Topic 11_Introduction to Organic Chemistry

7/29/2019 Topic 11_Introduction to Organic Chemistry

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Introduction To Organic

Chemistry


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Lecture 1

12.1 IntroductionLearning Outcomes:

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

1. List the elements that made up organic compounds C, H,O, N, P, S and halogens.

2. State the ability of carbon to form 4 covalent bonds withother carbons or elements.

3. Differentiate between saturated and unsaturated organic

compounds.4. Give examples of organic compounds used in medicine,

engineering, biotechnology and agriculture.


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WHAT IS ORGANIC CHEMISTRY?Organic chemistry is the chemistry of carboncompounds.

Organic compounds contain H as well as C, while

other common elements are O, N, the halogens, S andP.

There are many varieties of organic compounds

( more than 10 millions!!!)

They may exist as simple or complex molecules; asgases, liquids or solid and coloured or colourless.


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Examples:-

CH4

methane (a component of natural gas)

methyl salicylic acid (aspirin-a drug)

OCOCH3

COOH


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CH2 C

O

NH

O

S

N

COOH

penicillin (an antibiotic)

Cl CH Cl

CCl3

dichlorodiphenyltrichloroetane

(DDT- a pesticide component)


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All organic compounds consist ofcarbon atom.

Properties of carbon atom:

-has 4 valence electrons.

-can form 4 covalent bonds.

C C

Single bond

C C C C

Double bond Triple bond


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Hydrocarbons

saturated unsaturated

Contains only singlebonds ( -C-C- )

Examples: alkanes,

cycloalkanes

Contains at least onecarbon-carbon double

bond (-C=C-) or triplebond (-C C-).

Examples: alkenes,alkynes.


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Uses of organic compounds

Medicine Antibiotics are used to

fight bacterial and fungal

infections

Engineering Gasoline-as a fuel forinternal combustion

engines.

Biotechnology Genetic information like

DNA

Agriculture DDT-as insectisides to kill

harmful insects.


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Lecture 2:

12.2 Molecular and Structural Formulae

Learning Outcomes:

At the end of the lesson the students should

be able to :Define structural formula.

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

molecular formula.

Explain primary (1), secondary (2), tertiary (3)and quaternary (4) carbon.


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Structural formulashows how the

atoms in a molecule are bonded to

each other.

3 types of structural formula: condensed structure

expanded structure

skeletal structure


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2- Dimensional formula

Condensed StructureDoes not show single bonds between carbonand hydrogen atoms, but double and triple bondsare shown.

All atoms that are attached to a carbon arewritten immediately after that carbon.

C4H9Cl CH3CHCH2CH3

(Condensed structure)


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H2C

CH2

CH2

CH2

H2C

H2C

Examples:

ii) Cyclohexane, C6H12 iii) Aldehyde, CH3CHO

CH3CH

O


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Expanded Structure

Expanded structures indicate how atoms are

attached to each other but are not representations

of the actual shapes of the molecules.

C4H9ClMolecular

Formula

Expanded structure

C C C CH H

H H HCl

H H H H


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Examples:

i) Alcohol (C2H6O)

ii) Carboxylic acid (C3H6O2 )

C

H

H

H

C

H

H

OH

C

H

H

H

C

H

H

C

O

OH


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Skeletal Struc tu re

Shows only the carbon skeleton.Hydrogen atoms are not written.

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

i) CH3CH(Cl)CH2CH3 =

Cl

ii) =H2C CH2

H2C CH2


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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

or or

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

C

Br

HH

H

C

H

BrH

H

C

H

HBr

H
http://localhost/var/www/apps/conversion/tmp/4)%20intro.%20to%20org(shahnon)/link/3d.C3D


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A carbon atom can be classified as

primary carbon(1o)bonded to 1 C

secondary carbon(2o) bonded to 2 C

ter tiary carbon(3o) bonded to 3 C

quarternary carbon(4o) bonded to 4 C

Classification of C atoms:


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H C H

CH3

H

10 carbon10 carbon


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H C CH3

CH3

CH3

30 carbon


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11

1

11H C C C CH2 C CH3

H

H

H

H H

CH3 CH3

CH3


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2

2H C C C CH2 C CH3

H

H

H

H H

CH3 CH3

CH3


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43H C C C CH2 C CH3

H

H

H

H H

CH3 CH3

CH3


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CH3(CH2)CCl(CH3)2

Question

CondensedStructure

ExpandedStructure

SkeletalStructure

O

C

H

HH

C CH

H

H

CH3

CH3


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12.3 FUNCTIONAL GROUPS

AND HOMOLOGOUS SERIES


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Lecture 3

Functional Group and Homologous Series

Learning Outcomes:

At the end of the lesson the students should be ableto :

Define functional group.

Name functional groups and classify organiccompounds according to their functional groups.

Define homologous series and explain generalcharacteristics of its members.


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Functional group

is an atom or group of atoms in an organic

molecule which characterised the molecule

and enables it to react in specific wayswhich determines its chemical properties.


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Functional groups are important for three

reasons:

i. A basic by which organic compounds are

divided into different classes.

ii. A basic for naming organic compounds

iii. A particular functional group will always

undergo similar types of chemicalreactions.


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Homologous Series

is series of compounds where each member

differs from the next member by a constant

CH2 unitMembers of the same homologous series

are called homologs.


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Homologs Features

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


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4. Have same functional group

5. Have similar chemical properties

6. Can be prepared by similar general methods


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CH3C CCH3

carbon-carbontriple bond-C C-Alkyne

CH3CH=CH2

carbon-carbon

double bond-C=C-Alkene

CH3-CH3Alkane

Example

NameStructure

Class ofCompound

Functional Group

Classification of organic compound


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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- Alkoxide CH3-O-CH3


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Aldehyde

-C=O

H Carbonyl

CH3-C=O

H

KetoneR-C=O

R CarbonylCH3-C=O

CH3

Carboxylic

acid

-C=O

OH Carboxyl

CH3-C=O

OH

Ester

-C-O-C-

O Carboalkoxy

CH3-C=O

OCH3

Acyl chloride

-C=O

Cl

CH3-C=O

Cl


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Anhydride

O O

-C-O-C-

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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Exercises:

1. Identify the functional group in the following

molecules

a)(CH3)3CCH

2CH=CH

2

b)(CH3)3CCH=CHCH2-OH


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CH3 C

CH3

C C CH C NH2

O

NH2 CH

CH2

CH2

CO

OH

OH

C

O

O CH3

CH2

C

O

O C

O

CH3c)


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12.4 IsomerismLearning Outcomes:

At the end of the lesson the students should beable to :

Define isomerism.

Explain constitutional isomerism.

chain isomers

positional isomersfunctional group isomer


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Isomerism

Structural/

Constitutional IsomerismStereoisomerism

ChainIsomerism PositionalIsomerism

Functional Group

Isomerism

diastreomer enantiomer

cis- trans

isomerism

other

diastereomers


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I somerism

is the existence of different compounds with

the same molecular formula but differentstructural formulae.

Different structural formula that have the samemolecular formula are called isomers.


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1) Consti tutional isomers (Structural isomers)

are isomers with the same molecular formula

but differ in the order of attachment of atoms.2) Stereoisomers

are isomers with the same molecular formula

but different arrangement of atoms in space


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Consti tutional isomerism

Isomerism resulting from different order of

attachment of atoms.

Three types

a) Chain/skeletal isomer ism

b) Positional isomerism

c) Functional group isomerism


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CH3CH2CH2CH2CH3

C5H12

a) Chain/skeletal isomerism

The isomers differ in the carbon skeleton

(different carbon chain).

They possess the same functional group andbelong to the same homologous series.

Example:

CH3CHCH2CH3

CH3

CH3CCH3

CH3

CH3


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2)Positional isomerism

These isomers have a substituent group/ functional

group in different positions.Examples

C3H7Cl

CH3CH2CH2Cl1-chloropropane

2-chloropropane

CH3CHCH3

Cl


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C4H8

C8H10

CH3

CH

CH3

CH3

1,2-dimethylbenzene 1,3-dimethylbenzene

1-butene 2-butene

CH2=CHCH2CH3 CH3CH=CHCH3

CH3

C H 3

1,4-dimethylbenzene


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C6H13N

CH3

H N

CH2NH2

CH3

NH2

CH3

NH


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3)Functional group isomerism

These isomers have different functional groups and

belong to different homologous series with the samegeneral formula.

Different classes of compounds that exhibit

functional group isomerism :-

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

CnH

2n; n 3 alkene and cycloalkane

E l


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Examples

C2H6O

C3H6O

C3H6O2

CH3CCH3 CH3CH2CH

CH3COCH3CH3

CH2

COH

O

ethanoldimethyl ether

propanalpropanone

propanoic acid methyl ethanoate

CH3CH2OH CH3OCH3


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Exercise:

1. State how many are isomers with the following molecularformulae, 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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Lecture 5

12.4 IsomerismLearning Outcomes:

At the end of the lesson the students shouldbe able to :

Define stereoisomerism.

Describe cis-trans isomerism due torestricted rotation about C=C bond and CC

bond in cyclic compounds

Identify cis-trans isomerism of a givenstructural formula.


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Stereoisomerism / optical isomerism :

Isomerism that resulting from differentspatial arrangement of atoms

in molecules.

Two subdivisions of stereoisomers:i) Enantiomers (mirror image)

ii) Diastereomers (non-mirror image)


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Diastereomer

Cis-Trans Isomerism

The requirements for geometric isomerism

:i) restricted rotation about a C=C,double

bond in alkenes, or a C-C single bondin cyclic compounds.

ii) each carbon atom of a site of restrictedrotation has two different groupsattached to it.

E l


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Examples

C C

CH3H

H3CC C

CH3H3C

H

H

CH

H

CH

trans-2-butene cis-2-butene

cis-1,2-dimethylcyclohexane

trans-1,2-dimethylcyclohexane

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 cistrans isomer

C C

CH3H3C

H3C


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Lecture 6

12.4 IsomerismLearning Outcomes:At the end of the lesson the students should be able to:

Identify cis-trans isomerism of a given structural

formula. Define chirality centre and enantiomers.

Identify chirality centre in a molecule.

Explain optical activity of a compound.

Draw a pair of enantiomers using 3-dimensionalformula.

Define racemate.

State the applications of chiral compounds in daily

life.


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Optical I somerism Optically active compounds have the ability to

rotate plane-polarized light to the right

(dextrorotary) and to the left (levorotary)

The angle of rotation can be measured with an

instrument called polarimeter.

Enantiomer


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Polarimeter


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The requirements for optical isomerism :-

i) molecule contains a chiral carbon or chirality centre

or stereogenic centre (a sp3

-hybridized carbon atomwith 4 different groups attached to it)

ii) molecule is not superimposable with its mirror

image.

P

CQ

R

S*

PQRS*designates chiral centre

Enantiomers


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Enantiomers

a pair of mirror-image that are not superimposable.

Example:-

i) 2-butanol ,

C*

CH2CH3

H3C

OHH

C

CH2CH3

CH3HOH

CH3CHCH2CH3

enantiomers

ii) 2 hydroxypropanoic acid


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ii) 2-hydroxypropanoic acid,

enantiomers

COOH

HO H

CH

COOH

H OH

CH3
http://localhost/var/www/apps/conversion/tmp/scratch_6/link/imej%20cer2.C3D


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12.4.9 Racemate

A racemic mixtureorracemate isan

equimolar mixture of enantiomers which is

optically inactive because the two componentsrotate plane-polarized light equally (same

degree of rotation) but in opposite directions.

Hence it does not give a net rotation of plane-

polarized light.


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Applications of chiral compounds in

daily life.e.g:

() Dopa is used for treatment of Parkinsons

disease but (+) dopa is toxic to human. (S)-Ibuprofen the popular analgesic(the active

ingredient in motrin, advil, and many other

nonaspirin analgesics)


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REACTIONS OF ORGANIC

COMPOUNDS


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Lecture 7

12.5 Reactions of Organic Compounds

Learning Outcomes:

At the end of the lesson the students should

be able to :Explain covalent bond cleavage:

homolytic

heterolytic


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Types of Covalent Bond Cleavage/F ission

All chemical reactions involved bond

breaking and bond making.

Two types of covalent bond cleavage :-

Homolytic cleavage

Heterolytic cleavage


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a) Homolytic Cleavage

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.

Example:


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X : X

Example:

X + X

free radicals

X X

b) Heterolytic cleavage


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b) Heterolytic cleavage

Occurs in a polar bond involving unequal sharing of

electron pair between two atoms of differentelectronegativities.

A single bond breaks unsymmetrically.

Both the bonding electrons are transferred to the moreelectronegative atom.

Formed cation and anion.


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A is moreelectronegative.

B is more

electronegative.

A:- + B+anion cation

A : B

A+ + B:-cation anion


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Reaction I ntermediates

a) Carbocation

b) Carbanion

c) F ree Radical

They are unstable and highly reactive.


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a) Carbocation

Also called carbonium ion.

A very reactive species with a positive

charge on a carbon atom.Carbocation is formed in heterolyticcleavage.


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Example :

(CH3)3CCl (CH3)3C

+ + Cl-

carbocation anion

Chlorine is more electronegative than carbon and the

CCl bond is polar.

The CCl bond breaks heterolitically and both the

bonding electrons are transferred to chlorine atom to

form anion and carbocation.


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b) Carbanion

is an anion counterpart

a species with a negative charge on a carbonatom.

Carbanion is formed in heterolytic cleavage.


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example:

(CH3)3CLi (CH3)3C- +Li+ carbanion

kation


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b) F ree Radical

A very reactive species with an unpaired electron.

Formed in homolytic cleavage.

Examples:

i)

free radicals

Cl

Cl uv Cl

+ Cl


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C C C + C

H3C H H3CH+

ii)

iii)


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Lecture 7

12.5 Reactions of Organic Compounds

Learning Outcomes:

At the end of the lesson the students should beable to:

State the relative stabilities of primary, secondaryand tertiary free radicals, carbocations andcarbanions.

Explain the inductive effect of alkyl grouptowards the stability of carbocations andcarbanions.

Define electrophile and nucleophile.


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Relative Stabil i ties of Carbocations,

Carbanions and F ree Radicals

Carbocation, carbanion and free radical

can be classified into:

Primary

Secondary

Tertiary

Carbocat ion Stabi l i ty


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The alkyl groups (electron-releasing group)stabilise the positive charge on the

carbocation.

The stability of carbocation increases with the

number of alkyl groups present.

Carbocat ion Stabi l i ty


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Carbocation Stabil i ty:

H

C HH+

R

C RH+

R

C RR+

H

C RH+

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Topic 11_Introduction to Organic Chemistry