Chemistry Form 6 Sem 3 Chapter 1

CHEMISTRY FORM 6

ORGANIC CHEMISTRY

CHAPTER 1 :

1.1 The Chemistry of Carbon

Organic Chemistry ~ branch of chemistry concerning

compound of carbon (except CO, CO₂, CO₃2-₎

Aliphatic compounds Alicyclic compounds Aromatic compounds Heterocyclic compounds → _{Open chain} organic compounds. → _{Do not have} special stability like benzene. → _{Compound may} be unbranched or branched → _{Closed ring of} organic compounds. → _{Rings may} contain single or double bonds → _{Compound may} be branched or unbranched → _{Contain at least} one benzene ring (those with

in it) → _{Closed ring} contain element other than carbon in it (like N, S, O)
Alkane, alkene, alkyne, haloalkane
Cyclohexane, cyclobutene
Phenol, naphthalene, toluene
Pyridine

1. Hybridisation of alkane, alkene and alkyne

Carbon is a Group ___ element. It has the electronic

configuration of ______________ The orbital diagram

Ground state of C : _____ _____ _____ _____

2s 2p

Methane, CH₄ Type of hybridisation :

Excited state of C : _____ _____ _____ _____ 2s 2p
Hybridised state : _____ _____ _____ _____ sp3 14 1s2 _2s2 _2p2 sp3

Ethene, C₂H₄ Type of hybridisation :
Excited state of C : _____ _____ _____ _____ 2s 2p
Hybridised state : _____ _____ _____ _____

sp

_p

Molecular shape

Angle between bond

pair – bond pair

sp2

Trigonal planar

Ethyne, C₂H₂ Type of hybridisation :
Excited state of C : _____ _____ _____ _____ 2s 2p
Hybridised state : _____ _____ _____ _____

sp

p

p

Molecular shape

Angle between bond

pair – bond pair

sp

Linear

As a conclusion, the formation of double bond

(C=C) is due to ______sigma bond (σ) and

_____pi bond (π)

While the formation of triple bond (C≡C) is

due to ______sigma bond (σ) and _____pi

bond (π)

one

one

one

two

Hybridisation in benzene

Benzene (C₆H₆) is a flat and symmetrical molecule. All the

atom (6 C atom and 6 H atom) in a benzene molecule lie in the same plane. The Carbon atoms are arranged in the form of

hexagon as shown in diagram at the left.

The formation of benzene can be deduced using hybridisation

theory

Excited state of C : _____

_____ _____ _____

2s

2p

Hybridised state: _____ _____ _____ _____

sp

_p

Since each carbon contribute an unhybridised electron, so the

side touch between C–C atom will form a double bond. Hence

there are _____ double bond build between C–C in benzene ring. In another words, there are free delocalise electrons move about in benzene ring.

◦ The following notes are taken into consideration when drawing structure of benzene The unhybridised p orbitals do not overlap in pairs to form double bonds alternating with 3 single bonds as shown in Kekule structure.

◦ The structure of benzene is a ………... of 2 forms

◦ The resonance hybrid of benzene can be expressed as

◦ Unlike ethene, the double bond in benzene has a larger volume (space) to delocalise electron. Hence the more space provided, the lower the energy in benzene. This makes benzene posses an extra stability.

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1.2 Chemical formulae of Organic Compound

The formula of an organic compound can be represented by

the

(a) empirical formula (b) molecular formula (c) structural formula (d) skeletal formula

Empirical formula of a compound shows the simplest ratio

of the atoms of each element in the compound.

Molecular formula of a compound shows the actual number

1. Derive the empirical formula of a hydrocarbon that on analysis gave the following percentage composition: C = 85.63%, H = 14.37%. given the relative molecular mass of the hydrocarbon is 84, determine the molecular formula

2. A 1.367 g sample of an organic compound was combusted in a stream of dry oxygen to yield 3.002 g CO₂ and 1.640 g H₂O. If the original compound contained only C , H, and O, what is its empirical formula?

Element C H Mass 85.63 14.37 Mol 85.63 12 = 7.14 mol 14.37 1 =14.37 mol Ratio 7.14/7.14 = 1 14.37/7.14 = 2 Empirical formula = CH₂ (CH₂)n = 84 (12 + 2(1))n = 84 n = 6 Molecular formula = (CH₂)6 = C₆H₁₂ C_xH_yO_z + mO_{2  x CO2} + y/2 H₂O Since 1 C = 1 CO₂ x / 12 = 3.002 / 44 ; x = 0.8187 g Since 2 H = 1 H₂O y / 2= 1.640 / 18 ; y = 0.1822 g Mass of O, z = 1.367 – (0.8187 + 0.1822) = 0.3661 g Elemen C H O Mass 0.8187 0.1822 0.3661 Mol 0.8187 12 =0.0682 0.1822 1 =0.1822 0.3661 16 =0.0229 Ratio 0.0682/0 .0229 = 3 0.1822/0 .0229 = 8 0.0229/ 0.0229 = 1 Empirical formula = C₃H₈O

A 1.500 g sample of a compound containing only C, H, and O was burned completely. The only combustion products were 1.738 g CO₂ and 0.711 g H₂O. What is the empirical

formula of the compound?

Elementary analysis showed that an organic compound contained C, H, N, and O as its

elementary constituents. A 1.279-g sample was burned completely, as a result of which 1.60 g of CO₂ and 0.77 g of H₂O were obtained. A separately weighed 1.625 g sample contained 0.216 g nitrogen. What is the empirical

formula of the compound?

C_xH_yO_z + mO_{2  x CO2} + y/2 H₂O Since 1 C = 1 CO₂ x / 12 = 1.738 / 44 ; x = 0.474 g Since 2 H = 1 H₂O y / 2(1) = 0.711 / 18 ; y = 0.079 g Mass of O, z = 1.367 – (0.474 + 0.079) = 0.947 g Elemen C H O Mass 0.474 0.079 0.947 Mol 0.474 12 =0.0395 0.079 1 =0.079 0.947 16 =0.0592 Ratio 0.0395/0 .0395 = 1 0.079/ 0.0395 = 2 0.0592/ 0.0395 = 1.5 Empirical formula = C₂H₄O₃ Since 1 C = 1 CO₂ x / 12 = 1.60 / 44 ; x = 0.4364g Since 2 H = 1 H₂O y / 2(1) = 0.77 / 18 ; y = 0.08556 g Since 1.625 g of same sample produce 0.216 g of nitrogen Mass of N in sample, n n / 1.279 = 0.216 / 1.625 n = 0.17 g Mass of O, z = 1.279 –(0.4364 + 0.08556 + 0.17) = 0.587 g Empirical : C₃H₇NO₃

Structural formula of an organic compound is the formula which

shows how the atoms are bonded together as well as the numbers of each atom present. Structural formula can be expressed in a few ways.

Example : butane, C₄H₁₀
Example : 2-methylhexane a) Shorthand e) Simplified notation b) Displayed formula c) 3-D @ stereochemical d) Skeletal formula a) Shorthand b) Displayed formula c) 3-D @ stereochemical d) Skeletal formula CH₃CH₂CH₂CH₃ CH₃(CH₂)₂CH₃ CH(CH₃)₂CH₂CH₂CH₂CH₃

Example : 4,4-dimethylpent-2-ene a) Shorthand b) Displayed formula c) 3-D @ stereochemical d) Skeletal formula C(CH₃)₃CH=CHCH₃

Example : 3-ethyl-2,4-dimethylhexane
Example : 2,2,5-trimethylhex-3-yne a) Shorthand b) Displayed formula c) 3-D @ stereochemical d) Skeletal formula CH(CH₃)₂CH(CH₂CH₃)CH(CH₃)CH₂CH₃ a) Shorthand b) Displayed formula c) 3-D @ stereochemical d) Skeletal formula C(CH₃)₃CΞCCH(CH₃)₂

1.3 Classification of Organic Compounds based on the Functioning Group and its General Formulae

Homologous series ~ compounds with similar chemical

properties in which each member differs from the previous one by addition of –CH₂–. The characteristic of a homologous series are as follow.

All compounds in homologous series has the same functioning

group and chemical properties

Each member differ from the next series by a –CH₂– group, in

another words, molecular mass of each compound in series differ from next by …………

All the compounds in the series may be prepared by using the

similar methods.

Physical properties show a progressive change with increase

of molecular mass.

All the compounds in the series contain same elements and

functioning group, thus it can be represented by same general formula.

Homologous series General formula Examples + Structural formula Alkanes C_nH_2n+2 pentane heptane Cycloalkane C_nH_2n cyclobutane Cyclohexane Alkenes RCH=CHR C_nH_2n hex-1-ene but-2-ene Cycloalkene C_nH_2n–2 cyclohexene cyclobutene

R–H

Alkynes RCΞCH C_nH_2n–2 Propyne pent-1-yne Haloalkane R–X CnH2n+1X (X = Cl, Br, I) 1-chlorobutane 2-bromopentane Alcohol R–OH CnH2n+1OH / C_nH_2n+2O propan-1-ol pentan-3-ol Aldehyde R–COH C_nH_2nO Butanal hexanal Ketone R–CO–R C_nH_2nO Propanone heptan-3-one

Carboxylic acid

RCOOH

C_nH_2n+1COOH C_nH_2nO₂

butanoic acid ; pentanoic acid

Ester

RCOOR CnH2n+1COO–

C_mH_2m+1 / C_nH_2nO₂

ethyl ethanoate ; propyl butanoate

Primary amine R–NH₂ CnH2n+1NH2 ethylamine butylamine Amide RCONH₂ C_nH_2n+1CONH₂ propylamide ; pentylamide

1.4 Alkyl and Type of Alkyl groups

Alkyl group ~ obtained by removing a hydrogen atom from an

alkane.

Symbol of alkyl is R, where –R has the general formula of

C_nH_2n+1.

Alkane

Alkyl

Alkane

Alkyl

Methane

Ethane

Propane

Butane

Pentane

Hexane

Alkyl can be categorise into 3 groups

Type of alkyl

group Example Comment

Primary Only one alkyl group attached to carbon atom

Secondary Two alkyl group attached to carbon atom

Tertiary Three alkyl group attached to carbon atom

1.5 Isomerism in Organic Compound

Isomers ~ substances which have the same molecular formula

but different molecular structure

Isomerism Structural isomerism Stereoisomerism Chain isomerism Position isomerism Functional isomerism Geometrical isomerism Optical isomerism

1.5.1 Structural Isomerism

~ are isomers with same molecular formula but different

structural formula (link differently)

As mentioned above, structural formula can be separate into 3

different categories

◦ Chain isomerism

◦ Positional isomerism

1. Chain isomerism ~ isomers which have different carbon chain (straight or branched chain)

Pentane (C₅H₁₂) Butan-1-ol (C₄H₉OH)

1-chloropentane (C₅H₁₁Cl) Hexanal (C₆H₁₂O) CH₃CH₂CH₂CH₂CH₃ CH₃CH₂CH(CH₃)₂ C(CH₃)₄ CH₃CH₂CH₂CH₂OH CH(CH₃)₂CH₂OH C(CH₃)₃OH CH₃CH₂CH₂CH₂CH₂CHO CH(CH₃)₂CH₂CH₂CHO C(CH₃)₃CH₂CHO CH(CH₂CH₃)₂CHO CH₃CH₂CH₂CH₂Cl CH₃CH₂CH(CH₃)CH₂Cl C(CH₃)₃CH₂Cl

2. Position isomerism ~ isomers which the position of functioning group is different

Hexene (C₆H₁₂) Bromohexane (C₆H₁₃Br)

Pentanol (C₅H₁₁OH) Dichlorobenzene (C₆H₄Cl₂)

CH₂=CHCH₂CH₂CH₂CH₃ CH₃CH=CHCH₂CH₂CH₃ CH₃CH₂CH=CHCH₂CH₃ CH₃CH₂CH₂CH₂CH₂OH CH₃CH₂CH₂CH(OH)CH₃ CH₃CH₂CH(OH)CH₂CH₃ CH₂BrCH₂CH₂CH₂CH₂CH₃ CH₃CHBrCH₂CH₂CH₂CH₃ CH₃CH₂CHBrCH₂CH₂CH₃

3. Functional isomerism ~ isomers which has the same molecular formula but different molecule with different functioning group.

Alkene and cycloalkane – C₅H₁₀ Alcohol and ether – C₄H₁₀O

1.5.2 Stereoisomerism

Geometrical Isomerism ~ same structural formula but different

spatial arrangement.

~ also known as cis-trans isomer

~The essential requirement for the existence of geometrical

isomerism in organic compound must contains a carbon– carbon double bond (C=C)

~ A ring structure which hinders the rotation of a C–C single bond in a ring.

However, cis-trans isomers cannot occur if one of the carbon

atoms in the double bond has 2 identical atoms / groups.

pent-2-ene 1,2-dimethylcyclobutene

Physical Properties of Geometric Isomers

Cis-isomer usually has a lower melting point as the

structure of isomer is less symmetrical. Therefore, cis-isomer cannot be closely packed in the crystal lattice

resulting the intermolecular forces to become weaker than in trans-isomer.

On the other hand, cis-isomer has a higher boiling point

because the space arrangement in cis isomer caused the compound to become a polar molecule. As a result, the intermolecular forces fo cis-isomer in liquid is stronger, causing the temperature required to boil the substance become higher. Trans-isomer on the other hand, has 0 dipole moment.

(2) Optical isomerism

Optical activity is the ability of certain crystal or solution of

certain substances to rotate the plane of plane-polarised light. Such substances are said to be optically active (sometimes are known as chiral molecule)

Optical isomers are optically active substances which possess

the same molecular structure but different in their effect on plane polarised light.

For an optically active isomer, the object and mirrored image

cannot be superimposable to one and another. Example, in 1-bromo-1-chloro-1fluoromethane (CHClBrF), it shows optical

active when brought to plane-polarised light.

An organic molecule will exhibit optical isomerism if it contains

one chiral carbon atom.

Chiral carbon atom is a carbon atom which is attached by 4

different atoms / groups.

Also known as asymmetrical carbon often shown as C*

Example : State whether the organic compound below exhibit

optical isomerism. If yes, mark * at chiral carbon atom.

CH₃CH₂CH(Cl)CH₃ C(CH₃)₂(Br)(OH) CH₃CBr=CH₂ HOCH₂CH(NH₂)COOH CH₃CH₂CH(OH)CH₃

1.6 Lewis Acid & Lewis Base : Nucleophiles and Electrophiles

Lewis Acid : ………

Lewis Base : ………

Ammonia react with hydrogen ion to form ammonium ion

Boron trifluoride react with ammonia to form a complex

Aluminium trichloride react with chloride ion to form

tetrachloroaluminate ion

Water molecule

attached to cobalt ion to form

hexaaquacobalt (III) ion.

Methylamine react with water to form a basic solution

Ethanoic acid react with water to form ethanoate ion and hydroxonium ion

Substance which donate lone pair electron Substance which receive lone pair electron

In organic reaction, such reaction can be classified as either

nucleophile or electrophile, depending on whether they

attack regions of high electron density (δ –) or region with low electron density (δ +)

Nucleophile Electrophile

Definition

Nucleo = nucleus ; phile = love

Nucleophile mean love nucleus. In terms of Lewis acid-base theory, nucleophiles are often Lewis base, which donate lone pair electron.

Electro = electron ; phile = love

Electrophile mean love electron. In terms of lewis acid-base theory, electrophiles are often Lewis acid, which accept lone pair electron.

Examples :OH– _RO:– _:Cl– _H+ _(H 3O+) NO2+ Cl2 ; Cl+ :Br– _:I– _CN:– _Br 2 ; Br+ I2 ; I+ C⊕ carbonium :C– carbanion R–OH –C=C– RN2 + _R 3C+ BF3 :NH₂– _R–NH 2 H2O AlCl3 FeBr3 ZnCl2

1.7 Inductive Effect and Mesomeric Effect

Inductive effect is defines as the shift in electron density from one

atom to another to form a polar bond.

For example, in the bonding of –C–Cl. In the presentation of inductive

effect, the arrow shows that carbon atom repel electrons

Chlorine atom attract the electrons as it has a higher elctronegativity

1.7.1 Electron–withdrawing group

Most atoms and groups of atoms are more electronegative than carbon

atom, thus withdraw electron from carbon. When this occurs,

atoms/molecules are said to exert a negative inductive effect (–I effect).

Example of electron withdrawing group

–Cl ; –Br

; – I –NO2 –CN –COOH –COOR –C=O –SO3H –C6H5

1.7.2 Electron donating group

Some atoms/groups are less electronegative than carbon thus

donate electrons to carbon atom. These atoms/groups are also known as electron donating group and they exert a positive

inductive effect (+I effect). Alkyl group are known to have a +I effect. The effect increase with the increase of alkyl group

attached to the C atom.

The inductive effect produces a permanent dipole and this will

influence its physical properties such as boiling / melting point, acid and basic strength, reaction of functional group and rate of reaction.

Inductive effect may also occur when there is a permanent shift

of electron due to the polarisation of sigma bond (σ-bond). The shift of electron can also occur in a pi (π) bond. The shift of π

electron in multiple bonds toward a more electronegative atom is known as resonance effect (mesomeric [M] effect)

Example in methanal, oxygen atom is more electronegative

than carbon atom. Consequently, the π electron shifted to

oxygen atom. The structure can be represented as a resonance hybrid between structure I and II.

Same phenomenon can occur in carboxylate anion. M-effect

play an important role in the structure and stability of many compounds. The π electrons are spread and delocalised over the carbon and oxygen atoms. We can represent the

Type of Organic

Reaction

Substitution

Addition

Elimination

Definition : one or

more atom / group is

substitute by another

atom / group of atoms

Definition : 2 reactants

react together and

form 1 product

Definition : a small

atom / group is

removed (eliminated)

from the molecule

involve

Example :

CH

CH

Cl + OH

_

CH

CH

OH + Cl

Example :

CH

=CH

+ Br

_

CH

(Br) CH

(Br)

Example

CH

CH

_{OH }

CH

=CH

+ H

O

1.9 Bond Fission ~ Ways of Breaking Chemical Bond

In a covalent bond, a pair electron is shared between 2 atoms.

When the covalent breaks, these 2 electrons are redistribute between 2 atoms. There are 2 ways how these electrons are redistribute.

Bond fission

Homolytic fission Heterolytic fission • _{The 2 shared electrons distribute}

evenly between the 2 atoms • _{Free radicals are formed after}

homolytic fission and it is usually represented by •X

• _{Since it is a single electron, free} radicals tend to become unstable, thus are reactive.

• _{Example :}

Cl – Cl _ 2 Cl• (chlorine radical) H_{3C–H } CH₃• _{+ H•}

methyl radical hydrogen radical

o Shared electron in the covalent bond only goes to one of the atom. The pair of electron usually goes to the atom with higher electronegative

o Eventually, one will become negatively charged (as a result of receiving extra electron) while the other become

positively charged (lose electron to the other party) o Example : Cl – Cl _ :Cl– _{+ Cl}⊕ (CH₃)_{3C–Br } :Br– _{+ (CH} 3)3C ⊕

Practice

1.Base on the molecular structure, draw the hybridisation diagram of the following molecules, state the type of hybridisation and its shape, and state the angle of each hybridised structure

a) CH₃CH=CHOH H H H H _H

σ

σ

σ

σ

_σ

σ

σ

_π

σ

b) CHΞCCH₂NH₂ H H

σ

σ

σ

σ

σ

π

π

σ

σ

σ

Displayed formula Short hand / notation Skeletal formula

CH(CH₃)₂CH(OH)CH(CH₂CH₃)₂

CH(CH₃)₂C(CH₃)₂CH₂CH₂CH₃

C(CH₃)₃CH(CH₂CH₃)₂

CH₂(NH₂)CH(C₆H₅)CONH₂

COHCH(Br)C(CH₃)₂CH₂COOH

COHC(CH₃)₂CH(OH)CΞCCH₃

Name the functioning group presence CH₂ClCOCH₂CHNH₂CH₂CH₂CONH₂ Alkene_, alcohol, aldehyde Haloalkane_, Ketone ; Amine , Amide Aldehyde ; Alcohol ; Alkyne Carboxylic acid

Molecules Type of

isomerism Diagram of isomers

geometrical

geometrical

CH₂ClCH=C(CH₃)CH₂COOH

geometrical geometrical

optical

Particles Particles Particles

Br₂ Br+ _Br

-NH₃ CH₃NH₂ NH₂+

CH₃+ _CH

3- CH3COH

electrophile _{electrophile nucleophile}

nucleophile _{nucleophile electrophile}

electrophile nucleophile electrophile