DK024

UnitK

This Book belongs to *NoCH

H

tes

ZAMANURI FARHANA ZANARINA ASMAHAN KA
ASWARI MUHAMMAD AZIDA AFF

antanKimia, Kolej Matrikulasi Kel

ab*HEMISTRY 2

DK024 2021
2022

HANDBOOK

s*Tutorial Questions*Pre L

ASBAH ELMI ROSILAH SHARIFAH SUHAINA NAJIHAH
FENDY MELANI BADRISAM FATIMAH

ACKNOWLEDGEMENT

SPECIAL THANKS AND APPRECIATION TO:
FARHANA BINTI UMANAN

KASBAH BIN ANDI DAHRAN@ASMAD
DR ELMI SHARLINA BINTI MD SUHAIMI

KHAIRUL ASWARI BIN AB RAHMAN
NOOR ASMAHAN BINTI ABDULLAH

WAN ROSILAH BIN WAN LLAH
SUHAINA BINTI MOHD YATIM
SHARIFAH FADTHYAH BINTI SYED BAHARUDDIN
NOOR NAJIHAH BINTI KAMARUDDIN

ZANARINA BINTI SAPIAI
MUHAMMAD BIN SEMAN
NOOR AZIDA BINTI MOHD YUSOFF
AHMAD AFFENDY BIN ABDULLAH
MOHD ZAMANURI BIN MOHD SUHAIMI
MELANI SARI BINTI WAN ISMAIL
BADRISAM BIN SAINDIN
SITI FATIMAH BINTI MD SOLLHI
CHEMISTRY UNIT KELANTAN MATRICULATION COLLAGE
MATRICULATION DIVISION MINISTRY OF EDUCATION MALAYSIA

©chemistryunitkmkt2022

i

TABLE OF CONTENT

CONTENT TITLE PAGE
CHAPTER 1 INTRODUCTION TO ORGANIC CHEMISTRY
1
• NOTE ALKANES 20
• TUTORIAL
CHAPTER 2 ALKENES 28
• NOTE 57
• TUTORIAL AROMATIC COMPOUNDS
CHAPTER 3 62
• NOTE ALCOHOLS 74
• TUTORIAL
CHAPTER 4 CARBONYL COMPOUNDS 80
• NOTE 92
• TUTORIAL CARBOXYLIC ACID
CHAPTER 5 98
• NOTE QUALITATIVE ANALYSIS OF ORGANIC 121
• TUTORIAL CHEMISTRY
CHAPTER 6 SATURATED AND UNSATURATED 126
• NOTE HYDROCARBON. 143
• TUTORIAL REACTIONS OF ALCOHOLS
CHAPTER 7 ALDEHYDE AND KETONE 149
• NOTE 161
• TUTORIAL
PRE-LAB MODULE 166
• EXPERIMENT 1
169
• EXPERIMENT 2 171
174
• EXPERIMENT 3
• EXPERIMENT 4

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CHAPTER 1: INTRODUCTION TO ORGANIC CHEMISTRY | CHEMISTRY UNIT KMKt 2021/2022

CHAPTER 1 : INTRODUCTION TO ORGANIC CHEMISTRY

1.1 Introduction Identify the elements that made
up organic compounds.

Define structural formula

1.2 Structural Condensed structure, expanded
Formula structure, skeletal structure

Primary carbon, secondary carbon,
tertiary carbon, quarternary carbon

INTRODUCTION 1.3 Primary hydrogen, secondary
TO ORGANIC Classification of hydrogen, tertiary hydrogen
CHEMISTRY
Organic Define functional group
Compound Name functional group

Classify organic compound
according to their functional group.

Define and explain homologous series

Single bond, double bond and triple bond

1.4 Hydrocarbon Define hydrocarbons

Classify hydrocarbon into aliphatic,
aromatic, saturated and unsaturated

Differentiate between saturated and
unsaturated of organic compound

1.5 Classify and draw the constitutional isomers:
Isomerism chain, positional and functional group
isomers.

1.6 Basic reaction in Homolytic and heterolytic
organic compound cleavage

Define and stability of free
radical and carbocation.

Type of organic reactions:
substitution,addition and elimination

reaction.

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CHAPTER 1: INTRODUCTION TO ORGANIC CHEMISTRY | CHEMISTRY UNIT KMKt 2021/2022

1.1 Introduction to Organic Chemistry

 Organic chemistry is the branch of chemistry that deals with of carbon

compounds.
 Organic compounds are considered to be those that contain carbon and

hydrogen, while other common elements are O, N, the halogens, S and P.
 Examples of carbon compounds in our daily lives :-

Methane glucose

H OH

C H OH
H HH H

HH HO OH
• The simplest organic compound
• One of primary energy resource OH OH

• Principal sugar in blood

Urea DDT (Dichlorodiphenyltrichloroethane)

O

C Cl CH Cl

H2N NH2 Cl C Cl
Cl
• Use in urea–formaldehyde resins,
• Used as an insecticide
pharmaceuticals, and as a nitrogen

fertilizer

Penicillin Nicotine

N

CH2 CH3
CO N
NH • A colorless poisonous alkaloid

presents in tobacco

S CH3

N CH3

O
• An antibiotic derived from the mould

Penicillium notatum

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Cafeine

O CH3
H3C N N

ON

CH3
• Stimulant in coffee, tea, and many soft drinks

DNA (Deoxyribonucleic acid)

Genetic information is stored as a sequence of nucleotides

EXERCISE 1
Identify which of the following compounds are ORGANIC COMPOUND?

a) C6H5NH3Cl
b) (NH4)2SO4
c) Na2CO3
d) HCl
e) CH2═CHCH3
f) CH3CH2COCl
g) HNO3

1.2 Structural formula

Structural Formula is a formula that shows how the atoms of a molecule are bonded
to each other.

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Structural formula can be drawn in the form of condensed, expanded and skeletal
structure.

i) Condensed Structure

 In condensed formulae all the hydrogen atoms that are attached to a particular
carbon are usually written immediately after that carbon,

 In fully condensed formulae all atoms that are attached to a carbon are written
immediately after that carbon

 Example :

C4H9Cl CH3CH(Cl)CH2CH3

ii) 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 :

H H HH

C4H9Cl HCCCC H

H Cl H H

Expanded structure

iii) Skeletal Structure

 This structure shows only the carbon skeleton
• C chains are drawn zigzag, and rings are drawn as polygons

 The hydrogen atoms that are assumed to be present, are not written.
Other atoms such as O, Cl, N and etc. are shown.

 Examples :

i) CH3CH(Cl)CH2CH3 =

ii) Cl
H2C CH2 =

H2C CH2

iii) CH2 = CHCH2 OH = OH

EXERCISE 2
1. Convert each molecule into expanded structure.

a) (CH3)2CHCH2CH2CH(CH3)2
b) CH3CH(Cl)CH(OH)CH3

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CHAPTER 1: INTRODUCTION TO ORGANIC CHEMISTRY | CHEMISTRY UNIT KMKt 2021/2022
2. Write a condensed structural formula for each of the following compounds.

a)

b)

c)

d)
Classification Of Carbon And Hydrogen Atoms In Organic Molecules
 A carbon atom is classified as primary (1o), secondary (2o), tertiary (3o) or

quarternary (4o) depending on the number of carbon atoms bonded to it.
▪ a primary carbon is directly bonded to one other carbon atom (has 1

adjacent carbon atom).
▪ a secondary carbon is directly bonded to two other carbon atoms (has 2

adjacent carbon atoms).
▪ a tertiary carbon is directly bonded to three other carbon atoms (has 3

adjacent carbon atoms) .
▪ a quarternary carbon is directly bonded to four other carbon atoms (has 4

adjacent carbon atoms).

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 Similarly, a hydrogen atom is also classified as primary, secondary or tertiary
depending on the type of carbon to which it is bonded.
▪ 1o hydrogen atom is bonded to a 1o carbon atom.
▪ 2o hydrogen atom is bonded to a 2o carbon atom.
▪ 3o hydrogen atom is bonded to a 3o carbon atom.

KEEP IN MIND!
 No classification for C atoms in methane and in multiple bonds
 Only for sp3 C atom

EXERCISE 3
Identify the types of carbons (1o, 2o, 3o, or 4o) in the following molecules. Label at
each C atom.

CH3
a) CH3 CH CH2 CH3

CH3

b)

c)

d)
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1.3 Classification of organic compounds

Functional group is an atom or group of atoms that imparts characteristic
chemical properties to an organic compound.

Identifying and classifying functional groups of different class of compounds:

Class of Functional Group Example
compounds CH3CH3
Structure Name CH2=CH2
Alkane
CnH2n+2 None -
Alkene
CnH2n CC Carbon-carbon
double bond

Alkyne CC Carbon-carbon CHCH
CnH2n-2 triple bond
Aromatic/Arene
CnHn Aromatic ring

Alcohol -OH Hydroxyl CH3CH2OH
CnH2n+2O Halogen CH3CH2Cl
Haloalkane -X (F,Cl,Br,I) Alkoxy CH3OCH3
CnH2n+1X
C OC Carbonyl CH3CHO
Ether CH3COCH3
CnH2n+2O O Carbonyl CH3COOH
CH CH3COOCH3
Aldehyde O Carboxyl CH3COCl
CnH2nO RCR CH3COOCOCH3
O Carboalkoxy/ CH3CONH2
Ketone C OH ester
CnH2nO O Acyl CH3NH2
C OR
Carboxylic acid O Anhydride
CnH2nO2 C Cl
OO Carboxamide/
Ester CO C amide
CnH2nO2 O
CN Amino
Acyl chloride
CnH2n+1COCl CN Cyano CH3CN
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Anhydride CN
(CnH2n+1CO)O

Amide
CnH2n+1CONH2

Amine
CnH2n+1NH2

Nitrile
CnH2n+1CN

CHAPTER 1: INTRODUCTION TO ORGANIC CHEMISTRY | CHEMISTRY UNIT KMKt 2021/2022

EXERCISE 4

Identify the functional groups in the following molecules.

CH3 O

CH3 C CH2 C
a) CH3 NH2

CH3

b)
c) CH3 O CH2CH3

CH2CH3

d)

H3C
C O CH2CH3

e) O

H2N O OH
O Br
O
O CH2COOH

HO

f)

H2N CH2CH2OH
O
g)

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OH

C

h) OH OCH3

HN H HH H O
i) H C C N C C OH
H H H3C C O

Homologous series

 Homologous series is a series of compounds with the same functional group
where each member of the series differs from the next member by a constant –
CH2 unit.

 Members of the same homologous series are called homologs.

General Characteristics of homologous series.

There are four features of homologous series:

1. All the members of a particular homologous series share the same general
formula.

Example: Homologous series of alcohol

General formula: CnH2n+1OH, where n=1, 2, 3, etc.
Functional group: hydroxyl (OH)

Structure Name

n=1 CH3OH Methanol

n=2 CH3CH2OH Ethanol

n=3 CH3CH2CH2OH 1-propanol

*Methanol, ethanol and 1-propanol are said as members of homologous series.

2. All members of a particular homologous series have the same functional group
and same chemical properties.

Example:
▪ All alcohols contain the -OH group
▪ They react with carboxylic acids to give esters

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▪ The can be prepared by heating dilute sodium hydroxide solution with
anappropriate alkyl halide.

3. Successive member of a series differs by a –CH2 unit

Example:

ALKANES

CH4 Methane

CH3CH3 Ethane

CH3CH2CH3 propane

*each member differ by CH2

ALDEHYDE

HCHO Methanal

CH3CHO Ethanal

CH3CH2CHO Propanal

*each member differ by CH2

4. Each member of a homologous series gradually change in physical properties with
increasing number of carbon atoms.

Example:

PHYSICAL PROPERTIES

Compound Boiling point

CH4 -162 oC

CH3CH3 -88.2oC

As number of carbons increase, size of molecule becomes larger, boiling and melting
point increase.

General Formula Homologous Series

SERIES GENERAL ENDING NAME

FORMULA …ane
…ene
Alkane CnH2n+2 …yne
…ol
Alkene CnH2n …al
…one
Alkyne CnH2n-2 …oic acid
…amine
Alcohol R-OH …amide

Aldehyde R-CHO

Ketone R-COR’

Carboxylic acid R-COOH

Amine R-NH2

Amide R-CONH2

*R normally refer to alkyl group: CnH2n+1

*reduce 2 atom H for cyclic compound

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Homologous series of alkyl group

Alkane CnH2n+2 Alkyl CnH2n+1
Methane CH4 Methyl -CH4
Ethane CH3CH3 Ethyl -CH2CH3
Propane CH3CH2CH3 Propyl -CH2CH2CH3
Butane CH3CH2CH2CH3 Butyl -CH2CH2CH2CH3
Pentane CH3CH2CH2CH2CH3 Pentyl -CH2CH2CH2CH2CH3

Homologous series of cycloalkanes

Alkane CnH2n+2 Cycloalkane CnH2n
Methane CH4 - -
Ethane CH3CH3 - -
Propane CH3CH2CH3 Cyclopropane

Butane CH3CH2CH2CH3 Cyclobutane

Pentane CH3CH2CH2CH2CH3 Cyclopentane

Homologous series of alkenes and cycloalkenes

Alkane CnH2n Cycloalkane CnH2n-2
Ethene CH2=CH2 - -
Propene CH2=CHCH3 Cyclopropene
Butene Cyclobutene
CH2=CHCH2CH3
Pentene 1-butene Cyclopentene
CH3CH=CHCH3
2-butene
CH2=CHCH2CH2CH3
1-pentene

KEEP IN MIND!
Alkanes and cycloalkanes belong to the same class of compound but different
homologous series.

Ability of carbon to form 4 covalent bonds with other carbons or elements:

C atom can form four covalent bonds with other C or elements:

▪ Single bond (EXAMPLE: C‒C, C‒X)

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▪ Double bond (EXAMPLE: C=C, C=O)

▪ Triple bond (EXAMPLE: C≡C, C≡N)

1.4 Hydrocarbons
• Hydrocarbons are compound composed of only carbon and hydrogen.

Examples:

(aliphatic) (aromatic) (aliphatic)
(saturated)
(unsaturated) (saturated)

▪ Aliphatic hydrocarbon are composed of straight-chained, branched or cyclic
compounds and can be saturated (alkanes) or unsaturated (alkenes and
alkynes).

▪ Aromatic hydrocarbon have one or more conjugated, benzene or heterocylic
rings within their structures. “Conjugated” refers to the presence of delocalized
(shared) electrons within the chemical structure, such as benzene or arenes.

▪ Compounds such as the alkanes and cycloalkanes, whose molecules contain
only single bonds are referred to as saturated hydrocarbons. They contain the
maximum number of hydrogen atoms that the carbon compound can possess.

▪ Compounds with multiple bonds such as alkenes, cycloalkenes, alkynes and
aromatic hydrocarbons are called unsaturated hydrocarbons. They posses
fewer than the maximum number of hydrogen atoms.

Example: Aromatic hydrocarbon
benzene
Aliphatic hydrocarbon
CH4
methane

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CH2 CH3

cyclohexane ethylbenzene
CH3CH=CH2
propene naphthalene
Unsaturated hydrocarbon
Saturated hydrocarbon
-contains only single bond -contains double or triple bond
CH4 CH3CH=CH2
methane propene
CH3CCH
cyclohexane propyne
CH3CH2CH3
propane benzene

1.5 Isomerism

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

 Different compounds that have the same molecular formula are called isomers.

 Structural isomers are different compounds with the same molecular formula
but differ in the order of attachment of atoms.

 Three types of structural isomerism :-

a) Chain isomerism

 The isomers differ in the carbon skeleton (different carbon chain).
 They possess the same functional group and belong to the same

homologous series.
 Example:

How many structural isomers does pentane, C5H12, have?

• CH3 CH2 CH2CH2 CH3 n-pentane

CH3 CH CH2CH3 isopentane

• CH3

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CH3 neopentane
CH3 C CH3

• CH3

There are three isomers of C5H12. All the three isomers are chain isomers of
each other due to different in carbon skeleton.

b) b) Positional isomerism

 These isomers have a substituent group in different positions in the same
carbon skeleton.

 Example :
Molecular formula: C3H7Cl

CH3CH2CH2Cl Cl
1-chloropropane
CH3 CH CH3

2-chloropropane

Molecular formula: C4H8

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

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 :-

General formula Classes of compounds
CnH2n+2O alcohol and ether
CnH2nO aldehyde and ketone
CnH2n alkene and cycloalkane
CnH2nO2 carboxylic acid and ester

 Examples :

C2H6O: CH3CH2OH and CH3OCH3

ethanol dimethyl ether

C3H6O: CH3CH2CHO and CH3COCH3

propanal propanone

C3H6O2: CH3CH2COOH and CH3COOCH3

propanoic acid methyl ethanoate

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EXERCISE 5
Draw all structural isomers of compound with the molecular formula;
a) C4H10O
b) C4H8O
c) C4H8

1.6 Basic reaction in organic compounds

a) 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

i) 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.

free radicals

ii) 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 is more
anion cation electronegative
.
A:B

A+ + B:- B is more
cation anion electronegative

 Carbocations and free radicals are intermediates in organic reactions.
They are unstable and highly reactive.

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b) Reaction Intermediates

i) Carbocation

 Also called carbonium ion.
 A very reactive species with a positive charge on a carbon atom.
 Carbocation is formed in heterolytic cleavage.
 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.

ii) Free Radical

 A very reactive species with an unpaired electron.
 Formed in homolytic cleavage.
 Example: free radicals

X:X X. + X.

CC C+C

H3C H CH3 + H

c) Relative Stabilities of Carbocations and Free Radicals

 Carbocation and free radical can be primary, secondary or tertiary,
depending on the number of carbon atoms directly bonded to the :-
 positively charged carbon atom ( for carbocation )
 carbon atom with unpaired electron ( for free radical )

 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.

 As the number of alkyl groups attached to the positively charged carbon
atom increases, the stability of carbocation increases.

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i)Carbocation Stability:

HH RR

HCH < R C H < RC H< R C R
++ + +

methyl primary secondary tertiary
cation (1o)
(2o) (3o)

Increasing stability

 Likewise, the stability of free radical increases as more alkyl groups are
attached to the carbon atom with unpaired electron.

ii) Free Radical Stability :

HH RR
HCH < R C H
< RCH<R C R

methyl primary secondary tertiary
radical (1o) (2o) (3o)

Increasing stability

d) Types of Organic Reactions

 The four main types of organic reactions are:
i. Addition reaction
ii. Substitution reaction
iii. Elimination reaction
iv. Rearrangement Reaction

I) Addition Reaction

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

 Two types of addition :-

a) Electrophilic Addition Reaction

 Initiated by an electrophile, which attacks a nucleophilic site of a

molecule.
 Typical reaction of unsaturated compounds such as alkenes and

alkynes.
 Example :


room

CH3CH=CH2 + Br2 CH3CH(Br)CH2Br
electrophile
temperature

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b) Nucleophilic Addition Reaction

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

 Typical reaction of carbonyl compounds.
 Example :

−

O OH

CH3 C CH 3 + HCN CH3 C CH3
+ nucleophile CN

II) Substitution Reaction

 A reaction in which an atom or group in a molecule is replaced by
another atom or group.

 Three types of substitution :-

a) Free-radical Substitution Reaction

 Substitution which involves free radicals as intermediate species.
 Example :

uv light

CH3CH3 + Cl2 CH3CH2Cl + HCl

b) Electrophilic Substitution Reaction

 Typical reaction of aromatic compounds.
 The aromatic nucleus has high electron density, thus it is nucleophilic

and is prone to electrophilic attack.

 Example :

+ Br Br FeBr3 Br

electrophile + HBr

c) Nucleophilic Substitution Reaction

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

 Example : 
+ −
CH3CH2OH + Br-(aq)
CH3CH2Br + OH-(aq)
nucleophile

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III) 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.
 Example :

conc. H2SO4

CH3CH2OH CH2= CH2 + H2O



IV) 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 :

tautomerisme H CR
H C CR HC O

H OH H

EXERCISE 6
State the type of reaction for the following reaction.

a) CH3Br + KOH → CH3OH + KBr
b) CH3CH2OH → H2C═CH2 + H2O
c) H2C═CH2 + H2 → CH3CH3

Prepared by: Revised by:
ELMI SHARLINA BINTI MD SUHAIMI ZANARINA BINTI SAPIAI

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1. (a) Define functional group.

(b) Circle and classify organic compounds according to their functional group and class of
compound.

No Organic compounds Functional group Class of compound

1 CH3-CH3

2 H2C=CH2

3 HC≡CH

4 CH3

5 CH3CH2OH

6 CH3CH2Cl

7 H3COCH3

O
8

H3C C H

O

9

H3C C CH3

O

10

H3C C OH

11 O
H3C C O CH3

12 O

H3C C Cl

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OO
13

H3C C O C CH3

O
14

H3C C NH2

H
15 H3C N CH3

16 CH3 C N

2.
(a) Give the 8 structures of organic compounds.
(b) Define hydrocarbon and give one example.

3. Define homologous series and state four general characteristics of its members.
4. Define structural formula.
5. Draw the expanded structure of all compounds below and state the total number of single

bond, double bond and triple bonds.
(a) CH4
(b) C2H6
(c) C2H4
(d) C2H2
(e) C6H6
(f) CH3CH2OH
(g) CH3OCH3
(h) CH3CHO
(i) CH3COOH
(j) CH3COCH3

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(k) CH3CH2Cl
(l) CH3COOCH3
(m) CH3NH2
(n) CH3COCl
(o) CH3CH2COOCOCH2CH3
(p) CH3CONH2
(q) CH3CH2CN
6. (a) State the difference(s) between saturated and unsaturated organic compounds.
(b) Give an example for each saturated and unsaturated organic compound.
7. Draw the condensed structure for each of the compounds:
(a) OH
(b) O

(c)
(d) OH

Cl O

8. Draw the expanded structure for each compound given below, label and state the number of
primary (1o), secondary (2o), tertiary (3o) and quarternary (4o) carbons.

(a) O
CH3 CH2 CH2 C CH3

(b) CH3 CH3
CH3CH CH2CH2CH CH2CH3

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(c) (CH3)3CCH2CH2CH2OH

(d) Br
CH3CH2CHCH2C OH
O

9 For each of the following compounds, draw its isomeric pair that represents a chain,
positional and functional group isomerism.

(a) CH3CH2CH=CH2
(b) CH3CH(CH3)CH(CH3)CH3
(c) CH3CH2CH2CH2CH2OH

10 (a) Draw all possible isomers of compound with the molecular formula C4H10O and identify the
. types of isomerism present.

(b) Draw all possible isomers of compound with the molecular formula C3H6O and identify the
types of isomerism present.

(c) Draw all possible isomers of compound with the molecular formula C5H10O2 and identify
the types of isomerism present.

11 Draw the structural formula for
. i. an aldehyde (C5H10O)

ii. a ketone (C4H8O)
iii. a carboxylic acid (C3H6O2)
iv. an amine (C5H13N)
v. an amide (C2H5NO)
vi. an alcohol (C2H6O)
vii. an ester (C3H6O2)
viii. an ether (C2H6O)

12 Identify and name the functional groups present in the following molecules:
.

(a) CH2OH

NHCH3

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(b) CH2OCH3
NHCOCH 3
(c)
O Cl

CC COOH
O
C H
O C

O

13 Based on the reaction scheme below, state the type of reaction corresponding to steps A, B and C.
.

CH3 CH CH2 CH3 Br2

CH3 CH CH2 CH2 Br

CH3 uv CH3

A
B KOH/ethanol

Br HBr CH3 C CH CH3
CH3
CH3 C CH2 CH3 C
CH3

14. State the type of reactions: SOCl 2 O
(a) O C Cl
C OH

(b) CH2=CH2 + HCl ⎯→ CH3CH2Cl
(c) CH2=CH2 + H2O ⎯H⎯2SO⎯4 ,⎯cata⎯lyst → CH3CH2OH
(d) CH3CHBrCH3 ⎯K⎯OH⎯, alc⎯o⎯hol → CH2=CHCH3 + HBr
(e) CH4 + Br2 ⎯⎯uv→ CH3Br + HBr
(f) NO2

+ HNO3 H2SO4

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(g) Br + H
CH3CH CH CH3 Zn CH3COOH CH3C CH CH3
Br

15. Arrange the following carbocations in order of increasing stability.

OBJECTIVE QUESTIONS

1. A hydrocarbon is a compound consisting of

A. Carbon and oxygen only
B. Carbon and hydrogen only
C. Carbon, hydrogen and oxygen only
D. Carbon, hydrogen, oxygen and chlorine only

2. Which of the following does not contain a secondary carbon atom ?

A. (CH3)3CCH(CH3)2
B. CH3CH2CH3
C. C6H5CH2CH2CH3
D. (CH3)2CHCH2CH3

3. The compound R has the structural formula shown below.

OCH 3
CO

R is classified as C. Ester
D. Ketone
A. carboxylic acid
B. aldehyde

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4. The functional groups present in the amino acid with the structural formula as shown below are

HO CH2CHCOOH
NH2

A. alkoxy, carboxyl, amide and aromatic ring.
B. alkoxy, carboxyl, amino and aromatic ring.
C. hydroxyl, carboxyl , amino and aromatic ring.
D. hydroxyl, carbonyl, amino and aromatic ring.

5. The number of possible structural isomers for bromobutane, C4H9Br is
A. 2
B. 3
C. 4
D. 5

6. Which of the following compounds has a quaternary carbon atom?
A CH3CH2CH3
B CH3CH(OH)CH2OH
C (CH3)3CCH2CH3
D C6H5COCH(CH3)2

7. The functional groups found in the following compound E are

OH O O

NH2 OCH2CH3

A. amine, alkoxy, carbonyl, carboxyl

B. amino, hydroxyl, ketone, alkoxy

C. amino, hydroxyl, carbonyl, carboalkoxy

D. amino, alkoxy, carbonyl, carboalkoxy

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8. Which of the following is NOT grouped in the same homologous series?
A Propane, Butane, hexane
B Ethanol, methyl propanol, 1-butanol
C Ethanoic acid, propanoic acid, nitric acid
D Heptene, ethane, pentene.

9. What is the functional group of alcohol?
A Carboxyl
B Hydroxyl
C Hydroxide
D Carbonyl

10. Which of the following homologous series shows the CORRECT general formula and its
functional group?

Homologous series General formula Functional Group

A Alkane CnH2n C=C

B Alkene CnH2n+2 C-C

C Alcohol CnH 2n+1OH -OH

D Carboxylic acid CnH 2n+1 COOH -COO-

Revised by:
FARHANA BINTI UMANAN

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CHAPTER 2 : ALKANES

2.0 ALKANES 2.1 Structure of Alkanes as saturated
alkane hydrocarbons with the general

2.2 IUPAC formula CnH2n+2 for acyclic
Nomenclature alkanes and CnH2n for
cycloalkane
2.3 Sources of
alkane Straight chain and branched
alkanes(parent chain ≤ C10)

Cyclic alkanes (C3 ~ C6)

Alkyl groups (- R)

State the natural source of
alkanes from petroleum.

2.4 Physical Compare the boiling points
properties of i. alkanes based on
molecular weight
alkanes ii. isomeric alkanes

2.5 Chemical Solubility of alkanes in water
properties of and non-polar solvent

alkanes Unreactivity of alkane

Combustion of alkane in excess
oxygen and limited oxygen

Monosubstitution of alkanes
containing equivalent type of

hydrogen atoms

Free radical substitution
mechanism

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2.1 STRUCTURE OF ALKANE

✔ Alkanes are organic compounds that consist entirely of single-bonded carbon
and hydrogen atoms and lack any other functional groups.

✔ Alkanes have the general formula CnH2n+2, and can be subdivided into the

following three groups:
i. Linear straight-chain alkanes
ii. Branched alkanes
iii. Cycloalkanes

✔ Alkanes are also saturated hydrocarbons.
✔ Alkanes are the simplest and least reactive hydrocarbon species containing only

carbons and hydrogens.
✔ They are commercially very important, being the principal constituent of gasoline

and lubricating oils and are extensively employed in organic chemistry; though
the role of pure alkanes (such as hexanes) is delegated mostly to solvents.
✔ Have only sigma (σ) bonds and no functional groups.

✔ Two categories in alkane:
i. Acyclic alkanes CnH2n + 2
ii. Cycloalkanes CnH2n

EXAMPLE:

CH4 CH3CH2CH2CH2CH3 2-methylpropane Cyclopentane
Methane Pentane

EXAMPLE 1

Which molecular formulas correspond to acyclic alkane?
(a) C12H26
(b) C8H16
(c) C30H62

Note: acyclic = non cyclic = open chain

ANSWER:

Acyclic alkane: CnH2n+2
Cycloalkane: CnH2n
So, (a) & (c) is a non-cyclic alkane

EXAMPLE 2:

Which molecular formulas correspond to acyclic alkane?
(a) C7H14
(b) C22H24
(c) C10H22

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Note: acyclic = noncyclic = open chain
ANSWER:

Acyclic alkane: CnH2n+2
Cycloalkane: CnH2n
So, (c) C10H22 is acyclic alkane

2.2 NOMENCLATURE OF ALKANE

Alkanes are the simplest family of hydrocarbons - compounds containing carbon and
hydrogen only. Alkanes only contain carbon-hydrogen bonds and carbon-carbon
single bonds. The formula of any of the alkanes follow the general
formula CnH2n+2. The first ten alkanes are tabulated below:

Number of C atoms Molecular formula Name
1 Methane CH4
2 Ethane C2H6
3 Propane C3H8
4 Butane C4H10
5 Pentane C5H12
6 Hexane C6H14
7 Heptane C7H16
8 Octane C8H18
9 Nonane C9H20
10 Decane C10H22

Cycloalkanes

Cycloalkanes also only contain carbon-hydrogen bonds and carbon-carbon single
bonds, but the carbon atoms are joined in a ring. The smallest cycloalkane is
cyclopropane. Example of cycloalkanes:

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If you count the carbons and the hydrogens, you will see that they no longer adhere
to the general formula CnH2n+2. By joining the carbon atoms in a ring, two hydrogen
atoms are lost. The general formula for a cycloalkane is CnH2n and these are non-
planar molecules.
SYSTEMATIC NAMES
A systematic method of naming compounds was developed by International Union
of Pure and Applied Chemistry. It is referred to as IUPAC system of nomenclature.
Follow the accepted rules and indicates the compound’s chemical structure.

STEP TO DETERMINE THE IUPAC NAME:

Step 1 :
Find the parent carbon chain

Step 2:
Number the parent carbon chain

Step 3:
Identify and number the substituents

Step 4:
Write the name as a single word

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STEP 1 : PARENT NAME
Find the longest continuous carbon chain. It does not matter if the chain is
straight or bent.

EXAMPLE:

parent name: hexane

EXAMPLE:

If there are two chains of equal length, pick the chain with more
substituents

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STEP 2 : NUMBERING PARENT CHAIN

- Give the first substituent the lower number

STEP 3 : IDENTIFY AND NUMBER THE SUBSTITUENTS
- If the first substituent is the same distance from both ends, number the chain
to give the second substituent the lower number

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- If the location of two different substituents are same from both ends of the

main chain, assign the lower number alphabetically to the first substituent.

STEP 4 : WRITE THE NAME AS A SINGLE WORD
Writing the Name

- As a single word (no spacing)
Separate:

- Number from letter by hyphen “–”
- Numbers by comma “,”

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- If two or more different substituents are present, write them in alphabetical
order

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- When two or more substituents are identical, indicate these using prefixes

“di”, “tri”, “tetra”, “penta”, “hexa”, etc.

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- In deciding on alphabetical order, ignore prefixes such as “di”, “tri” and “tetra”

- Make sure each substituent has a number

Keep in mind!
If the structure given is too “condensed” and difficult to understand, convert it to C
skeleton

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EXAMPLE 1 :
Give the IUPAC name for each compound.

a) b)

c)

ANSWER:
a. 2,3–dimethylpentane
b. 2,4–dimethylhexane
c. 2,3,5–trimethylhexane

EXAMPLE 2 :
Give the IUPAC name for each compound.
a.

b.

ANSWER :
a. 3-ethyl-4–methylhexane
b. 3-ethyl-5–methylheptane

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CHAPTER 2: ALKANES | CHEMISTRY UNIT KMKt 2021/2022
EXERCISE 1 :
Give IUPAC name for the following molecules:

a. b.

d.
c.
EXERCISE 2 :
Give IUPAC name for the following molecules:

a. b.

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Some Common Substituent Groups

No. of Alkane Alkyl Group Structure Formula
Carbon Methane methyl

1 Ethane Ethyl
2
Propane propyl
3 Isopropyl

butyl

isobutyl

4 Butane sec-butyl

tert-butyl

pentyl
5 Pentane

neopentyl

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Substituent for cyclic structure

For another substituent, refer to the table below:

Name Substituent
bromo -Br
chloro -Cl
fluoro -F
iodo -I
hydroxyl -OH
amino -NH2
cyano -CN
nitro -NO2

- Prefix “iso” is considered part of alkyl group name for alphabetizing purpose,
but “sec” and “tert” are not considered

isopropyl, isobutyl listed under “i”

sec–butyl, tert–butyl listed under “b”

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EXAMPLE :

Substituent 4–isopropyl–2–methylheptane
Parent
4–isopropyl
2-methyl
heptane

Substituent 4–tert–butyl–2–methylheptane
Parent
4–tert–butyl
2–methyl
heptane

EXERCISE 3 b.
Name the following alkanes

a.

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NAMING CYCLOALKANES

Cycloalkanes are cyclic hydrocarbons, meaning that the carbons of the molecule are
arranged in the form of a ring. Cycloalkanes are also saturated, meaning that all of the
carbons atoms that make up the ring are single bonded to other atoms (no double or triple
bonds). There are also polycyclic alkanes, which are molecules that contain two or more
cycloalkanes that are joined, forming multiple rings.

EXAMPLE:

RULES FOR NAMING CYCLOALKANES
- If number of C atoms in ring is equal to or greater than the number in the

substituent, the compound is named as an cycloalkane.

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methylcyclopentane

1- cyclopropylbutane

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- When the compound contains important functional group, the cyclic portion is

sometimes named as cycloalkyl substituent.

- “cyclo” is considered part of alkyl–group name for alphabetizing purpose.

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- If only one substituent is present, no number is needed.

- When two same substituents are present, number the ring so that the
second substituent has as low number as possible.

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