Alkenes and Alkynes

Unsaturated Hydrocarbons
Unsaturated
hydrocarbons
 Have fewer hydrogen atoms attached to the carbon chain than alkanes.
 Are alkenes with double bonds.
 Are alkynes with triple bonds.

1

Structure of Alkenes
 Alkenes (and alkynes) are unsaturated hydrocarbons  Alkenes have one or more double bonds
 The two bonds in a double bond are different:
- one bond is a sigma () bond; these are cylindrical in shape and are very strong
- the other is a pi (π) bond; these involve sideways overlap of p-orbitals and are weaker than  bonds
 Alkenes are flat and have a trigonal planar shape around each of the two C’s in a double bond

Structure of Alkynes
 Alkynes have one or more triple bonds
 A triple bond consists of one  bond and two π bonds - the two π bonds are orthogonal (perpendicular)
 Alkynes are linear around each of the two C’s in the triple bond
 Because alkenes and alkynes have π bonds, which are much weaker than  bonds, they are far more chemically reactive than alkanes

Bond Angles in Alkenes and
Alkynes
According to VSEPR theory:
 Three groups in a double bond are bonded at 120° angles.
 Alkenes are flat: atoms in a C=C lie in the same plane.
 The groups attached to a triple bond are at 180° angles.

6

Naming Alkenes
The names of alkenes
 Use the corresponding alkane name.
 Change the ending to ene.

Alkene

IUPAC

Common

H2C=CH2

ethene

ethylene

H2C=CH─CH3

propene

propylene

cyclohexene

7

Ethene (ethylene)
Ethene or ethylene
 Is an alkene C2H4.
 Has two carbon atoms connected by a double bond.  Has two H atoms bonded to each C atom.
 Is flat with all the C and H atoms in the same plane.
 Is used to accelerate the ripening of fruits.
8

Naming Alkynes
The names of alkynes
 Use the corresponding alkane name.
 Change the ending to yne.
Alkyne

IUPAC

Common

HC≡CH

ethyne

acetylene

HC≡C─CH3

propyne

9

Guide to Naming Alkenes and
Alkynes

10

Naming Alkenes
Write the IUPAC name for CH2=CH─CH2─CH3 and
CH3─CH=CH─CH3
STEP 1 Name the longest carbon chain butene STEP 2 Number the chain from the double bond
CH2=CH─CH2─CH3
1

2

3

1-butene

4

CH3─CH=CH─CH3

2-butene
11

Comparing Names of Alkanes,
Alkenes, and Alkynes

12

Naming Alkenes with Substituents
CH3
│
Write the IUPAC name for CH3─CH─CH=CH─CH3

STEP 1 Name the longest carbon chain

pentene

STEP 2 Number the chain from the double bond
CH3
│
CH3─CH─CH=CH─CH3 2-pentene
5

4

3

2

1

STEP 3 Give the location of each substituent
4-methyl- 2-pentene

13

Naming Alkynes with Substituents
CH3
│
Write the IUPAC name for HC≡C─CH─CH3
1

2

3

4

STEP 1 Name the longest carbon chain

butyne

STEP 2 Number the chain from the double bond
1-butyne
STEP 3 Give the location of each substituent

3-methyl-1-butyne
14

Solution
Write the IUPAC name for each of the following:
1. CH2=CH─CH2─CH3

1-butene

2. CH3─CH=CH─CH3
CH3
|
3. CH3─CH=C─CH3

2-butene

4. CH3─CC─CH3

2-butyne

2-methyl-2-butene

15

Solution
Write the structural formula for each of the following:
A. CH3─CH2─C≡C─CH3

2-pentyne

CH3


B. CH3─CH2─C=CH─CH3

3-methyl-2-pentene

16

Cis and Trans Isomers
In an alkene, cis and trans isomers are possible because the double bond
 Is rigid.

 Cannot rotate.
 Has groups attached to the carbons of the double bond that are fixed relative to each other.
CH3
CH = CH

cis

CH3

CH3
CH = CH

trans

CH3
17

Cis-Trans Isomers
Cis-trans isomers

 Can be modeled by making a “double bond” with your fingers with both thumbs on the same side or opposite from each other. 18

Cis-Trans Isomers
Cis-trans isomers occur when different groups are attached to the double bond.
 In a cis isomer, groups on the same side of C=C

 In the trans isomer, the groups on opposite sides.

19

Cis-Trans Isomerism
 Cis-trans isomers do not occur if a carbon atom in the double bond is attached to identical groups.
Identical

HH

Identical

Br
H
Br

C C
HH

Br

H
C C

CH3

2-bromopropene
(not cis or trans)

H

Br
1,1-dibromoethene
(not cis or trans)
20

Naming Cis-Trans Isomers
 The prefixes cis or trans are placed in front of the alkene name when there are cis-trans isomers. cis trans

Br

Br
C C
H

H

Br
C C

H

cis-1,2-dibromoethene

H

Br

trans-1,2-dibromoethene
21

Pheromones
A pheromone

 Is a chemical messenger emitted by insects in tiny quantities.  Called bombykol emitted by the silkworm moth to attract other moths has one cis and one trans double bond.

22

Solution
Br

Br
C C

A.

cis-1,2-dibromoethene
H

H

H

CH3

trans-1,2-dimethylbutene

C C

B.
H

CH3
Cl

CH3
C.

C C
H

1,1-dichloro-2-methylpropene
Cl

Identical atoms on one C; no cis or trans

23

Addition Reactions
In addition reactions,
 Reactants add to the carbon atoms in double or triple bonds.  A double or triple bond is easily broken, which makes them very reactive. 24

Hydrogenation
In hydrogenation,
 Hydrogen atoms add to the carbon atoms of a double bond or triple bond.

 A catalyst such as Pt or Ni is used to speed up the reaction. H H
Pt
H2C CH2
H2C CH2 + H2

H H
HC CH + 2H2

Ni

HC CH
H H
25

Hydrogenation of Oils
Adding H2 to double bonds in vegetable oils produces
 Compounds with higher melting points.

 Solids at room temperature such as margarine, soft margarine, and shortening.

Copyright © 2007 by Pearson Education, Inc.
Publishing as Benjamin Cummings

26

Solution
Write an equation for the hydrogenation of 1-butene using a platinum catalyst.
Pt
CH2=CH─CH2─CH3 + H2
CH3─CH2─CH2─CH3

27

Trans Fats
In vegetable oils, the unsaturated fats usually contain cis double bonds.
 During hydrogenation, some cis double bonds are converted to trans double bonds (more stable) causing a change in the fatty acid structure

 If a label states “partially” or “fully hydrogenated”, the fats contain trans fatty acids.

28

Solution

Pt

CH3─CH=CH─CH3 + H2

+ H2

CH3─CH2─CH2─CH3

Pt

29

Halogenation
 In halogenation, halogen atoms add to the carbon atoms of a double bond or triple bond.

Br Br
H2C CH2 + Br2

H2C CH2
Cl Cl

HC C CH3 + 2Cl2

HC C CH3
Cl Cl

30

Solution
Write the product of the following addition reactions:
1. CH3─CH=CH─CH3 + Cl2

2.

Pt

Cl Cl l l
CH3─CH─CH─CH3

Br

+ Br2
Br

31

Hydrohalogenation
 In hydrohalogenation, the atoms of a hydrogen halide add to the carbon atoms of a double bond or triple bond.
H
CH3 CH CH CH3 + HCl

CH3

Cl

CH CH CH3

H

+ HBr
Br

32

Markovnikov’s Rule
 When an unsymmetrical alkene undergoes hydrohalogenation, the H in HX adds to the carbon in the double bond that has the greater number of
H atoms .
H

Cl

CH3 CH CH2 Does not form
CH3

CH CH2 + HCl
Cl
CH3

C with the most H
H

CH CH2

Product that forms

33

Hydration
In the addition reaction called hydration
 An acid H+ catalyst is required.
 Water (HOH) adds to a double bond.
 An H atom bonds to one C in the double bond.
 An OH bonds to the other C.
H OH
H+
│ │
CH3─CH=CH─CH3 + H─OH
CH3─CH─CH─CH3

34

Hydration
When hydration occurs with a double bond that has an unequal number of H atoms,
 The H atom bonds to the C in the double bond with the most H.
 The OH bonds to the C in the double bond with the fewest H atoms.
OH H
H+
│ │
CH3─CH=CH2 + H─OH
CH3─CH─CH2
(1H)

(2H)

35

Solution
1.

2.

H OH
│ │
CH3─CH2─CH─CH─CH2─CH3
CH3
│
CH3─C─CH─CH2─CH3
│ │
OH H
OH

3.
H
36

Solution
Cl
A.

CH3 CH CH2 + Cl2

CH3

CH CH2
H

+

B. CH3 CH CH CH3 + HOH

C.

+ H2

H

Cl

OH

CH3 CH CH CH3

Pt

H
H

37

Polymers from Addition Reactions

38

More Monomers and Polymers

39

Solution
F F
│

│

F─C=C─F tetrafluoroethene
F

F F

F F F F

│

│

│

│

│

│

│

F
│

─C─C─C─C─C─C─C─C─
│

│

│

F

F F

│

│

│

│

F F F F

portion of Teflon

│

F

40

CH2=CHC6H5
CH2 =CHCl
CH2=CH2
CF2=CF2
CH2=CCl2
CH2=CHCO2CH2CH3
CH3CH=CH2
CH2=CHCN
CH3CH2Cl
CH2=CCO2CH3

styrofoam
PVC, construction tubing ethelyne Teflon, non stick coatings
1,1 dichloroethylene poly ethyl acrylate, latex paint

propylene orlon, acrylics and acrylates chloroethane polymethyl methacrylate, glass substitutes


41

Recycling Plastics
Recycling is simplified by using codes found on plastic items.
1
2
3
4
5
6

PETE Polyethyleneterephtalate
HDPE High-density polyethylene
PV Polyvinyl chloride
LDPE Low-density polyethylene
PP Polypropylene
PS Polystyrene

Copyright © 2007 by Pearson Education, Inc
. Publishing as Benjamin Cummings

42

Learning Check
What types of plastic are indicated by the following codes? 3
PV
A.
B.

C.

5
PP
6
PS
43

Solution
What types of plastic are indicated by the following codes? A.

B.

C.
.

3
PV
5
PP

6
PS

Polyvinyl chloride

Polypropylene

Polystyrene
44

Chapter 12 Unsaturated
Hydrocarbons
12.5
Aromatic Compounds

Copyright © 2007 by Pearson Education, Inc.
Publishing as Benjamin Cummings

45

Benzene Structure
Benzene
 Has 6 electrons shared equally among the 6 C atoms.
 Is also represented as a hexagon with a circle drawn inside. 46

Aromatic Compounds in Nature and Health
O

Vanillin

Aspirin

CH

O
COH O
C O CH3

OCH3
O

OH

NH C CH3

Ibuprofen
CH3
H3C CH CH2

Acetaminophen
CH3 O
CH COH
OH
47

Naming Aromatic Compounds
Aromatic compounds are named

 With benzene as the parent chain.
 With one side group named in front of benzene.
CH3

methylbenzene

Cl

chlorobenzene

48

Some Common Names
Some substituted benzene rings

 Have common names used for many years.
 With a single substituent use a common name or are named as a benzene derivative.
CH3

toluene
(methylbenzene)

NH2

aniline
(benzenamine)

OH

phenol
(hydroxybenzene)
49

Aromatic Compounds with Two
Groups
Two naming systems are used when two groups are attached to a benzene ring.
 Number the ring to give the lowest numbers to the side groups.
 Use prefixes to show the arrangement: ortho(o-) for 1,2-

meta(m-) for 1,3para(p-) for 1,4-

50

Aromatic Compounds with Two
Groups
Cl

CH3

OH
Cl

Cl
Cl
3-chlorotoluene m-chlorotoluene 1,4-dichlorobenzene p-dichlorobenzene 2-chlorophenol o-chlorophenol 51

Learning Check
Select the correct name for each compound:
Cl
1) chlorocyclohexane
2) chlorobenzene
3) 1-chlorobenzene

1) 1,2-dimethylbenzene
2) m-xylene
3) 1,3-dimethylbenzene

CH 3

CH 3

52

Solution
Cl

2) chlorobenzene

CH 3

2) m-xylene
3) 1,3-dimethylbenzene
CH 3

53

Learning Check
Write the structural formulas for each of the following:
A. 1,3-dichlorobenzene

B. o-chlorotoluene

54

Solution
Cl

A. 1,3-dichlorobenzene
Cl

B. o-chlorotoluene

CH 3
Cl

55

Learning Check
Identify the organic family for each:
A. CH3─CH2─CH=CH2
B.
C. CH3─C≡CH
D.

56

Solution
Identify the organic family for each:
A. CH3─CH2─CH=CH2

alkene

B.

cycloalkane (alkane)

C. CH3─C≡CH

alkyne

D.

aromatic

57

Chapter 12 Unsaturated
Hydrocarbons
12.6
Properties of Aromatic
Compounds

Copyright © 2007 by Pearson Education, Inc.
Publishing as Benjamin Cummings

58

Properties of Aromatic
Compounds
Aromatic compounds
 Have a stable aromatic bonding system.
 Are resistant to many reactions.
 Undergo substitution reactions, which retain the stability of the aromatic bonding system.

59

Substitution Reactions
In a substitution reaction, a hydrogen atom on a benzene ring is replaced by an atom or group of atoms. Type of substitution

H on benzene replaced by

 Halogenation

chlorine or bromine atom

 Nitration

nitro group (—NO2)

 Sulfonation

—SO3H group

60

Halogenation
In a halogenation
 An H atom of benzene is replaced by a chlorine or bromine atom.
 A catalyst such as FeCl3 is needed in chlorination.
 A catalyst such as FeBr3 is needed in bromination.

H
+
Benzene

Cl
Cl2

FeCl3

+

HCl

Chlorobenzene
61

Nitration
In the nitration of benzene
 An H atom of benzene is replaced by a nitro (-NO2) group from HNO3.
 An acid catalyst such as H2SO4 is needed.

H
+ HNO3
Benzene

NO2
H2SO4

+

HOH

Nitrobenzene
62

Sulfonation
In a sulfonation
 An H atom on benzene is replaced by a —SO3H group from SO3.
 An acid catalyst such as H2SO4 is needed.
H
+ SO3
Benzene

SO3H
H2SO4
Benzenesulfonic acid
63

Learning Check
Write the equation for the bromination of benzene including catalyst.

64

Solution
Write the equation for the bromination of benzene including catalyst.
H
+ Br2

Br
FeBr3

+ HBr

65