Chapter 7Structure and Synthesis

of Alkenes

Organic Chemistry

Chapter 7 2

Functional Group

• Pi bond is the functional group.• More reactive than sigma bond.• Bond dissociation energies:

C=C BDE 146 kcal/molC-C BDE -83 kcal/molPi bond 63 kcal/mol

=>

Chapter 7 3

Orbital Description

• Sigma bonds around C are sp2 hybridized.• Angles are approximately 120 degrees.

• No nonbonding electrons.• Molecule is planar around the double bond.• Pi bond is formed by the sideways overlap of

parallel p orbitals perpendicular to the plane of the molecule. =>

Chapter 7 4

Bond Lengths and Angles

• Hybrid orbitals have more s character.

• Pi overlap brings carbon atoms closer.

• Bond angle with pi orbitals increases.Angle C=C-H is 121.7° Angle H-C-H is 116. 6° =>

Chapter 7 5

Pi Bond• Sideways overlap of parallel p orbitals.

• No rotation is possible without breaking the pi bond (63 kcal/mole).

• Cis isomer cannot become trans without a chemical reaction occurring.

=>

Chapter 7 6

Elements of Unsaturation

• A saturated hydrocarbon: CnH2n+2

• Each pi bond (and each ring) decreases the number of H’s by two.

• Each of these is an element of unsaturation.• To calculate: find number of H’s if it were

saturated, subtract the actual number of H’s, then divide by 2. =>

Chapter 7 7

Propose a Structure:

• First calculate the number of elements of unsaturation.

• Remember:A double bond is one element of unsaturation.A ring is one element of unsaturation.A triple bond is two elements of unsaturation. =>

for C5H8

Chapter 7 8

Heteroatoms

• Halogens take the place of hydrogens, so add their number to the number of H’s.

• Oxygen doesn’t change the C:H ratio, so ignore oxygen in the formula.

• Nitrogen is trivalent, so it acts like half a carbon.

C

H

H

C

H

H

N C

H

HH

=>

Chapter 7 9

Structure for C6H7N?

• Since nitrogen counts as half a carbon, the number of H’s if saturated is 2(6.5) + 2 = 15.

• Number of missing H’s is 15 – 7 = 8.

• Elements of unsaturation is 8 ÷ 2 = 4. =>

Chapter 7 10

IUPAC Nomenclature

• Parent is longest chain containing the double bond.

• -ane changes to -ene. (or -diene, -triene)• Number the chain so that the double

bond has the lowest possible number.• In a ring, the double bond is assumed to

be between carbon 1 and carbon 2. =>

Chapter 7 11

Name These Alkenes

CH2 CH CH2 CH3

CH3 C

CH3

CH CH3

CH3

CHCH2CH3H3C

1-butene

2-methyl-2-butene

3-methylcyclopentene

2-sec-butyl-1,3-cyclohexadiene

3-n-propyl-1-heptene =>

Chapter 7 12

Alkene Substituents

= CH2

methylene(methylidene)

- CH = CH2

vinyl(ethenyl)

- CH2 - CH = CH2

allyl(2-propenyl)

Name: =>

Chapter 7 13

Common Names

• Usually used for small molecules.

• Examples:

CH2 CH2

ethylene

CH2 CH CH3

propylene

CH2 C CH3

CH3

isobutylene=>

Chapter 7 14

Cis-trans Isomerism

• Similar groups on same side of double bond, alkene is cis.

• Similar groups on opposite sides of double bond, alkene is trans.

• Cycloalkenes are assumed to be cis.• Trans cycloalkenes are not stable

unless the ring has at least 8 carbons. =>

Chapter 7 15

Name these:

C CCH3

H

H

CH3CH2

C CBr

H

Br

H

trans-2-pentene cis-1,2-dibromoethene

=>

Chapter 7 16

E-Z Nomenclature

• Use the Cahn-Ingold-Prelog rules to assign priorities to groups attached to each carbon in the double bond.

• If high priority groups are on the same side, the name is Z (for zusammen).

• If high priority groups are on opposite sides, the name is E (for entgegen). =>

Chapter 7 17

Example, E-Z

C C

H3C

H

Cl

CH2C C

H

H

CH CH3

Cl1

2

1

2

2Z

2

1

1

2

5E

(2Z, 5E)-3,7-dichloro-2,5-octadiene =>

Chapter 7 18

Commercial Uses: Ethylene

=>

Chapter 7 19

Commercial Uses: Propylene

=>

Chapter 7 20

Other Polymers

=>

Chapter 7 21

Stability of Alkenes• Measured by heat of hydrogenation:

Alkene + H2 → Alkane + energy

• More heat released, higher energy alkene.

=>

30.3 kcal

27.6 kcal

Chapter 7 22

Substituent Effects

• More substituted alkenes are more stable.H2C=CH2 < R-CH=CH2 < R-CH=CH-R < R-CH=CR2 < R2C=CR2

unsub. < monosub. < disub. < trisub. < tetra sub.

• Alkyl group stabilizes the double bond.• Alkene less sterically hindered.

=>

Chapter 7 23

Disubstituted Isomers• Stability: cis < geminal < trans isomer

• Less stable isomer is higher in energy, has a more exothermic heat of hydrogenation.

27.6 kcalTrans-2-butene

28.0 kcal (CH3)2C=CH2Isobutylene

28.6 kcalCis-2-butene CH3C C

CH3

H H

HC C

CH3

CH3 H=>

Chapter 7 24

Cycloalkene Stability

• Cis isomer more stable than trans.• Small rings have additional ring strain.• Must have at least 8 carbons to form a

stable trans double bond. • For cyclodecene (and larger) trans

double bond is almost as stable as the cis. =>

Chapter 7 25

Bredt’s Rule• A bridged bicyclic compound cannot

have a double bond at a bridgehead position unless one of the rings contains at least eight carbon atoms.

• Examples:

Unstable.Violates Bredt’s rule

Stable. Double bondin 8-membered ring. =>

Chapter 7 26

Physical Properties

• Low boiling points, increasing with mass.• Branched alkenes have lower boiling points.• Less dense than water.• Slightly polar

Pi bond is polarizable, so instantaneous dipole-dipole interactions occur.

Alkyl groups are electron-donating toward the pi bond, so may have a small dipole moment. =>

Chapter 7 27

Polarity Examples

µ = 0.33 D µ = 0

=>

cis-2-butene, bp 4°C

C CH

H3C

H

CH3

trans-2-butene, bp 1°C

C CH

H

H3C

CH3

Chapter 7 28

Alkene SynthesisOverview

• E2 dehydrohalogenation (-HX)

• E1 dehydrohalogenation (-HX)

• Dehalogenation of vicinal dibromides (-X2)

• Dehydration of alcohols (-H2O) =>

Chapter 7 29

Removing HX via E2

• Strong base abstracts H+ as X- leaves from the adjacent carbon.

• Tertiary and hindered secondary alkyl halides give good yields.

• Use a bulky base if the alkyl halide usually forms substitution products. =>

Chapter 7 30

Some Bulky Bases

C

CH3

H3C

CH3

O_

tert-butoxide

(CH3CH2)3N :triethylamine

=>

N

H

CH(CH3)2

CH(CH3)2

diisopropylamine

N CH3H3C

2,6-dimethylpyridine

Chapter 7 31

Hofmann Product

• Bulky bases abstract the least hindered H+

• Least substituted alkene is major product.

CH3 C

H

H

C

CH3

Br

CH2

H

CH3CH2O

CH3CH2OH

_

C CCH3

CH3H

H3CC C

H

HH3C

CH3CH2

71% 29%

72%28%

C CH

HH3C

CH3CH2

C CCH3

CH3H

H3C_

CH3CH2OHCH3 C

H

H

C

CH3

Br

CH2

H

(CH3)3CO =>

Chapter 7 32

E2: Diastereomers

Stereospecific reaction: (S, R) produces only trans product, (R, R) produces only cis.

Ph

Br H

H CH3

Ph

≡

H

Ph CH3Br

PhHH

Ph

CH3

Ph

HBr

CH3Ph

PhH =>

Chapter 7 33

E2: Cyclohexanes

Leaving groups must be trans diaxial. =>

Chapter 7 34

E2: Vicinal Dibromides• Remove Br2 from adjacent carbons.

• Bromines must be anti-coplanar (E2).

• Use NaI in acetone, or Zn in acetic acid.

I-

Br

CH3H

Br

CH3H

C CCH3

H

H

H3C =>

Chapter 7 35

Removing HX via E1

• Secondary or tertiary halides

• Formation of carbocation intermediate

• Weak nucleophile

• Usually have substitution products too =>

Chapter 7 36

Dehydration of Alcohols

• Reversible reaction

• Use concentrated sulfuric or phosphoric acid, remove low-boiling alkene as it forms.

• Protonation of OH converts it to a good leaving group, HOH

• Carbocation intermediate, like E1

• Protic solvent removes adjacent H+ =>

Chapter 7 37

Dehydration Mechanism

C

H

C

O H

S

O

O

O HOH C

H

C

O H

H

HSO4 _

C

H

C

O H

H

C

H

CH2O:

C C H3O+ =>

Chapter 7 38

Industrial Methods

• Catalytic cracking of petroleumLong-chain alkane is heated with a catalyst to

produce an alkene and shorter alkane.Complex mixtures are produced.

• Dehydrogenation of alkanesHydrogen (H2) is removed with heat, catalyst.

Reaction is endothermic, but entropy-favored.

• Neither method is suitable for lab synthesis =>

Chapter 7 39

End of Chapter 7