Recall: Structure and Bonding

Alkenes

Introduction
Recall: Structure and Bonding
Degree of Unsaturation
Nomenclature of Alkenes
Cis/trans Isomerism of Alkenes
Cahn-Ingold-Prelog E/Z System
Physical Properties of Alkenes
Reactions of Alkene
Electrophilic Addition Reactions
Markovnikov’s Rule
Anti-Markovnikov’s Rule (Free-radical Reaction)
Carbocations as Intermediates in the addition of HX to alkenes
Rearrangements of carbocations
Addition of Halogens
Halohydrin Formation
Hydroboration-Oxidation
Oxymercuration-Reduction
Hydrogenation of Alkenes
Hydroxylation of Alkenes
Oxidative Cleavage of Alkenes – Ozonolysis
Addition of carbenes to Alkenes: Preparation of cyclopropane
Allylic bromination of Alkenes
Polymerization of alkenes
Preparation of Alkenes

Introduction

Unsaturated hydrocarbons have one or more carbon-carbon double or triple bonds and contain fewer hydrogen atoms than alkanes. There are three classes of unsaturated hydrocarbons, namely alkenes, alkynes and arenes (benzenes).

Alkenes have at least one carbon-carbon double bond and alkynes one triple carbon-carbon triple bond. Arenes, usually called aromatics, will be given a closer later.

Quite a number of alkenes are useful in nature. For example, ethylene is a plant hormone that induces the ripening of fruits; α-pinene is a major component of turpentine; β-carotene is the orange pigment responsible for the color of carrots and serves as a source of vitamin A.

Some pheromones – compounds produced by an organism for the purpose of communicating with other organisms of the same species, are alkenes. A pheromone may act as a sex attractant or set an alarm or mark a trail of food. For example, have you ever considered why ants follow each other in a line?

1. Recall: Structure and Bonding

sp2 +sp2 = σ bond

p + p = π bond

Bond angle H-C-H = 120 º

H-C-C = 120 º

Bond lengths C=C = 1.34 Å

C-H = 1.10 Å

The presence of double bonds in alkenes confers on them the restricted rotation of the carbon-carbon double bond. Thus conformational isomers are not easily recognizable in alkenes. Consequently, alkenes show cis/trans isomerism in which each carbon atom of the double bond has two different groups attached to it. For example, 2-butene shows cis/trans isomerism. If the two methyl groups are on the same side of the double bond, the molecule is said to exhibit cis isomerism and if the methyl are on opposite sides of the double bond the molecule is said to exhibit trans isomerism.

2.1 Degree of Unsaturation

The degree of unsaturation is the number of multiple bonds and/or rings in a molecule. This number can be calculated depending on the number and kinds of atoms in molecules.

Alkenes that contain only carbon and hydrogen

Degree of unsaturation (DU) = #(C+1) - #H/2

For example, cyclohexene, C6H10, DU = (6+1) – 10/2 = 2

The degree of unsaturation of cyclohexene is two which means that cyclohexene may have one triple bond, two double bonds, or a ring and a double bond. Cyclohexene has a ring and a double bond.

The presence of an oxygen atom does not affect the degree of unsaturation of a molecule. In other words, ignore the presence of the oxygen atom.

e.g., C6H11OH has a DU = (6+1) – 12/2 =1 (a ring or a double bond)

If a halogen is present, use the following formula

DU = #(C+1) - #(X+H)/2, where X is the halogen.

e.g., C4H6Br2 has a DU = (4+1) – (2+6)/2 = 1 (a double bond or ring)

In the presence of nitrogen atoms, the formula is modified as follows

DU = (C+1) + N/2 – (X+H)/2

e.g., C5H9N has a DU = (5+1) +1/2 – (0+9)/2 = 2 (a ring + double or 2 double bonds or 1 triple bond)

Problem 2.

Write out the structures for the following compounds: C4H6Br2’ C5H9N

Problem 2

Calculate the degree of unsaturation for each of the following compounds

C3H4Cl4 b. C5H8N2c. C8H14d. C5 H6e. C20H32f. C3H4

2.2 Nomenclature of Alkenes

Alkenes are named following a series of rules similar to those enumerated in alkanes with the suffix –ene instead of –ane. The basic rules are:

Name the longest parent chain containing the double bond using the suffix –ene.

The parent name is heptene (longest chain containing the double bond), not octene since it does not contain the double bond.

Number the carbon atoms in the chain beginning at the end nearest the double bond. If the double bond is equidistant from the two ends, begin numbering at the end nearer the first branch.

Write out the name in full naming the substituents according to their positions in the chain and listing them in alphabetical order. Indicate the position of the double bond by giving the number of the first alkene carbon atom.
If more than one double bond is present, indicate the position of each and use the suffixes –diene, -triene, -tetraene, etc.

Problem

Name the following compound.

Cycloalkenes are named in a similar way. We number the cycloalkenes such that the double bond is between C1 and C2 and the first branch point has as low a value as possible.

There are some low-molecular weight alkenes that do not conform to the IUPAC rules of nomenclature but they are of common usage. For example, ethylene (ethane) is accepted by the IUPAC because it has been used for so long. Other examples include groups like:

e.g.,

The table below indicates common names of some alkenes

Compound / Systematic Name / Common Name
CH2=CH2 / ethane / ethylene
CH3CH=CH2 / propane / propylene
(CH3)2C=CH2 / 2-methylpropene / isobutylene
CH2=C(CH3)CH=CH2 / 2-methyl-1,3-butadiene / isoprene
CH3CH=CHCH=CH2 / 1,3-pentadiene / piperylene
CH2=CH- / ethenyl / vinyl
CH2=CH-CH- / Propenyl / allyl
CH=C-CH2- / Propynyl / propargyl
CH2= / methylene
CH3CH= / Ethylidene

Both common and systematic names are recognized by IUPAC.

Problem

Give the IUPAC names for the following compounds.

CH2=CHCH(CH3)C(CH3)3
CH3CH2CH=CH(CH3)CH2CH3
CH3CH=CHCH(CH3)CH=CHCH(CH3)2

Problem

Draw the structures corresponding to these IUPAC names.

2-methyl-1,5-hexadiene
3-ethyl-2,2-dimethyl-3-heptene
2,33-trimethyl-1,4,6-octatriene
3,4-diisopropyl-2,5-dimethyl-3-hexene
4-tert-butyl-2-methylheptane

Problem

Provide systematic names for these cycloalkenes.

2.3 Sequence Rules: E/Z System of Nomenclature

Consider the following compound. Try to name this compound as cis or trans.

It can be labeled trans because the two identical groups are on opposite sides of the double bond. It can also be labeled cis because larger groups are on the same side of the double bond. Consequently, the naming of this compound is ambiguous as far as the cis/trans nomenclature is concerned. A set of rules were proposed to assign priorities to the groups attached to the carbon-carbon double bond of the alkenes. They postulated that if groups of high priority are on the same side of the double bond, the molecule is said to be in the Z (“zussamen” in German) conformation or geometry and if they are on opposite sides, the molecule is said to be in E (“entgegen” in German) conformation.

How do we prioritize the groups of substituents that are attached to the double bond? The following rules are proposed:

Atoms of higher atomic number have higher priority. Consider the atoms or groups of atoms attached to the following structure.

The carbon atom attached to the double bond has a higher priority than the H atom. The following examples illustrate this point.

For (Z)-2-pentene, the groups of higher priority are on opposite sides of the double bond, i.e., - CH3 and – CH2CH3 groups.

An isotope of higher atomic mass receives higher priority. For example, D (deuterium or 2H), an isotope of hydrogen has a higher priority than the 1H atom.

If atoms attached to the alkene carbon are identical, the next atom in each group is considered and so on.

Consider the structure above. Which of the groups have a higher priority? Which atom makes the difference?

For the purposes of assigning priority, triple bonds have higher priority than double bonds and double bonds have high priority than single bonds.

The E,Z nomenclature of compounds with more than one double bond is illustrated by the following molecule.

Study the name of this compound very well. Can you figure out how it was named? Easy. Right? Check out the priorities on the double bonds. Can you see them? Answer the question before you continue. If not, you are not ready for the next one. All right, pal, name the compound below.

Problem

Which member in each set is higher in priority?

– H or – Br
– Cl or – Br
– NH2 or – OH
– CH2OH or – CH3
– CH2OH or – CHO

Rank the sets of substituents in order of Cahn-Ingold-Prelog priorities.

– CH3, - CH2CH3, - CH=CH2, - CH2OH
– COOH, -CH2OH, -C=N (nitrile), - CH2NH2
–CH2CH3, - C=CH (triple bond), - C=N (nitrile), -CH2OCH3

Assign E or Z configurations for these alkenes.

Give the structures of

(2Z,4Z)-2,4-octadiene
(2Z,5Z)-2,5-octadiene

2.4Physical Properties of Alkenes

Physical properties of alkenes are similar to those of alkenes except for dipole moments and melting points. The low-molecular weight alkenes are flammable and nonpolar.

Let us compare some physical properties of 1-hexene and n-hexane.

Table showing some alkenes and physical properties

Property / Compound
Boiling point / 63.4 ºC / 68.7 ºC
Melting point / - 139.8 ºC / - 95.3 ºC
Density / 0.673 g/mL / 0.660 mg/mL
Solubility in water / negligible / negligible
Dipole moment / 0.46 D / 0.085 D

How we account for the higher dipole moments in alkenes? Electron density lies closer to the nucleus in sp2 orbitals than it does in sp3 orbitals. Alkyls groups attached to double bonds are polarized towards the double bond.

2.5Relative Stabilities of Alkene Isomers

Cis-alkenes are less stable than their trans counterparts because of steric (spatial) interference between two bulky substituents are on the same side of the double bond. Cis-and trans-2 butene is used to demonstrate the relative stability of alkenes.

In the cis isomer, the bulky methyl groups on the same side of the double bond tend to interfere with each other making the molecule unstable relative to the trans isomer where there less interference between the hydrogen atom and the methyl group.

Comparatively, the alkene with the greatest number of alkyl groups on the double bond is usually the most stable, i.e.,

An alkyl group stabilizes an electron-deficient carbocation (see later for explanation of this term) in two ways:

through the inductive effect
through the partial overlap of filled orbitals with empty ones.

The inductive effect is a donation of electron density through sigma bonds of the molecule. The positively charged carbon atom withdraws some electron density from the alkyl groups bonded to it.

An alkyl substituent with filled orbitals can overlap with π bond and a properly oriented C-H σ bond of a neighboring substituent. This type of overlap between a p-orbital and a sigma bond is called hyperconjugation.

2.6 Reactions of Alkenes

2.6a Addition Reactions of Alkenes

Alkenes act as nucleophiles, i.e., electron-rich species which donate a pair of electrons to an electron-deficient carbon atom called an electrophile. This reaction can be compared to an acid-base reaction in which the alkene acts as a Lewis base and the electrophile as the Lewis acid.

A reaction mechanism describes in detail how a reaction occurs. It attempts to elucidate which bonds are broken and formed and at what rate these processes occur. The typical mechanism for the addition reaction is as follows: When an electrophile, E+, approaches the weakly held pi electrons of the double bond of the alkene, the electron pair is donated to the electrophile forming a dative covalent sigma bond. A carbocation is formed, which is subsequently attacked by the nucleophile to form a neutral addition product.

There are two steps in this mechanism: Step 1 consists of the formation of the carbocation intermediate and in step 2, the carbocation is attacked by an electron-rich species. We will illustrate the steps in this mechanism by the reaction of 2-butene with a hydrogen halide.

Step 1.

eqn 2.1

Step 2.

eqn 2.2

The energetics of the reaction can be shown by the following diagram (Figure xx).

Define and indicate transition states on the diagram and show breaking and forming of bonds

This reaction depends on the presence of a strong electrophile to react with the π bond generating a carbocation. Since there is a rate-determining step, this kind of reaction is called an electrophilic addition reaction to an alkene. Here are some examples.

eqn 2.3

eqn 2.4

HI is usually generated in situ by the reaction of KI with H3PO4.

eqn 2.5

2.6b Orientation of Electrophilic Addition: Markovnikov’s Rule

In electrophilic addition reactions to unsymmetrical alkenes, it not obvious where the electrophile or nucleophile attaches itself to the alkene. For example, 2-methylpropene may react with HCl to yield 1-chloro-2-methylpropane in addition to 2-chloro-2-methylpropane as indicated above, but it did not. We say that the reaction is regiospecific when only one of the two possible products of an addition is obtained.

eqn 2.6

Markovnikov observed many of such addition reactions and proposed a rule which can be summarized today as follows: in an electrophilic addition reaction to an unsymmetrical alkene, the electrophile adds in such a way as to generate the most stable intermediate. In other words, the hydrogen atom becomes attached to the carbon atom with fewer alkyl groups. In biblical terms, it may be stated as “to he who hath more hydrogen atoms, more hydrogen atoms shall be given.” No offence intended.

When both ends of the double bond have the same degree of substitution, a mixture of products is formed.

Let us look at the mechanism for the following reaction which can be applied to others of a similar nature.

[A1]

Step 1: The alkene (nucleophile) attacks the halide to create a carbocation and a halide ion

Step 2: The halide ion (chloride ion) attacks the highly reactive carbocation to yield the final product.

Notice that the more highly substituted tertiary carbocation is formed. The primary carbocation is less stable and it is less likely to be formed.

Give the carbocations for the examples above. Indicate which of them are more stable.

Problems

Predict the products of the following reactions.

What alkenes would you start with to prepare

i. bromocyclohexane

ii. CH3CH2CH(Br)CH2CH3

iii. 1-iodo-1-ethylcyclohexane

2.6c Free-radical Addition of hydrogen halides to Alkenes: Anti-Markovnikov Addition

In the presence of peroxides, hydrogen halides add to alkenes to form anti-Markovnikov products.

Peroxides give rise to free radicals that confer a different mechanism to the addition reaction of alkenes. In the presence of peroxides, alkenes undergo a chain-reaction mechanism which consists of three steps: initiation, propagation and termination steps.

Initiation

In the presence of heat or ultraviolet light, the peroxide (commonly dibenzoyl peroxide) dissociates to yield free radicals.

eqn 2.0

free radical

eqn 2.

Propagation

The bromine radical thus generated reacts with the alkene to yield an alkyl radical which reacts with excess HBr to give a neutral product and a bromine radical and the reaction continues until one of the radicals is not sufficient enough for the reaction to continue.

eqn 2.

addition product

Termination

There are a number of ways the reaction can be terminated but essentially in this reaction, two radicals react to form a product.

eqn 2.

Can you guess other reactions that would be part of the termination step?

For example, 2-methyl-2-pentene reacts with hydrogen bromide in the presence of peroxides to yield 2-bromo-3-methylpentane instead of the 2-bromo-2-methylpentane.

Notice that the less stable secondary free radical would have formed the Markovnikov product. The reversal of regiochemistry in the presence of peroxides is called the peroxide effect.

2.6d Carbocations

Carbocations are reactive intermediates that have a positively charged carbon atom. They are often referred to as carbonium ions. They are classified as follows:

This classification depends on the number of alkyl groups attached to the positively charged carbon atom. In a primary carbocation, the positively charged carbon atom is attached to an alkyl group. Two alkyl groups are attached to the positively charged carbon atom in secondary carbocation and three alkyl groups are attached to the positively charged carbon atom in tertiary carbocations.

We have mentioned above that 3°carbocations are more stable than 2° carbocations which in turn are more stable than 1° carbocations. Free radicals follow the same pattern of stability.

2.6e Rearrangement

Consider the addition of HCl to 3-methyl-1-butene. The major product is not 2-chloro-3-methylbutane but the major product is 2-chloro-2-methylbutane.

What has happened? There has been a rearrangement. A hydrogen atom has moved with its electron pair from the carbon adjacent to the positively charged carbon atom. This transfer is called a hydride (H-) shift.

______

(i) Formation of a secondary (2°) carbocation

eqn 2.

(ii) Hydride shift (shift of a hydrogen atom with its electron pair to an adjacent deficient carbon atom) to form a more stable tertiary (3°) carbocation.

eqn 2.

(iii) The more stable tertiary carbocation accepts the chloride ion to yield the final product.

eqn 2.

______

Rearrangements are favored because the more stable tertiary carbocation is favored than the secondary carbocation.

Not only hydride shifts are possible but alkyl shifts do occur to. In this case, an alkyl group with its electron pair shifts from an adjacent carbon atom to the cation (eqn 2. ). Alkyl shifts are called Wagner-Meerwein rearrangements.

______

eqn 2.

2.7 Addition of Halogens to Alkenes

In the presence if a variety of solvents, halogens react with alkenes by electrophilic addition to yield vicinal dihalides

The halogens are usually chlorine or bromine and the reaction takes place rapidly at room temperature. The addition of bromine to an alkene serves as a simple laboratory test for unsaturation. A solution of bromine in carbon tetrachloride (CCl4) is red. On addition of a few drops of an alkene to this solution, the reddish brown color of bromine disappears, i.e., the solution becomes colorless.

Addititon of bromine to cycloalkenes illustrate anti addition stereochemistry: the two bromine atoms add on opposite faces of the double bond. In syn addition, the two bromine atoms add on the same face of the double bond.