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INTRODUCTION TO ORGANIC

CHEMISTRY

CHAPTER 4

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Chapter 4 : INTRODUCTION TO ORGANIC CHEMISTRY

4.1 Molecular and

Structural Formulae

4.2 Functional Groups and

Homologous Series

4.3 Isomerism

Learning Outcomes:

a) Define structural formula.

b) Show structural formula in the form of expanded, condensed and skeletal structures based on molecular formula.

c) Classify carbons into primary, secondary, tertiary or quaternary (1°,2°,3°,4°) and hydrogens into primary, secondary or tertiary (1°,2°,3°).

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4.1 MOLECULAR AND STRUCTURAL FORMULAE

•Structural formula shows how the atoms in a molecule are bonded to each other.

•3 types of structural formula:• expanded structure• skeletal structure• condensed structure

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STRUCTURAL FORMULA

•Expanded structures indicate how atoms are attached to each other but are not representations of the actual shapes of the molecules.

C4H9Cl(Molecular

Formula)

(Expanded structure)

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C C C CH H

H H HCl

H H H H

EXPANDED STRUCTURE

Examples:

Alcohol (C2H6O) C

H

H

H

C

H

H

OH

C

H

H

H

C

H

H

C

O

OH

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Examples:

Carboxylic acid (C3H6O2 )

(Expanded structure)

(Expanded structure)

• Shows only the carbon skeleton.

• Hydrogen atoms are not written.

• Other atoms such as O, Cl, N etc. are shown.

Examples:

i) CH3CH(Cl)CH2CH3Cl

ii) H2C CH 2

H2C CH 2

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SKELETAL STRUCTURE

• Does not show single bonds between carbon and hydrogen atoms, but double and triple bonds are shown.

• All atoms that are attached to a carbon are written immediately after that carbon.

C4H9Cl CH3CH(Cl)CH2CH3(Condensed structure)

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Examples:

CONDENSED STRUCTURE

H2C

CH2

CH2

CH2

H2C

H2C

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Cyclohexane, C6H12

Aldehyde, CH3CHO CH3CH

O

(Condensed structure)

(Condensed structure)

Examples:

A carbon atom can be classified as:

•primary carbon(1o) → bonded to 1 C

•secondary carbon(2o) → bonded to 2 C

•tertiary carbon(3o) → bonded to 3 C

•quarternary carbon(4o) → bonded to 4 C

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CLASSIFICATION OF CARBON ATOMS

A hydrogen atom can be classified as:

•Primary hydrogen(1o) → bonded to 1o C

•Secondary hydrogen(2o) → bonded to 20 C

•Tertiary hydrogen(3o) → bonded to 30 C

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CLASSIFICATION OF HYDROGEN ATOMS

THERE IS NO QUATERNARY(40) HYDROGEN

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H C H

CH3

H

10 carbon

10 carbon

10 hydrogen

Examples:

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H C CH3

CH3

CH3

30 carbon

30 hydrogen

Examples:

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11

1

11

H C C C CH2 C CH3

H

H

H

H H

CH3 CH3

CH3

Identify the 1o carbon:

Exercise 1:

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2

2

H C C C CH2 C CH3

H

H

H

H H

CH3 CH3

CH3

Identify the 2o carbon:

Exercise 2:

16

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H C C C CH2 C CH3

H

H

H

H H

CH3 CH3

CH3

Identify the 3o and 4o carbon:

Exercise 3:

CH3(CH2)CCl(CH3)2

C

H

HH

C CH

H

H

CH3

CH3

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Condensed

StructureExpanded

Structure

Skeletal

Structure

O

Exercise 4: Fill in the blank with the correct structure

Learning Outcomes:

a) Define functional group

b) State functional groups

c) Identify functional groups in a given compound

d) Define homologous series

e) Explain general characteristics of homologous series:

i. represented by a general formula

ii. Same functional group and chemical properties

iii.Gradual change in physical properties with increasing number of carbon atoms

iv. Successive member of a series differs by a –CH2-group

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4.2 FUNCTIONAL GROUPS AND HOMOLOGOUS SERIES

•Is an atom or group of atoms that determine the chemical properties of that compound.

• Organic compounds are classified based on the functional group present in their molecules.

• Molecules with the same functional group are considered to be in the same homologous series

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FUNCTIONAL GROUP

• Functional groups are important for three reasons:

1. A basic by which organic compounds are divided into different classes.

2. A basic for naming organic compounds

3. A particular functional group will

always undergo similar types of chemical reactions.

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•Series of compounds where each member differs from the next member by a constant – CH2 unit

•Members of the same homologous series are called homologs

•Compounds in the same functional group

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HOMOLOGOUS SERIES

1. Obey a general formula:

Examples:• Alkane: CnH2n+2• Alkene: CnH2n• Alcohol: CnH2n+1OH

2. Differ from the successive homolog by a CH2unit

3. Show a gradual change in the physical properties

4. Have same functional group

5. Have similar chemical properties

6. Can be prepared by similar general methods22

HOMOLOGS FEATURES

ALKANE (CnH2n+2)

Methane: CH4

Ethane: CH3CH3

Propane: CH3CH2CH3

Butane: CH3CH2CH2CH323

Examples:

CH2

CH2

CH2

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CH3C CCH3

carbon-carbon

triple bond-C C-Alkyne

CH3CH=CH2

carbon-carbon

double bond-C=C-Alkene

CH3-CH3Alkane

Example

NameStructure

Class of

Compound

Functional Group

CLASSIFICATION OF ORGANIC COMPOUND

Aromatic Benzene ring -CH3

Haloalkane X (F, Cl, Br, I) Halogen CH3Cl

Alcohol -OH Hydroxyl CH3-OH

Phenol -OH Hydroxyl -OH

Ether -C-O-C- Alkoxyl CH3-O-CH3

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Aldehyde

-C=O

H Carbonyl

CH3-C=O

H

Ketone

R-C=O

R Carbonyl

CH3-C=O

CH3

Carboxylic

acid

-C=O

OH Carboxyl

CH3-C=O

OH

Ester

-C-O-C-

O Carboalkoxyl

CH3-C=O

OCH3

Acyl chloride

-C=O

Cl

Acyl halide CH3-C=O

Cl

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Anhydride

O O

-C-O-C-

Anhydride O O

CH3C-O-CCH3

Amide -C=O

N-

Carboxamide CH3-C=O

NH2

Amine -NH2 Amino CH3-NH2

Nitrile -C N Cyano group CH3C N

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1. Identify the functional group in the following molecules

a)(CH3)3CCH2CH=CH2

b)(CH3)3CCH=CHCH2-OH

Exercises:

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CH3 C

CH3

C C CH C NH2

O

NH2 CH

CH2

CH2

C

O

OH

OH

C

O

O CH3

CH2

C

O

O C

O

CH3

c)

Learning Outcomes:

a) Define isomerism, constitutional isomerism and stereoisomerism

b) Identify and construct constitutional isomerism;

i. chain isomers

ii. positional isomers

iii. functional group isomers

c) Describe cis-trans isomerism due to restricted rotation about:

i. C=C bond

ii. C-C bond in cyclic compounds

d) Identify cis-trans isomerism of a given structural formula

e) Define chirality centre and enantiomers

f) Identify chirality centre(s) in a molecule

g) Determine optical activity of a compound

h) Draw a pair of enantiomers using 3-dimensional formula30

4.3 ISOMERISM

Isomerism

Structural/

Constitutional IsomerismStereoisomerism

Chain

Isomerism

Positional

Isomerism

Functional Group

Isomerism

Optical

isomerism

Geometrical /

cis-trans/

diastreomer

isomerism

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Active

(enantiomers)

Inactive

(racemate)

•is the existence of different compounds with the same molecular formula but different structural formulae.

•Different compound with different structural formula that have the same molecular formula are called isomers.

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ISOMERISM

•Are isomers with the same molecular formula but differ in the order of attachment of atoms.

•Also known as structural isomer

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Three types: a) Chain isomersb) Positional isomersc) Functional group isomers

CONSTITUTIONAL ISOMERISM

CH3CH2CH2CH2CH3

The isomers differ in the carbon skeleton (different carbon chain).

They possess the same functional group and belong to the same homologous series.

!Note : parent chain changed

Example: C5H12

CH3CHCH2CH3

CH3

CH3CCH3

CH3

CH3

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CHAIN ISOMERS

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

!Note: Parent chain unchanged.

Examples: C3H7Cl

CH3CH2CH2Cl

1-chloropropane

2-chloropropane

CH3CHCH3

Cl

POSITIONAL ISOMERS

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• C4H8

• C8H10

CH3

CH3

CH3

CH3

1,2-dimethylbenzene

1,3-dimethylbenzene

1-butene 2-butene

CH2=CHCH2CH3 CH3CH=CHCH3

CH3

CH3

1,4-dimethylbenzene

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• C6H13N

CH3

H2N

CH2NH2

CH3

NH2

CH3

NH2

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

Different classes of compounds that exhibit functional group isomerism.

!Note: different functional group

General formula Classes of compounds

CnH2n+2O ; n > 1 alcohol and ether

CnH2nO ; n ≥ 3 aldehyde and ketone

CnH2nO2 ; n ≥ 2 carboxylic acid and ester

CnH2n ; n ≥ 3 alkene and cycloalkane

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FUNCTIONAL GROUP ISOMERS

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Examples:

C2H6O

C3H6O

C3H6O2

CH3CCH3

O

CH3CH2CH

O

CH3COCH3

O

CH3CH2COH

O

ethanol dimethyl ether

propanalpropanone

propanoic acid methyl ethanoate

CH3CH2OH CH3OCH3

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

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Exercise:

Stereoisorism: Isomerism that resulting from different spatial arrangement of atoms in molecules.

Stereoisomers: Are isomers with the same molecular formula but different arrangement of atom in space.

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STEREOISOMERISM

Two subdivisions of stereoisomers:

i) Geometrical/cis-trans/diastreomers isomers

ii) Optical isomers:

a) Enantiomers (optically active)

b) Racemate(optically inactive)

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STEREOISOMERISM

The requirements for geometric isomerism :

i) Restricted rotation about a C=C double

bond in alkenes, or a C-C single bond

in cyclic compounds.

ii)Each carbon atom of a site of restricted

rotation has two different groups

attached to it.

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GEOMETRICAL / CIS-TRANS / DIASTEREOMERS ISOMERS

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Examples: 2-butane

C C

CH3

H

H

H3C

C C

CH3

H

H3C

H

trans-2-butene cis-2-butene

CH3CH=CHCH3

CH3-C=C-CH3

H H

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Examples: 1,3-dimethylcyclohexane

cis-1,3-dimethylcyclohexane trans-1,3-dimethylcyclohexane

CH3

CH3

H

H

CH3

CH3

CH3

CH3

H

H

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• If one of the doubly bonded carbons has 2 identical groups, geometric isomerism is not possible.

• Example

No cis – trans isomer

C C

CH3

H

H3C

H3C

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• Are isomers which are optically active compounds and able to rotate plane-polarized light to the right (dextrorotary) and to the left (levorotary),and exist as mirror images(enantiomers)

• The angle of rotation can be measured with an instrument called polarimeter.

OPTICAL ISOMERISM

3-Dimensional formula ( wedge – dashed wedge – line formula )

•Describes how the atoms of a molecule are arranged in space.

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Example : Bromoethane, CH3Br

Indication :

:bonds that lie in the plane

:bonds that lie behind the plane

:bonds that project out of the plane

C

Br

H

H

H

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C

Br

HH

HC

H

BrH

H

C

H

HBr

H

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Is a molecule which contains chiral carbon and not superimpose with its mirror images.

The requirements for optically active

isomerism/enantiomers :-

i) molecule contains a chiral carbon or chirality centre or stereogenic centre (a sp3-hybridized carbon atom with 4 different groups attached to it)

ii) molecule is not superimposable with its mirror image.

ENANTIOMERS / OPTICAL ISOMERS

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Example: 2-butanol

C *

CH2CH3

H3C

O HH

C

CH2CH3

CH3HO H

CH3CHCH2CH3

OH

enantiomers

CH3 C C CH3

H H

OH H

*

*

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Learning Outcomes:

a) Explain covalent bond cleavage:

i. homolytic cleavage

ii. heterolytic cleavage

b) Differentiate homolytic cleavage and heterolytic cleavage

c) State the relative stabilities of primary, secondary and tertiary free radicals, carbocations and carbanions.

d) Compare the stabilities of carbocations and carbanions by using the inductive effect of alkyl group.

e) Define

i. electrophile and nucleophile

ii. Lewis acid and Lewis base

f) State the main types of organic reactions

i. addition (nucleophilic and electrophilic)

ii. Substitution (free radical, nucleophilic, and electrophilic)

iii. Elimination

iv. Rearrangement

g) Identify the main type of organic reaction given a reaction equation.

4.4 REACTIONS OF ORGANIC COMPOUNDS

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Types of Covalent Bond Cleavage/Fission

All chemical reactions involved bond breaking and bond making.

Two types of covalent bond cleavage :-

Homolytic cleavage

Heterolytic cleavage

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• Occurs in a non-polar bond involving two atoms of similar electronegativity.

• A single bond breaks symmetrically into two equal parts, leaving each atom with one unpaired electron.

• Formed free radicals.

HOMOLYTIC CLEAVAGE

X : X X + X

free radicals

uv

Example:

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• Occurs in a polar bond involving unequal sharing of electron pair between two atoms of different electronegativities.

• A single bond breaks unsymmetrically.

• Both the bonding electrons are transferred to the more electronegative atom.

• Formed cation and anion.

HETEROLYTIC CLEAVAGE

H Br H+ + :Br-

Example:

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Reaction Intermediates

a) Carbocation

b) Carbanion

c) Free Radical formed from homolytic cleavage

They are unstable and highly reactive.

formed from heterolytic cleavage

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•Also called carbonium ion.

•A very reactive species with a positive charge on a carbon atom.

•Carbocation is formed in heterolytic cleavage.

•Is an e- deficient/poor.

•Usually left with 6 valence e-

CARBOCATION

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Example :

•(CH3)3C — Cl (CH3)3C+ + :Cl-

carbocation anion

•Chlorine is more electronegative than carbon and the C—Cl bond is polar.

•The C—Cl bond breaks heterolitically and both the bonding electrons are transferred to chlorine atom to form anion and carbocation.

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Is an anion counterpart

A species with a negative charge on a carbon atom.

Carbanion is formed in heterolytic cleavage.

Is an e- rich.

Usually left with 8 valence e-

Example:

(CH3)3C — Li (CH3)3C- + Li+

- + carbanion cation

CARBANION

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A very reactive species with an unpaired/single electron.

Formed from homolytic cleavage.

Is an e- deficient/poor.

Examples:

Cl – Cl Cl● + Cl ●

free radicals

FREE RADICAL

uv

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C C C ●+ C●

H3C H H3C● H●+

ii)

iii)

Examples:

uv

uv

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HOMOLYTIC CLEAVAGE HETEROLYTIC CLEAVAGE

Occur to a non-polar bond involving two atoms of similar electronegativity.

Occur to a polar bond involving unequal sharing of electron pair between two atoms of different electronegativity.

Bond breaks symmetricallyinto two equal parts leaving each atom with one unpaired/single electron.

Bond breaks unsymmetrically and both electron are transferred to the more electronegative atom.

Forms free radicals:

Cl-Cl Cl. + Cl.

Forms cation and anion:

H3C-Cl H3C+ + :Cl-

DIFFERENCES BETWEEN HOMOLYTIC AND HETEROLYTIC CLEAVAGE

uv

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•Carbocation, carbanion and free radical can be classified into:

Primary

Secondary

Tertiary

RELATIVE STABILITIES OF CARBOCATIONS,CARBANIONS AND

FREE RADICALS

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•The alkyl groups,R (e- releasing group @ ERG / e- donating group @ EDG) stabilisethe positive charge on the carbocation.

•As number of alkyl group present increases, the stability of carbocation increases.

CARBOCATION STABILITY

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H

C HH+

R

C RH+

R

C RR+

H

C RH+

< < <

Methylcation

Primary10

Secondary20

Tertiary30

Increasing stability

CARBOCATION STABILITY

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Alkyl group and other e- donating groups (EDG) destabilises carbanions.

As number of alkyl group present increases, the stability of carbanion decreases.

Electron withdrawing group / EWG (e.g: halogen) stabilises carbanions through the inductive withdrawal of electron density

CARBANION STABILITY

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H

C HH-

R

C RH-

R

C RR-

H

C RH-> > >

Methylanion

Primary10

Secondary20

Tertiary30

Decreasing stability

CARBANION STABILITY

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H R

C RH . C RH .

H

C HH . < < <

methylradical

Primary10

Secondary20

R

C RR .

Tertiary30

Increasing stability

As number of alkyl group present increases, the stability of free radical increases.

Radical carbon is an e- deficient/poor which stabilised by alkyl group (an e- donating group/EDG)

FREE RADICAL STABILITY

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Means ‘electron loving’.

An electron-deficient species and accepting electron from an attacking nucleophile.

Can be either neutral or positively charged

ELECTROPHILE

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Examples of electrophiles :

•cations such as H+, H3O+, NO2

+ etc.

•carbocations

•Alcohols (R-OH)

•Amines (R-NH2)

•Oxidizing agents (reduction, gain e-) such as Cl2, Br2 and etc.

•CO2•Lewis acids: a substance that can accept a pair of e- to form a coordinate covalent bond. Example: AlCl3, FeCl3, BF3etc.

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Electrophilic sites are molecules with low electron density around a polar bond.

Examples:

-C = O (carbonyl) ; -C – X (haloalkanes)

+ - + -

-C – OH (hydroxy compounds)

+ -

i) ii)

iii)

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means ‘nucleus loving’

An electron-rich species and donate electron to electrophile.

A nucleophile can be either neutral or negatively charged.

NUCLEOPHILE

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Examples of nucleophiles :

•anions such as OH-, RO-, Cl-, CN- etc.

•carbanions (species with a negative charge on carbon atoms ).

•Alkenes (–C=C-)

•Alcohols (R-OH)

•Amines (R-NH2)

•Lewis Base: a substance that can donate a pair of e- to form a coordinate covalent bond. Example: NH3 & H2O

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Addition

Substitution

Elimination

Rearrangement

4 TYPES OF REACTIONS

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A reaction in which atoms or groups add to adjacent atoms of a multiple bond.

Two types of addition :-a) Electrophilic Additionb) Nucleophilic Addition

ADDITION REACTION

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ELECTROPHILIC ADDTION

Initiated by an electrophile accepting electron from an attacking nucleophile.

Typical reaction of unsaturatedcompounds such as alkenes and alkynes.

Example :

CH3CH=CH2 + Br2 CH3CH-CH2‘Nu-’ ‘El+’ Br Br

Room temperature

CCl4

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NUCLEOPHILIC ADDITION

Initiated by a nucleophile, which attacks an electrophilic site of a molecule.

Typical reaction of carbonyl compounds.

CCH3 CH3

O

+ HCN CH3 C CH3

O-H

CN‘El+’ ‘Nu-’

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A reaction in which an atom or group in a molecule is replacedby another atom or group.

Three types of substitution :-a) free radical substitutionb) electrophilic substitutionc) nucleophilic substitution

SUBSTITUTION REACTION

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Substitution which involves free

radicals as intermediate species.

Typical reaction of ALKANES.

Example :

CH3CH3 + Cl2 CH3CH2Cl + HCl

FREE RADICAL SUBSTITUTION

uv

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The aromatic nucleus has high electron density, thus it is nucleophilic and is prone to electrophilic attack.

Typical reaction of aromatic compounds(benzene).

+ Br2 Br + HBrFe

catalyst

Example:

ELECTROPHILIC SUBSTITUTION

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Typical reaction of saturated organic compounds bearing polar bond as functional group, such as haloalkane and alcohol.

Example :

CH3CH2Br + OH- CH3CH2OH + Br

-

Δ

NUCLEOPHILIC SUBSTITUTION

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An atoms or groups are removed from adjacent carbon atoms of a molecule to form a multiple bond (double or triple bond).

Results in the formation of unsaturated molecules.

Example :

CH3CH2OH CH2=CH2 + H2O

ELIMINATION REACTION

conc. H2SO4

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

Occurs when a single reactant reorganizes the bonds and atoms.

Example :

C CC C

OHH

H R R

H

H

H

O

REARRANGEMENT REACTION

Exercises

1. Explain how the free radicals are formed in homolytic cleavage.

2. Write an equation for the bromine-bromine bond cleavage in the bromination of methane. State the type of bond cleavage.

3. Which would you expect to be the most stable free radical?

CH2CH3 , (CH3)2 CH , CH3 ,

CH3

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