HBr.

In the absence of any polar molecules or ions, the electron pair of the π-bond is equally shared by the two carbon atoms (Figure 4-39, A). Let us now mix ethylene with aqueous HBr, a strong acid that is dissociated into H+ and Br- in solution. As the positively charged H+ approaches the vicinity of the C=C bond, the negatively charged shared electron pair of the π-bond starts shifting toward the carbon that is closer to the proton (Figure 4-39, B). We say that the π-bond becomes polarized toward one of the two carbon atoms. Consequently, that carbon becomes partially negatively charged (δ-) and the other carbon partially positively charged (δ+).

Once the π-bond is polarized, the electrostatic attraction between its electron pair and the H+ grows stronger, forcing them to get closer and closer to each other. Eventually, a covalent bond is formed between the proton and the carbon atom toward which the π-bond was polarized (Figure 4-40). The electron pair that used to constitute the π-bond between the two C atoms is now shared by one of them and the hydrogen. Naturally, a full positive charge is formed on the other carbon atom as a result, because it has lost the electron it had originally contributed to the formation of the C-C π-bond.
Figure 4-40. Step-wise addition of HBr to ethylene.

Organic ions featuring a full positive charge on a carbon atom are called carbocations. Carbocations are highly unstable and reactive species that seek stabilization through chemical bond formation with a lone electron pair donor. In our case, the bromide anion that is left over from the dissociation of the HBr molecule serves as the lone pair donor to the carbocation to form the C-Br bond (Figure 4-40).

This step-wise mechanism of HBr addition to the C=C double bond explains the need for an acid catalyst to promote the reaction of alkenes with water (Figures 4-37 and 4-38). Unlike HBr, water is a poor electrolyte, which cannot generate protons in quantities sufficient for the reaction to occur. Therefore, H3PO4 is used to produce the H+ that is needed in the first step of the reaction. The sequence of events is shown in Figure 4-41.
Figure 4-41. Mechanism of acid-catalyzed addition of water to ethylene.

In the first step (A), a proton adds to ethylene to give the carbocation, exactly as in the reaction with HBr (Figure 4-41). The acid used in this reaction, however, is H3PO4, not HBr. Unlike the bromide anion, the dihydrogen phosphate anion, H2PO4-, produced on dissociation of H3PO4, is a very poor electron pair donor to the positively charged carbon atom. Water, which is plentiful in the reaction mixture, is a much better donor of a lone electron pair. The carbocation adds to a molecule of H2O in the second step of the reaction (B) just like a proton adds to water to produce the hydronium ion, H3O+. Finally (step C), the ethyl hydronium cation eliminates a proton to give the final product, ethanol, CH3CH2OH.

Notably, the proton needed for the first step of the reaction to happen (A) is regenerated in the last step (C) and can participate in the first step again. Consequently, the proton serves as a catalyst for the reaction of addition of water to ethylene.

The reaction of an alkene with bromine occurs via a similar mechanism, even though Br2 on its own does not dissociate to Br- and Br+. The shared electron pair that holds the two Br atoms together in the molecule of Br2 equally belongs to both atoms. In other words, the Br-Br bond is nonpolar covalent. Yet, this bond is poarizable. As a molecule of Br2 approaches a molecule of ethylene, the Br-Br bond and the π-bond of CH2=CH2 get mutually polarized, as shown in Figure 4-42 (A). The degree of this polarization of both bonds increases as the two species get closer to each other. Eventually, the π electron density is shifted altogether to the carbon closer to the Br2 molecule with the simultaneous heterolytic Br-Br bond cleavage and C-Br bond formation. In the final step (B), the carbocation and the bromide anion come together to give the final product.
Figure 4-42. Simplified mechanism of the reaction of ethylene with Br2.

4.3.6. Hydrogenation of Alkenes. Hydrogen can also be added to the C=C double bond of an alkene to give the corresponding alkane (Figure 4-43).
Figure 4-43. Hydrogenation of ethylene.

This alkene hydrogenation reaction requires a transition metal catalyst (Ni, Rh, Ru) to occur, and is governed by a distinct mechanism which is certainly beyond the scope of this course. Hydrogenation of alkenes is an industrially important reaction. For example, the process to make margarine is based on catalytic hydrogenation of the C=C double bonds in the molecules of vegetable oils.

4.3.7. Radical Polymerization of Ethylene. This reaction is of exceptional importance to the modern world. Ethylene undergoes polymerization to polyethylene, one of the three most important commodity polymers. Polymers are large molecules (also known as macromolecules) formed by linking together many small repeating molecular units. The repeating unit of polyethylene is [-CH2-CH2-]. In the general formula of polyethylene, [-CH2-CH2-]n, n is the degree of polymerization, the number of times the [-CH2-CH2-] unit repeats in the long chain. For different types of polyethylene, n varies from approximately 1,000 to 300,000.

One way to make polyethylene is by radical polymerization of ethylene. To initiate the polymerization, organic free radicals are generated. A good source of organic radicals is benzoyl peroxide, the very compound that is used as the active ingredient in many acne treatment creams and ointments. On slight heating, benzoyl peroxide decomposes via homolytic cleavage of the weak O-O bond (drawn in red in Figure 4-44). The resulting benzoyl radicals then decompose to CO2 and phenyl radicals.
Figure 4-44. Generation of phenyl radicals from benzoyl peroxide.

Do not be intimidated by the formulas of benzoyl peroxide and benzoyl and phenyl radicals. Toward the end of this course you will be fine with these structures. For the time being, however, all we need to know is that the phenyl radical (conventionally denoted as Ph•) is an extremely reactive species that instantly adds to the C=C bond of ethylene. In this addition, one of the two electrons of the π bond of the ethylene molecule pairs with the unpaired electron of the phenyl radical to form a Ph-C bond. The other electron of the π bond is now in full possession of the other carbon atom (Figure 4-45). The newly produced free radical then adds to another molecule of ethylene in the same manner. As this process repeats itself over and over again, the hydrocarbon chain grows to form the polymer, polyethylene.
Figure 4-45. Mechanism of radical polymerization of ethylene.

Does the chain grow until all of the ethylene has been consumed? Sometimes it may. However, there are pathways to termination of the polymerization process, such as when two radicals in the reaction mixture recombine to form a non-radical molecule (Figure 4-46).
Figure 4-46. Recombination of radicals leads to termination of the polymerization process.

Radical polymerization of ethylene furnishes so-called low-density polyethylene, which is rather flexible and is used to manufacture such products as thin grocery bags, clear sandwich bags, and squeeze bottles. Much harder and stronger high-density polyethylene for making plastic gallon milk jugs and thick glossy shopping bags (among many other things) cannot be produced by radical polymerization. High-density polyethylene is made by polymerizing ethylene in a non-radical catalytic process employing special Ziegler-Natta catalysts. These catalysts are named after Karl Ziegler (Germany) and Giulio Natta (Italy) who won the 1963 Nobel Prize in Chemistry for their groundbreaking developments of new polymerization reactions and catalysts.

4.3.8. Natural Occurrence, Production, and Applications of Alkenes. Ethylene is biosynthesized by plants and, as mentioned above, is an important hormone that accelerates the ripening and regulates many other processes in plants. Alkenes are present in crude oil, albeit in vastly smaller quantities than alkanes.

Alkenes and especially ethylene are paramount for the production of various chemicals and materials. Ethylene is manufactured on a scale that is unrivaled in the chemical industry of organic compounds and polymers. The petrochemical process called cracking is used to produce ethylene, propylene, and other light alkenes from longer chain alkanes and cycloalkanes distilled off crude oil.

Over one half of all ethylene produced worldwide is used to make polyethylene of various grades. Large quantities of ethylene are also consumed in the production of ethylene glycol (HOCH2CH2OH) and acetaldehyde (CH3CHO). Ethylene glycol is one of the two key compounds needed to make polyester. Acetaldehyde serves as a precursor to many industrially important chemicals. Propylene, the second most industrially important alkene after ethylene, is used to make polypropylene and many chemicals that are vital to several industries.

4.3.9. Exercises.

1. Ethylene (a) is dangerous because it is toxic while being odorless; (b) is a nontoxic odorless low-boiling liquid; (c) has a strong pungent odor; (d) is a flammable gas that is nontoxic and has a faint odor; (e) is a solid polymer used to make plastic bags and milk bottles. Answer

2. Ethylene is planar due to sp2 hybridization of the carbon atoms. True or false? Answer

3. The high reactivity of alkenes originates from (a) the C-H bonds at an sp2-hybridized C atom being highly polarizable and consequently reactive; (b) the σ-bond being polarizable and reactive; (c) the π-bond being polarizable and reactive. Answer

4. Using the hybridization concept, explain why ethylene is planar. [Answer: See 4.3.2]

5. Draw structural formulas for: 1-pentene; cis-3-hexene; trans-3-hexene; 2-methyl-1-hexene; 2,3-dimethyl-2-butene; cis-3,6-diethyl-4-octene; trans-3,6-diethyl-4-octene; cyclohexene; isobutene.

6. Name the following alkenes:

7. What is wrong with the following chemical names? (a) trans-3-methyl-1-pentene; (b) 2,2-dimethyl-3-pentene; (c) 3-methyl-3-ethyl-2-heptene. Draw the structures and give correct names. Answer

8. Are cyclohexane, 1-hexene, and 2,3-dimethyl-2-butene isomers? Answer

9. Finish the following equations by drawing structures of the products.
10. Write balanced chemical equations for combustion of (a) propene and (b) 2-methyl-1-hexene. Answer

11. What is Markovnikov's rule?

12. At temperatures above 60 oC, azobisisobutyronitrile (AIBN) decomposes to produce free radicals, as shown in the equation below.
The reaction produces nitrogen gas and isobutyronitrile radicals, which can be used to polymerize ethylene. Write a chemical scheme for ethylene polymerization, initiated by an isobutyronitrile radical. Answer

13. The general formula for cycloalkenes is (a) CnH2n; (b) CnH2n-2; (c) CnH2n-4. Answer

14. What are the three types of isomerism observed for alkenes? What is geometric isomerism in chemistry? Explain why geometric (cis-trans) isomerism is characteristic of alkenes but not alkanes. [Answer: See 4.3.3]