UNSATURATION OF ALKENES

Alkenes: Unsaturation

The defining characteristic of an alkene molecule is the double bond. This bond is composed of a σ and a π covalent bond. Because π bonds are formed exclusively by the side‐to‐side overlap of atomic p orbitals, any rotation along the σ‐bond axis requires disruption of the π‐bond system. Breaking the π‐bond system requires an energy input of 63 kilocalories per mole. Due to this energy barrier, free rotation about the carbon‐carbon double bond is not possible. Multiple bonds between two atoms always lead to restricted rotation.

Compounds containing double bonds are said to be unsaturated because they are capable of reacting with hydrogen in the presence of a catalyst. The resulting compounds, which contain no multiple bonds, are said to be saturated. For each π bond destroyed, two σ bonds are formed, and energy is normally released. Thus chemists refer to a π bond as an element of unsaturation. Each element of unsaturation corresponds to two fewer hydrogen atoms than are found in the formula of the saturated compound of the same carbon‐chain length.

MOLECULAR AND STRUCTURAL FORMULAS OF ALKENES

Alkenes: Molecular and Structural Formulas

The alkenes comprise a series of compounds that are composed of carbon and hydrogen atoms with at least one double bond in the carbon chain. This group of compounds comprises a homologous series with a general molecular formula of C _n H _2 n , where n equals any integer greater than one.

The simplest alkene, ethene, has two carbon atoms and a molecular formula of C ₂H ₄. The structural formula for ethene is

In longer alkene chains, the additional carbon atoms are attached to each other by single covalent bonds. Each carbon atom is also attached to sufficient hydrogen atoms to produce a total of four single covalent bonds about itself. In chains with four or more carbon atoms, the double bond can be located in different positions, leading to the formation of structural isomers. For example, the alkene of molecular formula C ₄H ₈ has two isomers.

Stereoisomers. In addition to structural isomers, alkenes also form stereoisomers. Because rotation around a multiple bond is restricted, groups attached to the double‐bonded carbon atoms always remain in the same relative positions. These “locked” positions allow chemists to identify various isomers from the substituents' locations. For example, one structural isomer of C ₅H ₁₀ has the following stereoisomers.

The isomer on the left, in which the two substituents (the methyl and ethyl groups) are on the same side of the double bond, is called the cis isomer, while the isomer on the right, with two nonhydrogen substituents on opposite sides of the double bond, is called the trans isomer.

If more than two substituents are attached to the carbon atoms of a double bond, the cis and trans system cannot be used. With such chemicals, E‐Z notation is used. In the E‐Z system, the molecule is first bisected vertically through the double bond. Second, the two atoms or groups on each carbon atom are ranked by atomic weight. The higher atomic weight is assigned priority. For example, in Figure , the carbon and chlorine atoms on the left side of the bisecting line are ranked. Chlorine has priority because it is heavier. On the right side, bromine outranks carbon. Third, the positions of the two atoms of higher rank are determined. If the two atoms are in the cis position, the arrangement is Z (for German zusammen, meaning “together”). If the atoms or groups are in the trans position, the arrangement is E (for German entgegen, meaning “opposite”).

The name of the chemical in Figure is ( E)‐2‐bromo‐3‐chloro‐2‐butene.

PREPARATION OF ALKENES

Alkenes: Preparations

Alkenes are generally prepared through β elimination reactions, in which two atoms on adjacent carbon atoms are removed, resulting in the formation of a double bond.

Preparations include the dehydration of alcohols, the dehydrohalogenation of alkyl halides, and the dehalogenation of alkanes.

Dehydration of alcohols. In dehydration reactions, a molecule of water is eliminated from an alcohol molecule by heating the alcohol in the presence of a strong mineral acid. A double bond forms between the adjacent carbon atoms that lost the hydrogen ion and hydroxide group.

The mechanism of this dehydration reaction consists of the following steps.

1. Protonation of the alcohol.

  This step is a simple acid‐base reaction, which results in the formation of an oxonium ion, a    positively charged oxygen atom.

2. Dissociation of the oxonium ion.

  Dissociation of the oxonium ion produces a carbocation, which is a positively charged carbon atom    and an unstable intermediate.

3. Deprotonation of the carbocation.

  The positively charged end carbon of the carbocation attracts the electrons in the overlap region that    bond it  to the adjacent a carbon. This electron movement makes the α carbon slightly positive,    which in turn attracts the electrons in the overlap regions of all other atoms bonded to it. This results    in the hydrogen on the α carbon becoming very slightly acidic and capable of being removed as a    proton in an acid‐base reaction.

Zaitsev rule. It may be possible in some instances to create a double bond through an alcohol dehydration reaction in which hydrogen atoms are lost from two different carbons on the carbocation. The major product is always the more highly substituted alkene, that is, the alkene with the greater number of substituents on the carbon atoms of the double bond, an observation called the Zaitsev rule. Thus, in the dehydration reaction of 2‐butanol, the following products are formed.

The Zaitsev rule predicts that the major product is 2‐butene. Notice that each carbon atom involved in the double bond of 2‐butene has one methyl group attached to it. In the case of 1‐butene, one carbon atom of the double bond has one substituent (the ethyl group), while the other carbon atom has no substituents.

Carbocation rearrangement. The carbocation in an alcohol dehydration may undergo rearrangement to form more stable arrangements. Dehydration of 2‐methyl‐3‐pentanol, for example, leads to the production of three alkenes. The mechanism for the reaction shows that the extra compound formation is due to rearrangement of the carbocation intermediate.

The 2‐methyl‐1‐pentene molecule is formed via rearrangement of the intermediate carbocation.

The movement of a hydride ion (H: ^‐) leads to the formation of a more stable carbocation. Carbocations are classified as primary, secondary, and tertiary, as are the carbon atoms. A primary carbocation has one alkyl group attached to it; a secondary carbocation is bonded to two alkyl groups; and a tertiary carbocation has three alkyl groups around it.

Alkyl groups theoretically have the ability to “push” electrons away from themselves. This phenomenon is called the inductive effect. The greater the number of alkyl groups “pushing” electrons toward a positively charged carbon atom, the more stable the intermediate carbocation will be. This increase in stability is due to the delocalization of charge density. A charge on an atom creates a stress on that atom. The more the stress is spread over the molecule, the smaller the charge density becomes on any one atom, reducing the stress. This lessening of stress makes the ion more stable. Thus, tertiary carbocations, with three alkyl groups on which to delocalize the positive charge, are more stable than secondary carbocations, which have only two alkyl groups on which to delocalize the positive charge. For the same reason, secondary carbocations are more stable than primary carbocations.

In reality, alkyl groups do not “push” electrons away from themselves, but rather they have electrons removed from them. When an atom picks up a positive charge and becomes an ion, its electronegativity changes. In the original σ bond between two carbon atoms, the location of the overlap region relative to each carbon atom is fixed in part by the electronegativity of the two atoms. With an increase in the electronegativity of one of the carbon atoms due to ion formation, the overlap region shifts closer to the more electronegative, positively charged carbon atom. This rearrangement of electron density produces a partial positive charge on the neighboring carbon. The amount of charge gained by the second carbon corresponds to the amount lost by the fully charged carbon atom. In this manner, the charge becomes delocalized over the two carbons.

Dehydrohalogenation of alkyl halides. The dehydrohalogenation of alkyl halides, another β elimination reaction, involves the loss of a hydrogen and a halide from an alkyl halide (RX). Dehydrohalogenation is normally accomplished by reacting the alkyl halide with a strong base, such as sodium ethoxide.

This reaction also follows the Zaitsev rule, so in the reaction of 2‐chlorobutane with sodium ethoxide, the major product is 2‐butene.

Dehydrohalogenation reactions proceed via the following mechanism.

    1. A strong base removes a slightly acidic hydrogen proton from the alkyl halide via an acid‐base reaction.

    2. The electrons from the broken hydrogen‐carbon bond are attracted toward the slightly positive carbon  atom attached to the chlorine atom. As these electrons approach the second carbon, the halogen atom  breaks free, leading to the formation of the double bond. The diagram below summarizes this  mechanism.

Dehalogenation. Vicinal dihalides, which are alkane molecules that contain two halogen atoms on adjacent carbon atoms, can form alkenes upon reaction with zinc.

PHYSICAL PROPERTIES OF ALKENES

Alkenes: Physical Properties

The physical properties of alkenes are very similar to those of alkanes. Alkenes also exist as gases, liquids, and solids at room temperature. Isomeric alkenes tend to have similar boiling points, which makes it difficult to separate them by boiling point differences.

Substituted alkenes show small dipole moments due to small electron distribution differences. These small differences allow cis and trans isomers to be distinguished from each other. The effects of substitution must be deduced for each molecule, based on the positions and the electronegativity of the atoms or groups attached to the carbon‐carbon double bond. Thus, in the case of cis‐ and trans‐2‐butene, the cis isomer shows a dipole moment of 0.33 debye units (D), while the trans isomer shows a dipole moment of 0 D, due to cancellation of the electronic effects.

NOMENCLATURE OF ALKENES

Alkenes: Nomenclature

Alkenes are normally named using the IUPAC system. The rules for alkenes are similar to those used for alkanes. The following rules summarize alkene nomenclature.

    1. Identify the longest continuous chain of carbon atoms that contains the carbon‐carbon double      bond. The parent name of the alkene comes from the IUPAC name for the alkane with the same  number of carbon atoms, except the ‐ane ending is changed to ‐ene to signify the presence of a double  bond. For example, if the longest continuous chain of carbon atoms containing a double bond has five  carbon atoms, the compound is a pentene.

    2. Number the carbon atoms of the longest continuous chain, starting at the end closest to the double  bond. Thus, is numbered from right to left, placing the double bond between the second and third  carbon atoms of the chain. (Numbering the chain from left to right incorrectly places the double bond  between the third and fourth carbons of the chain.)

    3. The position of the double bond is indicated by placing the lower of the pair of numbers assigned to the  double‐bonded carbon atoms in front of the name of the alkene. Thus, the compound shown in rule 2 is  2‐pentene.

    4. The location and name of any substituent molecule or group is indicated. For example, is 5‐chloro‐2‐` hexene.

    5. Finally, if the correct three‐dimensional relationship is known about the groups attached to the double‐ bonded carbons, the cis or trans conformation label may be assigned. Thus, the complete name of the  compound in rule 4 (shown differently here) is cis‐5‐chloro‐2‐hexene.