Kendriya Vidyalaya Sangathan
Lucknow Region
Study Material
Subject : CHEMISTRY
Class - XII

Prepared under supervision of
Mrs. NOMITA WILSON
PRINCIPAL, KENDRIYA VIDYALAYA IIM,LUCKNOW
Astt. Commissioner KVS (LR)
RANVIR SINGH
Under Guidance of
Dr. Satyendra pal
Principal
Kendriya Vidyalaya No. 1 , AFS, Agra
Prepared By
S.NO
NAME OF THE TEACHER
KENDRIYA VIDYALAYA
1
Dr. ABHYAS YADAV
BARABANKI
2
SANGITA RANI SINGH
GOMTI NAGAR, LUCKNOW
3
R.C.TIWARI
B.K.T. , LUCKNOW
4
B.K.SHARMA
ALIGANJ , LUCKNOW
5
A.K.DIXIT
SITAPUR
6
MANJULA DIXIT
I I M, LUCKNOW
7
RAJ SHEKHAR
I I M, LUCKNOW
CONTENTS
ONE PAPER
70 MARKS
UNIT NO.
TITLE
MARKS
I
SOLID STATE
4
II
SOLUTIONS
5
III
ELECTROCHEMISTRY
5
IV
CHEMICAL KINETICS
5
V
SURFACE CHEMISTRY
4
VI
GENERAL PRINCIPALES AND PROCESSES OF
3
ISOLATION OF ELEMENTS
VII
P-BLOCK ELEMENTS
8
VIII
D & F BLOCK ELEMENTS
5
IX
COORDINATION COMPOUNDS
3
X
HALOALKANES & HALORENES
4
XI
ALCOHALS, PHENOLS & ETHERS
4
XII
ALDEHYDES, KETONES & CARBOXYLIC ACID
6
XIII
ORGANIC COMPOUNDS CONTAINING NITROGEN
4
XIV
BIOMOLECULES
4
XV
POLYMERS
3
XVI
CHEMISRY IN EVERDAY LIFE
3
TOTAL
70
Unit-I THE SOLID STATE
Density of unit cell = Mass of unit cell =
z×M
Volume of unit cell
Na × a3
z- Number of atoms per unit cell M – Atomic mass or formula mass for ionic solids
Na – Avogadro number
a – Edge length
Radius ratio = r+ = Radius of the cation
rRadius of the anion
If R is the radius of spheres in the close packed arrangement then
I. Radius of octahedral void = r = 0.414 R
II. Radius of tetrahedral void = r = 0.225 R
Packing fraction =
Volume occupied by atoms in unit cell
Total volume of the unit cell

Different types of solids (table)
Unit cell
Distance b/w
nearest
neighbor (d)
Simple cubic
a
Face centered
a/
cubic (FCC)
Body centered
/2a
cubic (BCC)
Where, a = edge length of unit cell
Different Type of Solids
s.n. Type of
Solids
Constituent
Particles
Radius
Constituent
Atom
Coord
Ination
no.
Packing
fraction
Number of
Constituent
Atoms
a/2
a/2
6
12
0.52
0.74
8
0.68
1/8×8 = 1
(1/8×8) +
(1/2×6) = 4
(1/8×8) + 1 = 2
Bonding/
Attractive
Forces
/4a
Examples
Physical
Nature
Electrical
Conductivity
Melting
Point
1.
Molecular
Solids
(i)Non -polar
Soft
Molecules
Dispersion
or
London
forces
(ii)Polar
Ar, CCl4 ,
H2 , I2 , CO2
Insulator
Soft
Insulator
HCl , SO2
Dipoledipole
Interactions
(iii)Hydrogen
bonded
2.
Ionic solids
Ions
3.
Metallic
solids
4.
Covalent or
Network
solids
Positive
ions
In a sea of
delocalized
electrons
Atoms
Very
low
Low
Hard
Insulator
H2O(ice)
Hydrogen
bonding
Coulombic
NaCl, MgO
or
Zns, CaF2
electrostatic
Metallic
bonding
Fe, Cu, Ag,
Mg
Covalent
bonding
SiO2 (quartz),
SiC ,
C(diamond),
AIN,
C(graphite)
Low
Hard but Insulators in
brittle
solid state
but
conductors in
molten state
and in
aqueous
solutions
Hard but Conductors in
malleabl solid state as
e
well as in
and
molten state
ductile
Hard
Insulators
Soft
High
Fairly
high
Very
high
Conductor
(exception)
Important Questions;
1. Name the non-stoichiometric point defect responsible for colour in alkali halides.
Ans: F-centres
2. What happens when a ferromagnetic substance is heated to high temperature?
Ans: Ferromagnetic substance changes to paramagnetic substance due to randomization of
domains (spins) on heating.
3. A compound is formed by two elements X and Y. Atoms of the element Y (anions) make ccp
and those of the element X (cations) occupy all the octahedral voids. What is the formula of the
compound?
Ans: Suppose no. of atoms Y in ccp = N
:. No. of octahedral voids = N
:. No. of atoms X = N
Ratio of X : Y = N : N =1: 1
Formula of compound is XY
4. Explain the following terms with suitable examples:
(i) Schottky defect
(ii) F-centres
Ans (i) Stoichiometric defect when equal number of cations and anions are missing from the
lattice. Cations and anions are of almost same size. It lowers the density of solid e.g. NaCl , KCl
etc.
(ii)The electrons trapped at the anion vacancies are referred to as F-centres from German
word Farbenzenter meaning colour centre e.g. yellow colour imparted to NaCl when heated in an
atmosphere of sodium vapour.
5. Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87
g cm-3 Use this information to calculate Avogadro’s number. (At. Mass of iron = 56 g mol -1)
Ans :
d= zxM
a3 x N
A
For bcc lattice , z = 2
7.87 g cm-3 = 2 x 56 g mol-1
NA
(286.65 x 10-10 cm)3 x NA
= 6.04 x 1023 mol-1
Unit-II SOLUTIONS


Mass percent
=
Mass of component in solution x 100
Total mass of solution
Mole fraction of a component = Number of moles of the component
Total number of moles of all the components
χB =

Molarity (M)
=
Moles of solute
Volume of solution in litre
=
WB
x
1000
MB
Volume of solution in ml
WB – mass of solute
MB – molar mass of solute
 Molality (m) =
Moles of solute
Mass of solvent in Kg
=
WB x
1000
MB
WA (in grams)
WA – mass of
solvent
 Henry’s law
P = KH x χ1
P – partial pressure KH – Henry’s constant
χ1 - mole fraction
 Raoult’s law for volatile solute
PA = P°A χA
PA= partial pressure of component A in solution
P°A = vapour pressure of pure component
χA = mole fraction
P = PA + PB
= P°A χA + P°BχB
 Raoult’s law for non-volatile solute
P°A - PA = χB =
nB
= WB x MA
P°A
nA + nB
MB
WA (dilute solution)
 Elevation in boiling point
ΔTb = Kb x m = Kb x WB x
1000
MB
WA
 Depression in freezing point
ΔTf = Kf x m = Kf x WB
x
1000
An ideal solution should have:
ΔVmix = 0
ΔHmix = 0
Obeys Raoult’s law over a wide range of concentration
 Non-ideal solution
i.
ΔVmix ≠ 0
ii.
ΔHmix ≠ 0
iii.
Raoult’s law is not obeyed
Two types:
a) Positive Deviation : When partial pressure of each liquid and the resultant
total pressure is greater than the pressure expected on the basis of the
Raoult’s law
eg- Ether-acetone, Benzene-acetone , water-ethanol , n-hexane-ethanol etc.
ΔHmix = +ve
ΔVmix = +ve
PA > P°AχA
;
PB > P°BχB
i.
ii.
iii.
b) Negative deviation :When partial pressure of each liquid is less than the
vapour pressure expected on basis of Raoult’s law.
Eg- Methanol-acetic acid , water-nitric acid , acetone-chloroform , chloroformbenzene , chloroform-ether etc.
ΔHmix = -ve
1. How is the molality of a solution different from its molarity?
Ans- Molarity is the number of moles of solute per litre of solution whereas
molarity is the number of moles of solute in 1kg of the solvent.
2.State Henry’s law and mention two importants application.
Ans- Henry’s law : The pressure of gas over a solution is directly proportional to
the mol fraction of the gas dissolve in the mol fraction of the gas dissolve in tje
solution.
Applications :(i): To make soft drink bottle more rich in CO2 sealing is done at high pressure.
(ii):Deep sea divers use He and O2 mixture for respiration He is less soluble
in blood as compared to N2 at high pressure and create the pain ful effect
on the human body.
3. A 5 % sol (by mass) of Cane sugar in water has freezing point of 271K.
Calculate the freezing point of 5% glucose in water if freezing point of pure water
is 273.15K.
Ans- 5% glucose means 5gm in 100gm of sol
f=
For sugar 273.15 – 271 =
2.15 = kf × 0.154
For glucose f =
= Kf × 0.292
Comparing the two
=
f
Freezing point of solution will be 273.15K – 4.08K
= 269.07K
UNIT-III - ELECTROCHEMISTRY
Important formulae
• R=V/I
Where R is Resistance, V is Voltage, I is Current.
• Conductivity or conductance = 1/R = 1/V
The unit of conductance is
Ohm-1 = Siemens = S.
• Specific Conductance = 1/ Specific resistance
K=1/p
m = Specific Conductance/ Molarity
• ^
K/M
• Kohlrausch law - ^m = V+ ^+ + v-^-
•
Faraday’s First Law of Electrolysis
m=ZxQ
m=ZxIxt
Where is mass of substance deposited, I is current in ampere, t is time
in seconds,
Q=Ixt
Z = Eq. Wt./ 96500
Where z is electrochemical equivalent. Unit of electrochemical equivalent
is gram/coulomb
Faraday is charge on 1 mole of electrons.
Charge on 1 electron = 1.602 x 10 -19 C
Charge on 1 mole of electrons = 1.602 x 10 -19 x 6.023 x 10 23
= 96487 C
= 96500 C
•
•
Faraday’s Second Law of Electrolysis :
W1/E1 = W2/E2 = W3/E3
• Where W1 is mass of substance 1 deposited and E1 is
its equivalent weight,
• W2 is mass of substance 2 deposited and E2 is its
equivalent weight,
• W3 is mass of substance 3 deposited and E3 deposited
and E3 is its equivalent weight.
•
Nernst Equation
o
E oxidation = - Eo reduction
Eo cell = Eo cathode – Eo anode
Eo cell = Eo anode – Eo cathode
Mn+ + neM
o
E = E – 2.303 RT / nF log [M]/[Mn+]
where E = reduction electrode potential,
Eo = Standard Reduction electrode Potential,
n = No. of electrons
T = Temperature in Kelvin, F = 96500 C
Change in Gibb,s free energy ( ∆G ) = - n F E0 Cell
∆G0 = - 2.303RTlogKc
Important Questions
•
Why is the equilibrium constant K related to only Eocell and not Ecell ?
Ans: It is because Ecell at equilibrium is 0 volt .
•
Write the nernst Equation for the electrode reaction
M n+ + ne-
→
M(s)
Ans : Emn+/m = Eomn+/m +2.303RT/nF log[Mn+]
•
Predict the product of Electrolysis in each of the Following
a) An aq Sol. Of AgNO3 with silver electrodes.
b) An aq. Sol. Of AgNO3 with platinum electrodes.
Ans : At Cathode : Ag+ (aq) + eAg(s)
At anode
Ag(s)
Ag+ (aq)+ eAgNO3(aq)
Ag+(aq) + NO3-(aq); H2O(l) H+(aq) + OH- (aq)
b) At cathode
2Ag+(aq) + 2e2Ag(s)
At anode
2OH (aq)
O2(g) + 2H+(aq) + 4e● Can you store copper sulphate in zinc pot ?
Ans: No , because Zn is more reactive than Cu .
●If a current of 0.5 ampere flows through a metallic wire for 2 hours , then how
many electrons flow through the wire ?
Ans : Q = it = 5 Х 2 Х 60 Х 60 = 3600 C
No. of electrons flow = 3600 Х 6.022 Х 1023
9600
= 2.246 Х 1022
UNIT-IV Chemical Kinetics
Important formulae
•
Rate Of reaction : it is defined as the change in concentration of reactant (or
product) in a particular time interval. Unit of rate of reaction is mol L-1 s-1. If
time is in minutes, then units is mol L-1 min-1 and so on. R = K ﴾ Conc ﴿n
•
Rate Law or rate equation : It is the expression which relates the rate of
reaction with concentration of the reactants. The constant of proportionality
‘k’ is known as rate constant.
R = K [ A ]p [ B ]q Where , p and q are not stoichiometric coefficient but they
are order of the reaction
•
Rate constant : when concentration of both reactants are unity, then the rate
of reaction is known as rate constant. It is also called specific reaction rate.
Unit of K = ﴾ Conc ﴿ 1- n time -1
But in gaseous reaction , unit of K = atm 1-n time-1
•
Molecularity : Total number of atoms, ions or molecules of the reactants
involved in the reaction is termed as its molecularity. It is always in whole
number. It is never more than three. It cannot be zero.
Examples: Unimolecular reaction
NH4NO2
N2 + 2H2O
Bimolecular reaction
2HI(g)
H2(g)+I
Trimolecular reaction
2NO(g) + O2 (g)
2NO2(g)
•
•
•
Order of reaction : The sum of the exponents of the concentration of reactants
in the rate law is termed as order of the reaction. It can be in fraction. It can be
zero also.
Zero order reaction : The rate of reaction does not change with the
concentration of the reactants,
i.e., rate = k [A]0, where ‘k’ is rate constant
Unit of rate constant is mol L-1s-1
Half – life of a reaction : The time taken for a reaction when half of the starting
material has reacted is called half-life of a reaction. For first order reaction t1/2 =
0.693/k, where k is rate constant.
it is independent of initial concentration for first order reaction.
Important formulae
▪
K = 2.303 log a
t
a–x
For First order reaction
▪
t ½ = 0.693
K
▪ Slope = - K / 2.303
▪ General expression for the time taken for nth fraction of a reaction of 1st order
to complete :
t 1/n = 2.303 log 1
K
1- 1/ n
General expression for half life period of a reaction of nth order :
T 1/2 α [ Ao ] 1-n
▪ Amount of substance left after n half lives = [ Ao ] / 2n
No. of half lives = t / t ½
▪ Arrhenius equation :
K = A e –Ea / RT
Log K2 / K1 = Ea / 2.303 R ( T2 – T1 ) / T1T2
Slope = - Ea / 2.303 R
Important questions
•
A reaction is 50% complete in 2 hours and 75% complete in 4 hours. What is
the order of the reaction ?
Ans : First order because t75% = 2 t50%
•
Express the relation between the half life period of a reactant and its initial
concentration for a reaction of nth order.
Ans : t1/2 1/[R]on-1
•
Show that time required for 99% completion is twice the time required for
the completion of 90% reaction.
Ans :
t99% = 2.303/k log [R]o/([R]o – 99/100[R]0)
= 2.303/k log 100 = 2.303 x 2/k = 4.606/k
t 90% = 2.303/k log [R]0/([R]0 – 90/100[R]0 )
= 2.303/k x l
t99%/t90% = 4.606/k x k/2.303 = 2
t99% = 2t90%
UNIT-V- SURFACE CHEMISTRY
Adsorption : Phenomenon of attracting and retaining molecules of a substance on
the surface of a solid resulting in to a higher conc. On the surface than in the bulk .
Difference between Physical and Chemical adsorption :
Physical Adsorption
│
Chemical Adsorption
1. Arises due to van der waals forces
1.Chemical bond
2. Not specific
2. Highly specific
3. Reversible
3.Irreversible
4. Low enthalpy of adsorption
4.High enthalpy of adsorption
Adsorption Isotherm : Frendlich Adsorption Isotherm ( effect of Pressure )
X / m = K P 1/n
Where , x = mass of adsorbate
m = mass of adsorbent
P = pressure
K and n = Constants
Shape Selective Catalysis by Zeolites : The catalytic reaction that depends upon
the pore , structure of the catalyst and the size of the reactant and product molecules
. Zeolites are aluminosilicatees with three- dimensional network .
ZSM – 5 is used in the petroleum industry for the conversion of hydrocarbons
into gasoline
Sol = solid in liquid example starch sol
Gel = liquid in solid example butter
Emulsion = liquid in liquid example milk
Types of Colloids : 1.Lyophilic sol : Reversible
2.Lyophobic : Irreversible
3.Multimolecular : smaller to colloidal range
4.Macromolecular : bigger to colloidal range
5. Associated : soap solution
Peptisation : ppt + dispersion medium = colloid
Dialysis : Purification of colloid
Tyndall effect : Scattering of light by colloidal solution
Zeta Potential ; Potential diff . between the fixed layer and diffused layer of opposite
charges .
Electrophoresis : Movement of colloidal particles under the influence of electric
field .
Coagulation : Colloid + Electrolyte = ppt
Hardy Schulze Rule : Coagulation power of an ion α ( valency of the ion ) 4
Demulsification : E mulsions can be broken into the constituent liquids by heating ,
freezing , centrifuging .
Important Questions
Q1. What is meant by shape selective catalysis ?
Ans. The catalytic reaction yhat depends upon the pore structure and size of the
catalyst .
Q2. Define the following :
i.
Peptisation
ii.
Reversible sols
Q3.What is the diff between multimolecular and macromolecular colloids ?
Give one example of each .
Ans : Multimolecular Colloids -- smaller to colloidal range example gold sol
Macromolecular Colloids -- bigger to colloidal range example protein sol
UNIT:-VI- General Principles & Process of
Isolation of Elements
The chemical substances in the earth’s crust obtained by mining are called Minerals.
1.
2.
3.
4.
Minerals, which act as source for metal, are called Ore.
The unwanted impurities present in ore are called Gangue.
The entire process of extraction of metal from its ore is called Metallurgy.
Removal of gangue from ore is called Concentration, Dressing or Benefaction of
ore.
5. Concentration by Hydraulic washing is based on the difference in gravities of ore
and gangue particles.
6. Concentration by Magnetic separation is based on differences in magnetic
properties of ore components. If either of ore or gangue is capable of attracted
by a magnet field, then such separation is carried out.
7. Concentration by Froth Flotation Process is based on the facts that sulphide ore
is wetted by oil & gangue particles are wetted by water.
8. Concentration by Leaching is based on the facts that ore is soluble in some
suitable reagent & gangue is insoluble in same reagent. e.g. Bauxite ore contains
impurities of silica, iron oxide & TiO2 .The powdered ore is treated with NaOH
which dissolve Al & impurities remains insoluble in it.
Al2O3 +2NaOH + 3 H2O
2 Na [Al(OH)4].
10. Calcination involves heating of ore in absence of air below
melting
point of metal. In this process volatile impurities escapes leaving behind metal
oxide.
Fe2O3.xH2O
Fe2O3 +xH2O
ZnCO3
ZnO +CO2
CaCO3.MgCO3
CaO + MgO + 2CO2
11. Roasting involves heating of ore in presence of air below melting point of metal
in reverberatory furnace. In this process volatile impurities escapes leaving behind
metal oxide and metal sulphide converts to metal oxide.
2 ZnS + 3 O2
2ZnO+2SO2
2PbS + 3 O2
2 PbO +2 SO2
2 Cu2S + 3 O2
2Cu2O + 2 SO2
12. Reduction of metal oxide involves heating of metal in presence of suitable
reagent Coke or CO2.
13. Reactions taking place at different zones of blast furnace in extraction of iron:(i) Zone of reduction:- Temperature range 250oC-700oC
3Fe2O3+CO
2Fe3O4+CO2
Fe3O4+CO
3FeO+ CO2
FeO +CO
Fe+ CO2
(ii) Zone of slag formation:- Temperature range 800oC-1000oC
CaCO3
CaO+CO2
CaO+SiO2
CaSiO3,
P4O10+10C
4P+10CO,
SiO2+2C
Si+2CO,
MnO2+2C
Mn+2CO
o
o
(iii) Zone of fusion:- Temperature range 1150 C-1350 C
CO2 + C
2CO
(iv) Zone of fusion:- Temperature range 1450oC-1950oC
C +O2
CO2
ELLINGHAN DIAGRAM: For considering the choice of reducing agent in the reduction
of oxides
∆G ═ ∆H ─ T ∆S
∆G must be negative, ∆G═ ─ n E F , ∆G ═ ─ RT ln K
IMPORTANT QUESTIONS:
1.Is it true that under certain conditions Mg can reduce SiO2 and Si can reduce MgO? what
are these conditions ?
Ans.: By using Ellinghan Diagram
2. Diffrentiate between mineral and ores?
Ans: Mineral is a naturally occurring Chemical substance.
ore is a mineral from which we extract metal profitably.
3. How is cast Iron different from Pig Iron ?
Ans: Cast Iron = Iron + 3% Carbon
Pig Iron = Iron + 4% Carbon
4. Why is the reduction of a metal oxide is easier if the metal formed is in liquidstate at the
temperature of reduction/
Ans: Due to more ∆S ═ + ve
∆G become
more on negative side. And reduction become easier.
5.What is the role of depressant in froth flotation process?
Ans: An ore containing ZnS and PbS , the depressants usedis NaCN . It selectively
prevents ZnS from coming to the froth but allows PbS to come with the froth.
6.Describe Mond process and Van Arkel’s method .
Ans: Mond Process :refining of Nickel
330-350 K
450-470K
Ni + 4CO
→
Ni( CO)4
→
Ni + 4CO
Van Arkel Method : refining of Zr
Zr + 2I2
→
Zr I 4
1800K
→
Zr + 2 I 2
Unit-VII - p-BLOCK ELEMENTS
Nitrogen Family :
1) ns2np3
2) O.S. (N) -3,-2,-1,0,+1,+2,+3,+4,+5
Tendency to show +5 O.S. N>P>As>Sb>Bi
Tendency to show +3 O.S. N<P<As<Sb<Bi
Disproportionation
HNO3 →
HNO3+H2O+2NO
H3PO3 → H3PO4 +PH3
3) Formation of hydrides
angle
NH3>PH3>AsH3>SbH3>BiH3
Reducing Power Nh3<PH3…………
Boiling Point
NH3
SbH3
PH3
4) Anomalous behavior of (Nitrogen, Oxygen,Fluorine)
→Small Size, High Electronegativety,Non availability of d-orbitals & High I.E.
5) 4NH3+5O2
→
4NO+6O2
P4+3NAOH+3H2O →
PH3+3NaH2PO2(sodium hypophosphite)
PCL3+3H2O
→
H3PO3+3HCl
Sulphur Family
1)
2)
3)
4)
ns2np4
O.S.
-2
Hydrides
NO + O3
(+2+4+6)
H2E(H2O,H2S,H2Ss……………)
NO2+O2(Depleting of oxygen)
Halogens
1)
2)
3)
4)
ns2np5
O.S. <Fluorine (-1)>
(+1,+3,+5,+7)
Halogen oxides halide ions of higher atomic number.
Hydride :
Acidic strength HF<HCl<HBr<HI
Dissociation enthalpy HF>HCI>HBr>HI
5) Inter Halogen Compounds :Halogens combine among themselves to form a number of compounds
known as interhalogens of the types
XX’,XX’3,XX’5,XX’7
6) Cl2 is a powerful bleaching agent
7) Due to high electronegativity and small size. Fluorine forms only one oxoacid(HOF)
8) Acidic strength HClO<HCIO2<HCIO3<HCIO4
GROUP 18
1) ns2np6
2) Nobel gases have low b.p. because being mono atomic they have no inter atomic forces except
weak dispersion forces .
3) Neil Bartett prepared a compound (O2PtF6)
Ionization enthalpy of Xe (1170 KJ/Mol)
O2 (1175 KJ/Mol)
+
4) XeF2 + PF5→ [XeF] [PF6]
XeF4 + SbF5→ [XeF]+[SbF6]XeF6 + MF→ M+[XeF7]{M=No,K,Rb or Cs)
5) Hydrolysis
6XeF4+12H20 →
4Xe + 2Xe03 + 24HF + 3O2
XeF6+ 3H2O
→
XeO3 +6HF
Questions :
1) How does ammonia reacts with a solution of Cu++?
2) Why does No2 dimerise?
3) Are all five bonds in PCl5 molecule equivalent?
4)Are all five bonds in PCl5 molecule equivalent ?Justify your answer
5) H3PO3 is diprotic and why?
6) Which form of sulphur shows paramagnetic behavior ?
7) Complete the following relations
1. Cl2 + 2X- →
2. Br2 + 2I- →
8) Why is helium used in diving apparatus?
9)What is covalency of nitrogen inN2O5 ?
10) Why does R3P=0 exits but R3N=0 does not exists ?
11) PCl5 Is ionic in nature in the solid state .Give reason?
12) ClF3 exists but FCl3 does not .Why?
d-Block Elements
1. The transition metals are very much hard and have low validility, because greater number of
electrons form (n-1)d in addition to the ns electrons in the interatomic metallic bonding .
2. High enthalpies of atomization .
3. Variable o.s. – The enthalpies difference b/w (n-1)d orbitals is small.
4. Formation of colored ions : When an electron from the lower energy d-orbital is exited to a higher
energy d-orbital the energy of excitation corresponding to the frequency of light absorbed .the
color observed corresponding to the complementary color of the light absorbs .
5. Formation of complex compound : Due to the small size of metal ions their high ionic charges and
availability of vacant d-orbitals for bond formation .
6. Catalytic Properties :
a. Variable O.S.
b. Metal surfaces have free vacancies.
c. They can form complexes.
7. Alloy Formation : d-block transition metals have similar atomic size due to which one metal ion
can be readily replaced by another metal atom.
8. Interstitial state : Transition metals accommodate atoms of small size in the voids of there crystal
lattice and form interstitial compounds.
Questions :
A. Why is highest O.S. of a metal exhibited in its oxide or fluoride only ?
B. What is lanthanoid contraction ? what are its consequences ?
C. Why La(OH)2 is more basic than Lu(OH)3
D. Write preparation of K2Cr2O7?
E. What is meant by disproportionation? Give two examples?
F. What are interstitial compounds? Why are such compounds well known for transition
metal?
G. Explain why transition elements have many irregularities in their electronic configurations?
H. Which metal in the first series of transition metals exhibits +1 oxidation state most
frequently and why?
UNIT IX-- COORDINATION COMPOUNDS
1. A co-ordination compound is a molecular compound in which the central metal atom or
metal ion is linked toa number of ions or neutral molecules by coordinate bonds e.g.
K4[Fe(CN)6]
2. Central metal atom & ligands are collectively written inside square bracket known as
COORDINATION SPHERE.
3. The ligands whose only one donor atom is bonded to metal atom/ion are called mono
dentate ligand. CO, NH3 ,H2O.
4. The ligand, which contain two-donor atom through which it is bonded to central atom, is
called bidentate ligands. e.g. C2O4-- , NH2-CH2-CH2-NH2 .
5. A complex in which there is a close ring of atoms caused by attachment of ligand to a metal
atom at two points is called CHELATES.
6. A molecule or ion that can form close ring with metal atom is called chelating ligand. e. g.
C2O4--, NH2-CH2-CH2-NH2.
7. 10. Ethylene Di Amine Tetra Acetate is hexa dentate chelating ligand.
11. Ionization isomerism occurs when there is interchanging of groups between coordination
sphere to ionization sphere.
e.g. [ Co (NH3)5Br] SO4 and [Co (NH3)5 SO4] Br
13. Linkage isomerism exhibited by that compound in which a ligand can form linkage with metal
through different atoms i.e. ambidentate ligand form this isomerism.
e.g. [Co (NH3)5ONO]Cl2 and [Co (NH3)5NO2]Cl2
14. Hydrated isomerism occurs due to interchanging of H2O ligand from coordination sphere to
ionization sphere.
e.g. [Co (H2O)6]Cl3 [Co (H2O)5Cl]Cl2.H2O
15. Tetrahedral complex does not show geometrical isomerism.
16. In square planar complex cis (when same group are on same side) and trans (when same
group on are opposite side) isomers are possible.
17. In octahedral complex cis –trans isomers are as following:-
18. According to crystal field theory as ligands approaches to metal atom then splitting of dorbitals of metal atoms takes place. These d-orbitals splits into two categories:(i) t2g orbital contains dxy, dyz, dzx
(ii) eg orbital contains d X2- Y2 , d Z2.
21. For octahedral complex energy of eg is more than that of t2g orbitals.
22. For tetrahedral complex energy of eg is less than that of t2g orbitals.
23. Coordination compounds find use in followings field:(i) In qualitative analysis K4[Fe(CN)6] gives blue and brown colour with Fe3+
&Cu2+
respectively.
(ii) In quantitative analysis EDTA is used to remove hardness of water.
(iii) In extraction of metals like gold silver complex is formed.
(iv) In biological system Chlorophyll & Haemoglobin is a complex of Mg2+ &Fe3+ respectively.
Important Questions:
Q.1- Write the IUPAC name of linkage isomer of [Cr (en)2(ONO)2]Cl.
A.1-Linkage isomer is [Cr (en)2 (NO2)2] Br.
IUPAC name- bis(ethane-1,2-diamine)dinitrito-N-chromium(iii) bromide
Q.2- What is meant by the chelate effect? Give an example.
A.2- A complex in which there is a close ring of atoms caused by attachment of ligand to a metal
atom at two points is called chelate effect. e.g.
Q.3- Which complex is used in the treatment of cancer?
A.3- cis-platin or cis- [Pt (NH3)2 Cl2]
Q.4-Draw optical isomers of [Co(en)3]2+
A.4-
Q.5- Draw a sketch to show the splitting of d-orbitals in an octahedral crystal field.
A.5-
Unit-X HALOALKANES & HALOARENES
POINTS TO BE REMEMBERED
A Haloalkane:1. Halogen derivative of alkanes are called as Haloalkanes & of aromatic
hydrocarbon are called Haloarenes.
2. In Haloalkane C-atom of C-X bond is sp3 Hybridized while in Haloarene it is
sp2 hybridized.
3. Alcohols can be converted into Haloalkanes by reacting with HalogenAcids,
+H-X
PCl5
R-OH
SOCl2
R-X+H2O
R-Cl+POCl3+HCl
R-Cl+SO2+HCl
Phosphorous halide &Thionyl chloride (SOCl2).
4. Haloalkanes can also be prepared by electrophilic addition of Halogen Acids
(HX) to the alkanes.
R2O2
R-CH=CH2 + HBr
R-CH2-CH2Br
Markownikoff’s Rule
R-CH=CH2+HBr
R-CHBr-CH3
5. The Physical Properties of Haloalkanes are attributed to dipole-dipole and
dipole-induced –dipole interaction.
6. Haloalkanes undergo Nucleophilic Substitution reaction (SN) due to polar
character of Cδ+—Xδ- bond.
7. There are two types of SN reaction –
(a). SN1-Unimolecular Nucleophilic Substitution reaction proceeds in two
steps (slow and fast).
(b)
SN2-Bimolecular Nucleophilic substitution reaction proceeds in a single
steps and its rate depends on the concentration of both Haloalkane &
Nucleophile.
8. The reactivity of Haloalkane in SN1reaction follows the order ; Tertiary >
Secondary > Primary. While reactivity of Haloalkane in SN2 reaction follows the
order ; Primary > Secondary > Tertiary.
9. Optically Active compounds have the capacity to rotate the plane of polarized
light either to Right (Dextorotatory) or to the left (Laevorotatory).
10. Optical Isomerism in compounds is due to the presence of Chiral carbon
Atom as well as Dissymetry.
11. SN1 Mechanism in optically active alkyl Halide leads to Racemisation .
12. SN2 Mechanism in optically active alkyl halide leads to inversion of
configuration .
13. Alkyl Halide on reaction with metals form Organometallic compounds such
as Grignard Reagent (a useful compound) .
B. HALOARENES:-
1. Haloarenes are prepared from their parent arenes and from diazonium salts.
C6H6+X2
FeCl3,Dark
C6H5X + HX
Cu2X2 / HX
C6H5N2+X-
C6H5X + N2 +HX
2. Haloarenes are more stable & hence less reactive than Haloalkane due to
(a) Sp2- hybridized C atom of C-X bond .
(b) delocalization of Π- electron cloud.
3. Halogen present on the benzene is Ortho and Para directing but deactivating
in nature.
4. Halogenation of benzene proceeds by Electrophilic substitution Mechanism.
5. m.p.of p-dichlorobenzene is higher than that of O & m isomers.
C. polyhalogen compound :1.
Organic compound having two or more than two halogen atom are
called as polyhalogen compounds.
2.
polyhalogen compounds such as dichloromethane(CH2Cl2),
trichloromethane(CHI3), tetrachloromethane(CCl4),p-dichlorobenzen,
Freons(CFC,CCl2F2)BHC,(Benzene hexachloride C6H6Cl6), DDT&PFC
have many industrial application.
3.
chloroform is no longer used as aneasthatic due to it’s poisionous
nature &tendency to form phosgen gas(COCl2)when exposed to air .
4.
BHC is a better insecticide as well as pesticide than DDT as DDT is
non-biodegradable. γ-Isomers of BHC is called Gammexane or lindane
or 666.
5.
Freon used as a refrigerants but cause depletion of ozone layer.
VERY SHORT ANSWER TYPE QUESTIONS
(1MARK)
Q.1. Write the structural isomers of C3H6Cl2 which can exihibit
enantiomerism ?
H
Ans.
CH3— C—CH2Cl
Can exihibit enantiomerism
Cl
(i.e optical isomerism).
Q.2 which compound (CH3)3 –C-Cl or ;CH3Cl Will react faster in SN2 Reaction
with –OH?
Ans.CH3Cl is more reactive than (CH3)3-C-Cl because of steric Hinderance of
three bulky group in (CH3)3CCl.
Q.3. Why is vinyl chloride less reactive than ethyl chloride?
Ans. Conjugation of electron pair with π-electron pair of double bond in vinyl
chloride results to a partial double bond character in C-Cl &it’s bond strength
increases. CH2=CH-Cl — CH2-CH=Cl+
Q.4.Write the formula & chemical name of DDT?
Ans. DDT is 2 , 2-bis(p-chlorophenyl)-1,1,1-trichioroethane
Cl
CH
Cl
C.Cl3
Q.5.Write down the structure of the following compounds;
(a) 1-chloro-4-ethyl cyclohexane
(b) 1, 4-dibromo but-2-ene
(c) 4-tert.butyl-3-iodoheptane
(d) 1-bromo-4-secbutyl-2-methylbenzene
(e) perfluorobenzene
Cl
Ans. (a)
(b) Br-CH2-CH=CH-CH2-Br
Cl
s
C2H5
(c) H3C-CH2-CH-CH-CH2-CH2-CH3
I
F
(d)
CH3
C-(CH3)3
F
(e)
Br
H3C-CH-C2H5
F
F
F
F
Q.6. An alkyl halide having molecular formula C4H9Cl is optically active. What is
its structure ?
Ans. CH3
*CH
CH2 CH3
Cl
Q.7. Out of C6H5CH2Cl & C6H5CHCl C6H5 which is more easily hydrolysed with
aq. KOH & why?
Ans. C6H5 CH C6H5 will get hydrolysed easily because
Cl
Carbocation formed will be stabilized by resonance effect of two phenyl group.
Q.8. Why is sulphuric acid not used during the reaction of alchohols with KI?
Ans. Because HI formed will get oxidized to I2 by conc. H2SO4 which is an
oxidizing agent.
Q.9. A hydrocarbon C5H10 does not react with chlorine in dark but it gives a
single monobromo compound in bright sunlight. Identify the compound.
Cl
Ans.
sun
+
Cl2
light
+ HCl
Cyclopentane
Chlorocyclopentane
Q.10.Chloroform is stored in dark coloured & sealed bottle. Why ?
Ans. As Chloroform reacts with O2 of air in presence of sunlight & forms
Phosgene gas which is poisonous .
light
2 CHCl3 + O2
2COCl2
+ 2HCl
Phosgene gas
SHORT ANSWER TYPE QUESTIONS
(2MARK)
Q.1. Give the IUPAC names of the following compounds ?
a) ClCH2C
CCH2Br
b) (CCl3)3CCl
c) CH3CH(Cl)CH(Br)CH3
d)
Br
Cl
Ans. (a)1-Bromo-4-chlorobut-2-yne
(b) 2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane
(c) 3-bromo-3-chlorobutane
(4) p-bromo chlorobenzene
Q.2. Predict the product of the following reactions;
(a) CH3
CH
C(CH3)2
+ HCl
(b) CH3
CH
I
CH3
(c)
u.v.light
+ Cl2
(d) CH3
Na/Dry ether
CH
CH2
Br KOH(alc)
CH3
(e)
CH3
Markonikoff’s rule
+
Ans. (a) CH3
HI
CH3
CH
C CH3
Cl
(b)
CH3
H
CH
2-chloro-3-methylbutane
I
+ 2Na
+
I
CH
CH3
CH3
ether
CH3
CH3-CH-CH-CH3+2NaF
(c)
Cl
Cl
Cl
CH3 CH3
(2,3-dimethylbutane)
Cl
Cl
Cl
Benzene Hexachloride (BHC)
CH
CH3
(d). CH3
CH
CH2
Br
CH3
C
CH2 +KB r+H2O
2-methyl propene
CH3
+ HI Markonikoff’s
rule
(e).
I
CH3
1-Iodo-1-methyl
Cyclohexane
Q.3. Starting from Methyl Iodide, how will you prepare :
(a) Nitromethane.
(b) Methyl Nitrite.
O
Ans. (a) CH3I
+ AgNO2
(b) CH3I + K
O
CH3----N=O
CH3
N
+ AgI
O
O
N
O KI
Q.4. How can Iodoform be prepared from ethanol ?
Ans. C2H5OH + 6NaOH +4I2
CHI3 + HCOONa + 5NaI + 5H2O
Q.5.Write the reactions involved in :
(a). The Isocyanide test.
(b). Iodoform test .
Ans.(a). R-NH2 + CHCl3 + 3KOH(alc.)
RN=C + 3KCl+3H2O
o
1 -Amine chloroform
O
(b) CH3-C-CH3+3I2 +4NaOH
CHI3 +CH3COONa +3NaI+3H2O
Q.6.How will you distinguish between
(i) CH3NH2 and (CH3)2NH
(ii) Ethanol & 1-propanol
Ans. (i) CH3NH2 (Methyl amine) gives Isocyanide test while (CH3)2NH do not.
CH3-NH2 + CHCl3 + 3KOH
CH3-N=C +3KCl +3H2O
(ii) Add I2 & NaOH, ethanol will gives yellow ppt. of Iodoform whereas 1-propanol
will not .
Q.7. Propose the Mechanism of the following reaction ;
CH3-CH2 –Br +CH3OCH3-CH2-OCH3 + BrAns.
CH3 H
H3C
H
CH3
CH3-O +C-Br
[ CH3O
C
Br ]
CH3O C H +BrH
H
(Nucleophile)
H
Q.8.Give the uses of (a) CCl4 (b) Iodoform .
Ans. (a). Ccl4 is used as a solvent and also in fire Extinguisher.
(b). Iodoform is used as Antiseptic.
Q.9.Which will have a higher boiling point 1-chloropentane or
2-chloro-2-methylbutane ?
Ans. CH3 CH2 CH2
CH2
CH2
Cl has higher boiling point
1-chloropentane
than
CH3
CH3
C
CH2
CH3 (2-chloro-2-methyl butane).Because branc-
Cl
-hing of C-atom chain decreases the molecular size and vander
waal’s force of attraction leading to decrease in boiling point.
Q.10. Rearranging the following in order of increasing ease of dehydro
-halogenation CH3CH2CH2Cl , CH3CHClCH3 ,
CH3
C
Cl(CH3)2.
Ans. CH3CH2CH2Cl < CH3CHClCH3 < CH3
C
Cl(CH3)2.
As per saytzeff’s rule more alkylated alkene is more stable and hence is
formed.
SHORT ANSWER TYPE QUESTIONS
(3MARK)
Q.1. How will you bring the following conversion?
(a) Propene to Propyne
(b) Toluene to Benzyl Alcohol
(c) Aniline to Phenylisocyanide
Ans.
Br2 / CCl4
KOH(alc)
(a) CH3-CH=CH2
CH3-CH—CH2
CH3-C=CH
(Propyne)
Br
Br
(b)
CH2OH
CH3
CH2Cl
Cl / hν
aq.KOH
2
-HCl
-KCl
N≡C
→
NH2
(c)
+ CHCl3 + 3KOH(alc.)
+3KCl +
3H2O
Q.2. What happen when;
(a) n-butyl chloride is treated with alc. KOH.
(b)ethyl chloride is treated with aq.KOH.
(c)methyl chloride is treated with KCN.
Ans.
+KOH(alc.)
(a) CH3-CH2-CH2-CH2-Cl
CH3CH2CH=CH2 + KCl + H2O
(b) CH3-CH2-Cl+KOH(aq.)
CH3CH2OH+KCl+H2O
(c) CH3-Cl +KCN
CH3CN+KCl
Q.3. Complete the following reaction;
(a) Cl2CH-CH-Cl2+Zn
∆
(b)(CH3)2-CH-CH-CH2-CH3
C2H5ONa
Ethanol
Br
(c) C6H5ONa +C2H5Cl
Ans. (a). Cl2CH-CHCl2+Zn
(b). (CH3)2-CH-CH-CH2-CH3
Br
(C). C6H5O-Na+ + C2H5Cl
Sodium
Phenoxide
∆
Cl-CH=CHCl+ZnCl2
C2H5ONa
Ethanol
(CH3)2-CH-CH-CH2-CH3
OC2H5
C6H5OCH2CH3+NaCl
(Phenetole)
Unit-XI Alcohols, Phenols and Ethers
Points to be remembered:Alcohols:
Compounds containing one or more -OH group directly attached to C- atom of an
aliphatic system.
Phenol:
Compounds containing – OH group directly attached to C-atom of an aromatic
system.
Ethers:
Compounds obtained by substitution of a H-atom in a hydrocarbon by an alkoxy or
aryloxy group.
Structure of functional group:
H
142 pm sp3
sp3
O
C
108.9
96 pm
H
H
H
Alcohol
sp2
C
H
O
136 pm
Phenol
C-O bond length (136pm) is less than that in alcohol due to –
1. Partial double bound character on account of the conjugation of unshared e pair of
oxygen with aromatic ring.
2. sp2 state of C to which O-atom is attached
141pm
sp2
H
H
O
C
111.7
H
H
C
H
H
Ether
The C-O-C bound angle is slightly greater than angle due to repulsion interaction
between two bulky (-R)gps
Preparation of Alcohols:
8. Prom Alkene:
(i) By Acid Catalysed hydration:
H+
>C = C
+ H2 O
> C -C<
H
OH
(ii) By hydroboration – oxidation:
H2O2, OH3CH3 - CH = CH2 + BH3 → (CH3-CH2-CH2)3B
3CH3-CH2-CH2-OH
Propan -1-ol
+ B (OH) 3
(b) From Carbonyl Compounds:
(i) By reduction of aldehydes and ketones Pd/Pt/Hi
RCHO + H2
RCH2OH
(Primary alcohols)
NaBH4
RCOR1-
R-CH-R1
(Sec. Alcohols)
1 LiACH4
OH
(ii) By reduction of Carboxylic acids and esters –
(i) LiAl H4
RCOOH
R CH2 OH (Pri alcohol)
(ii) H2O
Or
R1OH
H2
1
RCOOH
RCOOR
RCH2OH + R1OH.
+
H
Catalyst
(c) From Grignard Regards H2O
(i) HCHO + RMg X
RCH2OMgX
RCH2OH + Mg(OH)X
Formaldehyde
Pri. Alcohol
1
H2O R
(ii) RCHO + R1MgX
R-CH – OMgX
R-CH-OH+Mg(OH)X
Other aldelydc
R1
R1
H2 O
(iii) RCOR + R1MgX
R–
C – OMgX
R – C – OH + Mg (OH) X
Ketone
R1
R
Test alcohol
Sec. Alcohol
Preparation of phenols –
(a) From Haloarenes Cl
|
623k
OH
ONa+
HCl
+NaOH
9. From Benzene 300atm
Sulphoric acid –
SO3H
Oleum
OH
NaOH
H+
10. From diazonium Salts –
NH2
|
N2+Cl
273278K
NaNo2
+HCl
300atm
Aniline
OH
H2O
warm
N2 +HCl
Benzene
diazonium
Chloride
(d) From Cumene (Manufacture)
CH3
CH3
H3C-CH
CH3---C-OO-H
O2
OH
H+
+ CH3COCH3
Cumene
H2O
Cumene
Hydro peroxide
Physical Properties:
Boiling Point –

The B.P. of alcohols and phenols increase with increase in the no. of C-alones
(increase in Vander waal forces).

In alcohols B.P. decrease with increase in branching (decrease in vander waals
forces with decrease in surface area).
(iii) Due to intermolecular H-bonding, B.P. of alcohols & phenols are higher than
hydrocarbons, ethane, haloalkanes and haloarenes of comparable messes.
Solubility:
Solubility of alcohols and phenols in water due to them ability to form H-bond with water
molecules. Solubility decrease with increase in size of alkyl / aryl group.
Na
C2H5ONa + H2
NaH
C2H5 ONa + H2
PCl5
C2H5CL
PX3
C2H5X
Alcohol
C2 H5 OH
(X=Cl, Br, 1)
SOCl2
O
C2 H5 Cl
CH3 – C – OH / H+
O
CH3 – C OC2H5
CH3 – C – Cl
Ch3 – C – OC2H5
O
K2 Cr2 O7 / H+
CH3 – C – OH
O
Cu / 575 K
CH3 – C – H
ONa
OH
NaOH
COOH
CO2
Kolbe’s reaction
H+
OH
OH
NO2
Dil.HNO3
+
Nitration
OH
OH
NO2
OH
O2N
NO2
Conc.HNO3
OH
OH
Br2
CHCl3
+
aq.NaOH
OH
CS2
Br
+
273
OH
Reimer Tiemrnn
Reaction
Br
Br
Br
aq.
Br2
ONa
Br
ONa
CHCl2
[
]
OH
CHO
H+
NaOH
O
Na2Cr2O7
H2SO4
O
O
Acidity of Phenols:
Phenol is acidic due to resonance stabilized phenoxide ion.
CHO
Acidity – Phenol Vs Alcohol:
Phenols are stronger acid than alcohol and water because –
(i) In phenol –OH gp is attached to sp2 C-atom which being more electronegative
decreases c-dually on o-atom and increase of ionisation of phenols than that of alcohols.
(ii) In alkoxide in, the negative charge is localized on O-atom while in phenoxide ion,
change is delocalized as phenoxide in is resonance stabilized. Hence phenoxide ion is more
stabilized than alkoxide ion.
Distinction between primary, secondary and tertiary alcohols –
Lucas test – Alcohols are soluble in Lucas reagent (conc. HCland ZnCl2) while than
halides are insoluble and produce turbidity in solution.
If turbidity is produced immediately it is test alcohol, if after 5 min. it is sec alcohol and if
turbidity is not produced at roar keep. It is pri. Alcohol.
Test for phenol:
(i) FeCl3 test :- Phenol gives purple color with FeCl3 solution.
(ii) Br2 test :- Phenol decolorizes Br2 water forming tribromphenol as while ppt.
Preparation of Ethane:(i)
By dehydration of alcohols.
H2SO4
CH2 = CH2
443K
CH3 CH2 OH
H2 SO4
C2 H5O C2H5
413K
(ii) Williamson Syntheses – (lab method)
RX + R1O Na
R OR1 + NaX
Physical Properties of Ethane:
(i) Boling point – C-O bond in ethane polar and ethers have a net dipole moment. But in
weak and not affect then B.P. which they have lower B.P. than alcohol due to absence of
intermolecular H – bonding in ethers.
(ii) Solubility – The miscibility of ethers with reasonable those of alcohols of same
molecular ware as O-atom of ether can form H-bond with water.
Chemical Properties of ethers:
(a) Cleavage of C-O bond in ethers R – O – R + HX
R – OH + HX
R
O
+ HX
RX + R – OH
RX + H2O
OH
+ RX
(b) Electrophonic Substitution In Anisole –OCH3 group attached to benzene group is ortho para directing and ring
activating and thus it gives ortho and para substituted products.
Very Short Answer type Questions
(1mark)
(i) How do you account for themiscibility of ethoxyethane with water?
Ans:- Due to intermolecular hydrogen bonding between ether and water molecule.
(ii) Write the I.U.P.A.C. name of the following compound :
H3C – CH – CH2 – CH – CH – CH2OH
I
I
I
CH3
OH CH3
Ans:- 2,5 – dimethylhexane – 1,3 – diol
(iii)Give the I.U.P.A.C. name of the following compound :
H3C – C = C – CH2OH
I I
CH3 Br
Ans:- 2 – bromo – 3 – methylbut – 2eu – 1ol
(iv) Among HI, HBr, HCl, HI is most reactive towards alcohols. Why ?
Ans:- HI has lowest bond dissociation enthalpy due to longer bond lengh, that is why it is most
reactive.
(v) Give the I.U.P.A.C. name of the following compound :
H2C = CH – CH – CH2 – CH2 – CH3
I
OH
Ans:- Hex – 1 – en- 3 – ol
(vi) C2H5ONa + CH3Br
C2H5OCH3 + NaBr, What is the name of the
reaction?
Ans:- Williamson’s synthesis of ethers.
(vii) Convert ethyl alcohol to diethyl ether.
Ans:- 2C2H5OH
C2H5OC2H5 + H2O
(viii) Give chemical test to distinguish between phenol and ethanol in seemingly similar
conditions.
Ans:Test
Phenol
Ehtnaol
Lucas Test:
No positive test
Cloudiness appear on heating. Comp. + Lucas
Reagent
(conc. HCl + Anhyd. ZnCl2)
(ix) Ortho nitrophenol is more acidic than ortho – methoxy phenol. Why?
Ans:- Because nitro group being electron withdrawing, increases acidic character whereas
methoxy group ( - OCH3) being electron releasing group decreases acidic character.
(x) Complete the following reaction –
CH3 – CH – CH3 – PCl5
I
OH
Ans:- CH3 – CH – CH3 – PCl5
CH3 – CH – CH3 – POCl3 – HCl
I
OH
|
Cl
Short Answer type Questions
(3mark)
Give a chemical test to distinguish between the following pairs of compounds.
(b) CH3 – CH – CH3
And
I
OH
(i)
Why is phenol more acidic than ethanol?
Ans:- (i) (a)
Test
1. Bromine Water test.
2. Ferric Chloride Test
Phenol
Cyclohexanol
Pale yellow ppt. of 2,4,6 Does not give this test
Tri bromo phenol is formed
Give Red to purple color.
Do not give any color
(b)
Test
2 – Propanol (Sec. alcohol)
Lucas Test :
produces lubridity within 5 min.
No turbidity
Comp. + Lucas Reagent
Cl
(HCl + ZnCl2)
I
(ii)
CH3 – CH – CH3 + HCl ----- CH3 – CH – CH3 + H2O
I
OH
(1) O – H group in this group is much more polar than that in alcohol.
(2) O – H group in phenol is directly attached to the sp2 – C of benzene which acts
as an electron withdrawing group. So, in resonance structure of phenol the oxygen
of – OH is +ve, which increases the polarity of O – H bond.
(3) In alkoxide ion the –ve change is localized on oxygen while in phenoxide ion,
the charge is delocalized. So, phenoxide ion is more stable (resonance establised)
(2) Write the reaction of Williamson’s synthesis of 2-ethoxy-3methylpentane starting from
ethanol and 3-methylpentan-2-ol.
Ans:- (i) H3C – CH2 – CH – CH – CH3 ----Na--- H3C – CH2 – CH - ONa
I
I
I
I
CH3 OH
CH3 CH3
3 methylpentan-2ol
(ii)
(iii)
C2H5OH ------------P,Br2-------
C2H5Br
H3C – CH2 – CH – ONa + C2H5Br ---------- H3C – CH2 – CH – CH – OC2H5
I
I
I
I
CH3 CH3
CH3 CH3
Chapter 12 ALDEHYDES, KETONES AND CARBOXYLIC ACID
ALDEHYDES: Compounds in which the carbonyl group (>C=O) is bonded to a carbon & hydrogen
atom. (R-CHO)
KETONES: Compounds in which the carbonyl group (>C=O) is bonded to 2 C-atoms.
CARBOXYLIC ACID: Compounds in which the carbonyl group (>C=O) is bonded to O-atom.
COOH)
(R-
STRUCTURE OF CARBONYL GROUP:The carbonyl C-atom is sp2 hybridised forming 3 sigma bonds & remaining electron in p-orbital
forms pie bond with O-atom. The O-atom has 2 L.p. The bond angle is 120 degree (trigonal
coplaner).
The C O double bond is polar with net dipole moment & are more polar than ethers.
PREPARATION OF ALDEHYDE:11.
From acyl chloride (acid chloride) or Rosenmind reaction.
O
CHO
H2
C
Cl
Pd /
BaSO4
12.
From nitrile or Stephen reaction.
RCN + SnCl2 + HCl
RCH=NH
RCHO
H3O+/H2O
13.
From esters.
1. DIBAL-H
CH3(CH2)9 --- C --- OC2H5
CH3(CH2)9 --- C --- H
+C2H5OH
2. H2O
14.
From hydrocarbons.
1. By oxidation of methylbenzene :=
[i] use of chromyl chloride (CrO2Cl2) or Etard reaction :::)
CH3
CH (OCrOHCl2)2
+ CrO2Cl2
CHO
[ii] Use of chromic oxide (CrO3) :::)
CH3
+ CrO3 + (CH3CO)2O
273278K
CHO
CH(OCOCH)2
H3O+
2. By side chain chlorination followed by hydrolysis :=
CH3
CHCl2
Cl2 / hv
CHO
H2O
373K
3. By Gatterman – Koch reaction :=
CHO
CO\HCl
Anh. AlCl3 / CuCl3
PREPARATION OF KETONES:
From acyl chlorides .
2R –C – Cl + R2Cd

2R – C – R + CdCl2
From nitriles.
NMgBr
CH3—CH2— C-- N
+ C6H5MgBr
CH3—CH2— C
C6H5
H30+/H2O
C6H5 -- C
C6H5

From benzene or substituted benzene or Friedal crafts acylation
reaction.
C – Ar / R
+ Ar/R— CO—Cl
PHYSICAL PROPERTIES OF ALDEHYDE & KETONES:(ii) Boiling point:: B.P of aldehyde & ketones are higher than hydrocarbons & ethers of
comparable masses due to weak molecular association in aldehydes & ketones arising
due to dipole-dipole interaction . But the B.P are lower than alcohols of similar molecular
masses due to absence of intermolecular H-bonding.
(iii) Solubility:: Lower members are soluble in water due to intermolecular H-bonding but
solubility decreases with increasing molecular mass.
3.
Test To Distinguish Aldehydes And Ketones
(i)
Tollens test:
Aldehydes give this test but ketones do not.
+
RCHO + 2[Ag(NH3)2] + 3OH
RCOO + 2H2O +4NH3 +2Ag
Tollens reagent : Tollens reagent is Ammonical Silver nitrate solution
(ii)
Fehlings Test: Aldehydes give this test but aromatic aldehydes and ketones do not.
Fehlings reagent- Fehling solution A(aq. Copper sulphate solution) and Fehling
solution
B(alkaline sodium potassium tartarate or Rochelle salt)
+2.
RCHO + 2Cu
+ 5OH
RCOO + 3H2O + Cu2O (Red brown ppt.)
Haloform (Indoform) test: Aldehydes and ketones having methyl ketones CH3CO group
give this test. Compounds containing CH3CH(OH) group give this test of ethanol.
X2/NaOH
R CO CH3
R COONa + CHX3 (X=Cl, Br,I)
(Haloform)
Reaction due to α hydrogen
Aldol condensation :
Given by aldehydes and ketones containing atleast one α – H- atom
2CH3CHO
CH3-CH-CH2-CHO
H20
CH3- CH - CHDil.NaO
- aldol condensation product
OH
H
H2O
Aldol
CHO
CH3
Ba(OH)2
2CH3COCH3
CH3-C-CH2-COCH3
OH
-H2O
(CH3)2- C= CH-COCH3
Cross Aldol condensation: Aldol condensation is carried out
between two different or ketenes
2CH3CH0+CH3CH2CHO
CH3CH =CHCHO+CH3-CH2-CH = CH(CH3)CHO +CH3CH=CH(CH3)CHO +CH3CH2CH= CHCHO
Other reactions:
1.
Cannizzaro’s reaction: this is a disproportion reaction given by only aldehydes
having no α-H atom. Ketones do not give this reaction
2.
Electtrophilic substitution reaction: Given by both Aldehydes and ketones .
in this
C
O group acts as a deactivating and meta-directing group.
CARBOXYLIC ACID
1. Structure of carboxylic group: carboxyl C-atom is less electrophilic than carbonyl C-atom
because of resonance.
Preparation of carboxylic acid:
1. From primary Aldehydes and alcohols:
Alk. KMnO4
RCH2OH
RCOOH
H3O
2.
+
From alkyl benzene:
aromatic carboxylic acid are prepares by this method.
Entire side chain is oxidized to the carboxyl group irrespective to the of its length. Test
group is not affected.
CH2CH2CH3
CHO
KMnO4KOH
3.
From nitriles and amides:
H2O/dil
RCN
H20
R-CO-NH2
HCL
H20/dil HCL
CH3CONH2
CH3COOH + NH3
RCOOH
H3O
+
COOH
4.
From Grignard reagent:
RMgX + O
5.
C
O
R-COOMgX
RCOOH + Mg(OH)X
From alkyl halides and hydrides:
H20
RCOOH + Cl
-
RCOCl
H 20
(C6H5CO) 2O
6.
RCOOH + HCl
-
RCOOH
2C6H5COOH
From esters:
COOC2H5
COOH
+
C2H5OH
NaOH
CH3CH2CH2COOC2H5
CH3CH2CH2COONa + C2H5OH
Basic
hydrolysis
Physical properties:
Boiling point: have higher boiling point than Aldehydes and ketones and even
7.
1.
alcohols of comparable molecular mass due to more extensive association of carboxylic
acid molecules through intermolecular H-bonding, which is not broken completely even in
vapour phase as most of them exist as dimer in the vapour phase and in aprotic solvents.
2. Solubility: decreases with increased mass
1. Acidic nature: cleavage of O-H bond
2RCOOH+2Na
2R-COO-Na+ + H2
R-COOH +NaOH
RCOONa+ + H2O
RCOOH + NaHCO3
RCOO-Na+ + H2O + CO2
2. Cleavage of C-OH bond:
i.
Formation of anhydride:
2RCOOH
ii.
RCOOCOR
Esterification :
RCOOH +R’OH
iii.
RCOOR’ + H2O
Reaction with PCl5,PCl3 and SOCl2
PCl5
RCOCl + PCl3 + HCl
PCl3
RCOOH
RCOCl +H3PO4
SOCl2
RCOCl +SO2 +HCl
iv.
Reaction with ammonia:
CH3COOH + NH3
CH3COO-NH4+
CH3CONH2
Substitution reaction in the hydrogen part:
1. Hell – Volhard Zelinsky reactionHalogenations at the α position of carboxylic acid having an α - H atom to give α halogen
carboxylic acid
Cl2/P(Red)
R-CH2-COOH
RCH-COOH
X
2.Ring substitution-
(X=Cl,Br)
Electrophilic substitutionin which the – cooh group acts as
deactivating and m-directing group
COOH
COOH
Conc. HNO3
+
Conc. H2SO4
NO2
ONE MARK QUESTIONS –
(i)
Write the structural formula of
(a) 1-phenylpentan-1-one
(b) 3-oxopentanal
Ans: - (a)
(ii)
O
O
II
II
(b) H-C-CH2-C-CH2-CH3
Give the I.U.P.A.C names of the following compound –
Ans: - 2-ethylcyclopent-3enecarboxylic acid
(iii) Give chemical test to distinguish between phenol and benzoic acid.
Ans:- Test
Phenol
Benzoic Acid
NaHCO3 Test:
No Reaction.
With effervescence CO2
Add a few drops of
is evolved.
aq. sol. (saturated) of
C6H5COOH + NaHCO3 --
NaHCO3.
C6H5COONa + H2O + CO2
(iv)
Why do acryl chlorides have lower boiling point than corresponding acids ?
Ans:- Acyl chlorides do not have intermolecular H-bonding while corresponding
acids have H-bonding.
(v)
Complete the following reactions and name the major products
(a) CH3CHO
------------
(b) CHCl3 + C2H5NH2 + 3KOH(alc)
-------------
OH
O
.
I
II
Ans:-(a) CH3CHO ------- CH3-CH-CH2-C-H (3-Hydroxy butanal or aldol)
(b) CHCl3 + C2H5NH2 + 3KOH (alc.) --------- C2H5NC + 3KCl + 3H2O
Ethyl isocyanides
(vi)
How will you convert Ethanol to Acetaldehyde?
O
II
Ans:- CH3CH2OH --------- CH3-C-H + H2
Ethanol
Acetaldehyde
(vii) Arrange the following in increasing order of acidic character :
HCOOH, CH2ClCOOH , CF3COOH , CCl3COOH
Ans:- HCOOH < CH2ClCOOH < CCl3COOH < CF3COOH
(viii) Write the IUPAC name of the compound :
CH3-CH-CO-CH-CH3
I
I
CH3
CH3
Ans:- 2,4 – Dimethylpentan-3-one
(ix)
Name two important uses of formalin
Ans:-1. Used as preservative for biological specimens.
2. Used for preparation of Bakelite.
(x)
Write the structure of phenyl-2-(phenyl) ethanoate.
Ans:-
5 – Marks Questions –
1. Identify the unknown organic compounds. (A) to (E) in the following series of
chemical reactions –
2. An organic compound (A) having molecular formula C9H10O forms an orange red ppt (B) with
2,4 –DNP reagent. Compound (A) gives a yellow ppt (C) when heated in the presence of Iodine
and NaOH along with a colourless compound (D). (A) does not reduce Tollen’s reagent or
Fehling’s sol. Nor does it decolourise bromine water. On drastic oxidation of (A) with Chromic
acid, a carboxylic acid (E) of molecular formula C7H6O2 is formed. Deduce the structure of the
organic compounds (A) to (E).
3. A compound (A) with molecular formula, C5H11O, on oxidation forms compound (B) with
molecular formula, C5H10O. The compound B gives iodoform test but does not reduce ammonical
silver nitrate. (B) on reduction with amalgamated Zinc & HCl gives compound (C) with molecular
formula C5H12. Identify A, B & C. Write down the chemical reaction involved.
5.
(a)
(i)
Ket
ones are less reactive towards nucleophiles than aldehydes.
(ii) Benzoic acid is stronger acid than ethanoic acid.
(iii) Which acid is stronger- Phenol or Cresol? Explain.
Ans.(a)
(i)
(ii)
The Alkyl group in Ketones decreases the positive charge on the Carboxyl carbon due
to + I effect, makes > C= O less polar than Aldehyde and decreasing the attacking
tendency of the nucleophile.
(ii) Phenyl group in Benzoic acid has weak +R effect and thus destablises the
Carboxylate anion, makes it a stronger base. +I effect of Methyl group in Acetic acid is
still greater than +R effect of Phenyl group. This further destabalises the Acetate ion
comparatively, makes it a stronger base than Benzoic acid. So, acetic acid is weaker
than Benzoic acid.
Cresol is a weaker acid than Phenol because +I effect of –R group at the ortho and para position
increases electron density on the Benzene nucleus, decreasing stability of ArO- ion making it a
stronger base as O-H bond becomes less polar. Hence, substituted Phenol is less acidic than
Phenol.
Unit-13 Organic Compounds containing nitrogen
Points to be remembered:1. Alkyl or Aryl derivatives of ammonia
10 Primary amine
R-NH2
0
2 Secondary amine
R2—NH
30 Tertiary amine
R3-N
2. IUPAC nomenclature
Alkane amine in place of ‘e’ of alkane for secondary and tertiaryN-alkyl or N<N-dialkyl
alkanamium.
3. Structure is pyramidal
4. Preparation
(a)Reduction of nitro compounds
H2/ Pd or Ni or Pt
Sn + HCl
Ar-NO2
Ar-NH2
Ar-NH2
(b)Ammonolysis of alkyl halides
R-X + NH3
R-NH2 + NH4X
0
0
If Alkyl halide is in excess 2 and 3 Amines can be prepared
(c)Reduction of nitrites
H2 /Ni
R-C
N
R-CH2-NH2
(d)Reduction of amides
R-CONH2
R-CONH2
LiAlH4
R-CH2-NH2
R-CH2-NH2
(F)Hoffman bromamide degradation reaction
R-CONH2 + Br2 + 4NaOH
R-NH2 + Na2CO3 + 2NaBr + 2H2O
PHYSICAL PROPERTIES:
5.(a)Boiling point increase with the increase in molecule weight. Primary and secondary
amines have higher boiling point than the tertiary amines of same molecular weight due to
possibility of intermolecular hydrogen bonding in
(b)Solubility: - lower members are readily soluble in water due to H-bonding.Solubility in
water decreases and in organic solvent increase with increase in molecular weight. As the
alkyl group predominates over the amino group thus less tendency of H-bond with water.
6. Basic characters of amines
R-NH2 + H2O
R-N+H3 + OHLarger value of kb or smaller the value of pkb stronger is the base.
In gases phase basis strength is as follows:
Tertiary amine > secondary > primary > ammonis
Because of the presence of electron release alkyl group which stabilizes ammonium cations
and makes nitrogen unshared or lone pair of electron available for sharing.
In solution
Secondary > tertiary > primary
Stabilization due to: electron-releasing effect of alkyl group
: salvation trough H-bonding
: steric factor/hindrance
(C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3
(CH3)2NH > CH3NH3 > (CH3)3N > NH3
Aryl/Aromatic amines are weaker than aliphatic amines and ammonia.
Resonance stabilization of aryl amines makes unshared electron pair on nitrogen less
available for protonation.
In case of substituted aniline electron-releasing groups like -0CH3,
-CH3 etc increases basic strength
Whereas electron with drawing groups like –NO2,-SO3,-COOH,-X decreases basic strength.
E.g. increasing basic strength
7. Alkylation:
+R-Br
+R-Br
R-NH2
8. Acylation:
-HBr
R2NH
-HBr
R3N
O
R-NH2 + R’C-Cl
acid chloride
R-N-H2 + Cl – C-R’
R-N-C-R’ + HCl
R-N-C-R’ + HCl
O
C6H5-N-H + R-C-O-C-R
H
O O
acid anhydride
RNH2 + C6H5COCl
BENZOYL CHLORIDE
9. Carbylamines reaction (isocyanide test):
HEAT
R-NH2 + CHCl3 + 3KOH
H O
C6H5-N-C-R + RCOOH
H O
RNHCOC6H5 +HCl
R-NC +3KCl +3H2O
alkyl carbylamines
10. Reaction with nitrous acid:
R-NH2 + `HNO2
C6H5-NH2
NaNO2+H [R-N2+Cl-]
ROH +N2 +HCl
H2O
Cl
NaNO2+2HC
C6H5N2+Cl- + NaCl + 2H2O
l
273-278 K
11. Reaction with arylsulphonyl chloride (Hinsberg’s test):
O
O
Ar-S-Cl + H-N-C2H5
Ar-S-N-C2H5 + HCl
O
Benzenesuphonylchloride
O
O(N- ETHYLBENZENE SULPHONAMIDE)
O
Ar-S-Cl + H-N-C2H5
O
Ar-S-N-C2H5 + HCl
C2H5
12. Electophilic substitution:
a) Bromination
NH2
Br
O C2H5
(N,N-DIETHYLBENZENESULPHONAMIDE)
NH2
Br
Br2/H2O
+
3HBr
+
Br
(2,4,6-TRIBROMOANILINE)
O
NH2
H-N-C-CH3
H-
O
N-C-CH3
OH- OR
Br
PYRIDI
NE
NH2
H+
CH3CO
OH
Br
b) Nitration:
NH2
NH2
Br
NH2
NH2 NO2
HNO3, H2SO4
+
NH2
NHCOCH3
NHCOCH3
HNO3,
H2SO4
288K
+
+
OH- OR
H+
(CH3CO)2O
288K
NH2
(D) FRIEDAL CRAFT REACTON
Aniline does not undergo this reaction due to salt formation with AlCl3, the Lewis acid
: NH2
NH2AlCl3-
+AlCl3
DIAZONIUM SALTS
( RN+2X-)
Preparation by diazotization.
C6H5NH2 + NaNO2 + 2HCl
C6H5N2Cl + NaCl + 2H2O
1. Reaction involving displacement of nitrogen.
(a) Replacement by halide or cyanide ion
Sand Meyer Reaction:
ArCl + N2
CuCl/HCl
ArN+2XArBr + N2
CuBr/HBr
ArCN + N2
Gatterman Reaction:
ArN+2X-
CuCN/KCN
Cu/HCl
Cu/HBr
(b) Replacement of iodine ion:
ArCl + N2 +CuX
ArBr + N2 +CuX
ArN+2Cl- + KI
ArI + KCl + N2
( c) Replacement by Fluoride ion :
ArN+2Cl- +HBF4
ArN+2 BF-4
(d) Replacement By H :
H3PO2 + H2O
ArN+2ClArH + N2 +H3PO3 +HCl
+CH3CH2OH
Ar-F + BF3
ArH +N2 +CH3CHO + HCl
(e) Replacement by Hydroxyl group:
ArN+2Cl- + H2O
ArOH + N2 + HCl
(f)
Replacement by --NO2 group:
+N2Cl+N2BF-4
NO2
NaNO2
+ HBF4
Cu
Flouroboric acid
2. Reaction involving retention of Diazo Group
Coupling reaction:
OHN+= NCl- + H-
-N=N-
-OH
p-hydroxyozobenzene
(Orange dye)
-N+=NCl-
+
H-
OH--
-NH2
-N=N--
+ N2 + NaBF4
-OH
+H2O+Cl-
NH2+
P-Amioazobenzene
+Cl-+H2O
(Yellow dye)
Very Short Answer type Questions
(1mark)
Q.1.How will you convert aniline to benzoic acid?
Ans.
NH2
N +Cl2
ANILINE
CN
COOH
H2O
Q.2.What product is obtained by Hoffman degradation of m-bromobenzamide?
Ans. m-bromoaniline is obtained
Q.3.What happen when benzonitrite is treated with LiAlH4?
Ans. Benzylamine is obtained
CN
CH2NH2
LiAl
H4
Q.4.What for are quaternary ammonia salts widely used?
Ans. Quaternary ammonium salts are used as detergents.
Q.5.How is aniline converted into flourobenzene ?
+ N2 Cl
NH2
N2BF4
HONO+HQ
A
0-5oc
F
HBF4
LiAl
H4
Aniline
benzenediazonium chloride
Flourobenz
Q.4.Arrange in increasing orderof their basic strength
(a)C2H5OH, (CH3)2NH, C2H5NH2.(Increasing order of b.p.)
(b)C6H5NH2,(C2H5)2NH,C2H5NH2.(INCREASING ORDER OF SOLUBILITY IN WATER)
Ans.
(a). (CH3)2NH< C2H5NH2< C2H5OH.
(b). C6H5NH2<(C2H5)2NH<C2H5NH2.
Q.5.Explain (a).coupling reaction.
(b). Sandmeyer’s reaction.
Ans.(a). Benzene diazonium chloride couples with electron rich aromatic compounds like
phenols and amines to give azocompounds. The azo compounds contains –N=N- bond and
the reaction coupling reaction.
C6H5N2+Cl- + C6H5OH
→ C6H5-N=N-C6H5-0H (p)
p-HYDROXYAZOBENZENE
(b)When benzene diazonium salts is treated with cuprous chloride or cuprous bromide at
roomtemrature, chlorobenzene or bromobenzene is formed .this reaction is known as
sandmeyer’s reaction.
C6H5N2+Cl- +
CuCl/HCl → C6H6Cl + N2
Short Answer type Questions
(3mark)
Q1.(a) Give possible explanation for each of the following:(i) the presence of a base is needed in the ammonolysis of alkyl halides.
(ii)aromatic primary amines cannot be prepared by Gabriel phtaliminde synthesis
(b)Write the IUPAC name of CH3-N-C-CH3 C2H5
(a)
(i) To remove the HX produced during the rn.
(ii)
Aryl halides do not undergo nucleophilic substitutionWITH POTASSIUM
PHTHALIMIDE.
(b) N-Ethyl-N-methylethanamide.
2.Account for the following :(i) Pkb of aniline is morethan that of methylamine.
(ii) Ethylamine is soluble in water whreas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
Ans- (i) aniline has an electron withdrawing phenyl group so it is the weaker base than
ammonia. Methyl group in methylamine is electron donating group so it is stronger base than
ammonia.
(ii)ethyl group in ethylamine is comparatively a small group and causes no hindrance in
formation of hydrogen bonding and hence it is soluble in water.
(iii) methylamine behaves like aqueous ammonia and in similar way it precipitates metal as
hydroxides from metal salts
CH3NH2 + H2O  CH3N+H3OH- 3CH3NH3+OH- + FeCl33(CH3NH3)+Cl- + Fe(OH)3
Ques) arrange the following
(i)
in decreasing order of the pKb value:
C2H5NH2, C6H5NHCH3,(C2H5)2NH AND CH3NH2
(ii)
in decreasing order of basic strength: C6H5NH2, C6H5N(CH3)2,(C2H5)2NH and
CH3NH2
(iii)
in increasing order of basic strength:
(a)
aniline,p-nitroaniline and p-toylidine
(b)
C6H5NH2,C6H5NHCH3,C6H5CH2NH2
Ans) (i) C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH
(ii) (C2H5)2NH>CH3NH2>C6H5NH2>C6H5N(CH3)2
(iv) (a) p-nitroaniline<aniline<p-toulidine
(b)C6H5NH2<C6H5NCH3<C6H5CH2NH2
Q-4 Give plausible explanation of each of the following:
i)
Why are amines less acidic than alcohols of comparable molecular masses?
Why primary amines are having higher boiling point than tertiary amines?
ii)
Why aliphatic amines are having stronger bases than aromatic amines?
iii)
(i)Amines are less acidic than alcohols of comparable molecular masses because the
anion formed is not is not stabilized enough due to presence of unshared electro pair on the
nitrogen atom.
iv)
R-NH2
RNH- + H+
v)
(ii) In primary amines , nitrogen bears two hydrogen atoms whereas in tertiary
amines nitrogen bears no hydrogen atom hence the hydrogen bonding is more prominent.
This is why primary amines have higher boiling point than amines.
vi)
(iii) In aromatic amines, aryl group has an electron withdrawing Phenyl group on the
nitrogen atom. It decreases electron density on nitrogen atom and makes it a weaker base
than ammonia.
CHAPTER-14 BIO MOLECULES
POINTS TO BE REMEMBERED:1. CARBOHYDRATES:-There are optically active poly hydroxyl aldehydes or ketones or the
compounds which produce such units on hydrolysis they have general formula Cx(H )y.
2. CLASSIFICATION On the basis of them behaviours on hydrolysis they are classified as:
(a).Monosaccharide:- The carbohydrates which cannot hydrolysied further .
Eg. Glucose , Fructose.
(b).Oligosaccharides:- The carbohydrates which give 2 to10 monosaccharids units on
hydrolysis. Eg. Sucrose, maltose.
( c).Polysaccharides:- The carbohydrates which give large number of monosaccharides on
hydrolysis. Eg. Starch, cellulose.
3. SUGARS: - The carbohydrates which are sweets in taste are called sugar .eg. Sucrose,
lactose.
4. NON SUGARS: - The carbohydrates which are not sweets in taste.eg. Starch, celloluse.
5. REDUCING SUGAR:- Sugars which reduse fehling solution and tollens regent.
Eg. Glucose, fructose.
6. NON REDUCING SUGER:-which do not reduce Fehling or tollen’s reagent. Eg;-sucrose.
Structure of Glucose
7. CYCLIC STRUCTURE OF GLUCOSE:The straight chain structure of glucose is unable to explain the following reaction.
(a) It does not give the2, 4-DNP test, Schiff test and doesn’t form hydrogensulphide
addition product with NaHSO3.
(b) The pentaactate of glucose does not react with NH2OH indicating the absence of
CHO-group.
(c) It exits in two different crystalline ά and β forms .They are called anomers.They
differ in optical rotation .They have different melting point.
HAWORTH STRUCTURE
ά- D Glucose
β- D Glucose
8. STARCH:- Main storage polysaccharide of plants has two components amylase and
amylopectin.
9. PROTIENS: - Protiens are long polymers of amino acids linked by peptide bands. They are
essential for proper growth and maintaince of body.
10. AMMINOACIDS:- It contain amino and carboxylicacid functional group.
R-CH-COOH
|
NH2
11. ESSENTIAL AMMINO ACID:-Amino acid which cannot be synthesized in the body and
are obtained from diet.eg.Valine, Lysine.
12. NON-ESSENTIAL AMINO ACID:- The amino acid which can be synthesized in the body
are called non essential amminoacid.eg.Glysine, serine, proline.
13. ZWITTER ION: - In aqueous solution the carboxyl group can lose a proton and amino
group can accept a proton, giving rise a dipoler ion called as zwitter ion.
14. CLASSIFICATION:(A) FIBROUS PROTEIN:1. Polypeptide chains run parallel or antiparllel & held together by hydrogen and
disulphide bonds.
2. Generally in soluble in water. Eg-collagen, myorin, fibroin.
(B) GLOBULAR PROTIEN:1. Chains of polypeptide coil around to give spherical shape.
2. Usually soluble in water. E.g. Insulin, hemoglobin.
15. DENATURATION OF PROTEIN: - The proteins in its native form, when subjected to
physical change like temperature, PH etc undergoes uncoiling and loses its biological activity.
This is called denaturation of proteins.
The secondary and tertiary structures are destroyed, only primary structure retained.
eg: coagulation of egg while on heating.
16. VITAMINS:The organic compound required in the diet in small amounts to perform
specific biological function and is not synthesized by our body.
17. CLASIFICATION: - ON THE BASIS OF THEIR SOULIBLITY IN WATER OR FAT.
(a)FAT SOLUBLE VITAMIN: - They are insoluble in water but soluble in fat.eg vitaminA, B, E,
K.
(b)WATER SOLUBLE VITAMIN: - Vitamin B & C
Vitamin
Source
A
CARROT, BUTTER & MILK
B
YEAST, LIVER, GREEN VEGETABLE
deficiency disease
XEROPHTHAL MIA, NIGHT
BLIDNESS
BERI-BERI
C
CITRUS FRUIT.AMLA
SCURVY
D
SUNLIGHT ,FISH
EGG YALK
RICKETS
18. NUCLEIC ACIDS: - These are bimolecules which are long chain polymers of nucleotides.
They are DNA& RNA.
19. COMPOSITION O F NUCLEIC ACID: - They are made-up of a pentane sugar (B-D-2
dexoyribose in DNA d B-D-ribose in RNA), phosphoric acid and nitrogen containing
heterocyclic compound. (BASE)
(a) DNA: - Base adenine (A), Thymine (t), guanine (g) and cytosine(C).
(b)RNA:- Contain adenine(A),guanine(G),cytosine(C)and Uralic(U).
20. ENZYMES: - Enzyme are bio-catalyst and are generally globular protein e.g. zymase,
maltase.
VERY SHORT ANSWER TYPE QUESTION
(1 marks)
Q.1 what are reducing sugar?
A.1 The sugar which reduces Fehling solution and tollen’s regent .eg.glucose.
Q.2 Name the deficiency diseases resulting from lack of vitamin A and C?
A.2 Xerophthalmia & scurvy.
Q.3 What is name given to the linkage which hold together to monomeric units in
polysaccharide?
A.3 Glycosidie linkage.
Q.4 Name the enzyme which convert sucrose into glucose and fructose?
A.4 Invertase.
Q.5What are the components of starch?
A.5 Amylase and amylopectin..
Q.6 Why is cellulose in our diet not nourishing?
A.6 Our body does not have enzyme which help in digestion of cellulose.
Q.7 Name the purines present in DNA?
A.7Adenine and guanine.
Q.8Why vitamin C cannot be stored in our body?
A.8 It is water soluble & readily excreted in urine.
Q.9Give two examples of non- essential amminoacid?
A.9 Glycine and alamine.
Q.10 What are different types of RNA found in the cell?
A.10 1.m. RNA 2. t. RNA 3.r. RNA
SHORT ANSWER TYPE QUESTION
(2 marks)
Q.1STATE two difference between globular protein and fibrous protein?
A.1.(a)They are α-helix structure. While they have β-pleated structures.
(b)They are soluble in water while they are insoluble in water.
Q.2What are essential amino acids? Give two examples.
A.2 The amino acid which cannot be synthesized in the body we get from our
Diet e.g. Valine and leuline.
Q.3 What is glycogen? How is it different from starch?
A.3 It is a polysaccharide found in all animal cells mainly in muscles and liver.
Carbohydrates are stored in animal body as glycogen where in plant body as starch.
Q.4How are vitamins are classified? Name the vitamin responsible for the
cogulation of blood.
A.4Vitamins are classified on the basis of their solubility as fat soluble and water soluble.
Vitamin K
Q.5 What are nucleic acids? Mention their two important function.
A.5 They are polymers of nucleotides having ribose or deoxyribose sugar, hetrocyclic base
like A,G,C,T and U and phosphoric acid.
Q.6 What is difference between DNA &RNA on the basis of bases they contain?
A.6 Both have two bases derived from purine, adenine and guanine and two bases from
pyramidine, thymine and cytocine but in RNA there is Uralic in place of thymine of DNA.
Q.7 (a). Which forces are responsible for the stability of x-helix? (b). What is
denaturation of protein?
A.7 a. Hydrogen bonding
b. When 2 &3 degree structure are ruptured then the proteins are said to be denatured.
It is done by heating or change in pH.
Q.8 Where does the water present in the egg go after boiling the egg?
A.8 When egg is boiled the water present in the egg is used in denaturation of protein
through H-bounding.In this the globular protein in the egg changes to rubber like insoluble
mass
Q.9What are the enzymes? Give two examples.
A.9 The enzymes are biocatalyst . Eg maltose and urease.
Q.10 What is difference between nucleoside and a nucleotide?
A.10 A nucleotide contain only two basic components of nucleic acid pentose sugar and
nitrogenous base .But nucleotide contain all three components phosphoric acid
group,pentosesugar and nitrogenous base.
SHORT ANSWER TYPE QUESTION
(3 marks)
Q.1Define the following as related to proteins
A.Peptide linkage
B. Primary structure
C.Denaturation
A.1A. Peptide linkage: - The bond formed by condensation reaction between amminoacid in
protein and polypeptide is called peptide linkage.
B. Primary structure:- The sequence in which amminoacid are linked with each other in
polypeptide chain forms primary structure .
C. Denaturation:- The process in which 2&3 degree structure of protein are destroyed but 1
degree structure remain as the same is called denaturation.
Q.2 What are essential and non-essential amminoacid? Give two examples of each
type.
A.2 Essential amminoacid: - Those amminoacid which are not synthesized in the body.
Eg. Valine, leucine.
Non-essential amminoacid: - those amminoacid which are produced in our body.
Eg.glycine,alaine.
Q.3 Enumerate the reaction of D-glucose which can not be explained by it open chain
structure?
A.3 Open chain structure can not be explained by following reasons:a. Despite having the aldehydegroup, glucose does not give suhiffs test . 2, 4-DNA test
b. Glucose does not react with NaHSO3 to form addition product.
c. The pent acetate of glucose does not react with NH2OH showing the absence of –CHO
group.
Q.4 Define the following and give one example of each?
(a) ISOELECTRIC POINT (b) MUTAROTATION (c) ENZYMES
(a) ISOELECTRIC POINT:- The PH at which no net migration of amminoacid take place under
the influence of an applied electric field is called isoelectric point.
Ex-Amminoacid exits as zwitter ion at PH 5.5 to 6.3
(b) MUTAROTATION:- Spontaneous change in optical rotation when an optical active
substance is dissolved in water. e.g. ά-D-glucose when dissolved in water, its optical rotation
changes from 111 to 52.5 degree.
(C) ENZYMES: - They are biological catalyst which catalises specific biochemical reaction.
-----------------------------------------------------------------------------------------------
UNIT 15 POLYMERS
Points to be remembered:Polymers – Polymer is a very large molecule having high molecular mass 103 to 107 g/mol.
They are formed by joining together repeating simple unit.
Polymerizations- The process of formation of polymers from respective monomers is called
polymerization.
CLASSIFICATION OF POLYMERS
1) Based on source –
a) Natural – found in plants and animals e.g proteins, cellulose, starch,
rubber.
b) Semi-synthetic – Derived from naturally occurring polymers by chemical
modifications e.g cellulose acetate, cellulose nitrate.
c) Synthetic – These are prepared in laboratories .They are also called man
made polymers. e.g Nylon, polythene, PVC , Buna-S.
2) Base on structure –
a) Linear polymers – They consists of long and straight chain of repeating
units e.g polythen,PVC.
b) Branched chain – They contain linear chain having some branches. e.g.
amylopectin, glycogen.
c) Cross linked polymers – They have strong covalent bond between
various linear polymer chains e.g Bakelite, melamine.
c) Based on mode of Polymerization:(i) Addition Polymers: - These are formed by the repeated addition of monomer molecules
having multiple bonds. e.g. polythenes, polypropene, polystyrene.
(ii) Condensation Polymers: - These are formed by repeated condensation reaction of
different bi-functional or tri-functional monomers with the elimination of small molecules
like water, HCl, NH3, etc. e.g.: Polythene, Bakelite, nylon.
(d) Based on molecular forces:(i) Elastomers: - Forces of interaction between polymer, chain is weakest. e.g.: Natural
rubber, neoprene.
(ii) Fibres: - Strong hydrogen bonds are present between the polymers chain .e.g.: Nylon,
Bakelite.
(iii) Thermoplastics: - They are linear or slightly branched chain molecule capable of
repeated softening of heating and hardening on cooling. E.g.: Polythene, PVC, polypropene.
(iv) Thermosetting Polymers: - They are cross linked or heavily branched molecules which
on heating undergo extensive cross linking and become infusible. e.g.: Bakelite.
(e) Based on growth of polymerization:- Depending upon the mechanism of polymerization
polymers are classified as…
(i) Addition Polymer OR Chain Polymer:-
They follow mostly free radical mechanism. The process starts with phenyl free radical
formed by peroxide which adds with the double bond of ethene to form larger free radical.
This step is called Chain Initiating Step. This reacts with another molecule to form bigger
radical. This is Chain Propagation Step. This reacts with another radical to form polymerized
product. This is called the Chain Terminating Step.
CHAIN INITIATING STEP:
C6H5-C-O-O-C-C6H5
2C6H5C-O*
2*C6H5
O
O
O
Phenyl Radical
Benzoyl peroxide
*C6H5 + CH2=CH2
C6H5-CH2-*CH2
CHAIN PROPAGATION:
C6H5CH2-*CH2 + CH2=CH2
C6H5-(CH2-CH2)n-CH2-*CH2
CHAIN TERMINATING:
C6H5-(CH2-CH2)n-CH2-*CH2
C6H5-(CH2-CH2)-CH2-CH2
+
C6H5-(CH2-CH2)-CH2-*CH2
C6H5-n(CH2-CH2)-CH2
Polythene
(ii) Condensation Polymers or Step Growth Polymers:-They are formed by gradual
steps.
Some Important Polymers:
NAME OF POLYMERS
MONOMERS
STRUCTURE
Polythene
Ethane
-(CH2-CH2)n-
Teflon
Terylene or Daeron
CF2=CF2
CH2OH
|
CH2OH
+
HOOC-COOH
-(CF2-CF2)n-[O-CH2-CH2-O
|
-n[C-C=O
H2N-(CH2)6-NH2
+
HOOC-(CH2)4COOH
-[N-(CH2)6-N-C=O
|
|
H
O H
Nylon-6,6
Nylon-6
O
H2C
|
H2C
C
H
N
CH2-
CH2
CH2
USES
Insulating wire ,
dustbin, toys
Oil seal & gaskets.
Ropes, tyre cord
,Safely belts, sarees &
dress materials
O
-n[-C--(CH2)4
O
H
||
|
-[C-(CH2)5-N]n-
Socks, ropes , bristle
for brushes
Tyre cord, Fabrics,
Ropes
Buna-S
CH2=CH-CH=CH2
+
-CH=CH2
Buna-N
Natural Rubber
Neoprene
Nylon-2-Nylon-6
CH2=CH-CH=CH2
+
CN
|
CH2=CH2
CH2=C-CH=CH2
|
CH3
Cl
|
CH2-C-CH=CH2
H2NCH2COOH
Glycine
+
H2N-(CH2)5-COOH
Amino caproic acid
Floor tiles, foot
wears
-[CH2-CH=CH-CH2-CH-CH2]n-[CH2-CH=CH-CH2-CH2-CH]n
|
CN
Oil seal tank lining
CH3
|
-[CH2-C=CH-CH2]nCl
|
-[CH2-C=CH-CH2]n-
Used for tyres
O
-[C-CH2-NH-CO(CH2)5-NH]n-
Conveyor belts,
gasket.
Biodegradable
Polymer
RUBBER:1.
Natural Rubber: - It is a linear polymer of isoprene. It is prepared from rubber latex.
CH3
CH3
|
|
CH2=C-CH=CH2
-[CH2-C=CH-CH2]nIsoprene
Natural rubber
2.
Vulcanization of Rubber:- Natural rubber is soft , soluble in non-polar solvents. To
improve its physical properties it is vulcanized. The process of heating raw rubber with
sulphur and additive to make it hard, more resistant and elastic is called Vulcanization of
rubber. Sulphur form cross- links at the reactive site of double bonds.
3.
Synthetic Rubber:- These are obtained by polymerization of butadiene or its
derivative are called Synthetic rubber. e.g. Buna-S, Buna-N.
4.
Biodegradable Polymers: - The polymers which are decomposed by the microorganism into simple molecule are called Biodegradable Polymers. They do not pollute the
environment. e.g. PHBV, Nylon-2-nylon-6.
HO-CH-CH2-COOH + HOCH-CH2-COOH
CH3
CH2
CH3
3-Hydroxy butanoic acid
3 hydroxy pentanoic acid
-[O-CH-CH2-C-O-CH-CH2-C]nCH3
O CH2
O
CH3
PHBV
Very Short Answer type Questions
(1mark)
Q1: What is polymerisation?
Ans: The process of formation of polymers from respective monomer is called
Polymerisation.
Q2: Is (-NH-CHR-CO-)n, a homopolymer or a copolymer?
Ans: It is a homopolymer because it is obtained from a single monomer.
Q3: Write the name and structure of one of the common initiators used in free radical
addition polymerisation?
Ans: Benzoyl peroxide (C6H5CO-O-O-CO-C6H5).
Q4: Write the structures of monomer used in the preparation of
i. Teflon
ii. PVC.
Ans: (i) CF2=CF2
(ii) CH2=CHCl.
Q5: What are the monomers of bakelite?
Ans: Phenol and formaldehyde HCHO.
Q6: Write the equation used for the synthesis of neoprene.
Ans: nH2C=CCl-CH=CH2
(H2C-CCl=CH-CH2-)n –
Q7: What is mean by vulcanisation of rubber?
Ans: The process of heating natural rubber with sulphur to improve its properties.
Q8: What are homopolymer?
Ans: The polymers formed by the polymerisation of a single monomeric species are called
homopolymer.
Q9: On the basis of forces between their molecules in a polymer to which class does Nylon 6,6 belong?
Ans: It belongs to the class of fibres.
Q10: Write the chemical equation for the preparation of terylene.
Ans:
nHO-CH2-CH2-OH + nHOOC-------_______-COOH
-(O-CH2-CH2-O-OC-
-COO)n--
Short Answer type Questions
Q1: What are natural and synthetic polymers? Give two examples of each type.
(2mark)
Ans: Polymers of high molecular mass obtained from plants and animals are called natural
polymers.
Ex. Cellulose, starch.
Polymers prepared in laboratories are called synthetic polymers.
Ex. Nylon, PVC, Buna-S etc.
Q2: Explain the difference between Buna-N and Buna-S.
Ans: Buna-N is copolymer of 1,3 butadiene & acrlonitrile.
Buna-S is copolymer of 1,3 butadiene & styrene.
Q3: What are biodegradable polymers? Give two examples.
Ans: Polymers which are decomposed by the micro-organism. eg. PHBV & Nylon-2, Nylon-6.
Q4: In which classes, the polymers are classified on the basis of molecular forces?
Ans: On the basis of molecular forces they are classified as
1 Elastomers –natural rubber
2 Fibres –nylon-6,6
3 Thermoplastics –Polythene
4 Thermosetting plastics- Bakelite.
Short Answer type Questions
(3mark)
Q1) What are the synthetic and natural polymers? give two examples?
Ans:-polymers of high molecular mass obtained from plant and animals are called natural
polymers e.g.;-cellulose and starch.
Polymers prepared in the laboratories are called synthetic polymers e.g:-Nylon, Buna-S and
PVC.
q2) Explain the difference between Buna-N and Buna –S.
Ans:-Buna –n is copolymers of 1,3 butadiene and acrlonitrile Buna-s is copolymers of 1,3
butadiene and styrene.
Q3) what are biodegradable polymers? give two examples.
Ans:- polymers which are decomposed by micro-organism dg:- PHBV and nylon -2 and
nylon-6.
Q4) In which the polymers are classified on the basis of molecular forces?
Ans- on the basis of molecular forces they are classified as
1. Elastomers ------ natural rubber
Fibres------- nylon-6 and 6
2. Thermoplastics----- polythene
3. Thermosetting plastics-----Bakelite.
Q5) write the structural formula for the monomers of the following polymers;
(i) Nylon-,6,6
(ii) Natural rubber
Ans: (i) Nylon-6,6: Adipic acid HOOC-(CH2)4 –COOH Hexamethylene diamine H2N-(CH2)6NH2
(ii) Natural rubber: isoprene CH2C-CHCH2CH3
Q6) what is the step growth polymerisation give 1 example for it
Ans: It is a process in which isomers undergo condensation polymerisation step wise.
Ex: Nylon and Terylene are step growth polymers.
Q7) Write the chemical equation for the synthesis of glyptal Mention one use of it.
Ans:
+
{-O-CH2-CH2OOC
HOCH2-CH2OH
HOOC
COOH
Phthalic acid
ethylene glycol
Glyptal
It is used in the manufacture of paints and lacquers.
Q8) Comment on the natural difference b/w thermoplastics and thermosetting polymers
Ans: Thermoplastics have less intermolecular forces of attraction because they are not
cross linked and become soft on heating and can be moulded. Thermosetting polymers
have strong forces of attraction due to cross linkage. They do not become soft on heating.
Q9)(a) How does vulcanization change the character of NATURAL RUBBER
(b)Why 66 and 6 put in the name NYLON-66AND NYLON -6!
Ans:(a) Natural rubber become more hard and elastic on vulcanization due to formation of
cross linkage with sulphur.
(b) Nylon-6,6 is a polymer of adipic acid and Hexamethylene diamine, both the monomers has 6
carbon atom each Nylon-6 is polymer on caprolactum which contain 6 carbon atom.
Q10) write the chemical reaction involved in the synthesis of Bakelite.
Ans:
OH
OH
CH2OH
+ HCHO
Phenol
Formaldehyde
O-hydroxy benzyl alcohol
OH
+
CH2OH
-H2O
OH
-CH2-
OH
OH
-CH2CH2
OH
-CH2-
Polymerisation
-CH2-
CH2
-CH2-
-CH2 CH2
-CH2-
OH
OH
Bakelite
(2) Write the name and structure of the monomers of the following polymers. (a) Buna-S
(b) Buna-N (c) Dacron.
Ans. Buna-s 1, 3 butadiene CH2=CH-CH=CH2
Styrene C6H5 CH=CH 2
(b)Buna-n
1,3butadiene CH2=CH-CH=CH2
CO }-
Acronitrile CH2=CH-CH
ethylene glycol HO-CH2-CH2O+
Terephthalic acid
HOO---COOH
(3) What are biodegradable polymers? Give an example
and write its preparation.
O
Ans. The polymers which get decomposed by micro organisms. e.g: PHBV and Nylon-2,
nylon-6.
Preparation of Nylon-2-nylon-6:O
HOOC-CH2-NH2 + HOOC-(CH2)5NH2 -------→ [-C-CH2-NH-CO-(CH2)5-NH-]
Glycerine
amino caproic acid
nylon-2-nylon-6
(4) How does the presence of double bond in rubber molecules influences their structure and
reactivity?
Ans. The natural rubber is cis-1,4 polyisoprene. The double bonds located b/w C2 & C3 of
isoprene uniprene units. The cis-configuration about double bonds do not allow chains to
come closer for effective attraction due to weak intermolecular forces of attraction .That is
why the natural rubber has a coiled structure and shows elasticity.
(5) How can you differentiate between addition & condensation polymerisation . Give one
example of each?
Ans.
ADDITION POLYMERISATION
CONDENSATION POLYMERISATION
(c)Dacron
=
1.It takes place in unsaturated
monomer
2. Loss of small molecules like HCl,
H2O, NH3 does not take place .
e.g: Formation of polythene &
polypropene
1.It takes place in monomer having multi
functional groups
2. Loss of small molecules likeH2O,NH3, NaCl takes
place.
e.g: Formation of nylon-6,6 terylene
(6) What are monomers? Write monomers of PHBV, nylon-6, 6. Give there uses.
Ans. The simplest molecules which combine to give polymers are called Monomers.
PHBV: -3-hydroxybutanoic acid & 3-hydroxy pentatonic acid
Nylon-6, 6:- Hexamethylene diamine adipic acid.
Uses: 1.PHBV is used in specialty packaging, and in controlled release of drug.
2. Nylon-6, 6 is used in making sheets, bristles for brushes and in textile industry.
--------------------------------------------------------------------------------------------------------
Unit-16
CHEMISTRY IN EVERYDAY LIFE
POINTS TO BE REMEMBERED
1. DRUGS – Drugs are chemical of low molecular masses, which interact with
macromolecular targets and produce a biological response.
2. CHEMOTHERAPY- The use of chemicals for therapeutic effect is called
chemotherapy.
3. CLASSIFICATION OF DRUGS –
(a)
ON THE BASIS OF PHARMACOLOGICAL EFFECT-drugs for a particular type
of problem as analgesics-----for pain relieving.
(b)
ON THE BASIS OF DRUG ACTION-Action of drug on a particular biochemical
process.
(c)
ON THE BASIS OF CHEMICAL ACTION-Drugs having similar structure .egsulpha drugs.
(d)
ON THE BASIS OF MOLECULAR TARGETS- Drugs interacting with
biomolecules as lipids , proteins.
4. ENZYMES AS DRUG TARGETS
(i)
CATALYTIC ACTION OF EN ZYMES(a)
enzymes have active sites which hold the substrate molecule .it can be attracted by
reacting molecules.
(b)
Substrate are bonded to active sites through hydrogen bonds , ionic bonds ,Vander
Waal or dipole –dipole interactions.
(ii)
DRUG- ENZYME INTERACTIONS(a)Drug complete with natural substrate for their attachments on the active sites of
enzymes .They are called competitive inhibitors.
(b)Some drugs binds to a different site of the enzyme called allosteric sites which changes
the shape of active sites.
5. ANTAGONISTS- The drugs that binds to the receptor site and inhibit its natural
function.
6. AGONISTS-Drugs mimic the natural messenger by switching on the receptor.
7. ANTACIDS-These are compounds which neutralize excess acid of stomach.egAluminium hydroxide, Magnesium hydroxide.
8. ANTI HISTAMINES-The drugs which interfare with the natural action of
histamines and prevent the allergic reaction. eg-rantidine,tegarnet, avil.
9. TRANQULIZERS-The class of chemical compounds used for the treatment of
stress,mild or even severe mental diseases. Eg-idardil, iproniagid,luminal,second
equaqnil.
10. ANALGESICS-They reduce pain without causing impairment of consciousness,
mental confusion or some other disturbance of the nervous system.
Eg-aspirin,seridon,phenacetin.
11. ANTIMICROBIALS-They tend to prevent/destroy or inhibit the pathogenic action of
microbes as bacteria,virus,fungi etc.They are classified as
(i)ANTIBIOTICS-Those are the chemicals substances which are produced by microorganisms.
Eg-Pencillin, ofloxacin.
NARROW SPECTRUM ANTI-BIOTICS-These are effective mainly against gram positive
or gram negative bacteria. Eg-Penicillin,streptomycin.
BROAD SPECTRUM ANTI-BIOTICS-They kill or inhibit a wide range of microorganisms.
eg- chloramphenicol,tetracydine
(ii)ANTISEPTICS OR DISINFECTANT-These are which either kill/inhibit the growth of
micro-organisms
Antiseptics are applied to the living tissuses such as wounds,cuts,ulcers etc. egfuracine,chloroxylenol & terpinol(dettol).Disinfectant are applied to inanimate objects
such as floors , drainage , system.
Eg- 0.2% solution of phenol is an antiseptics while 1% solution is an disinfectant.
12. ANTIFERTILITY DRUGS- These are the chemical substances used to control the
pregnancy.They are also called oral contraceptives or birth control pills.
Eg-Mifepristone, norethindrone.
13. ARTIFICIAL SWEETNING AGENTS-These are the chemical compounds which
give sweetening effect to the food without adding calorie.
They are good for diabatic people eg- aspartame, saccharin,alitame,sucrolose.
14. FOOD PRESERVATIVES- They prevent spoilage of food to microbial growth.eg-salt,
sugar, sodium benzoate.
15. CLEANSING AGENTS(i) SOAPS- They are sodium or potassium salts of long chain fatty acids.They are
obtained by the soapnification reaction , when fatty acids are heated with aqueous
sodium hydroxide.
They do not work well in hard water.
(iii)
TOILETS SOAP-That are prepared by using better grade of fatty acids and
excess of alkali needs to be removed .colour & perfumes are added to make
them attractive.
(iv)
MEDICATED SOAPS- Substances of medicinal value are added.egButhional,dettol.
16. SYNTHETIC DETERGENTS-They are cleaning agents having properties of soaps
,but actually contain no soap.They can used in both soft and hard water .They are(i)ANIONIC DETERGENTS-They are sodium salts of sulphonated long chain alcohols
or hydrocarbons.eg-sodium lauryl sulphonate. They are effective in acidic solution.
CH3(CH2)CH2OH → CH3(CH2)10CH2OSO3H
(lauryl
alchol)
+
→CH3(CH2)10CH2SO3 Na
(sodium lauryl sulphonate)
(ii)CATIONIC DETERGENTS- They are quarternary ammonium salts of amines with
acetates , chlorides, or bromides.They are expensive used to limited extent.egcytyltrimethylammoniumbromide
(iii)NON-IONIC DETERGENTS- They do not contain any ions. Some liquid dishwashing
detergents which are of non-ionic type .
17. BIODEGREDABLE DETERGENTS-The detergents which are linear and can be
attacked by micro-organisms are biodegradable.
Eg-sodium 4-(1-dodecyl)benzene\ sulphonate.
18. NON-BIODEGREDABLE DETERGENTS-The detergents which are branched and
cannot be decomposed by micro-organisms are called non-biodegdradable.eg-sodium 4(1,3,5,7 tetramethyloctl)-benzene sulphonate.It creates water pollution.
VERY SHORT ANSWER TYPE QUESTION
(1 marks)
Q-1 Define the term chemotherapy?
Ans-1 Treatment of diseases using chemicals is called chemotherapy.
Q-2 Why do we require artificial sweetening agents?
Ans-2 To reduce calorie intake.
Q-3 What are main constiuent of dettol?
Ans-3 Choloroxylenol & Terpinol.
Q-4 What type drug phenaticinis?
Ans-4 It is an antipyretics.
Q-5 Name the drug that are used to control allergy?
Ans-5 Antihistamines.
Q-6Why is the use of aspartame limited to cold food and drinks?
Ans-6 It is unstable at cooking temperature and decompose.
Q-7What are tranquilizers? Give an example?
Ans-7 They are the drug used in stress , mild severe mental disease.
Q-8 What type of drug chloramphenicol?
Ans-8 It is broad spectrum antibiotic.
Q-9Why is biothional is added to the toilet soap?
Ans-9It acts as antiseptics.
Q-10 What are food preservatives?
Ans-10 The substances that prevent spoilage of food due to microbial growth. eg- sodium
benzonate.
SHORT ANSWER TYPE QUESTION
(2 marks)
Q-1 Mention one important use of the following(i) Equanil
(ii)Sucrolose
Ans-1 (i) Equanil- It is a tranquilizer.
(ii) Sucrolose-It is an artificial sweetener.
Q-2 Define the following and give one example(i)Antipyretics
(ii)Antibiotics
Ans-2 (i) Antipyretics- Those drugs which reduce the temperature of feveral body are called
Antipyretics.
Eg - Paracetamol
(ii) Antibiotics-The drugs which prevent the growth of other micro-organisms. Eg- Pencillin.
Q-3 Name the medicines used for the treatment of the following(i) Tuberculosis
(ii) Typhoid
Tuberculosis- Sterptomycin
Typhoid- Cholororophenicol
Q-4 What are tincture of iodine?
Ans-4 2-3% iodine solution of alcohol water is called tincture of Iodine. It is a powerful
antiseptics and is applied on wounds.
Q- 5 What are artificial sweetening agent? Give two examples?
Ans-5 The substances which give sweetening to food but don’t add calorie to our body .
Eg- Saccharin , alitame.
Q-6 How are synthetic detergents better than soaps?
Ans- 6 (i)Detergents can be used in hard water but soaps cannot be used.
(ii) Detergents have a stronger cleansing action than soaps.
Q-7 What are sulpha drugs? Give two examples?
Ans-7 A group of drugs which are derivatives of sulphanilamide and are used in place of
antibiotics are called sulpha drugs.
Eg- sulphadizine, sulphanilamide.
Q-8 What forces are involved in holding the active sites of the enzymes?
Ans-8 The forces are involved in holding the active sites of the enzymes are hydrogen
bonding , ionic bonding , dipole-dipole attractions or Vander waals force of attractions.
Q-9 Describe the following giving an example in each
case- (i) Edible colours
(ii) Antifertility drugs
(i)
Edible colours- They are used for dying food.
Eg- saffron is used to colour rice.
(ii)
Antifertility drugs- Those drugs which control the birth of the child are called
Antifertility drugs.
Q-10 Give two examples of organic compounds used as antiseptics?
Ans-10 Phenol(0.2%) , iodoform
SHORT ANSWER TYPE QUESTION
(3 marks)
Q-1 What are Biodegredable and non-biodegdredable detergents? Give one example of each.
Ans-1 Detergents having straight hydrocarbon chain and are easily decomposed by microorganisms are called Biodegredable detergents.The detergents having branched hydrocarbon
chain and are not easily decomposed by micro-organisms are called Non-Biodegredable
detergents.
Q-2 What are barbiturates? To which class of drugs do they belong? Give two examples.
Ans-2 Derivatives of barbituric acid are called barbiturates.They are tranquilizers . They
also act as hypnotics. eg- luminal , seconal.
Q-3 What is the use of –
(i) Benadryl (ii) sodium benzoate (iii) Progesterone
Ans-3 (i) Antihistamines
(ii) Preservatives
(iii) Antifertility drug
Q-4 Identify the type of drug(i) Ofloxacin (ii) Aspirin (iii) Cimetidine
Ans- 4 (i) Antibiotic (ii) Analgesics & Antipyretics
(iii) Antihistamines & antacid
Q-5 Describe the following with suitable example(i) Disinfectant (ii) Analgesics
(iii) Broad spectrum antibiotics
(i)
Disinfectant- chemicals used to kill the micro-organisms can applied on non
living articles.
(ii)
Analgesics- They are the drugs which are used to relieve pain . eg – Aspirin ,
Ibuprofen.
(iii)
Broad spectrum antibiotics- They kill the wide range of gram positive and
gram negative bacteria.
Eg- Chloramphenicol , ofloxacin.
-----------------------------------------------------------------------------------------------------------------------
BIOMOLECULES
Carbohydrates: These are optically active poly hydroxyle or ketones
or molecule which provide such unit on hydrolysis.
Classification of Carbohydrates
Carbohydrates are classified into three classes depending on well
behaviour towards hydrolysis:
 Monosaccharide:-The carbohydrates which cannot hydrolysied
futher. Eg: glucose, fructose.
 Oligosaccharides: The carbohydrates which have 2 to 10
monosaccharides units on hydrolysis. Eg: sucrose,cellulose.
 Polysaccharides: the carbohydrates which give large no. of
monosaccharides on hydrolysis. Eg: starch,cellose
Sugar: The carbohydrates which are sweet in taste are called
sugar. Eg: sucrose ,lactose
Non-Sugars:The carbohydrates which are not sweet in
taste.Eg:starch,cellose.
Reducing Sugar: Sugars which reduce fehling solution and
tollens reagent. Eg: glucose, fructose.
NON Reducing Sugar: which do not reduce Fehling or tollen’s
reagent. Eg: sucrose.
CHO
H
OH
OH
H
H
OH
H
OH
CH2COOH
Structure Of Glucose
Cyclic Structure Of Glucose:
The straight chain structure of glucose is unable to explain the
following reaction.
1. It does not give the 2,4-DNP test, Schiff test and dosen’t
form hydrogensulphide addition product with NaHSO3.
2. The penta acetate of glucose does not react with NH2Oh
indicating the absence of CHO group.
3. It exist in two different crystline α and β forms. They are
called anomers. They differ in optical rotation. They
have different melting point.
CH2OH
Starch: Main storage polysaccharide of plants has two components
amylase and amylopectin.
Protiens: Protiens are long polymers of amino acids linked by peptide
bands.they are essential for proper growth and maintaince of body.
Amino Acids: It contains amino and carboxylicacid functional group
R-CH-COOH
NH2
Essential Amino Acids: Amino acid which cannot be synthesized in
the body and are obtained from diet.eg:Valine,Lysine.
Non-Essential Amino Acid: The amino acid which can be synthesized
in the body are called non essential amino
acid.eg:glysine,serine,praline.
Zwitter Ion:In aqueous solution the carboxyl group can lose a
protone and amino group can accept a protone,giving rise a dipoler
ion called as zwitter ion.
NH3+----CH----COOR
Classification:
1. Fibrous Protein:
 Polypeptide chains run parallel or antiparalle and held
together by hydrogen and disulphide bonds.
 Generally in soluble in water.eg:collagen,myorin,fibroin.
2. Globular Protien:
 Chains of polypeptide coil around to give spheical shape.
 Usually soluble in water.eg:insulin,haemoglobin.
Denaturation Of Protein: The proteins in its native form, when
subjected to physical change like temperature,PH etc undergoes
uncoiling and loses its biological activity. This is called denaturation
of proteins.
The secondary and tertiary structure are destroyed, only primary
structure retained.eg:coagulation of eggs while on heating.
Vitamins:The organic compound required in the diet in small
amounts to perform specific biological functions and its not
synthezised by our body.
Classification: On the basis of their souliblity in water or fat.
(a) Fat Soluble Vitamin: They are insoluble in water
but soluble in fat.eg: vitamin A,B,E,R.
(b) Water Soluble Vitamin: Vitamin B & C.
VITAMINS
A
B
C
D
SOURCE
DEFICIENCY
DISEASE
CARROT,BUTTER & XEROPHTHAL
MILK
MLA,NIGHT
BLINDNESS
YEAST,LIVER,GREEN BERI-BERI
VEGETABLE
CITRUS FRUIT,AMLA SCURVY
SUNLIGHT,FISH,EGG RICKETS
YALK
Nucleic Acids:These are bimolecules which are long chain polymers
of nucleotides.They are DNA & RNA.
Composition Of Nucleic Acid: they are made-up of a pentane sgar (BD-2 dexoyribose in DNA d B-D-ribose in RNA), phosphoric acid and
nitrogen containing
(a) DNA: Base adenine
(A),Thymine(t),guanine(g),and cytosine(C).
(b) RNA: Contain adenine(A), guanine(g),
cytosine(C)and Uralic(U).
Enzymes:enzyme are bio-catalyst and are generally lobular potein eg:
zymase,maltase.
VERY SHORT ANSWER TYPE QUESTIONS
Q1 What are carbohydrates?
ans- These are optically active poly hydroxyle or ketones or molecule
which provide such unit on hydrolysis.
Q2 Name the deficiency disease resulting from lack of vitamin C and
D?
ans- SCURVY and RICKETS.
Q3 what are the name given to the linkage which hold together to
monomeric units in polysaccharide?
Ans-Glycosidie linkage.
Q4 Name the enzyme which convert sucrose into glucose and
fructose?
Ans-Invertase
Q5 What are the components of starch?
Ans-Amylase and amylopectin.
Q6 Why is cellulose in our dite not nourishing?
Ans-Our body does not have enzyme which help in digestion of
cellulose.
Q7 Name the purines present in DNA?
Ans-Adenine and Guanine.
Q8 Why vitamine C cannot be stored in our body?
Ans-It is water soluble and readily excreated in urine.
Q9 Give two examples of non-essential ammino acid?
Ans-Glycine and alamine
Q10 What are different types of RNA found in the cell?
Ans-1.m.RNA 2.t.RNA 3.r.RNA.