Organic chemistry and inorganic chemistry .pdf

Course No: CHE 101N Credit: 3.0 Year: First Semester: Second
Course Title: General Chemistry PHY Course Status: Theory
Course Content:
1) Atoms, molecules and ions: Atomic Theory, components of atoms.
Electronic Structure: The quantum theory, the atomic spectrum of
hydrogen and the Bohr model, Quantum numbers, Energy levels and
orbitals, Electronic configuration, Chemical bonding and molecular
structure.
2) The periodic Table: Development of the periodic table, Electron
arrangements and the periodic table, Summarized chemical properties of
s-block, p-block, d-block and f-block elements.
3) Chemical formulas and equations: Types of formulas, Percent
composition from formula, Formulas from experiment, Formulas of ionic
compounds, Names of compounds, Writing and balancing chemical
equations, Mass relations in reactions, Limiting reagent and theoretical
yield. Concept of mole, Solution: different concentration units.
4) Acids and Bases: Theories and Modern definition of acids and bases,
Dissociation constant, strength, pH, Buffer solution etc.

5) Gaseous State: Measurement on gases, the ideal gas law, Volumes of
gases involved in reactions, Gas mixtures, Partial pressure, Kinetic theory
of gases, Real gases.
6) Introduction to Chemical Kinetics: Rate laws, rate constant,
equilibrium constant, order of reaction etc.
7) Organic Chemistry: Introduction, Classification of Organic
compounds, Nomenclature, Synthesis, Physical & Chemical properties
and application of (i) Aliphatic and aromatic hydrocarbons, (ii) alcohols
and amines, (iii) Carbonyl compounds, (iv) Carboxylic acids and their
derivatives, (v) Carbohydrates (mono- and disaccharides) etc.
8) Modern Perspective of Chemistry: (a) Fuels e. g. Hydrocarbon,
Hydrogen (b) Fertilizer (c) Medicine (d) Electronic Industries e.g. LCD,
pure Silicon for IC, Semiconductor, insulator, etching materials etc.

What is organic chemistry?
• study of carbon, the compounds it makes, and
the reactions it undergoes
• over 16 million carbon-containing compounds
are known
• because the C-C single bond (348 kJ mol-1) and
the C-H bond (412 kJ mol-1) are strong, carbon
compounds are stable
• carbon can form chains and rings

Homologous series/compounds (10.1.1)
• related compounds that have the same
functional group (groups of atoms found
within molecules that are involved in the
chemical reactions characteristic of those
molecules)

General members of a homologous series
general formula is CnH2n+2 - for non-cyclic alkanes
saturated hydrocarbons - all carbon-carbon bonding is single
bonds are spaced tetrahedrally about carbon atoms.
Isomerism the first example of structural isomerism occurs with C4H10
BUTANE 2-METHYLPROPANE
Structural isomers have the SAME MOLECULAR FORMULA BUT
DIFFERENT STRUCTURAL FORMULA
They possess different physical properties such as boiling point,
melting point and density
ALKANES

HYBRIDISATION OF ORBITALS
The electronic configuration of a
carbon atom is 1s22s22p2
1 1s
2
2s
2p

HYBRIDISATION OF ORBITALS
The electronic configuration of a
carbon atom is 1s22s22p2
1 1s
2
2s
2p
If you provide a bit of energy you
can promote (lift) one of the s
electrons into a p orbital. The
configuration is now 1s22s12p3
1 1s
2
2s
2p
The process is favourable because the of arrangement of
electrons; four unpaired and with less repulsion is more stable

HYBRIDISATION OF ORBITALS IN ALKANES
The four orbitals (an s and three p’s) combine or HYBRIDISE to give four
new orbitals. All four orbitals are equivalent.
Because one s and three p orbitals are used, it is called sp3 hybridisation
2s22p2 2s12p3 4 x sp3

In ALKANES, the four sp3
orbitals of carbon repel each
other into a TETRAHEDRAL
arrangement with bond angles
of 109.5º.
Each sp3 orbital in
carbon overlaps with
the 1s orbital of a
hydrogen atom to form
a C-H bond.
THE STRUCTURE OF ALKANES
109.5º

Boiling point increases as they get more carbon atoms in their formula
more atoms = greater induced dipole-dipole interactions
greater intermolecular force = more energy to separate the molecules
greater energy required = higher boiling point
CH4 (-161°C) C2H6 (-88°C) C3H8 (-42°C) C4H10 (-0.5°C)
difference gets less - mass increases by a smaller percentage
PHYSICAL PROPERTIES OF ALKANES

Boiling point increases as they get more carbon atoms in their formula
more atoms = greater induced dipole-dipole interactions
greater intermolecular force = more energy to separate the molecules
CH4 (-161°C) C2H6 (-88°C) C3H8 (-42°C) C4H10 (-0.5°C)
difference gets less - mass increases by a smaller percentage
Straight chains molecules have greater interaction than branched
“The greater the branching, the lower the boiling point”
HIGHEST BOILING POINT LOWEST BOILING POINT
STRUCTURAL ISOMERS OF C5H12

Melting point general increase with molecular mass
the trend is not as regular as that for boiling point.
Solubility alkanes are non-polar so are immiscible with water
they are soluble in most organic solvents.

Introduction - fairly unreactive; (old family name, paraffin, meant little reactivity)
- have relatively strong, almost NON-POLAR, SINGLE covalent bonds
- they have no real sites that will encourage substances to attack them
Combustion - make useful fuels - especially the lower members of the series
- react with oxygen in an exothermic reaction
complete CH4(g) + 2O2(g) ——> CO2(g) + 2H2O(l)
combustion
incomplete CH4(g) + 1½O2(g) ——> CO(g) + 2H2O(l)
combustion
the greater the number of carbon atoms, the more energy produced
BUT the greater the amount of oxygen needed for complete combustion.
Handy tip When balancing equations involving complete combustion, remember...
every carbon in the original hydrocarbon gives one carbon dioxide and
every two hydrogen atoms gives a water molecule.
Put the numbers into the equation, count up the O’s and H’s on the RHS
of the equation then balance the oxygen molecules on the LHS.
CHEMICAL PROPERTIES OF ALKANES

Reagents chlorine and methane
Conditions UV light or sunlight - heat is an alternative energy source
Equation(s) CH4(g) + Cl2(g) ——> HCl(g) + CH3Cl(g) chloromethane
CH3Cl(g) + Cl2(g) ——> HCl(g) + CH2Cl2(l) dichloromethane
CH2Cl2(l) + Cl2(g) ——> HCl(g) + CHCl3(l) trichloromethane
CHCl3(l) + Cl2(g) ——> HCl(g) + CCl4(l) tetrachloromethane
Mixtures free radicals are very reactive - they are trying to pair their electron
with sufficient chlorine, every hydrogen will eventually be replaced.
CHLORINATION OF METHANE

Reagents chlorine and methane
Conditions UV light or sunlight - heat is an alternative energy source
Equation(s) CH4(g) + Cl2(g) ——> HCl(g) + CH3Cl(g) chloromethane
CH3Cl(g) + Cl2(g) ——> HCl(g) + CH2Cl2(l) dichloromethane
CH2Cl2(l) + Cl2(g) ——> HCl(g) + CHCl3(l) trichloromethane
CHCl3(l) + Cl2(g) ——> HCl(g) + CCl4(l) tetrachloromethane
Mixtures free radicals are very reactive - they are trying to pair their electron
with sufficient chlorine, every hydrogen will eventually be replaced.
Mechanism Mechanisms portray what chemists think is going on in the reaction,
whereas an equation tells you the ratio of products and reactants.
Chlorination of methane proceeds via FREE RADICAL SUBSTITUTION
because the methane is attacked by free radicals resulting in
hydrogen atoms being substituted by chlorine atoms.
The process is a chain reaction.
In the propagation step, one radical is produced for each one used

Initiation Cl2 ——> 2Cl• RADICALS CREATED
The single dots represent UNPAIRED ELECTRONS
During initiation, the WEAKEST BOND IS BROKEN as it requires less energy.
There are three possible bonds in a mixture of alkanes and chlorine.
412 348 242
Average bond enthalpy kJ mol-1
The Cl-Cl bond is broken in preference to the others as it is the weakest and
requires requires less energy to separate the atoms.

Propagation Cl• + CH4 ——> CH3• + HCl RADICALS USED and
Cl2 + CH3• ——> CH3Cl + Cl• then RE-GENERATED
Free radicals are very reactive because they want to pair up their single electron.
They do this by abstracting a hydrogen atom from methane; a methyl radical is formed
The methyl radical is also very reactive and attacks a chlorine molecule
A chlorine radical is produced and the whole process can start over again

Termination Cl• + Cl• ——> Cl2 RADICALS REMOVED
Cl• + CH3• ——> CH3Cl
CH3• + CH3• ——> C2H6
Removing the
reactive free
radicals brings an
end to the reaction.
This is not very
likely at the start of
the reaction
because of their low
concentration.

Initiation Cl2 ——> 2Cl• radicals created
Propagation Cl• + CH4 ——> CH3• + HCl radicals used and
Cl2 + CH3• ——> CH3Cl + Cl• then re-generated
Termination Cl• + Cl• ——> Cl2 radicals removed
Cl• + CH3• ——> CH3Cl
CH3• + CH3• ——> C2H6
OVERVIEW
Summary
Due to lack of reactivity, alkanes need a very reactive species to persuade them to react
Free radicals need to be formed by homolytic fission of covalent bonds
This is done by shining UV light on the mixture (heat could be used)
Chlorine radicals are produced because the Cl-Cl bond is the weakest
You only need one chlorine radical to start things off
With excess chlorine you get further substitution and a mixture of chlorinated products

Initiation
Propagation
Termination
RADICALS
PRODUCED
RADICALS USED
AND REGENERATED
RADICALS
REMOVED

Further
propagation If excess chlorine is present, further substitution takes place
The equations show the propagation steps for the formation of...
dichloromethane Cl• + CH3Cl ——> CH2Cl• + HCl
Cl2 + CH2Cl• ——> CH2Cl2 + Cl•
trichloromethane Cl• + CH2Cl2 ——> CHCl2• + HCl
Cl2 + CHCl2• ——> CHCl3 + Cl•
tetrachloromethane Cl• + CHCl3 ——> CCl3• + HCl
Cl2 + CCl3• ——> CCl4 + Cl•
Mixtures Because of the many possible reactions there will be a mixture of products.
Individual haloalkanes can be separated by fractional distillation.

General are members of a homologous series
hydrocarbons - contain only C and H
general formula is CnH2n - for non-cyclic alkenes
unsaturated - atoms can be added to their formula
contain a C=C double bond somewhere in their structure
THE STRUCTURE OF ALKENES
Because carbon atoms can link up with other carbon atoms in straight and
branched chains, carbon can form more compounds than any other element.
Q. Explain why carbon is able to form so many more compounds than
any other element.

General are members of a homologous series
hydrocarbons - contain only C and H
general formula is CnH2n - for non-cyclic alkenes
unsaturated - atoms can be added to their formula
contain a C=C double bond somewhere in their structure
Structure spacial arrangement around the C=C is planar
the bond angles are 120°

HYBRIDISATION OF ORBITALS - ALKENES
Alternatively, only three orbitals (an s and two p’s) combine or
HYBRIDISE to give three new orbitals. All three orbitals are
equivalent. The remaining 2p orbital is unchanged.
2s22p2 2s12p3 3 x sp2 2p
In ALKENES, the three
sp2 orbitals repel each
other into a planar
arrangement and the 2p
orbital lies at right angles
to them

Covalent bonds are formed
by overlap of orbitals.
An sp2 orbital from each carbon
overlaps to form a single C-C bond.
The resulting bond is called
a SIGMA (δ) bond.

 The two 2p orbitals also overlap to form a second bond. This is
known as a PI (π) bond.
 For maximum overlap and hence the strongest bond, the 2p
orbitals are in line.
 This gives rise to the planar arrangement around C=C bonds.

two sp2 orbitals overlap to form a sigma
bond between the two carbon atoms
ORBITAL OVERLAP IN ETHENE - REVIEW
two 2p orbitals overlap to form a pi
bond between the two carbon atoms
s orbitals in hydrogen overlap with the sp2
orbitals in carbon to form C-H bonds
the resulting shape is planar
with bond angles of 120º

Boiling point trends are similar to those shown in alkanes
increases as they get more carbon atoms in their formula
more atoms = greater intermolecular Van der Waals’ forces
greater intermolecular force = more energy to separate molecules
the lower members are gases at room temperature and pressure
cyclohexene C6H10 is a liquid
for isomers, greater branching = lower boiling point
C2H4 (- 104 °C) C3H6 (- 48°C) ....... C6H10 (83°C)
Melting point general increase with molecular mass
the trend is not as regular as that for boiling point.
Solubility alkenes are non-polar so are immiscible (don’t mix with) with water
miscible with most organic solvents.
PHYSICAL PROPERTIES OF ALKENES

CHEMICAL PROPERTIES OF ALKENES
ELECTROPHILIC ADDITION MECHANISM
The main reaction of alkenes is addition
These species are called electrophiles; they
possess a positive or partial positive charge
somewhere in their structure.
Examples include... hydrogen halides
concentrated H2SO4
Because of the extra electron density
in a C=C double bond, alkenes are
attacked by species which ‘like’
electrons.

CHEMICAL PROPERTIES OF ALKENES
The addition of bromine dissolved in tetrachloromethane (CCl4) or
water (known as bromin. If the reddish-brown colour is removed
from the bromine solution, the substance possesses a C=C bond.e
water) is used as a test for unsaturation.
ELECTROPHILIC ADDITION OF BROMINE
TEST FOR UNSATURATION
PLACE A SOLUTION OF BROMINE
IN A TEST TUBE
ADD THE HYDROCARBON TO BE
TESTED AND SHAKE
IF THE BROWN COLOUR
DISAPPEARS THEN THE
HYDROCARBON IS AN ALKENE
A
B
C
A B C
Because the bromine adds to the alkene, it no longer exists as
molecular bromine and the typical red-brown colour disappears

When a hydrogen halide react with an unsymmetrical alkenes, there are
two possible addition products as the hydrogen atom is placed on one or
other carbon of the double bond. Thus propylene with HBr can form n-
propylbromide and isopropyl bromide. But it has been established
experimentally that isopropyl bromide is obtained predominantly.
Addition to unsymmetrical alkenes
Isopropyl bromide
(MAJOR PRODUCT)
N-propyl bromide
(MINOR PRODUCT)
CH
CH2
Br
CH3
CH
CH3
CH3
HBr
CH2
CH2
CH3
Br

Russian chemist Markovnikov studied a number of addition reactions
and formulated an empirical rule:
In the ionic addition of an unsymmetrical reagent (H-G), the
hydrogen or positive end of the reagent becomes attached to the
carbon atom of the double bond bearing the larger number of
hydrogen atom.
Markovnikov Rule
Where, HG=
Unsymmetrical
reagent; R-alkyl
H
R
H
H
H G
+

Carbocation Stabilities
A carbocation is an ion with a positively-charged carbon atom.

The mechanism of addition reaction of halogen halides to unsymmetrical
alkenes involves the formation of carbonium ion (carbocation).
Explanation : Mechanism
Step 1: Formation of a cyclic intermediate and Br -
CH
CH
H
CH
HBr
CH
CH
CH 3
2
3
2 









Br





 Br
CH
CH
CH 3
2

H
Step 2 : the cyclic intermediate is capable of forming two carbonium ions,
one primary and one secondary.

CH
CH
CH 3
2 


H
Step 3: The more stable carbonium ion react with the nucleophile Br- to
give the isopropyl bromide , the major product.
CH
H
C
CH 3
3 


3
2
2 H
C
CH
H
C 


-
3
3 Br
CH
H
C
CH 



 CH
H
C
CH 3
3 

Br
2 carbonium ion (more stable)
1 carbonium ion (less stable)
Isopropyl bromide
(major product)

Modern Statement of Markovnikov’s Rule
With this understanding of the mechanism for the ionic addition of
hydrogen halides to alkenes, we can now give the following modern
statement of Markovnikov’s rule.
“In the ionic addition of an unsymmetrical reagent to a double bond,
the positive portion of the adding reagent attaches itself to a carbon
atom of the double bond so as to yield the more stable carbocation
as an intermediate”.
An Exception to Markovnikov’s Rule
This exception concerns the addition of HBr to alkenes when the addition
is carried out in the presence of peroxides (i.e., compounds with the
general formula ROOR).

When alkenes are treated with HBr in the presence of peroxides, an
anti-
Markovnikov addition occurs in the sense that the hydrogen atom
becomes attached to the carbon atom with the fewer hydrogen
atoms.
This anti-Markovnikov addition occurs only when HBr is used in
the presence of peroxides and does not occur significantly with
HF, HCl, and HI even when peroxides are present.

The vast majority of organic molecules contain elements in addition to carbon
and hydrogen. However, most of these substances can be viewed as
hydrocarbon derivatives, molecules that are fundamentally hydrocarbons
but that have additional atoms or groups of atoms called functional groups.
Functional Groups:

The vast majority of organic molecules contain elements in addition to carbon
and hydrogen. However, most of these substances can be viewed as
hydrocarbon derivatives, molecules that are fundamentally
hydrocarbons but that have additional atoms or groups of atoms called
functional groups.

 Alcohols have higher solubility and boiling points than
hydrocarbons of the same molecular weight.
 This is due to the hydrogen bonding of alcohol. Alcohols form
intermolecular hydrogen bonding.
Alcohol
Sources of alcohols:
(a) Petroleum
b) Fats
(c) Sugars
(d) natural gas
(e) Coal
(f) Biomas

Industrial preparation of alcohols:
General Preparation of alcohols:

Test for –OH group:
1) Sodium metal is added to alcohol: If bubbles of H₂ gas.
given off, the compound contains an -OH group.
2 ROH + 2Na 2 RONa + H2(g)

2) Phosphorus pentachloride is added to alcohol: If the
mixture becomes warm with evolution of hydrogen
chloride gas, the given substance is hydroxy compound.
ROH + PCl5 → RCI + POCl2 + HCl ↑

Test for –OH group:
3) Acetyl chloride or benzoyl chloride is added to alcohol:
The separation of an oily layer of ester and evolution of HCl gas
indicates the presence of аn OH group. It may be noted that if the
substance was taken in an organic solvent, the oily layer may
dissolve in it and thus limiting the test to the evolution of HCl gas
only.
+ HCl

How to distinguish between 1°,2° and 3° alcohols?

Phenols are acidic in character, while alcohols are almost neutral:

Carbonyl compounds
The carbonyl carbon atom is sp2
hybridized; the bond angles
approximately 120:

Carbonyl compounds
Preparation of aldehydes:

Reactions of carbonyl compounds
One of the most important reactions of carbonyl compounds is the
nucleophilic addition to the carbonyl group.
 The carbonyl carbon bears a partial positive charge ⇒ the carbonyl
group is susceptible to nucleophilic attack.
 The electron pair of the nucleophile forms a bond to the carbonyl
carbon atom.
 The carbonyl carbon can accept this electron pair because one pair of
electrons of the carbon-oxygen group double bond can shift out to
the oxygen.
A nucleophilic addition reaction is a chemical addition reaction in which a
nucleophile forms a sigma bond with an electron-deficient species.

 The carbon atom undergoes a change in its geometry and its
hybridization state during the reaction.
 It goes from a trigonal planar geometry and sp2 hybridization to a
tetrahedral geometry and sp3 hybridization.
Nucleophilic addition to the carbon–oxygen double bond occurs in
either of two general ways.

The important aspect of this step is the ability of the carbonyl oxygen atom to
accommodate the electron pair of the carbon–oxygen double bond.
2. When an acid catalyst is present and the nucleophile is weak:
The carbonyl oxygen with the acid enhances electrophilicity of the carbonyl
group

Relative reactivity: Aldehydes versus Ketones
In general, aldehydes are more reactive in nucleophilic additions
than are ketones. Both steric and electronic factors favor aldehydes.
Steric Factors :
 In aldehydes, where one group is a hydrogen atom, the central carbon of
the tetrahedral product formed from the aldehyde is less crowded and the
product is more stable. Formation of the product, therefore, is favored at
equilibrium.
 With ketones, the two alkyl substituents at the carbonyl carbon
cause greater steric crowding in the tetrahedral product and make
it less stable. Therefore, a smaller concentration of the product is present
at equilibrium

Because alkyl groups are electron releasing, aldehydes are more
reactive on electronic grounds as well.
 Aldehydes have only one electron-releasing group to partially
neutralize, and thereby stabilize, the positive charge at their
carbonyl carbon atom.
 Ketones have two electron-releasing groups and are stabilized
more. Greater stabilization of the ketone (the reactant) relative to
its product means that the equilibrium constant for the
formation of the tetrahedral product from a ketone is smaller
and the reaction is less favorable:
Electronic Factors:

How would you differentiate Aldehydes from Ketones

Preparation of carboxylic acids:

Reactions of carboxylic acids:

STRUCTURE & CLASSIFICATION
Structure Contain the NH2 group
Classification
primary (1°) amines secondary (2°) amines
tertiary (3°) amines quarternary (4°) ammonium salts
Aliphatic methylamine, ethylamine, dimethylamine
Aromatic NH2 group is attached directly to the benzene ring (phenylamine)
R N:
H
H
R N:
R
H
R N:
R
R
R
+
R N R
R
Amines

NOMENCLATURE
Nomenclature Named after the groups surrounding the nitrogen + amine
C2H5NH2 ethylamine
(CH3)2NH dimethylamine
(CH3)3N trimethylamine
C6H5NH2 phenylamine (aniline)

BASIC PROPERTIES
Measurement the strength of a weak base is depicted by its pKb value
the smaller the pKb the stronger the base
the pKa value can also be used;
it is worked out by applying pKa + pKb = 14
the smaller the pKb, the larger the pKa.
Compound Formula pKb Comments
ammonia NH3 4.76
methylamine CH3NH2 3.36 methyl group is electron releasing
phenylamine C6H5NH2 9.38 electrons delocalised into the ring
strongest base methylamine > ammonia > phenylamine weakest base
smallest pKb largest pKb

• In early 19th century, the term aromatic was used to describe
some fragrant compounds
– Not correct: later they are grouped by chemical behavior
(unsaturated compounds that undergo substitution
rather than addition).
coal distillate cherries, peaches and almonds Tolu balsam
Benzene and Aromatic Compounds

• Currently, the term aromatic is used to refer to the class of
compounds related structurally to benzene
– They are distinguished from aliphatic compounds by
electronic configuration
steroidal hormone analgesic tranquilizer

• Aromatic compounds have many common names that have
been accepted by IUPAC:
• Toluene = methylbenzene
• Phenol = hydroxybenzene
• Aniline = aminobenzene
Nomenclature of Benzene Derivatives

• Monosubstituted benzenes, like hydrocarbons, are
systematically named with –benzene as the parent
name
C6H5Br C6H5NO2 C6H5CH2CH2CH3
Monosubstituted benzenes

• Arenes are alkyl-substituted benzenes
– If # Csubstituent < or = 6, then the arene is named as an
alkyl-substituted benzene
– If # Csubstituent > 6, then the arene is named as a phenyl-
substituted alkane
Arenes

Aryl groups
• “Phenyl” refers to C6H5
-
– It is used when a benzene ring is a substituent
– “Ph” or “f” can also be in place of “C6H5
-”
• “Benzyl” refers to “C6H5CH2
-”

• The ortho- (o), meta- (m), and para- (p) nomenclature is useful
to describe reaction patterns
Example: “Reaction of toluene with Br2 occurs at the para
position”

• Relative positions on a disubstituted benzene ring:
– ortho- (o) on adjacent carbons (1,2 disubstituted)
– meta- (m) separated by one carbon (1,3 disubstituted)
– para- (p) separated by two carbons (1,4 disubstituted)
Disubstituted benzenes

• Multisubstituted benzenes (more than two substituents) are
named as follows:
– Choose the sequence when the substituents have the lowest
possible number
– List substituents alphabetically with hyphenated numbers
– Use common names, such as “toluene”, as parent name (as in
TNT)
Multisubstituted benzenes

– Use common names, such as “toluene”, as parent name
 The principal substituent is assumed to be on C1

Practice Problem: Tell whether the following compounds are
ortho-, meta-, or para-disubstituted
(a) Meta (b) Para (c) Ortho

Practice Problem:
Give IUPAC names for the following compounds
(a) m-Bromochlorobenzene (b) (3-Methylbutyl)benzene
(c) p-Bromoaniline (d) 2,5-Dichlorotoluene
(e) 1-Ethyl-2,4-dinitrobenzene (f) 1,2,3,5-Tetramethylbenzene

a) p-Bromochlorobenzene
b) p-Bromotoluene
c) m-Chloroaniline
d) 1-Chloro-3,5-dimethylbenzene
Practice Problem:
Draw structures corresponding to the following compounds

93
• The resonance description of benzene consists of two equivalent Lewis
structures, each with three double bonds that alternate with three single
bonds.
• The true structure of benzene is a resonance hybrid of the two Lewis
structures, with the dashed lines indicating the position of the  bonds.
• Because each  bond has two electrons, benzene has six 
electrons.
• Benzene (C6H6) is the simplest aromatic hydrocarbon.

94
• In benzene, the actual bond length (1.39 Å) is intermediate
between the carbon—carbon single bond (1.53 Å) and the
carbon—carbon double bond (1.34 Å).

Aromatic Character: The (4n + 2 ) π Rule
Hückel’s Rule
1) A molecule must be cyclic.
2) A molecule must be planar.
3) A molecule must be completely conjugated.
4) A molecule must satisfy Hückel’s rule, and contain
a particular number of  electrons.
4n+2 π electrons (n= 0, 1, 2, 3, ….= 2, 6, 10, 14, ...)

Electrophilic Aromatic Substitution
 Benzene does not undergo addition reactions like other
unsaturated hydrocarbons, because addition would yield a
product that is not aromatic.
 Substitution of a hydrogen keeps the aromatic ring intact.
Halogenation, Alkylation, Nitration, and Sulfonation are the typical
electrophilic aromatic substitution reactions.
1- Specific Electrophilic Aromatic Substitution Reactions

General Mechanism-Electrophilic Aromatic Substitution

• A heterocycle is a cyclic compound that contains an atom
or atoms other than carbon in its ring, such as N, O, S, P
• There are many heterocyclic aromatic compounds and
many are very common.
• Example: Pyridine and Pyrrole
Aromatic Heterocycles: Pyridine and Pyrrole

• Pyridine is a six-membered heterocycle with a nitrogen atom in its
ring
•  electron structure resembles benzene (6 electrons)
• The nitrogen lone pair electrons are in sp2 orbital, not
part of the p aromatic system (perpendicular orbital)
• Pyridine is a relatively weak base compared to normal amines
but protonation does not affect aromaticity
Pyridine

• Pyrrole is a five-membered heterocycle with a nitrogen
atom in its ring
Pyrrole
  electron system is similar to that of cyclopentadienyl
anion
 Four sp2-hybridized carbons with 4p orbitals
perpendicular to the ring and 4p electrons

 It has 6p electrons: 4n + 2 = 6, thus n = 1 (an integer)
 It has a lone pair of electrons in a p orbital perpendicular to the
plane
Problem: Thiophene, a sulfur-containing heterocycle, undergoes
typical aromatic substitution reactions rather than addition
reactions. Explain why thiophene is aromatic.

 It has 6 p electrons: 4n + 2 = 6, thus n = 1 (an integer)
 It has a lone pair of electrons in a p orbital perpendicular to the plane
Problem: Draw an orbital picture of furan to show how the
molecule is aromatic.

Carbohydrates [Cx(H2O)y] are usually defined as polyhydroxy aldehydes and
ketones or substances that hydrolyze to yield polyhydroxy aldehydes and ketones.
• Simple carbohydrates are known as sugars or saccharides (Latin saccharum,
sugar) and the ending of the names of most sugars is -ose. For example:
1. Glucose (for the principle sugar in blood)
2. Fructose (for a sugar in fruits and honey)
3. Sucrose (for ordinary table sugar)
4. Maltose (for malt sugar)
A carbohydrate is an organic compound that consists only of carbon, hydrogen, and
oxygen. Three groups of carbohydrates are: monosaccharides, disaccharides, and
polysaccharides

 Monosaccharide is a simple sugar consist only of one unit ("mono" means one) and
they cannot be broken down into simple sugar units. They serve as building blocks for
more complex carbohydrate forms.
Examples: glucose, galactose (found in milk), and fructose (found in many
fruits)
These three monosaccharides are combined in various ways to make more complex
carbohydrates.

Disaccharides are formed when two monosaccharides are joined together and a
molecule of water is removed, a process known as dehydration reaction.
Examples: maltose and sucrose (table sugar)

Polysaccharides are complex carbohydrates composed of numerous monosaccharides
combined through the loss of water molecules.
Examples: cellulose (found in plant cell walls), chitin (found in animal exoskeletons),
starches
Cellulose

A reducing sugar is a sugar that has a free aldehyde or ketone and that can act as a reducing
agent. A non-reducing sugar does not have a free aldehyde or ketone, so it cannot act
as a reducing agent.
 Why is it called reducing sugar?
 Why is glucose called a 'reducing sugar'?
They are called 'reducing sugars' because the presence of the aldehyde group makes
them undergo oxidation readily to form carboxylic acid and in the process the
reactive reagents are reduced easily.
All monosaccharides are reducing sugars because all monosaccharides have an aldehyde group (if they are
aldoses) or can tautomerize in solution to form an aldehyde group (if they are ketoses). This includes
common monosaccharides like galactose, glucose, glyceraldehyde, fructose, ribose, and xylose. They are
called 'reducing sugars' because the presence of the aldehyde group makes them undergo
oxidation readily to form carboxylic acid and in the process the reactive reagents are reduced easily.
 What is reducing and non-reducing sugar?

116
Oxidation reactions of glucose
(i) Tollens’ reagent (Ag+)/ Fehling’s solution (Cu2+)
Glucose reduces Fehling's solution to reddish brown cuprous oxide. Fehling's
solution is an aqueous solution of copper sulfate, sodium hydroxide, and potassium
sodium tartrate, used to test for the presence of sugars and aldehydes in a substance.

117
3. Oxidation with strong oxidizing agent: dil. HNO3
CHO
OH
H
H
HO
OH
H
OH
H
CH2OH
COOH
OH
H
H
HO
OH
H
OH
H
COOH
dil. HNO3
D-glucaric acid)
D-glucose
CHO
OH
H
H
HO
OH
H
OH
H
CH2OH
COOH
OH
H
H
HO
OH
H
OH
H
CH2OH
Br2, H2O
D-gluconic acid)
D-glucose
2. Oxidation with mild oxidizing agent: bromine water

118
Osazone formation:
Q. Why 3 molecules of phenylhydrazine is required? 2 for making osazone and the 3rd one is
turned into aniline and ammonia.
Reduction of glucose:
CHO
OH
H
OH
H
OH
H
OH
H
CH2OH
CH2OH
OH
H
OH
H
OH
H
OH
H
CH2OH
NaBH4
D-sorbitol (glucitol)
D-glucose

119
Sucrose
Table sugar/ Ordinary sugar
Obtained from sugar cane and sugar beets
Consist of α-D-glucose and β-D-fructose
Has an α,β-1,2-glycosidic bond
Disaccharides are carbohydrates that yield two monosaccharide molecules on hydrolysis.
For example: Sucrose, lactose, maltose Or
A disaccharide (double sugar) is the sugar formed when two monosaccharides are
joined by glycosidic linkage.
Disaccharides
α-D-glucose
β-D-fructose

120
Which is invert sugar and why?
Sucrose(which is dextrorotatory) is known as invert sugar because upon hydrolysis it breaks down
into Fructose (laevorotatory) and Glucose (dextrorotatory). The specific rotation value of
Fructose is more than glucose due to which the mixture overall is laevorotatory.
Why is sucrose a non reducing sugar?
Sucrose is a non-reducing sugar because
the two monosaccharide units are held
together by a glycosidic linkage between
C1 of α-glucose and C2 of β-fructose.
Since the reducing groups of glucose and
fructose are involved in glycosidic bond
formation, sucrose is a non-reducing sugar.
It gives negative tests with Fehling’s
solution.
α-D-glucose
β-D-fructose

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