Alkenes: Structure & Properties

Alkane (acyclic): CnH2n+2 > saturated.

Alkene (acyclic): CnH2n > unsaturated.

eg ethylene (IUPAC: ethene), C2H4: H2C=CH2

The carbon-carbon double bond is the distinguishingfeature of alkenes.

It is formed between two sp2 carbons. One bond is σ,formed by head to head overlap of sp2 orbitals; the secondis π, formed by overlap of parallel 2p orbitals. The π bondhas two halves, one above and the other below the planedefined by the two carbons and the four other atoms towhich they are attached.

The bond angles around the sp2 carbons are � 120o.

σ-bond requires ca. 360kJ (90 kcal)/mole to break.

π-bond requires ca. 260kJ (65 kcal)/mole to break.

Owing to the shape of the π-bond and its strength, there isnot free rotation around a double bond as there is aroundmost single bonds.

C C

CH3

H

H3C

H

C C

H

CH3

H3C

H

cis-2-butene trans-2-butene

The butenes: C4H8

CH3-CH2-CH=CH2 : 1-butene,(CH3)2C=CH2 : 2-methylpropeneCH3-CH=CH-CH3 : 2-butene

Appear to be 3 isomers, but again!look

( (. .

There are two isomeric 2-butenes, for 4 butene isomers inall.

The cis/trans isomers are configurational (notconformational) stereoisomers and are sometimes calledgeometric isomers.

These isomers have different properties:cis- bp = 4oC, mp = -139oC; trans- bp = 1oC, mp = -106oC.

The potential for geometric isomerism will exist in alkenesunless at least one of the carbons joined by the doublebond carries two identical groups.

H

C

H3C

C

CH3

H3C

C

H

C

CH3

H3C

C

H

C CH3

Traversing this ring in clockwise direction from =C: CH(CH3), CH2, CH2; counterclockwise: CH2, CH2, CH(CH3).

Since we geta differentresult dependingon the direction of our traverse, and since the other doubly

bonded carbon has two different groups (H, CH3) attached, these aregeometric isomers.

Traversing this ring in eitherdirection leads to:CH2, CH(CH3), CH2. Therefore, there are no geometric isomers here even though there are two different groups on the other =C.

geometric isomers

nogeometricisomers

If one of the carbons is part of a ring to which the doublebond is exocyclic, the carbon is considered to carry twoidentical groups if, starting at the trigonal carbon,traversing the ring in the clockwise direction is the same astraversing it in a counterclockwise direction.

C C

�high

�low

high�

low�

C C

�low

�high

high�

low�Z E

C

high�

low�

Some examples in whichthe higher priority groupis on top, as shown tothe right.

C

H

H3C

C

H3CCH2O

H3CCH2S

C

H3C12

H3C14

E, Z System for Naming Geometric Isomers ---

1) At each atom joined by the double bond determinewhich of the groups attached by single bonds has thehigher priority (vide infra). If the groups of higherpriority are on the same side of the double bond: Z; if on opposite sides: E.

2) To establish priorities at each doubly bonded atom,consider the atoms directly attached by single bondsto the atom bearing the double bond, and rank themin order of decreasing (highest first) atomic number(atomic mass for isotopes).

C

high�

low�

Some examples in whichthe higher priority groupis on top, as shown tothe right.

C

CH3CH2

CH3

C

H3CCH2O

HOCH2O

C

H3CCH2

(H3C)2CH

C

(H3C)3C

H2NCH2

3) If a decision cannot be made by the above rule, workoutward to the first point of difference, then use rule 2comparing atoms of highest atomic number, then, if adecision cannot be made, next lower atomic number,etc.

4) If a group has multiple bonds, the multiple bonds areconverted to single bonds for purposes of E/Z naming;this is done in the following way. A multiple bond fromatom A to atom B is considered to be equivalent to thesame number of single bonds from atom A to atom B. This results in “dummy” bonds and atoms – shown inred below.

Once the conversion is complete, priority assignmentis done in the usual way.

C

C C

H H

H

C

C C

H H

HC

Cbecomes

C

C O

H

C

C N

C

C O

H

C

O

C

C N

N

N

C

C

becomes

becomes

C C

H

CH3

H3C

H

C C

CH3

H

H3C

H

H+

24% Z 76% E

(E)-2-butene is more stable than (Z)-2-butene.

Evidence:

1)

2) (Z)-2-butene + H2 > butane + 28.6 kcal/mole

(E)-2-butene + H2 > butane + 27.6 kcal/mole

Why?

Steric repulsion of methyl groups in Z isomer.

C C

R

R

R

R

C C

R

H

R

R

>small

most stable,least energy

> C C

R

H

H

Rsmall>

C C

H

H

R

R

~ C C

R

H

R

H

C C

H

H

R

H

> > C C

H

H

H

H

least stable,most energy

∆H for hydrogenation of alkenes also indicates thefollowing order of stability of alkenes (compared to theproduct alkanes).

Why?

Some controversy; two possibilities ---

1) Orbital interaction between π*-orbital and σ-orbital ofthe first carbon in R (not the σ-orbital joining R to thetrigonal carbon) which lowers the energy of theoccupied σ-orbital and raises the energy of theunoccupied π*-orbital.

2) The more highly substituted alkenes have a greaternumber of sp3 - sp2 C-C bonds: these σ-bonds arestronger than those between two sp3 carbons.