Alkene: Structure and Reactivity

Chapter 6

Alkenes

• An alkene (also called an olefin) is a hydrocarbon with a carbon-carbon double bond.

• Alkenes are present in most organic and biological molecules.

Examples of Alkenes

H3C

H3C

H H

H

OH

O

-Carotene

Testosterone

Calculating Degrees of Unsaturation

• Because a double bond has fewer hydrogens than an alkane it is refereed to as unsaturated. One example is that of ethylene.

H

HH

H H

HH

H

HH

Ethylene

C2H4

Ethylene

C2H6

Saturated Alkane Chains

• To calculate the amount of hydrogens in a alkane chain all you have to do is take the number of carbons (make that n) and multiply it by 2 then add 2 more for the additional hydrogens at the terminals.

CnH2n+2

Degrees of Unsaturation

• There are several ways that a molecule can be unsaturated. The first is by an double or triple bond. The second is by the formation of a ring. Or a combination of both.

All of these are C6H10

How to Calculate Unsaturation• Take the original number of hydrogens found.

• Use the calculation of saturated alkanes.

• Subtract the original hydrogens from saturated alkenes.

• For every two hydrogens (divide by 2) that are missing 1 degree of unsaturated is found.

C6H14

C6H10

C6H2(6)+2 = 14

14 – 10 = 4 hydrogens

4 hydrogens / 2 = 2 degrees of unsaturation (answer)

Calculating Unsaturation in other Molecules

• What if the molecules contains halogens, oxygen, and nitrogen? There are rules for these possibilities.

• For halogens (F, Cl, Br, I) since they bond only once similar to hydrogen, you can consider them a hydrogen.

• Since oxygen forms 2 bonds it doesn’t effect the formula. IGNORE IT.

• For nitrogen, subtract the number of nitrogens from hydrogens to get the proper saturated number.

Summarize These Rules

• Add the number of halogens to the number of hydrogens.

• Ignore the number of oxygens.

• Subtract the number of nitrogens from hydrogens.

Naming Alkenes

• 1) Name the Parent hydrocarbon (THAT INCLUDES THE ALKENE).

• 2) Number the carbon atoms in the chain with either the alkene or branching with the lowest number.

• 3)Write the full name. Name it accordingly with substituents properly named and numbered. Also remember it there are more than 1 alkene it is diene (2), triene (3), and so forth.

Examples

H3CH2CH2C

H

H

CH3CH2CH3

CH3123

4 5

1-cyclohexene4-methyl-1,3-cyclopentadiene

2-ethyl-1-pentene1,5-dimethyl-1-cyclopentene

Older Names Still in Use

• Methylene (group)

• Vinyl (group)

• Allyl (group)

• Ethylene instead of Ethene

H2C

H2C CH

H2C CH CH2

Cis –Trans Isomers in Alkenes

• Because double bonds do not rotate freely they are similar to ring structure because they can be stereoisomers. They can have both a cis and trans configuration. This occurs when the double bond is disubstituted (not when 3 hydrogens are bound to the double bond).

H

CH3

H

H3C

H

CH3

H

H

H

CH3

H3C

H

Cis TransNot Stereoisomers

E, Z Designation

• Cis and Trans only describes disubstituted alkanes. It does not describe when other atoms (halogen, oxygen, nitrogen) attaches to the double bond. The E,Z system is used to describe when this occurs.

• E = Trans

• Z = Cis (for me I remember CIZ)

E, Z Rules

• Rule 1. Consider Each Carbon Separately. Look at the atoms directly attached to each and rank them accordingly BY ATOMIC NUMBER.

• Br > Cl > S > P > O > N > C > 2H > 1H

• Rule 2. If the first atom don’t give you an answer then move down the chain until you find a difference.

• Rule 3. Multiple bonds atoms are equivalent to the same number of single bonds.

C OH C O

O

H

CC = H,O,O O = C,C

Stability of Alkenes

• Because of steric strain the energy of cis isomers are higher than that of the trans isomers. This does not mean that you can not find the cis isomer but often an equilibrium can form with the trans more abundant.

Alkene Stability (Lower Energy)

The Heats of Hydrogenation is a determination of the relative stabilities of cis and trans isomers without looking at equilibrium positions. Alkyl groups are better than H.

R

R

R

R

R

R

R

H

R

R

H

H

H

R

H

R

H

R

H

H

> > >~tetrasubstituted trisubstituted disubstituted monosubstituted

Stability of Alkenes

• Stability order of alkenes is due to a combination of two factors. The most important is the stabilizing interactions between the C=C p bond and adjacent substituents. This is called hyperconjugation.

Comparing Stabilities of Alkenes

• Evaluate heat given off when C=C is converted to C-C.

• More stable alkene gives off less heat.– Trans butene generates 5 kJ less heat than cis-

butene.

H

CH3

H

H3C H2

Pd H

H3C

H

CH3H H

H

CH3

H3C

HH2

Pd H

H3C

H

CH3H H

Cis

Trans

Electrophilic Addition of HX to Alkenes

• General reaction mechanism: electrophilic addition.• Attack of electrophile (such as HBr) on bond of

alkene.• Produces carbocation and bromide ion.• Carbocation is an electrophile, reacting with

nucleophilic bromide ion.

R

R

H

H

H Br

Br-Br

HH

H

HH

NucleophileR

R

H

HH

• Two step process.• First transition state is high energy point.

Electrophilic Addition Energy Path

Example of Electrophilic Addition

• Addition of hydrogen bromide to 2-Methyl-propene.

• H-Br transfers proton to C=C• Forms carbocation intermediate.

– More stable cation forms.• Bromide adds to carbocation.

Energy Diagram for Electrophilic Addition

• Rate determining (slowest) step has highest energy transition state.– Independent of

direction.– In this case it is the first

step in forward direction.

– “rate” is not the same as “rate constant”.

Carbocation Structure and Stability

• Therefore stability of carbocations: 3º > 2º > 1º > +CH3

C

R

R

R C

H

R

R C

H

R

H C

H

H

H> > >Tertiary (3o) Secondary (2o) Primary (1o) Methyl

Orientation of Electrophilic Addition: Markovnikov’s Rule

• In an unsymmetrical alkene, HX reagents can add in two different ways, but one way may be preferred over the other.

• If one orientation predominates, the reaction is regiospecific.

• Markovnikov observed in the 19th century that in the addition of HX to alkene, the H attaches to the carbon with the most H’s and X attaches to the other end (to the one with the most alkyl substituents).– This is Markovnikov’s rule.

Example of Markovnikov’s Rule

• Regiospecific – one product forms where two are possible

• If both ends have similar substitution, then not regiospecific.

CH2H Br

CH2

H

Br-

CH2

H

Br

CH2

Br

HCH2

H

Br

Not Favored Favored

Energy of Carbocations and Markovnikov’s Rule

• More stable carbocation forms faster.• Tertiary cations and associated transition states

are more stable than primary cations.

Mechanistic Source of Regiospecificity in Addition Reactions

• If addition involves a carbocation intermediate.– and there are two

possible ways to add.– the route producing the

more alkyl substituted cationic center is lower in energy.

– alkyl groups stabilize carbocation.

The Hammond Postulate

• If carbocation intermediate is more stable than another, why is the reaction through the more stable one faster?– The relative stability of the intermediate is related to

an equilibrium constant (Gº).– The relative stability of the transition state .(which

describes the size of the rate constant) is the activation energy (G‡).

– The transition state is transient and cannot be examined.

Transition State Structures

• A transition state is the highest energy species in a reaction step.

• By definition, its structure is not stable enough to exist for one vibration.

• But the structure controls the rate of reaction.• So we need to be able to guess about its properties

in an informed way.• We classify them in general ways and look for trends

in reactivity – the conclusions are in the Hammond Postulate.

Statement of the Hammond Postulate

• A transition state should be similar to an intermediate that is close in energy.

• Sequential states on a reaction path that are close in energy are likely to be close in structure. - G. S. Hammond

carbocation

G

Reaction

In a reaction involving a carbocation, the transition states look like the intermediate.

Competing Reactions and the Hammond Postulate

• Normal Expectation: Faster reaction gives more stable intermediate.

• Intermediate resembles transition state.

“Non-Hammond” Behavior

• More stable intermediate from slower reaction• Conclude: transition state and intermediate must not

be similar in this case – not common.

Take Home Message

• Alkene Structures

• Calculating Degrees of Unsaturation

• Naming Alkenes

• Alkene Stability

• Markovnikov’s Rule

• Transition States