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## intro organicPresentation Transcript

• CHAPTER 11 : INTRODUCTION TO ORGANIC CHEMISTRY
• CHAPTER 11 : INTRODUCTION TO ORGANIC CHEMISTRY 11.1 Introduction 11.2 Empirical molecular and structural formulas 11.3 Functional groups and homologous series 11.4 Classification of carbon atoms in organic molecules 11.5 Isomerism 11.6 Reactions in organic compound
• Organic and Inorganic Compound Organic compound Inorganic compound
• were defined as
• compounds that
• could be obtained
• from living
• organisms
• were those that
• came from
• nonliving sources
• Some examples of carbon compounds in our daily lives :- methane (a component of natural gas) Methyl salicylic acid (aspirin-a drug) alanine (amino acid-a protein component) cocaine (a pain killer)
• 11.2 EMPIRICAL, MOLECULAR AND
• STRUCTURAL FORMULAE
• Empirical formula is the simplest formula that
• shows the relative numbers of the different kinds
• of atoms in a molecule.
• Molecular formula is a formula that states the
• actual number of each kind of atom found in the
• molecule. Example : C 2 H 4 , C 2 H 4 O 2
• Quantitative example :- A sample of hydrocarbon contains 85.7 % carbon and 14.3 % hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound.
• Solution :- = 7.142 = 14.3  Empirical formula = CH 2 n (empirical molar mass)= molar mass n ( 12 x 1 + 1 x 2 ) = 56 n = 4 Molecular formula = C 4 H 8 14.43 7.142 = 2 7.142 7.142 = 1 Smallest ratio 14.3 1.0 85.7 12.0 Moles (n) 14.3 85.7 Mass (g) H C Element
• Practice Exercise:
• 1) A complete combustion of 10.0 g of compound X,
• C x H y O z forms 12.0 g H 2 O and 22.0 g CO 2 . Its
• molar mass is 60.
• a) Determine the percentage composition of
• C, H and O.
• b) Determine the molecular formula of
• compound X.
• A complete combustion of 0.6 g of hydrocarbon,
• C x H 12 forms 1.98 g CO 2 . Determine the percentage
• of C and H in the hydrocarbon and write the
• molecular formula.
• Structural formula shows the order in which
• atoms are bonded together
Representation of structural formula :- a) Condensed Structure b) Expanded Structure c) Skeletal Structure d) 3-Dimensional formula e) Ficher Projection
• a) Condensed Structure
• In condensed formulae all the hydrogen atoms
• that are attached to a particular carbon are
• usually written immediately after that carbon
• Example : C 4 H 9 Cl CH 3 CHCH 2 CH 3 or CH 3 CH(Cl)CH 2 CH 3 Condensed structure Cl
• b) Expanded Structure
• Expanded structures indicate the way in which
• the atoms are attached to each other and are not
• representations of the actual shapes of the
• molecules.
Example : C 4 H 9 Cl Expanded structure
• c) Skeletal Structure
• This structure shows only the carbon skeleton
• The hydrogen atoms that are assumed to be
• present, are not written.
Other atoms such as O, Cl, N and etc. are shown Example : CH 3 CH(Cl)CH 2 CH 3 Cl = 1.
• 2. = CH 2 =CHCH 2 OH 3. = OH
• Practice Exercise : Rewrite each of the following structures using skeletal formula :- (CH 3 ) 2 CHCH 2 CH 2 CH(CH 3 )CH 2 CH 3 CH 2 = CHCH 2 CH 2 CH = CHCH 3 1. 2. 3. 4.
• d) 3 - Dimensional formula (wedge - dashed wedge - wedge) Describes how the atoms of a molecule are arranged in space Example : (Bromoethane) Indication :- bonds that lie in the plane of the page bonds that lie behind the plane bonds that project out of the plane of the paper OR OR
• e) Fischer Projection
• Vertical lines represent bonds that project behind
• the plane of paper
• Horizontal lines represent bonds that project out of
• the plane of paper
• The intersection of vertical and horizontal lines
• represent a carbon atom, that is stereocentre
• Example : 2 – butanol , CH 3 CH(OH)CH 2 CH 3 OR
• 11.3 FUNCTIONAL GROUPS AND HOMOLOGOUS SERIES
• A functional group is an atom or group of
• atoms in an organic molecule which characterized
• the molecule and enables the molecule to react
• in specific ways (determines its chemical properties)
• Some important functional groups in organic compounds :- CH  CH ethyne -yne C n H 2n-2 C  C (triple bond) alkynes CH 2 =CH 2 ethene -ene C n H 2n C = C (double bond) alkene CH 4 methane -ane C n H 2n+2 none alkane Example IUPAC nomenclature Prefix- -suffix General Formula Functional Group Homologous Series
• aromatic ring methylbenzene CH 3 CH 2 Cl chloroethane haloalkane C n H 2n+1 X – X (halogen) haloalkane CH 3 OCH 3 methoxymethane alkoxyalkane C n H 2n+2 O – OR (alkoxy) ether CH 3 CH 2 OH ethanol alkanol C n H 2n+1 OH – OH (hydroxyl) alcohol -benzene C n H 2n-6 arene Example IUPAC nomenclature Prefix- -suffix General Formula Functional Group Homologous Series
• carbonyl H ethanal carbonyl CH 3 propanone C OH O carboxyl OH ethanoic acid CH 3 C=O alkanoic acid CnH 2n O 2 carboxylic acid CH 3 C=O alkanone C n H 2n O ketone CH 3 C=O alkanal C n H 2n O aldehyde Example IUPAC nomenclature Prefix- -suffix General Formula Functional Group Homologous Series
• acyl Cl ethanoyl chloride ester ethyl ethanoate amide ethanamide amino methanamine CH 3 NH 2 -amine C n H 2n+1 NH 2 -NH 2 amine CH 3 CONH 2 -amide C n H 2n+1 CONH 2 amide CH 3 COOCH 3 alkyl alkanoate C n H 2n O 2 ester CH 3 C=O alkanoyl chloride C n H 2n+1 COCl acyl chloride Example IUPAC nomenclature Prefix- -suffix General Formula Functional Group Homologous Series
• 11.4 CLASSIFICATION OF CARBON AND HYDROGEN ATOMS IN ORGANIC MOLECULES
• Carbon atom classified
primary (1 o ) secondary (2 o ) tertiary (3 o ) quarternary (4 o ) depending on the number of carbon atoms bonded to it
• A primary carbon – directly bonded to one other
• carbon atom
• (has 1 adjacent carbon atom)
Example : 1 o carbon 1 o H
• A secondary carbon – directly bonded to two other
• carbon atoms
• (has 2 adjacent carbon atoms)
Example : 2 o carbon 2 o H
• A tertiary carbon – directly bonded to three other
• carbon atoms
• (has 3 adjacent carbon atoms)
Example : 3 o carbon 3 o H
• A quarternary carbon – directly bonded to four other
• carbon atoms
• (has 4 adjacent carbon atoms)
Example : 4 o carbon
• Similarly, a hydrogen atom is also classified as
• primary, secondary or tertiary depending on the
• type of carbon to which it is bonded.
1° hydrogen atom bonded to a 1° C atom 2° hydrogen atom bonded to a 2° C atom 3° hydrogen atom bonded to a 3° C atom
• Classification of haloalkanes (alkyl halides)
• Alkyl halides are classified based on the carbon atom
• to which the halogen is directly attached.
1 ° alkyl halide – the halogen atom is bonded to a primary carbon atom 2 ° alkyl halide – the halogen atom is bonded to a secondary carbon atom 3 ° alkyl halide – the halogen atom is bonded to a tertiary carbon atom
• 1 ° alkyl chloride 1 ° C 2 ° alkyl chloride 2 ° C 3 ° alkyl chloride 3 ° C
• Classification of alcohols
• Alcohols are classified based on the carbon atom
• to which the hydroxyl group is directly attached.
1 ° alcohol – the hydroxyl group is attached to a 1 ° carbon atom 2 ° alcohol – the hydroxyl group is attached to a 2 ° carbon atom 3 ° alcohol – the hydroxyl group is attached to a 3 ° carbon atom
• 1 ° alcohol 1 ° C 2 ° C 2 ° alcohol 3 ° alcohol 3 ° C
• Classification of amines
• Amines are classified based on the number of alkyl
• groups or carbon atoms that are directly attached
• to the nitrogen atom
1 ° amine – N is bonded to one alkyl group 2 ° amine – N is bonded to two alkyl groups 3 ° amine – N is bonded to three alkyl groups
• N bonded to one alkyl group A primary (1°) amine N bonded to two alkyl group A secondary (2°) amine N bonded to three alkyl group A tertiary (3°) amine
• ISOMERISM Structural/Constitutional Isomerism Stereoisomerism Isomerism Chain isomerism Positional isomerism Functional group isomerism Geometric isomerism Optical isomerism
• Isomerism is the existence of different compounds
• with the same molecular formula but different
• structural formulae
• Isomers – different compounds that have same
• molecular formula
Two types of isomerism structural isomerism stereoisomerism different order of attachment of atoms different spatial arrangement of atoms in molecules
• Structural isomerism Chain/skeletal isomerism
• Structural isomers are different compounds with
• the same molecular formula but differ in the order
• of attachment of atoms
Positional isomerism Functional group isomerism
• The isomers differ in the carbon skeleton
• ( different carbon chain )
a) Chain/skeletal isomerism
• They possess the same functional group and
• belong to the same homologous series
Example : C 5 H 12 : CH 3 CH 2 CH 2 CH 2 CH 3 CH 3 CHCH 2 CH 3 CH 3 CH 3 -C-CH 3 CH 3 CH 3
• b) Positional isomerism
• These isomers have a substituent group in
• different positions in the same carbon skeleton
Example : C 3 H 7 Cl i) CH 3 CH 2 CH 2 Cl 1-chloropropane CH 3 CHCH 3 Cl 2-chloropropane C 4 H 8 ii) CH 2 =CHCH 2 CH 3 CH 3 CH=CHCH 3 1-butene 2-butene
• 1,2-dimethylbenzene iii) C 8 H 10 CH 3 CH 3 CH 3 CH 3 1,3-dimethylbenzene CH 3 CH 3 1,4-dimethylbenzene
• c) Functional group isomerism
• 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 :-
carboxylic acid and ester C n H 2n O 2 alkene and cycloalkane C n H 2n aldehyde and ketone C n H 2n O alcohol and ether C n H 2n+2 O Classes of compounds General formula
• Example : i) C 2 H 6 O CH 3 CH 2 OH ethanol CH 3 OCH 3 dimethyl ether ii) C 3 H 6 O CH 3 CH 2 C-H O propanal CH 3 C-CH 3 O propanone iii) C 3 H 6 O 2 CH 3 CH 2 C-OH O propanoic acid CH 3 C-O-CH 3 O methyl ethanoate
• Stereoisomerism Geometric Isomerism Optical Isomerism
• Geometric isomerism
• occurs only in two classes of compounds :
Alkenes & cyclic compound (because of rigidity in molecules)
• Geometric isomers (also called cis-trans isomers)
• are stereoisomers that differ by groups being
• on the same side (cis-isomer) or opposite
• sides (trans-isomer) of a site of rigidity in a molecule
• The requirements for geometric isomerism :
• 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
• Examples : H 3 C CH 3 H H C C = i) cis -2-butene H 3 C C = C CH 3 H H trans -2-butene ii) H 3 C CH 2 CH 3 C = C H CH 3 trans -3-methyl-2-pentene H 3 C CH 3 C = C H CH 2 CH 3 cis -3-methyl-2-pentene
• iii) H H CH 3 CH 3 cis -1,2-dimethylcyclohexane H H CH 3 CH 3 trans -1,2-dimethylcyclohexane Cl Cl H H iv) Cl Cl H H cis -1,3-dichlorocyclopentane trans -1,3-dichlorocyclopentane
• If one of the doubly bonded carbons has 2 identical
• groups, geometric isomerism is not possible.
Examples : CH 3 CH 2 i) C = C H H H 3 C 2-methyl-2-butene H ii) C = C CH 3 CH 3 Cl 1-chloro-2-methylpropene
• cis-trans isomers have similar chemical properties
• but different physical properties
• They differ in melting and boiling points and
• solubility due to different polarity of the molecules
cis -isomers polar molecules trans -isomers non-polar
• Melting point: trans - isomer > cis -isomer
• Boiling point: cis -isomer > trans - isomer
• Stability: trans -isomer > cis -isomer
• b) Optical isomerism
• If a beam of light is passed through a piece of
• polarizer prism, the emergent light vibrates in a
• single plane, hence it is called a plane-polarized
• light
• Optically active compounds have the ability to
• rotate plane-polarized light
• The angle of rotation can be measured with an
• instrument called polarimeter
• Schematic representation of a polarimeter containing an optically active sample : clockwise rotation – plus sign (+) / dextrorotarory anticlockwise rotation – minus sign (-) / levorotorary
• The requirements for optical isomerism :-
• molecule contains a chiral carbon or chiral centre
• (carbon atom with 4 different groups attached to it)
ii) molecule is not superimposable with its mirror image
• A representation of a chiral molecule with
• 3-dimensional formula :-
P  Q  R  S *designates chiral centre
• Enantiomers are a pair of mirror-image molecules
• that are not superimposable (must have one or more
• chiral carbons)
Examples : i) 2-butanol, CH 3 CHCH 2 CH 3 OH enantiomers :-
• ii) 2-hydroxypropanoic acid, CH 3 CHCOOH OH :- COOH COOH OH HO H H CH 3 CH 3 enantiomers
• A racemic mixture or racemate is an equimolar
• mixture of enantiomers which is optically inactive
• because the two components rotate plane-polarized
• light equally (same degree of rotation but in opposite
• direction– so they can cancel each other’s rotation)
• A pair of enantiomers have identical chemical and
• physical properties but differ in the direction of
• rotation of plane-polarized light
• A compound with n chiral centers can have a
• maximum of 2 n stereoisomers
• If a molecule contains two or more chiral centers,
• diastereomers may exist
• Diastereomers are stereoisomers that are not
• mirror images of each other
• All physical properties of diastereomers are usually
• different from one another
• Example : The 4 stereoisomers of 2-amino-3-hydroxybutanoic acid CH3CH-CHCOOH are shown below using OH NH 2 Fischer projection formula:- COOH COOH CH 3 CH 3 H H H H NH 2 H 2 N OH HO enantiomers COOH NH 2 H H HO CH 3 COOH H H OH H 2 N enantiomers CH 3 A B C D Four pairs of diastereomers are identified : A and C ; A and D ; B and C ; B and D
• Meso compound is a stereoisomer that has more
• than one chiral centres and that is superimposable
• on its mirror image because of the presence of an
• internal plane of symmetry , hence it is optically
• inactive (does not cause a rotation of plane-polarized
• light)
• Example : Tartaric acid , HOOCCH(OH)CH(OH)COOH COOH OH OH H H COOH COOH COOH HO HO H H COOH COOH H H OH OH plane of symmetry plane of symmetry rotate 180 o P Q identical
• At first glance, P and Q are assumed to be enantiomers
• But if compound Q is rotated 180 o in the plane of
• the paper, it is actually identical to compound P ,
• therefore P and Q are superimposable mirror images
• P and Q are the same compound
• It is a meso compound
• COOH OH OH H H COOH COOH COOH R S COOH COOH H HO H OH * not a plane of symmetry * not a plane of symmetry rotate 180 o different H OH H HO
• R and S are related as mirror images and are not
• superimposable even if rotated 180 o
• Thus R and S constitute an enantiomeric pair
• There are 2 pairs of diastereomers :
P and R & P and S
• Further examples of meso compounds:
CH 3 CH 3 Cl Cl H H CHO CHO HO HO H H H OH plane of symmetry
• 11.6 REACTIONS OF ORGANIC COMPOUNDS
• 11.6.1 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
• 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
• Free radicals are formed in homolytic cleavage
X X X + X ≡ 2X • • • • • free radicals
• b) Heterolytic Cleavage
• Occurs in a polar bond involving unequal sharing
• of electron pair between two atoms of different
• electronegativities
• A single bond breaks unsymmetrically and both the
• bonding electrons are transferred to the more
• electronegative atom
• Cation and anion are formed in heterolytic cleavage
A B • • A • • - + B + A is more electronegative A + + B • • - B is more electronegative cation anion anion cation
• Carbocations and free radicals are intermediates in
• organic reactions.
• They are unstable and highly reactive
• 11.6.2 Reaction Intermediates
a) Carbocation
• Also called carbonium ion
• A very reactive species with a positive charge
• on a carbon atom
• Carbocation is formed in heterolytic cleavage
• Example : (CH 3 ) 3 C — Cl   (CH 3 ) 3 C + carbocation + Cl - 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
• A very reactive species with an unpaired electron
• Formed in homolytic cleavage
Cl – Cl Example : uv Cl • free radicals + Cl • C C C • + C • H 3 C H CH 3 • + H •
• 11.6.3 Relative Stabilities of
primary secondary tertiary depending on the number of carbon atoms directly bonded to the :-
• positively charged carbon atom (for carbocation)
• The stability of carbocation increases with the
• number of alkyl groups present
• The alkyl groups are electron-releasing relative to hydrogen, thus help to stabilize the positive charge
• on the carbocation
• Carbocation Stability:
methyl cation primary (1°) secondary (2°) tertiary (3°) Increasing stability
• As the number of alkyl groups attached to the
• positively charged carbon atom increases, the
• stability of carbocation increases
• Likewise, the stability of free radical increases as
• more alkyl groups are attached to the carbon atom
• with unpaired electron
methyl radical primary (1°) secondary (2°) tertiary (3°) Increasing stability
• 11.6.4 Reagents and Sites of Organic Reactions a) Electrophile (E + )
• Means ‘ electron loving ’
• An electron-deficient species and electron-pair
• acceptor that attacks a part of a molecule where
• the electron density is high
• An electrophile can be either neutral or positively
• charged
• Examples of electrophiles :-
• cations such as H + , H 3 O + , NO 2+ , Br + etc.
• carbocations.
• Lewis acids such as AlCl 3 , BF 3 etc.
• oxidizing agents such as Cl 2 , Br 2 and etc
• Examples of electrophilic sites in organic molecules :-
• molecules with low electron density around a
• polar bond such as :-
 +  -  +  -  +  - C = O C – X C – OH carbonyl haloalkanes hydroxyl compound
• b) Nucleophile (Nu - )
• Means ‘ nucleus loving ’
• An electron-rich species and electron-pair donor
• that attacks a part of a molecule where the electron
• density is low
• A nucleophile can be either neutral or negatively
• charged
• Examples of nucleophiles :-
• anions such as OH - , RO - , Cl - , Cn - etc.
• carbanions. (species with –ve charge on C atoms)
• Lewis bases which can donate lone pair electrons
• such as NH 3 , H 2 O etc.
• Examples of nucleophilic sites in organic molecules :-
molecules with high electron density around the carbon-carbon multiple bond such as :- -C=C- (alkenes) , -C  C-(alkynes), (benzene ring) and etc.
• 11.6.5 Types of Organic Reactions
• The four main types of organic reactions are:
• Substitution
• Elimination
• Rearrangement
a) Electrophilic addition b) Nucleophilic addition A reaction in which atoms or groups added to a multiple bond
• Initiated by an electrophile , which attacks a
• nucleophilic site of a molecule
• Typical reaction of unsaturated compounds such
• as alkenes and alkynes
Example : CH 3 CH=CH 2 + Br 2 CH 3 CHBrCH 2 Br room temperature electrophile
• Initiated by a nucleophile , which attacks an
• electrophilic site of a molecule
• Typical reaction of carbonyl compounds
Example :   + H + CN -
• 2. Substitution Reaction A reaction in which an atom or group in a molecule is replaced by another atom or group a) Free-radical Substitution b) Electrophilic Substitution c) Nucleophilic Substitution
• Substitution which involves free radicals as
• intermediate species
Example : CH 3 CH 3 + Cl 2 CH 3 CH 2 Cl + H Cl uv light
• b) Electrophilic Substitution
• Typical reaction of aromatic compounds
• The aromatic nucleus has high electron density,
• thus it is nucleophilic and is tend to electrophilic
• attack
Example : + Br 2 Fe catalyst Br + H Br electrophile Br  Br 
• c) Nucleophilic Substitution
• Typical reaction of saturated organic compounds
• bearing polar bond as functional group, such as
• haloalkane and alchohol
Example : CH 3 CH 2 Br + OH - (aq) CH 3 CH 2 OH + Br - (aq) nucleophile  
• 3. Elimination Reaction
• A reaction in which atoms or groups are removed
• from adjacent carbon atoms of a molecule to form
• a multiple bond (double or triple bond)
• Elimination reaction results in the formation of
• unsaturated molecules
CH 3 C H 2 OH CH 2 = CH 2 + H 2 O Conc. H 2 SO 4  Example :
• 4. Rearrangement Reaction
• A reaction in which atoms or groups in a molecule
• change position
• Occurs when a single reactant reorganizes the
• bonds and atoms
Example : tautomerisation enol keto (more stable) H C C H OH R H C C R H H O