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Structure

Alkenes are hydrocarbon compounds which contain less than the maximum quantity of hydrogen. Thus, they are referred to as unsaturated hydrocarbons. The simplest member of the alkene family is ethylene gas, or ethene, C2H4 (B.P.: - 104 °C). We begin by connecting the two C atoms with a covalent bond, and then attach two H atoms to each C atom. With only 6 atoms in the valance shell of each C atom, we find that the molecule needs an additional pair of electrons in order to be neutral.  

The double bond which exists between the two carbon atoms is a result of the equal sharing of 2 electron pairs between the 2 carbon atoms.    In order to form bonds with 3 other atoms, carbon makes use 3 equivalent sp2 hybridized orbitals (mixing one s and two p orbitals).

                                                         Ethylene - C2H4

As we have seen in methane, the sp2 hybridized orbitals lie in a single plane and are directed towards the corners of an equilateral triangle. All bond angles are thus 120 degrees. This trigonal planar arrangement permits the hybrid orbitals to be as far apart as possible. Just as mutual repulsion among sp3 hybridized orbitals gives 4 tetrahedral bonds, so among sp2 hybridized orbitals it gives 3 trigonal planar bonds. 

Thus each carbon atom lies a the center of a triangle, at whose corners are located the 2 hydrogen atoms and the other C atom. The individual bonds are very similar to he bonds in methane, and are given the same designation: sigma bond.

The molecule is not complete however. In forming the sp2 orbitals, each C atom has used only two of its three p orbitals. The remaining p orbitals consist of two equal p lobes -- one lying above and one lying below the plane of the three sp2 hybridized orbitals. The p orbital of one C atom overlaps the p orbital of the other C atom. The electrons pair up and the additional (pi) bond is formed.

The alkene double bond therefore consists of the combination of a strong sigma bond (w/ bonding electrons) and a weak pi bond (w/ non-bonding electrons). The total bond energy of 146 kcal is nearly twice that that of the C - C single bond of ethane. One result of this that the bond distance is shorter since the atoms are more tightly bound. The high electron density in the vicinity of the double bond makes it a natural location and/or center for a number of various types of chemical reactions.

The next member of the alkene family is propylene, C3H6. Note the numbering of the carbon atoms, and the isomeric equivalence of all 4 potential positions of the methyl group. Thus, there are no isomers for this molecule.

Moving on to the butylenes, C4H8, we find that there are a number of possible geometric arrangements possible which satisfy the chemical formula. For example, we may have straight chain skeleton (as in n-butane) or a branched-chain structure (as in isobutane). In the first case, we have: 

1) A straight chain terminating in a double bonded carbon atom. 

2) A straight chain with a double bond in the middle. In this case, we have two diastereomers (cis vs. trans) of the same molecular compound 

3) We also have a branched structural isomer of this same compound which we call Isobutylene or 2-methylpropene.

On hydrogenation, the branched compound yields isobutane, while all 3 of the straight chain compounds yield n-butane. [ Note: There are other organic compounds which have the formula C₄H₈, namely cyclobutane and methylcyclopropane, but are not alkenes and are not discussed here. There are also four-carbon cycloalkenes such as cyclobutene and methylcyclopropene, but they do not have the formula C₄H₈ and are not discussed here.]

The alkenes form another homologous series, with the increment being the same as for alkanes: CH2. The general formula for this family is:

                                                                                    C_nH_2n 

As we ascend the series of higher alkenes, the number of isometric structures for each member increases even more rapidly than in the case of the alkane series. In addition to variations in the carbon skeletons, there are variations in the positions of the double bonds for a given skeleton, and the possibility of structural isomerism.

Properties & Functions

As a class, the alkenes possess physical properties similar to those of the alkanes. They are insoluble in water, but quite soluble in non-polar solvents like benzene, ether, chloroform or ligroin. They are less dense than water. The boiling point increases with increasing molecular weight, rising 20 - 30 degrees with each added carbon (except for very small homologs). Branching lowers the boiling point. The simplest alkenes, ethylene, propylene and butylene are gases. Linear alkenes of approximately five to sixteen carbons are liquids, and higher alkenes are waxy solids.

Ethylene acts physiologically as a growth hormone in plants. It stimulates the ripening of fruit, the opening of flowers, and the abscission (or shedding) of leaves. Its biosynthesis starts from methionine with 1-aminocyclopropane-1-carboxylic acid (ACC) as a key intermediate. It is also found in many lip gloss products.

At room temperature and pressure, propylene is a gas. It is colorless, odorless , and highly flammable. It is found in coal gas and can be synthesized by cracking petroleum. Propylene is a major commodity in the petrochemicals industry. The main use of propylene is as a monomer, mostly for the production of polypropylene.

Polypropylene is a thermoplastic polymer, used in a wide variety of applications, including food packaging and textiles for high performance outdoor clothing.  Propylene is extracted from petroleum and then transformed into polypropylene which is melted and then spun into fibers, which are solidified by cooling and stretched. Plastic parts and reusable containers of various types can be formed, as wel as laboratory equipment, loudspeakers, automotive components, and polymer banknotes. The material is rugged and unusually resistant to many chemical solvents, bases and acids. Propylene is also used as a fuel gas for various industrial processes.

For a more detailed description of the relationship between the molecular structure of polymerized alkenes, methods of organic synthesis, and practical uses as engineering materials, please see the Unit 22: Synthetic Polymers

Production & Synthesis

Ethylene is the organic compound produced on the largest scale. Global production of ethylene exceeded 75,000,000 metric tons per year in 2005. Ethylene is produced in the petrochemical industry by steam cracking.

In this process, gaseous or light liquid hydrocarbons are briefly heated to 750–950 °C, inducing numerous free radical reactions. Generally, in these reactions, large hydrocarbons break down in to smaller ones and saturated hydrocarbons become unsaturated. The result of this process is a complex mixture of hydrocarbons in which ethylene is one of the principal components. The mixture is separated by repeated compression and distillation. Thus, alkenes up to four carbon atoms can be obtained in pure form from the petroleum industry. Pure samples of higher alkanes must be prepared by methods like those outlined below.

The introduction of a C - C double bond into a molecule containing only single bonds must necessarily involve the elimination of atoms or constituent groups from two adjacent carbons.

Dehydrohalogenation of Alkyl Halides

When isopropyl bromide is treated with a hot concentrated alcoholic solution of a strong base (e.g. potassium hydroxide, KOH) the result is propylene, potassium bromide and water.

This is an example of dehydrohalogenation: 1,2 elimination of the elements of hydrogen halide. Dehydrohalogenation involves the loss (or elimination) of the halogen atom and of  a hydrogen atom form a carbon adjacent to the one losing the halogen. The reagent required is a base, whose function is to abstract the hydrogen as a proton.

1

In order to elucidate the reaction mechanism, we note that the halogen atom leaves the molecule as a halide ion, and thus mast take its electron pair along with it. Hydrogen is abstracted by the base as a proton, and hence must leave its electron pair behind. This is ht electron pair that is available to form the second bond - the pi bond - between the carbon atoms.

We have called this 1, 2 elimination. In order for the double bond to form, the hydrogen must come form a beta-carbon that is adjacent to the alpha-carbon holding the halogen atom.

In some case, this reaction yields a single alkene, and in other cases yields a mixture of alkenes. In order to predict which products can be formed in a given reaction, we have only to examine the structure of the substrate. We can expect an alkene corresponding to the loss of any one of the beta hydrogens -- but no other alkenes.

1) n-butyl bromide can lose hydrogen only form C-2, and yields only 1-butene. 

2) sec-Butyl bromide can lose hydrogen either form C-1 or C-3, and yields a mixture of both 1-butene and 2-butene. Where the two alkenes can be formed, 2-butene is the chief product. (We will discuss this more later in this Unit).

Duality of Mechanism:  E1 vs. E2

The theory of elimination reactions developed in a way remarkably similar to the way the theory of nucleophilic substitution developed. We begin by examining the reaction kinetics. Normally, the reaction will follow second-order kinetics. That is the rate of alkene formation depends on the concentrations of two different substances: alkyl halide and base. The 2nd order reaction is observed for all classes of alkyl halides.

                                                                  Rate  =  k [ RX ] [ :B ]

Now, if one proceeds along a series of substrates, 1° to 2° to 3°, and if one reduces the strength or concentration of the base, a second type of behavior begins to appear which is dominated by first-order reaction kinetics. The rate of elimination depends only upon the concentration of alkyl halide, and is independent of the concentration of the base.

                                                                      Rate  =  k [ RX ] 

In general, this 1st-order reaction is encountered only with secondary or tertiary substrates, and in solutions where the base is either weak or in low concentration.

For the reaction that proceeds by 2nd-order kinetics, we refer to what has been named the E2 mechanism (Hughes and Ingold). This reaction involves only one single step.

The base pulls a proton away form the carbon. At the same time, a halide ion (the leaving group, X) departs and the double bond is formed. The halogen ion takes its electron pair with it. The hydrogen leaves its electron pair behind to form the double bond.

In the transition state, two bonds are being broken: C-H and C-X. So where does the energy for the bond breaking come from ? As usual, it comes from bond-making:

a) Formation of the bond between the proton and the base

b) Formation of the pi bond.

The rate-determining step - the only step - involves reaction between a molecule of alkyl halide and a molecule of base. The rate is proportional to the concentration of both reactants. This is a bimolecular elimination reaction, because in the rate-determining step two molecules undergo covalency changes.

The primary evidence for this reaction mechanism is that:

a) The reaction follows 2nd-order reaction kinetics.

b) The reactions are not accompanied by molecular rearrangements.   

We have already observed that htis reaction often yields a mixture of isomeric alkenes. So which isomer, if any, will predominate ? Studies have shown that one isomer generally will predominate, and that it is possible to predict which isomer this will be -- that is to predict the orientation of elimination -- on the basis of molecular structure.

E.G. Consider sec-butyl bromide.

Attack by base at any one of the three beta-hydrogens on C-1 can lead to the formation of 1-butene. Attack at either of the two beta-hydrogens on C-3 can lead to the formation of 2-butene. We see that 2-butene is the preferred product despite a probability factor of 3:2 working against its formation. 

If we focus our attention, not on the hydrogen being lost, but rather on the alkene being formed, we see the following. The preferred product, 2-butene, is a di-substituted alkene, whereas 1-butene is a mono-substituted alkene. I.E. In 2-butene, there are two alkyl CH3 groups attached to the doubly bonded carbons, and in 1-butene there is only one alkyl C2H5 group.

In the other examples we see that a di-substituted alkene is preferred over a mono-substituted alkene, and a tri-substituted alkene is preferred over a di-substituted alkene.

The preferred product is the alkene that has the greater number of alkyl groups attached to the double bonded carbon.

Keep in mind that these are all irreversible reactions, so that orientaiotn is determined by the relative rates of competing reactions. More 2-butene is formed because 2-butene is formed faster than 1-butene under these conditions. The alkene with the greater number of alkyl groups is the preferred product because it is formed faster than the alternative alkenes. Thus we have a sequence for the relative rates of alkene formation as follows:

We shall see in Unit 10 that the stability of alkenes follows exactly the same sequence.

Thus, the more stable the alkene, the faster it is formed. Predominant formation of the more stable isomer is called Saytzeff orientation.

Consider the transition state for this E2 reaction. Bonds to both hydrogen and the leaving group are partially broken, and the double bond is partially formed. The transition state has thus acquired considerable alkene character.  

Factors that stabilize the alkene -- alkyl groups in this case -- also stabilize the incipient alkene in the transition state. The activation energy is lowered, and the alkene is formed more rapidly. Thus, the nature of the transition state is a major factor in determining its stability, and hence the rate of reaction.  

Alkene stability not only determines orientation of the reaction, but also is an important factor in determining the reactivity of an alkyl halide elimination. Relative rates of reaction (per hydrogen) increase as the alkene becomes more highly substituted. As one proceeds along a series of alkyl halides from 1° to 2° to 3°, the structure by definition becomes more branched at the carbon carrying the halogen. This increased branching has two results: 

a)  It provides a greater number of beta-hydrogens for attack by base, and hence a more favorable probability factor towards elimination.

b)  It leads to a more highly branched, more stable alkene, and hence a more stable transition state and a lower activation energy. 

The result is that in E2 dehydrohalogenation, the order of reactivity of alkyl halides is: 

Reactivity of RX towards E2    

3°  >  2°  >  1° 

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In the proposed E1 mechanism, the electronic changes (including the bond-breaking and bond- making) are the same as in E2. However, instead of occurring simultaneously, they are occurring one after the other. Where E2 involves a single step, E1 involves 2 steps. 

1) The substrate undergoes a slow heterolysis to form the halide ion and a carbocation.

2) The carbocation rapidly loses a proton to the base and forms the alkene. 

We recognize step (1) as identical ot the first step in SN1. In the second step of SN1, the carbocation combines with a nucleophile to yield the substitution product. In step (2) of E1 the carbocation reacts with the base to yield the elimination product. 

Again, the reactions of the carbocation provide a pair of electrons to complete the octet of the electron-deficient carbon. In  SN1these electrons are provided as an unshared pair on the nucleophile. In E1 they are the pair originally shared by the proton, and made available - thru pi bond formation -- by departure of the proton.  Thus, we know now that a carbocation can:

a)  Combine with a a nucleophile.

b)  Rearrange to a more stable carbocation.

c)  Eliminate a proton to form an alkene.

In the E! reaction follows 1st-order reaction kinetics. I.E. the overall rate of reaction is determined solely by a single step: the first slow / rate-determining step. Except for the solvent molecules, this rate-determining step involves only one substrate, and its rate depends only on the concentration of that single substrate. The rate of an E1 reaction is independent of base concentration. 

   Rate  =  k [ RX ] 

because the reaction whose rate we are measuring does not involve the base. Thus it is the rate of formation of carbocations that determines how fast a reaction goes. Once formed, the carbocations rapidly react to yield a product -- an alkene

The primary evidence for the E1 reaction mechanism is that:

1)  The reaction follows 1st-order reaction kinetics

2)  The reactions show the same effect of structure on reactivity as SN1 reactions do.

3)  Where structure permits, the reactions are accompanied by rearrangement. 

In order to understand the effect of structure on reactivity, we need only recall that E1 involves exactly the same first step as SN1. Since this first step is rate-determining, it follows that the order of reactivity of alkyl halides in E1 must be the same as in SN1. Experiment has proven this to be the case. Thus, 

Reactivity in E1    

3°  >  2°  >  1° 

In E1, as in SN1, reactivity is determined by the rate of formation of the carbocation, which depends on the stability of the carbocation.

Where the structure permits, these first-order elimination reactions are accompanied by rearrangement. Since the first step is identical to SN1 and yields carbocations, it follwos that E1 should also be susceptible to carbocation rearrangements. This is indeed, confirmed by experiment. 

The double bonds appear in places remote from the carbon that held the leaving group. 

Sometimes the carbon skeleton is changed.

Alkenes are often obtained from substrates that do not contain a beta-hydrogen.

In each case, it is evident that if, indeed, the alkene is formed form a carbocation, it is not the same carbocation that was initially formed from the substrate. 

In each of these examples, the initially formed carbocation can rearrange by a 1,2 shift to form a more stable carbocation. When this can happen, it does.

It is this new carbocation that loses the proton -- in a perfectly straightforward way from the beta position -- to yield the new alkenes.

 Elimination by E1 shows strong Saytzeff orientation. Thus, when more than one alkene can be formed, the more highly branched -- the more stable -- alkene is the preferred product. Thus a di-substituted alkene is preferred over a mono-substituted one.  

Also, a tri-substituted alkene is preferred over a di-substituted one. 

Interestingly, in the E1 reaction mechanism, orientation and reactivity are still determined by the relative rates of reaction -- but of different steps. How fast the substrate reacts is determined by the rate of step (1). But which alkene is produced is clearly determined by which beta-proton is lost faster from the carbocation in step (2). 

Let us examine the transition state, then, for this product-determining step. 

The C-H bond is partially broken, and the double bond is partially formed. Thus, as before, the transition state has acquired alkene character. As in E2, factors that stabilize the alkene also stabilize the incipient alkene in the transition state. The activation energy is lowered, and the alkene is formed faster. 

When rearrangement occurs in E1, we still predict orientation by Saytzeff's rule. But we must now consider the loss of beta-protons form the rearranged cations as well as from the cations initially formed. 

So which mechanism is preferred ?? How can we tell which mechanism, E2 or E1, is likely to be implemented under a particular set of conditions. 

Let's start by looking at the effect of the nature of the substrate's alkyl group. As one proceeds along the sequence from 1° to 2° to 3°, the reactivity by both mechanisms increases, although for different reasons. 

1) Reactivity by E2 increases largely because of the greater stability of the more highly branched alkenes being formed. 

2) Reactivity by E1 increases because of the greater stability of the carbocation being formed in the rate-determining step.

Thus, except that it is very difficult for primary substrates even to form carbocations, we can expect no abrupt shift in mechanism due simply to changes in the alkyl group.

But if we turn now to the role played by the base, we find a striking difference. In E2, the base participates, or plays an active role, in the rate-determining step. In E1, it does not. Thus, the rate of E2 depends on the concentration of the base, while that for E1 does not. The rate of E2 also depends on the nature of the base. In E2, A stronger base pulls proton away from the substrate more rapidly. In E1, the base always waits until the carbocation is formed. 

Thus, for a given substrate, the more concentrated or powerful the base, the more E2 is favored over E1. Under the conditions typically used to bring about dehydrohalogenation -- a concentrated solution of a strong base -- the E2 mechanism is the path taken by elimination. In general, the E1 mechanism is encountered only with secondary or tertiary substrates, and in solutions where the base is dilute or weak  -- typically where the base is the solvent. 

Substitution vs. Elimination

Another quandary appears when considering the competition between substitution and elimination. The nature of the alkyl group is perhaps the major factor in this competition. For example, both bimolecular reactions result from attack on the substrate by the reagent :Z (acting as a nucleophile or base). 

Trendwise, primary substrates undergo elimination slowest and substitution fastest. Tertiary substrates undergo elimination fastest and substitution slowest. 

In summary, the proportion of elimination increases as the structure of the substrate is changed from primary to secondary to tertiary. Many tertiary substrates yield almost exclusively alkenes under these conditions. 

In addition, many nucleophiles are, like the hydroxide ion, very strong bases. Here elimination competes strongly with substitution. There are other reagents, however, that are good nucleophiles but comparatively weak bases, where substitution tends to be favored.

A less polar solvent tends to favor elimination, as does a higher temperature (e.g. hot KOH). Alternatively, a lower temperature and a more polar solvent (e.g. water) tend to increase the yield of the substitution product: the alcohol. 

In the competition between the unimolecular reactions, since the first step (heterolytic cleavage) is identical on both cases, it is the second step that the reaction path forks.   

In this second step, there is attack by the nucleophilic, basic reagent :Z, which is typically the solvent. This time, the attack is not on the substrate itself, but rather on the carbocation. Attack at the electron-deficient carbon brings about substitution, whereas attack at he hydrogen brings about elimination.

The proportion of products are determined by the relative rates of reaction. How fast a carbocation loses a proton depends upon the stability of the alkene being formed (consider branching). Thus the rate of elimination follows the sequence 3°  >  2°  >  1° while the rate of substitution is opposite: 1°  >  2°  >  3°.

In summary, when we want the product of a substitution reaction, elimination is a nuisance to be avoided. With some nucleophiles, an acceptable yield of a substituted product can be obtained only with primary (and possibly secondary) substrates. But when we want an alkene, in order to achieve elimination we encourage the bimolecular reaction E2 by using a solvent of low polarity and a high concentration of strong base.    

Dehydration of Alcohols

In contrast to the elimination reactions we have considered so far, the dehydration of alcohols is an acid catalyzed elimination reaction in the presence of heat. 

Dehydration is accomplished by the elimination of a molecule of water, H20. It is generally accomplished in two different ways:  

1)  By heating the alcohol with sulfuric acid or phosphoric acid. 

2)  By passing the alcohol vapor over a catalyst, typically alumina (Al2O3), at high temperatures. 

(Note: The alumina functions either as an electrophilic Lewis acid, or as a classical acid via OH groups on the surface).  

The ease of dehydration of alcohols proceeds as:

                                                                              3°  >  2°  >  1° 

The following examples show how these differences in reactivity affect the conditions of dehydration. Certain tertiary alcohols are so prone to dehydration that they can be distilled only if precautions are taken to protect the system form all acidic fumes.

For dehydration of secondary and tertiary alcohols, the following mechanism is generally accepted. 

Step (1): Fast acid-base reaction between the alcohol and the catalyzing acid. This gives the protonated alcohol and the conjugate base.

Step (2): The protonated alcohol undergoes heterolysis to form the carbocation and water.

Step (3): The carbocation loses a proton to the base to yield the alkene. 

In steps 2 and 3, we recognize a kind of E1 elimination with the protonated alcohol as substrate. Step 1 is simply the fast, reversible prelude that produces the actual substrate. Strong evidence has been logged to substantiate the validity of this proposed mechanism.

Dehalogenation of Vicinal Dihalides

Dehalogenation of vicinal (neighboring) dihalides is severely limited by the fact that these dihalides are themselves generally prepared from the alkenes. However, it is sometimes useful to convert an alkene into a dihalide while we perform some operation on another part of the molecule, and then to regenerate the alkene by treatment with zinc. This procedure is referred to as protecting the double bond.