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Chapter Outline

Chapter 20

Introduction to General, Organic, and Biochemistry, 10e John Wiley & Sons, Inc

Morris Hein, Scott Pattison, and Susan Arena

Unsaturated Hydrocarbons

The aromas of many fragrant plants are mixtures of unsaturated organic molecules.

Chapter Outline

2

Chapter Outline

20.1 Bonding in Unsaturated Hydrocarbons

20.2 Nomenclature of Alkenes

20.3 Geometric Isomerism in Alkenes

20.4 Cycloalkenes

20.5 Preparation and Physical Properties of Alkenes

20.6 Chemical Properties of Alkenes

20.7 Alkynes: Nomenclature and Preparation

Chapter Outline

3

20.8 Physical and Chemical Properties of Alkynes

20.9 Aromatic Hydrocarbons: Structure

20.10 Naming Aromatic Hydrocarbons

20.11 Polycyclic Aromatic Hydrocarbons

20.12 Sources and Physical Properties of Aromatic Hydrocarbons

20.13 Chemical Properties of Aromatic Hydrocarbons

Chapter Summary

Chapter Outline

Chapter Outline

Unsaturated hydrocarbons enhance our standard of living. They are used to make:

1. Polyethylene plastic bags and bottles. 2. Polystyrene Styrofoam cups. 3. Plastic wraps. 4. Cosmetics, medicines, flavorings, perfumes. 5. Detergents, insecticides, and dyes.

Bonding in Unsaturated Hydrocarbons

4

Chapter Outline

Types of Unsaturated Hydrocarbons

• Alkenes contain carbon-carbon double bonds.

• Alkynes contain carbon-carbon triple bonds.

• Aromatic compounds

contain benzene rings.

C C

C C


5

Chapter Outline

Figure 20.1 Schematic hybridization of 2s2 2p 2p orbitals of carbon to form three sp2 electron orbitals and one p electron orbital.


6

The carbon atoms connected to double bonds in alkenes and aromatic compounds are sp2 hydridized.

Chapter Outline

Figure 20.2 (a) A single sp2 electron orbital and (b) a side view of three sp2 orbitals all lying in the same plane with a p orbital perpendicular to the three sp2 orbitals.

sp2


7

Chapter Outline

Figure 20.3: Pi (π) and sigma (σ) bonding in ethene.


8

Chapter Outline


The carbon atoms connected to triple bonds in alkynes are sp hybridized as shown in Figure 20.3 on the following slide . . .

9

Chapter Outline


Figure 20.3: Pi (π) and sigma (σ) bonding in acetylene.

10

Chapter Outline

IUPAC Rules for Naming Alkenes

1. Identify the longest chain containing the C=C bond.

2. Name the parent alkane and change the –ane ending to –ene.

CH3CH2CH3 is propane

CH3CH=CH2 is propene

Nomenclature in Alkenes

11

Chapter Outline

3. Number the carbon chain to give the double bonded carbons the lowest numbers.

4. Number and name branched alkyl groups as shown below.

CH3CH2CH=CH2 1-butene1234

CH3CH=CHCH3 2-butene1234

CH3 CH CH2 CH CH2

CH3

4-methyl-1-pentene

12345CH2 CH2 CH CH CH2

12345CH3

CH2 CH2 CH3

3-propyl-1-hexene

6

Nomenclature in Alkenes

12

Chapter Outline

Your Turn!

What is the structural formula of 4-methyl-2-pentene?

Chapter Outline

The name indicates: • Five carbons in the

longest chain containing the double bond.

• The double bond is between carbons #2 and #3

• A methyl group is on carbon #4.

CH3

CHCH

CHCH3

CH3

12

3

4

5

Your Turn!

What is the structural formula of 4-methyl-2-pentene?

Chapter Outline

CH3

CH2

CCH2

CH2

CH3

CH2

15

Your Turn!

What is the name of this compound?

Chapter Outline

16

Your Turn!

• Five carbon atoms in the longest chain containing the double bond.

• The double bond is between carbons #1 and #2. • The ethyl group is attached to the #2 carbon atom. • The name of this compound is 2-ethyl-1-pentene.

CH2

CH3

CH2

CCH2

CH2

CH3

1

23

4

5

Chapter Outline

Cl

C

H

C

H

Cl H

C

Cl

C

H

Cl

cis-1,2-dichloroethene(bp = 60.1 C)

trans-1,2-dichloroethene(bp = 48.4 C)

17

Geometric Isomerism in Alkenes

Alkenes that have the same molecular formula and the same connectivity between atoms but different spatial orientation of the atoms are called geometric isomers or cis-trans isomers.

Chapter Outline

Alkenes with the a/b pattern shown here will exhibit cis-trans isomerism.

a

C

b

C

b

a a

C

b

C

a

b

cis isomer trans isomer18


Chapter Outline

If a C=C carbon has two identical groups as shown here, then cis-trans isomerism will not occur in the alkene.

H

C

H

C

H

H CH3

C

CH3

C

H

CH3

two groupsare the same

19


Chapter Outline

Draw the chemical structure of cis-5-chloro-2-hexene.

20

Your Turn!

Chapter Outline

H

CH3

C C

CH2 CH

CH3

H

Cl

1 4

6

2 3

5

21

Your Turn!

Draw the chemical structure of cis-5-chloro-2-hexene. This molecule contains six carbons with a C=C

between carbons #2 and #3, and a Cl atom on carbon #5.

Chapter Outline

H

CH3

C C

CH2 CH

CH3

H

Cl

1 4

6

2 3

5a

C

b

C

b

a

22

Your Turn!

Draw the chemical structure of cis-5-chloro-2-hexene.

This molecule is also cis because the carbon atoms in the longest chain containing the double bond are on the same side of the double bond.

Chapter Outline

23

Your Turn!

Is this the cis or trans isomer of 3-methyl-2-pentene?

Chapter Outline

This is trans-3-methyl-2-pentene because the carbon atoms in the longest chain are on opposite sides of the double bond. 24

Your Turn!

Is this the cis or trans isomer of 3-methyl-2-pentene?

a

C

b

C

a

b

Chapter Outline

25


Many compounds have more than one C=C. Compounds with two C=C are called dienes as shown below. Compounds with three C=C are called trienes.

Chapter Outline

Cycloalkenes

Cycloalkenes are cyclic compounds with a C=C bond in the ring.

The cyclo- in the name indicates that the molecule is cyclic

and the –ene ending indicates that there is a double bond in the molecule.

cyclopentene cyclohexene 26

Chapter Outline

CH3

CH3

CH3

12

1

23

45

4

5

6

3

1-methylcyclopentene 1,3-dimethylcyclohexene 27

Cycloalkenes

Naming cycloalkenes Number the carbon atoms in the ring. The carbon atoms

with the double bond are given the lowest numbers.

Chapter Outline

CH3

CH2CH3

12

34

5

61

2

3

45

6

1, 3-cyclohexadiene 4-ethyl-1-methyl-1,3-cyclohexadiene 28

Naming cycloalkenes with two double bonds. The double bonds are given the lowest numbers. Diene in

the name indicates that each molecule contains two double bonds.

Cycloalkenes

Chapter Outline

Br

CH3

H3C

29

Name the following compounds.

Your Turn!

Chapter Outline

Br

CH3

H3C1

2

3

4

5

6

1

2 3

4

5

30


Your Turn!

1-bromo-4-methylcyclohexene

2-methyl-1,3-cyclohexadiene

Chapter Outline

31

Preparation and Physical Properties of Alkenes

Common preparation methods for alkenes start with saturated organic molecules.

Atoms must be removed to form the double bonds. Alkene

synthesis commonly means “getting rid” of some atoms in elimination reactions.

Chapter Outline

Two examples of alkene preparation are cracking and dehydration of alcohols.

• Cracking (splitting of large hydrocarbon molecules to

form smaller ones) • Dehydration of alcohols (elimination of H2O from an

alcohol molecule)

32


Chapter Outline

catalyst

33


Alkenes can be prepared by cracking petroleum (i.e. crude oil) using a catalyst like silica-alumina as shown in the reaction below.

Chapter Outline

CH3 C

H

H

C

OH

H

CH3

conc. H2SO4

heatCH3 C

H

C

H

CH3+ H2O

34


Alkenes can also be prepared by dehydration of alcohols. The reaction is catalyzed by an acid .

Chapter Outline

35


The physical properties of alkenes are similar to alkanes. Alkenes are nonpolar and insoluble in water but soluble in

organic solvents.

Chapter Outline

What type of reaction might be expected for an alkene (or an alkyne)?

Both alkenes and alkynes have fewer than the maximum

of four atoms bonded per carbon. These molecules are more reactive than the corresponding alkanes and readily undergo addition reactions.

36

Chemical Properties of Alkenes

Chapter Outline

Alkenes undergo addition reactions to the C=C bond with these reactants.

• Hydrogen (H2)

• Halogens (Br2, Cl2)

• Hydrogen halides (HBr, HCl, HI) • Sulfuric acid (H2SO4)

• Water (H2O)

37


Chapter Outline

38

Consider the reactions of ethene first. Addition of hydrogen (H2) to ethene forms ethane. This

is type of addition reaction is called a hydrogenation reaction.


Chapter Outline

39


Addition of halogen to ethene (Br2 in this case) forms 1,2-dibromoethane (an alkyl halide).

Chapter Outline

During the reaction of bromine with an alkene, the red-orange color of bromine (flask on the left) dissipates to form a colorless alkyl halide (flask on the right).

40


Chapter Outline

41

Addition of sulfuric acid (H2SO4) to ethene forms ethyl hydrogen sulfate.


Chapter Outline

H C

H

C

H

H H C

H

C

H

H+ H2O

H OH

H+

42

Addition of water (H2O) to ethene forms ethanol.


Chapter Outline

43

Addition of alkyl halide (HCl, HBr, HI) to ethene forms chloroethane, bromoethane and iodoethane.


H C

H

C

H

H + HCl H C

H

C

H

H

H Cl

H C

H

C

H

H + HBr H C

H

C

H

H

H Br

H C

H

C

H

H + HI H C

H

C

H

H

H I

Chapter Outline

44

The preceding examples dealt with ethene, but reactions of this kind occur on almost any molecule that contains a carbon–carbon double bond.

If a symmetrical molecule such as Cl2 is added to a

larger alkene like propene only one product, 1,2-dichloropropane, is formed.


H2C CH CH3 + Cl2 HC C CH3

Cl Cl

Chapter Outline

45

If an unsymmetrical molecule such as HCl is added to propene, two products are theoretically possible, depending on the carbon atom that the hydrogen atom connects to.

The two possible products are 1-chloropropane and 2-

chloropropane . . .


Chapter Outline

However only one product of the two possible products is produced. This occurs because addition reactions involving unsymmetrical alkenes follow Markovnikov’s rule.

CH3 CH CH2 + H-Cl

CH3 CH CH2

HCl

CH3 CH CH2

ClH

(about 100% yield)

(trace) 46


Chapter Outline

What is Markovnikov’s rule? It is a rule that states the H in HX adds to the C=C carbon that has the largest number of hydrogen atoms.

This rule can be explained by a reaction mechanism (i.e.

the specific steps from reactants to products).

47


Chapter Outline

CH3 CH CH2 + H-Cl CH3 CH CH2

H123

+ Cl-

isopropyl carbocation48

HX addition to alkenes is a two-step reaction mechanism. A carbocation is produced in step one. (A carbocation is an ion where a carbon atom has a positive charge.)


1. In step one a secondary carbocation (2o) is produced when the pi electrons of the C=C are attacked by HCl.

Chapter Outline

CH3 CH CH2

H

CH3 CH CH2

HCl

+ Cl-

2-chloropropane

Note: A 2° carbocation is more stable than a 1° carbocation so the 2° carbocation forms preferentially over the 1° carbocation. This difference in carbocation stabilty is the basis for Markovnikov’s rule.

49

2. In step two the chloride ion produced in step one adds to the carbon atom with the positive charge to produce the product.


Chapter Outline

There are four types of carbocations and these can be arranged by their relative stability

H

CH

H

H

CC

H

C

CC

H

C

CC

C

methyl carbocation

primary (1o)carbocation

secondary (2o)carbocation tertiary (3o)

carbocation

increasing stability50


Chapter Outline

Predict the major products formed when 2-methyl-1-butene reacts with:

a) H2, Pt/25°C b)Cl2

c) HCl d)H2O, H+

51

Your Turn!

Chapter Outline

2-methyl-1-butene + H2, Pt/25 °C Hydrogen adds to the double bond to form an alkane (2-methylbutane).

CH2 C

CH3

CH2CH3 + H2 CH2 C

CH3

CH2CH3

H H

Pt

52

Your Turn!

Chapter Outline

CH2 C

CH3

CH2CH3 + Cl2 CH2 C

CH3

CH2CH3

Cl Cl

53

2-methyl-1-butene + Cl2 Chlorine adds to the double bond to form an alkyl halide (1,2-dichoro-2-methylbutane).

Your Turn!

Chapter Outline

CH2 C

CH3

CH2CH3 + HCl CH2 C

CH3

CH2CH3

H Cl54

2-methyl-1-butene + HCl The hydrogen atom adds to the carbon atom with the larger number of hydrogen atoms and the chlorine adds to the other carbon atom (Markovnikov’s rule) to form an alkyl halide (2-chloro-2-methylbutane).

Your Turn!

Chapter Outline

CH2 C

CH3

CH2CH3 + H2O CH2 C

CH3

CH2CH3

H OH

H+

55

2-methyl-1-butene + H2O The hydrogen atom adds to the carbon atom with the larger number of hydrogen atoms and the –OH group adds to the other carbon atom (Markovnikov’s rule) to form an alcohol.

Your Turn!

Chapter Outline

CH2 CH2 + KMnO4 (aq) + H2O CH2 CH2

OH OH

+ MnO2 + KOHethene(ethylene)

(purple) (brown)56


Another typical reaction of alkenes is oxidation of the double bond.

For example when an alkene is shaken with a cold, dilute

solution of potassium permanganate, KMnO4, the alkene is converted to a glycol (glycols are dihydroxy alcohols).

Chapter Outline

CH2 CH2 + KMnO4 (aq) + H2O CH2 CH2

OH OH

+ MnO2 + KOHethene(ethylene)

(purple) (brown)

57

The reaction below is used in the Baeyer test which is a test for the presence of double or triple bonds in an unknown sample.


Chapter Outline

The rules for naming alkynes are the same as those used to name alkenes except the suffix –yne is used to indicate the presence of the C≡C bond. The carbon atoms with the triple bond get the lowest numbers.

IUPAC Rules for Naming Alkynes

CH3 CH2 C C H

1-butyne

1234

58

Alkynes: Nomenclature and Preparation

Chapter Outline

59


Chapter Outline

Although triple bonds are very reactive, it is relatively easy to synthesize alkynes. Acetylene, the simplest alkyne, can be prepared inexpensively from calcium carbide and water or by the cracking of methane.

60


Chapter Outline


61

Your Turn!

H C C C

CH3

CH3

H

H3C C C CH2

Cl

Chapter Outline


62

Your Turn!

H C C C

CH3

CH3

H

H3C C C CH2

Cl

1 2 3 4 1234

3-methyl-1-butyne 1-chloro-2-butyne

Chapter Outline

Physical properties of acetylene …… • a colorless gas. • little odor when pure. • insoluble in water. • a gas at normal temperature and pressure. • subject to explosive decomposition.

63

Physical and Chemical Properties of Alkynes

Chapter Outline

Chemical properties of alkynes Alkynes undergo addition reactions similar to those

of alkenes. They react with: • Cl2 and Br2 • HCl and HBr • KMnO4 to give a positive result from Baeyer’s

test.

64


Chapter Outline

HC≡CH + Br2 → CHBr=CHBr

HC≡CH + 2 Br2 → CHBr2-CHBr2

65

Reaction of acetylene with bromine (bromination).


Chapter Outline

HCl addition to unsymmetrical alkynes follows Markovnikov’s rule like the reactions with alkenes.

The hydrogen atom adds to the carbon atom that has the largest number of hydrogens.

CH3C≡CH + HCl → CH3CCl=CH2

CH3C≡CH + 2 HCl → CH3CCl2-CH3

66


Chapter Outline

Representations of Benzene

67

Aromatic Hydrocarbons: Structure

Benzene is an aromatic compound. Aromatic compounds are those that resemble benzene in structure and reactivity.

Chapter Outline

68

Benzene is aromatic because it has an unusually stable electronic structure created by six delocalized pi electrons.


Chapter Outline

Figure 20.5 (a) sp2-sp2 orbital overlap to form the carbon ring structure.

69


Chapter Outline

Figure 20.5 (c) pi electron clouds above and below the plane of the carbon ring.

71


Chapter Outline

72

Naming Aromatic Compounds

Substituted benzenes are the most common benzene derivatives because substitution is the most common reaction type for benzene.

A substituted benzene is derived by replacing one or more

hydrogen atoms of benzene by another atom or group of atoms.

Chapter Outline

nitrobenzene ethylbenzene chlorobenzene bromobenzene

NO2 Cl BrCH2CH3

73


A monosubstituted benzene has the formula C6H5G, where G is the group replacing a hydrogen atom.

Monosubstituted benzenes can be named by adding the

substituent prefix in front of the word benzene as shown below.

Chapter Outline

CH3

C

O

HC

O

OH

benzoic acid(benzene carboxylic acid )

benzaldehyde(benzene carboxaldehyde)

styrene(vinylbenzene)

phenol(hydroxybenzene)

aniline

toluene(methylbenzene)

OHCH=CH2

NH2

(aminobenzene) 74


There are several monosubstituted benzenes that have special names.

Chapter Outline

CH

CH3

CHCH2CH3

CH2

diphenylmethane

Cl

3-chloro-2-phenylpentane

1

23 4 5

This is the phenyl group which is a benzene ring minus ahydrogen atom .

These are examples of benzene derivatives that are easier to name using the phenyl group.

75

The word phenyl represents the C6H5- group. It is used to name benzene derivatives that would otherwise be difficult to name.

Chapter Outline

G

ortho

meta

para

meta

ortho

76

Disubstituted benzenes The prefixes ortho-, meta-, and para- (abbreviated o-, m-,

and p-) are used to name disubstituted benzenes where G is given here as a reference point on the ring.


Chapter Outline

G

ortho

meta

para

meta

ortho

Cl

Cl

77


For example, this would be ortho-dichlorobenzene because the chlorine atoms are on adjacent carbon atoms on the ring.

Chapter Outline

78

The three isomers of dichlorobenzene have different physical properties. Notice how their melting and boiling points are different.

Chapter Outline

Cl

ortho-bromochlorobenzene meta-ethylnitrobenzene

Br

NO2

CH2CH3

79


When naming disubstituted benzenes the substituents are given in alphabetical order followed by the word benzene.

Chapter Outline

The dimethylbenzenes have the special name xylene.

CH3

ortho-xylene

CH3

CH3

CH3para-xylenemeta-xylene

CH3

CΗ3

80


Chapter Outline

ortho-nitrophenol

NH2 CH3

meta-nitrotoluenepara-bromoaniline

OH

NO2 Br

NO2

81

A disubstituted benzene is named as a derivative if one of the substituents corresponds to a benzene derivative with a special name.


Chapter Outline

82

Naming Polysubstituted Benzenes If there are three groups on the ring, start with number one

and number in the direction that gives the lowest number to each group as shown on the next slide . . .


Chapter Outline

CH3

O2N NO2

NO2

2,4,6-trinitrotoluene (TNT)

12

3

45

6

OH

Cl

5-bromo-2-chlorophenol

12

35

6

Br 4

83


Naming Polysubstituted Benzenes

Chapter Outline

84

Polycyclic Aromatic Compounds

There are many other aromatic ring systems besides benzene. Their structures consist of two or more rings in which two carbon atoms are common to two rings.

These compounds are known as polycyclic or fused

aromatic ring systems.

Chapter Outline

naphthalene phenanthrene

anthracene 85

Polycyclic Aromatic Compounds

Three of the most common hydrocarbons in this category are naphthalene, anthracene, and phenanthrene.

Chapter Outline

86

Your Turn!


CH2CH3

F

CH3

NO2

Cl

Br

NO2

Chapter Outline

87

Your Turn!


CH2CH3

F

CH3

NO2

Cl

Br

NO2

1

2

3

4

5

6

1

2

3

4

5

6 1

2

3

4

5

6

1-ethyl-2-fluorobenzene 3-nitrotoluene

1-chloro-2-bromo-4-nitrobenzene

Chapter Outline

Coal Coke + Coal gas + Coal tar∆

88

Sources and Physical Properties of Aromatic Hydrocarbons

Aromatic hydrocarbons (to include benzene, toluene, napthalene etc.) are produced from petroleum.

A limited amount can be prepared from coal tar. Coal tar is

only a by-product of coke. Consequently this method is no longer the principal source of aromatic hydrocarbons.

Chapter Outline

89

Sources and Physical Properties of Aromatic Hydrocarbons

Aromatic hydrocarbons have properties common to all hydrocarbons. They are nonpolar substances, they are insoluble in water but soluble in most organic solvents, and have densities less than the density of water

They also burn readily, usually with smoky yellow flames.

Chapter Outline

Aromatic hydrocarbons undergo the following substitution reactions.

• Halogenation – net addition of –Br or –Cl

• Nitration – net addition of –NO2

• Alkylation – net addition of –R (alkyl group)

90

Chemical Properties of Aromatic Hydrocarbons

Chapter Outline

+ X2

X

+ HXFeX3

benzenebromine orchlorine bromobenzene or

chlorobenzene

91

Halogenation: Benzene is halogenated in the presence of a mixture of X2 and an iron(III) halide catalyst.


Chapter Outline

+ HO-NO2

NO2

+ H2OH2SO4

benzenenitric acid

nitrobenzene

92

Nitration: Benzene is nitrated in the presence of a mixture of concentrated nitric acid/sulfuric acid at ~50°C.


Chapter Outline

+ CH3CH2Cl

CH2CH3

+ HCl

benzene ethylbenzene

AlCl3

chloroethane

93

Alkylation: The reaction below is an example of the Friedel-Crafts reaction. An alkyl group from an alkyl halide (RX) is substituted on the ring for a hydrogen atom. The reaction is catalyzed by AlCl3.


Chapter Outline

CH2CH3

ethylbenzene

K2Cr2O7/H2SO4heat

COOH

+ CO2

benzoic acid

94

Oxidation of alkyl groups: Alkyl groups on a benzene ring are easily oxidized to benzoic acid.


Chapter Outline

Unsaturated hydrocarbons consist of alkenes, alkynes, cycloalkenes, and aromatic compounds. Each molecule in this class contains one or more pi (π) bond. Pi (π) bonds result from sideways overlap of p-orbitals on two sp- or sp2-hybridized atoms. Alkenes undergo addition reactions. For unsymmetrical alkenes the product obtained is predicted by Markovnikov’s rule.

95

Chapter 20 Summary

Chapter Outline

Markovnikov’s rule is explained by the variation in stability among carbocations. Aromatic compounds are relatives of benzene and undergo substitution reactions because of the special electronic stability these molecules derive from electron delocalization. Aromatic compounds undergo substitution reaction due the stability of the benzene ring.

96

Chapter 20 Summary

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