chapter11 introduction of organic chemistry

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3

At the end of this lesson, students should be able to:

Define structural formula.

Identify classes of carbons (10,20,30,40) and hydrogens

(10,20,30).

Draw structural formula in the form of expanded condensed and skeletal structures based on molecular formula.

Chemical Formula represented by :

EMPIRICALFORMULA

MOLECULARFORMULA

STRUCTURALFORMULA

shows all the covalent bonds

between the atoms

does not represent the actual shapes of the molecules

EXPANDED structure

Examples

Covalent bondC

H

H

H H

C

H

H

HC

H

H

H

Methane, CH4

Ethane, C2H6

C C

H H

H H

Ethene, C2H4

C CC

H

H

H C

H

H

H

2-propene, C4H8

H H

does not show single bonds between C and H atoms

BUT double and triple bonds are shown

all atoms which attached to a particular C

are written immediately after the C

condensed structure

Example 1

C C

H H

H H

Ethene, C2H4

H2C=CH2

C CC

H

H

H C

H

H

H

2-propene, C4H8

H H

CH3CH=CHCH3

Example 2

Example 3

2-Chlorobutane

C CC

H

H

H C

H

H

H

H Cl

CH3CH2CHCH3

H H @CH3CH2CH(Cl)CH3

Cl

if there are 2 or more similar groups

bonded

to the same C atom,

the group is enclosed in brackets and a subscript numeral

is used to indicate its number

Example

Pentane, C5H12

C CC

H

H

H C

H

H

H

H H

CH3(CH2)3CH3

H H

C

H

H

bonds are represented as lines C atoms assumed to be at the

beginning and end of a line

and where 2 or more lines meet does not show C and H atoms BUT

functional groups and atoms other than C and H

MUST be shown

SKELETAL structure

CH3CH(Cl)CH2CH3

Cl

C1

C2

C3

C4

H2C CH2

H2C CH2

CH2 = CHCH2OH

OH

C1 C2

C3

(i)

(ii)

(iii)

1 2 3 4

1 2

34

Examples

describes how atoms of the molecules are

arranged in space

e.g. Bromomethane

C

Br

H

H

H 2-dimensional structure

3-Dimensional structure

C

Br

HHHC

H

BrHHC

H

HBrH

Represent a bond on the plane of the paper

Represent a bond comingout of the paper

Represent a bond pointingto the back of the paper

3–Dimensional Structure of Bromoethane

CLASSIFICATION OF CARBON AND HYDROGEN ATOMS

Depending on the number of C atom bonded to it

Type of C atom No. of C atoms bonded to it Example

None12

3

4

MethylPrimary (10)Secondary (20)

Tertiary (30)

Quarternary (40)

OHH3CCH3H3C

CH2 CH3H3CCH CH3

CH3

H3C

CCH3

CH3

CH3

H3C

Classification of C atoms

H atom is also classified as 1o, 2o and 3o

depending on the class of C bonded to it e.g.

CH

CH2

CH3

H3C CH3

1o

2o

3o

Classification of H atom

1o2o

3o

1o

1o

However, there is no 40 H

21

1. Draw skeletal structure for each condensed structure given below:

CH3 CH2CH2CH3

=1 2 3 4

CH3 CH2 C N

= N

CH2 = CHCH2OH

OH

=1 2 3

22

=CH3 CH2CH CH2CH3CH3

1 2 3 4 5

CH3 CH2C-CH3 O O

=

CH2CHClH2C

H2C CH2

Cl

=

23

2.A hydrocarbon Q has the molecular formula of C4H10 . Write all the possible structural formulae of Q in the form of :

i. condensed structures

ii. skeletal structuresCH3 CH2CH2CH3

CH3 CH(CH3 )2

CH3 (CH2)2 CH3

i.

ii.

or

CH C C CH2 C CH3

H

H

H

H H

CH3 CH3

CH3

10 20

4030

1o carbon - 5 2o carbon - 23o carbon - 14o carbon - 1

10 10

10

10

20

24

26

At the end of this lesson, students should be able to:

Define functional group.Identify functional group

Explain general characteristics of homologous serries.

Classify organic compounds according toTheir functional groups.

Define homologous serries.

29

Homologous Series Alkene Alkyne Arene

Functional Group

C = C (carbon-carbon double bond)

C ≡ C (carbon-carbon

triple bond)

(aromatic ring)

General Formula CnH2n CnH2n-2 CnH2n-6

IUPAC Nomenclature -ene -yne -benzene

Example CH2= CH2 ethene

CH ≡ CH ethyne methylbenze

ne

CH3

Benzene ring

30

Homologous Series Haloalkane Alcohol Ether

Functional Group

C–X halogen

–OH (hydroxyl)

–O– (alkoxy)

General Formula CnH2n+1X CnH2n+1OH CnH2n+2O

IUPAC Nomenclature

haloalkane alkanol alkoxy

alkane

Example CH3CH2Clchloroethane

CH3CH2OH ethanol

CH3OCH3 methoxy methane

31

Homologous Series Aldehyde Ketone Nitrile

Functional Group

–C–H

carbonyl

–C–

carbonyl

–C≡N(cyano)

General Formula CnH2nO CnH2nO CnH2n+1CN

IUPAC Nomenclature alkanal alkanone alkanonitrile

Example CH3C-H

ethanal

CH3COCH3

propanone

CH3CH2C≡Npropanonitril

e

O O

OO

32

Homologous Series Carboxylic Acid Ester

Functional Group

–C–OH (carboxyl)

–C–O–C

General Formula CnH2nO2 CnH2nO2

IUPAC Nomenclature alkanoic acid alkyl alkanoate

Example CH3C–OH

ethanoic acid

CH3C–OCH3

methyl ethanoate

O (carboalkoxy)

O

O

O

33

Homologous Series Acyl halide Anhydride

Functional Group –C–X

–C–O–C–

General Formula CnH2n+1COX (CnH2n+1CO)2O

IUPAC Nomenclature alkanoyl halide alkanoic anhydride

Example

CH3C–Cl

ethanoyl chloride

CH3C–O–C-CH3

ethanoic anhydride

O OO

O OO

34

Homologous Series Amine Amide

Functional group

–C–N– (amino)

–C–N–

(carboxamide)

General Formula CnH2n+1NH2 CnH2n+1CONH2

IUPAC nomenclature -amine -amide

Example CH3NH2 methanamine

CH3CONH2

ethanamide

35

C

O

CC

CHCH2C-CH3

OO

1. Circle and name all the functional groups present in the following molecules.

a.

aromatic ring

carbonyl

Carbon-carbon double bond

carboalkoxy

36

b.

CH2CH2CHCH2C-OCH3

O

NH2

O

Br

carbonyl

Carbon-carbon double bond

amino

carboalkoxy

halogen

37

2. With reference to the following structural formula:

CH3CH2CH=C-CH2CH2C-NH2 B

rO

a. Give the molecular formula.

b. Write its skeletal formula.

C7H12ONBr

Br

O

NH2

38

CH3CH2CH=C-CH2CH2C-NH2 B

rO

Carbon-carbon double bond

halogen

carboxamide

c. State 3 functional groups that exist in the molecule.

39

1. Write and name the functional groups that exist in the following molecule :

O

CH2CH2CHCH2COCH3

O

NH2

carbonyl

Carbon-carbon double bond

carboalkoxy

amino

40

OH

O CH3

CO H O H O

║ │ ║NH2 – CH-C-N-CH-C-OH │ │ H CH3-C=O

2. Identify the functional groups in the following compounds. Write and name the class of the compounds.

FIRST THREE MEMBERS OF THE SERIES

CH3OH Methanol

Example: ALCOHOLS

(Homologous Series)

C2H5OH Ethanol

C3H7OHPropanol

(Homolog)

A homologous series has four features :

share the same general formulaExample:

saturated aliphatic ALCOHOL CnH2n+1OH, where n = 1, 2, 3, etc.

(1)

n = 1 CH3OH methanoln = 2 C2H5OH ethanol

n = 3 CH3CH2CH2OH propanol

Have same functional group. Thus, have the same chemical properties and can prepared by same methods contain the –OH group (Hydroxyl

Group) react with carboxylic acids to give

esters can be prepared by heating dilute

sodium hydroxide solution with an appropriate alkyl halide

Example: ALL Alcohols

(2)

Each homolog differs from the homolog above or below it in the series by – CH2

CH4 methaneCH3CH3

ethaneCH3CH2CH3 propane

(3)

Example 1

HCHO methanal

Example 2

CH3CH2CHO propanal

CH3CHO ethanal

As the size of the molecules increases, boiling point and melting point increases

show a gradual change in physical properties as molar mass increases

(4)

Example

Compound Boiling pointCH4 - 162CCH3CH3 -

88.2C

48

At the end of this lesson, students should be able to:Define isomerism.

Explain constitutional isomerism.

Define and identify chirality centre and enantiomers.

Explain optical activity of a compound.

Draw a pair of enantiomers using 3-D formula.

Define stereoisomerism.

Describe and identify cis-trans isomerism.

ISOMERISM

Structural/Constitutional Isomerism Stereoisomerism

Chain Isomerism

Positional Isomerism

Functional Group Isomerism

DiastreomerEnantiomer

cis-transisomerism

otherdiastereomers49

ISOMERISM

Structural isomerism

Stereoisomerism

50

the phenomenon of the existence of 2 or more compounds with the same molecular formula

but different in structures

The different compounds : called isomer

The isomerism resulting from different order of attachment of atoms

The isomers differ in their bonding sequence; their atoms are connected differently

Chain Isomerism

Positional Isomerism

Functional Group Isomerism

51

same functional group same homologous series same chemical properties but different physical properties linear or branch chain

a) Chain isomerism different carbon chain i.e continous-chain or branched-chain.

Example : C5H12

CH3 CH2 CH2 CH2 CH3

CH2 CH3CH3 CHCH3

CH3 C

CH3

CH3

CH3

Linear form

Branch form

3 Isomers

Characteristic of Chain Isomers

Chemical properties isomers show same chemical properties

because it has same functionalPhysical properties

density and boiling point show trends according to the degree of branching

“straight” chain isomers have higher boiling points

than branched chain isomers the greater the degree of branching, the lower

the boiling point

branching decreases the effectiveness of intermolecular attractive forces, so less

energy has to be put in to separate the molecules

Boiling PointCharacteristic of Chain Isomers

- 0.5°Cstraight chain

greater branching = lower boiling point

-11.7°C branched

• same skeletal structure• different position of the functional group or

substituent groupsExample : C3H7Cl

CH3CH2CH2Cl 1-chloropropane CH3CHCH3 2-chloropropane Cl

Example : C4H8

CH2 = CHCH2CH3 1-butene CH3CH = CHCH3 2-butene

b) Positional isomerism

C8H10

1,2-dimethylbenzene

1,3-dimethylbenzene

1,4-dimethylbenzene

CH3

CH3

CH3

CH3

CH3

CH3

58

same molecular formula have different functional groups belongs to different homologous series different chemical and physical properties

c) Functional Group isomerism

Alcohols and Ethers

Aldehydes and Ketones

Carboxylic Acids and Esters

Compounds which exhibit functional group isomerism

Alkene and Cycloalkane

ExampleALCOHOLS and ETHERS

Name ETHANOL METHOXYMETHANEClassification ALCOHOL ETHER

Functional Group -OH -OR

Physical polar O-H bond gives rise no hydrogen bondingproperties to hydrogen bonding. low boiling point

get higher boiling point insoluble in water and solubility in water

General Formula CH3CH2OH CH3OCH3

Examples

C2H6O

C3H6O

C3H6O2

ethanol dimethyl ether

propanal propanone

propanoic acid methyl ethanoate

CH3CH2OH CH3OCH3

CH3 CH2C

O

H

CH3CH2C

O

OH

CH3 CO

CH3

CH3 CO

O CH3

62

Exercise:1.State how many are isomers with the

following molecular formulae, identify the type of isomerism and draw the structural formula of the isomers.

a) C5H10

b) C5H10O2

c) CH3CH=C(Cl)CH3

d) C4H6Cl2 e) CH3CH2CH(OH)CH(Br)CH2CH3

63

Chain Isomerism

Positional Isomerism

The isomers differ in the carbon skeleton (different carbon chain i.e continous-chain or branched-chain).

The isomers have a substituent group/ functional group in different positions.

The isomers have different functional groups and belong to different homologous series with the same general formula.

Functional Group

Isomerism

Stru

ctur

aliso

mer

ismSUMMARY

64

The isomerism resulting from different arrangement of atoms in molecules.

Isomers have the same bonding sequence, but they differ in the oriention of their atoms in space.

Enantiomer Diastereomer65

cis-isomer trans-isomer

C CCl

H H

ClC C

Cl

H Cl

H

Same side

opposite side

same atom or groups of atom on the same

side

same atom or groups of atom on the opposite side

a) DiastereomerCis-Trans Isomerism/geometric isomerism

66

♫ restricted rotation about a C=C, double bond in alkenes, or a C-C single bond in cyclic compounds.

♫ each carbon atom of a site of restricted rotation has two different groups attached to it.

The requirements for geometric isomerism :

67

trans-2-butene cis-2-butene

C CH

CH3 H

CH3

C CH

H3C CH3

H

cis-1,2-dimethylcyclohexane

trans-1,2-dimethylcyclohexane

H

CH3

H

CH3

H

CH3

CH3

H

Examples :

68

H

Cl

H

Cl

H

Cl

Cl

Hcis-1,3-dichlorocyclohexane

trans-1,3-dichlorocyclohexane

69

cis-trans isomers are two different compounds different physical properties :

melting point : trans-isomer > cis-isomer boiling point : trans-isomer < cis-isomer stability : trans-isomer > cis-isomer

similar chemical properties (have the same functional group) cis-isomer : polar molecules; trans-isomer : non-polar molecules

Characteristic of Cis-Trans Isomers

If there are 3 different groups attached to the doubly bonded C,the following rule applies:

higher-priority groups are on the same side

higher-priority groups are on the opposite site

cis-isomer trans-isomer

71

C CH

CH3 CH2CH3

CH3

C CH

CH3 CH3

CH2CH3

Higher priority groups

Atom with a higher atomic number is in higher priority than an atom with lower atomic number

Higher priority groups

cis-3- methyl-2-pentenetrans-3- methyl-2-pentene72

♫ If one of the doubly bonded carbons has 2 identical groups, geometric isomerism is

not possible.

2-methyl - 2- butene

C CH

CH3 CH3

CH3

C CH

H CH2CH3

CH3

2-methyl - 1- butene

No cis – trans isomer

73

® The chiral molecule and its mirror image are

Enantiomers

Molecules/objects that are not superimposable with their mirror images are said to be

chiral.

hands glovesshoes

74

have one or more chiral carbons. A chiral carbon has four different

atoms or group atoms attached to it.

P

CQR

S* PQRS

*designates chiral centre

b) Enantiomers Are a pair of stereoisomers that are mirror-image but not superimposable.

75

Enantiomers i) 2-butanol

C*CH2CH3

H3COH

HC

CH2CH3

CH3HOH

enantiomers

CH3 CH CH2CH3

OH

76

*

ii) 2-hydroxypropanoic acid

CCOOH

H3COH

HCCOOH

CH3HOH

enantiomers

CH3 CH COOH

OH

77

*

Enantiomer

Diastereomer

a pair of mirror-image molecules that are not superimposable.

A pair of enantiomers have identical chemical & physical properties but differ in the direction of rotation of plane- polarized light. Stereoisomers that are not mirror images of each other

All physical properties of diastereomers are usually different from one another.

Stre

reoi

som

erism

78

OPTICALLY ACTIVE molecules which are optically active are

measured using : polarimeter able to rotate the plane-polarised light to be optically active molecules must contain at least one chiral C

Plane – polarised Light is composed of waves that vibrate in only one plane.

Plane – Polarised Light

A beam of light is passed through a piece of polariser prism,the emergent light vibrates in a single plane.The light is called plane-polarised light.

80

Substances that could rotate the plane-polarized light are called Optically active compounds . 81

PLANE-POLARISED LIGHT

Dextrorotatory isomer(d or +)

Laevorotatory isomer(l or -)

Clockwise rotation

Anti-Clockwise rotation

i) molecule contains a chiral carbon/chirality centre

ii) molecule is not superimposable with its mirror image.

The requirements for optically active compounds :-

A pair of enantiomer rotate the plane of polarised light by exactly the same degree

but in opposite directions.83

PRACTICE EXERCISEFor the following questions match each definition to a term from the list below.

a. Optically activeb. Diastereomer

c. Chirality centre

2. Stereoisomers that are not mirror images3. Is an atom in a molecule that is bonded to 4 different atoms or groups of atoms

1. Describes organic molecules which rotate plane-polarised light Optically

active Diastereomer

Chirality centre

86

At the end of this lesson, students should be able to:Explain covalent bond

cleavage:homolytic and heterolitic.

State relative stability of 10,20 and 30 free radicals, carbocations nad carboanions.

Define electrophile and nucleophile.

State the main types of organic reactions.

Explain the inductive effect of alkyl group

Towards the stabilities of carbocations

and carboanions.

87

• involve the breaking and making new covalent bonds

• two means of electron transfer during the breaking of covalent bond

represents the transfer of one electron.

represents the transfer of two electrons or a pair of electrons.

:

:

88

• Occurs in a non-polar bond

• involve two atoms of similar electronegativity.

• single bond breaks symmetrically,

• leaving each atom with one unpaired electron.

• Forms free radicals.

Occurs in a polar bond involve two atoms of

different electronegativities.

single bond breaks unsymmetrically,

both the bonding electrons are transferred to the more electronegative atom.

Forms cation and anion.

89

Homolytic Cleavage

Heterolytic Cleavage

A – B A B+ A – B A– B++

A – B A+ B–+

A is > electro-ve

B is > electro-ve

90

Cl – Cla.

H – CH3b.

2 Cl

H CH3+

H – Cl

(CH3)3C – Br

+ –

+ –

+H+ Cl

+ Br(CH3)3C+

uv

uv

Homolytic Cleavage

Heterolytic Cleavage

91

Unstable and highly reactive species.Formed during a reaction as a result of bond cleavage.

Three types:

Carbocation or carbonium ion

Carbanion Free radical

92

Reactive species with unpaired electronAn e deficient speciesResults from homolytic cleavageExamples

Reactive species with + charge on C atom An e deficient speciesResults from heterolytic cleavageExamples

Reactive species with - charge on C atom An e rich speciesResults from heterolytic cleavageExamples(CH3)3C

+ CH3+

(CH3)2CH+(CH3)3C- CH3

-

(CH3)2CH-

(CH3)3C CH3

(CH3)2CH

93

Carbocations, carbanions & free radicals are classified as 1o , 2o or 3o based on the number of adjacent C atoms of the:

positively charged C atom (for carbocation)

negatively charged C (for carbanion)

carbon atom with an unpaired electron(for free radical)

94

Carbocations and free radicals are electron-deficient (lack an octet around the C atom)

Both are stabilised by the electron-donating effect of alkyl groups, –R, The more highly substituted carbocations or free radicals are more stable.

95

(3o

)(2o

)(1o

)methylium

Decreasing stability

RRCR

+ H CH

+H

CH

+R

RHCH

+R

(CH3)3C

+

(3o

)(1o

)(2o

)

+(CH3)2CH

+CH3CH2

+CH3

Stability of carbocation

96

The stability of benzylium ion (a 1o aromatic carbocation) is comparable to that of 3o aliphatic carbocation.

CH2+

CH2+ CH2

+CH2+

CH2

+

Reason The positive charge on C is resonance-stabilised:

97

3o 2o 1o

(CH3)3C

(CH3)2CH

CH3CH2

CH3

RRCR

H CH

H

CH

R

RHCH

R

Decreasing stability

methyl radical

3o1omethyl radical

2o

98

Carbanions are high electron density species. (electron-rich)

Electron-donating groups ( e.g. alkyl group, –R , etc) destabilise a carbanion.

Electron-withdrawing groups (e.g. Halogen) stabilise carbanions ( reduces the electron density of carbanions)

The order of stability is the opposite to that of carbocations and free radicals

99

3o 2o 1o methyl anion

(CH3)3C

– –(CH3)2CH

–CH3CH2

–CH3

Increasing stability

RRC:R

– C:H

RR –

HC:H

R – H C:H

H–

3o 1o methyl anion

2o

100

Arrange the following species in order of increasing stability:

CH2+ CH3+ CH3

+

(CH3)2CCH2CH3

+ (CH3)3CCHCH

3

+

(CH3)3CCH2

+

a.

b.

1o 3o 2o

3o

2o

1o

101

CH2+ CH3

+c. CH2

+

(CH3)2CCH2CH3

+d.

CH3CHCH2CH2

+

CH3

CH2CHCH2CH3

+

CH3

1o

3o

CH3CHCHCH3

+

CH3 2o

1o

1o aromatic

2o

1o

102

There are 2 types of chemical reagents

Electrophile (E+) Nucleophile (Nu-)

“ loves electron ” “ loves nucleus ”

An electron-deficient species and electron-pair acceptor that attacks part of a molecule where the electron density is high (nucleophilic site)

An electron-rich species and electron-pair donor that attacks a part of a molecule where the electron density is low (electrophilic site)

103

Cations H+, H3O+, NO2

+, X + All carbocations R+

Lewis acidsAlCl3 , FeCl3 , BF3 , ZnCl2 , H2O Oxidising agents Cl2 , Br2 etc.

Electrophile Nucleophile

AnionsOH–, RO–, Cl– , CN– All carbanionsR – (species with a negative charge on carbon atoms).Lewis basesspecies which can donate lone pair electrons : NH3 , H2O, H2S, RNH2

104

Molecules with low electron density around a polar bond such as:

Electrophilic Sites in Organic Molecules

carbonyl group

haloalkane

hydroxyl group

C Oδ+ δ-

C Xδ+ δ-

C OH+δ δ-

105

Molecules with high electron density around the C-C multiple bond:

C = C

C C

Alkenes

Alkynes

Aromatic compounds

Nucleophilic Sites in Organic Molecules

106

Identify the following species as nucleophile or electrophile:

CH3OCH2CH3 AlCl3

CH3CH2OH SO3

CO2

CH3CH=CH2

CH3CH2O– (CH3)2CH +

C6H5N2+

Nu.E

E / Nu

Nu. E E

Nu. E

Nu.

107

SubstitutionRearrangementElimination

4 main types of chemical reactions

Addition

electrophilic

nucleophilic

electrophilic

nucleophilic

Free radical

108

A reaction in which atoms or groups add to adjacent atoms of a multiple bond (double or triple bond). • involve breaking of one bond

to form two sigma bonds

109

atoms or groups of atom are added to adjacent atoms of a multiple bond

atoms or groups of atom are added to adjacent atoms of a multiple bond

Initiated by electrophileTypical rxn for alkene

Initiated by nucleophileTypical rxn for carbonyl

110

Reagent : electrophileSubstrate : unsaturated compounds with

nucleophilic sites such as C=C or C≡C.Note : typical reaction of alkenes

CH2= CHCH2 CH3CHBrCH2+ HBra.

CH2=CH2 CH2(Br)CH2 (Br)+ Br2b.

CH3 + H2O CH3

OHc.

111

Reagent : nucleophileSubstrate : unsaturated compounds

with electrophilic sites such as –C=O

Note : typical reactions of aldehydes & ketones

a.

CH3CHO

+ NH3 CH3CHNH2

OH

b. CH3CCH3

O+ HC

NCH3C-CH3

OH

CN

112

A reaction in which an atom or a group in a molecule is replaced by another atom or group of atoms from the reagent.

113

+ Br2Fe Br + HBra

.

CH3+NO2+ CH3

NO2

b.

Reagent : electrophileSubstrate : aromatic nucleus, stable

with high electron density.Note : typical reaction of

aromatic compounds.

114

CH3CH2Cl + NaOH

a.

CH3CH2OH + NaCl

Brb. + KCN

CN + KBr

+ SOCl2

c. + H2O

CH3CHOHCH3

CH3CHClCH3

Reagent : nucleophilicSubstrate : saturated compounds

with polar bonds as functional group

Note : typical reactions of haloalkanes and alcohols.

115

CH3CH2CH3 + Cl2

a.

+ HCluv CH3CHCH

3 Cl

b. CH3 + Br2

uv

CH3

Br + HBr

Substitution which involves free radical as intermediate species.

Typical reaction of alkanes.

116

A reaction in which atoms or group of atoms are removed from adjacent carbon atoms of a molecule to form a multiple bond (double or triple bond)

Typical reaction of alcohols & haloalkanes.

Product: Unsaturated molecules.

117

CH3CHOH CH3

conc. H2SO4 CH2=CHCH

3

+H2Oi.

Br conc. KOHethanol, ∆

+HBrii.

118

A reaction in which atoms or groups in a molecule change position.

Occurs when a single reactant reorganises the bonds and atoms

Occur in reactions involving the formation of a carbocation intermediate

119

c. OH-C CH2C

H3

CH3

Cl

CH3

a. CH3CCH3

OCH3C=CH2

OH

tautomerism

b. CH3CHCH2OH CH3

conc. H2SO4

C CH3CH3

OH

CH3

CH3CH CHCH3CH3

Cl

120

State the type of reaction for each of the following reactions:

(CH3)3C-Br + NaOH

(CH3)3C-OH+ HBr

a.

CH3C–CH2OH CH3C=CCH3 CH3CH3

CH3 CH3

+ H2O

conc. H2SO4

b.

Elimination

Nucleophilic substitution

121

OH

HO

CH3CH3Cl

AlCl3+ HCl

+ Br2(aq)

CH3CH=CCH3 CH3

CH3CHCCH3 CH3

OH

Br

c.

d.

Electrophilic substitution

Electrophilic addition

122

CH3C–CH3 O

e.

CH3C–CH3 OH

CN+ HCN

+ Br2

Br

+ HBr

f.u

Nucleophilic addition

Free radical substitution

123

State the type of reaction for each of the following

reactions :

b. CH2 = CH2 + HBr CH3CH2Br

c. C6H6 + CH3Cl C6H5CH3 + HCl

e. CH3CH2Cl + NaOH CH3CH2OH + HCl

a. CH2 = CHOH CH3CHO

d. CH3CH2OH CH2 = CH2 + H2O

[5 marks]

rearrangement

Electrophillic addition

Electrophillic substitution

Elimination

Nucleophillic substitution

124

Nucleophilic substitution

Nucleophilic addition

Electrophilic substitution

+ HNO3

Conc H2SO4

NO2

CH3COCH3 + HCN CH3 C CH3

CN

OH

CH3CH(OH)CH2CH3

CH3CH(Cl)CH2CH3

+ SOCl2

125

Substitution

RearrangementElimination

Addition

electrophilic

nucleophilic

electrophilic

nucleophilic

Free radical

(alkene/alkyne)

(aldehyde/ketone)

(alkane)

(aromatic ring)

(R-OH, R-X)

(R-OH, RX alkene)

(Rxn involve C+ / C intermediates)

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