MAHARASHTRA STATE BOARD12th Class

CHEMISTRYMOCK TEST PAPER

withSOLUTIONS

Presents

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1. Select and write the most appropriate answer from the given alternatives for each subquestion :

[7]

(i) Paramagnetism is a property of : (1)

(a) Completely filled electronic sub-shells (b) Unpaired electrons

(c) Non-transition elements (d) Melting point and boiling point of elements

(ii) The molal freezing poing constant for water is 1.86 /C mol° . If 342 gms of cane sugar ( )12 22 11C H O

is dissolved in 1000g of water, the solution will freeze at .................... (1)

(a) 3.92 C− ° (b) 1.86 C− ° (c)1.86 C° (d) 2.42 C°

(iii) For a first order reaction A → products, if k is the rate constant, then the half life period of the

reaction is equal to (1)

(a) [ ]0.693

k A (b) [ ]1 1

.k A (c)

[ ][ ]

1ln

A

k A ° (d) ln 2

k

(iv) 31 10 g× atoms of Ag can be oxidised to Ag

+ ions by passing the electricity of ................ (1)

(a) 0.001 Faraday (b) 0.1 Faraday (c) 1 Faraday (d) 1000 Faraday

(v) Chemical formula of copper pyrites is ............... (1)

(a) 2CuFeS (b) 2Cu S (c) 2Cu O (d) 3 2. ( )CuCO Cu OH

(vi) Two neutrons are introduced in the nucleus of an atom. The resulting atom and the parent atom form

a pair of .............. (1)

(a) isobars (b) isotopes (c) isomers (d) isotones

(vii) Which of the following properties is intensive ? (1)

(a) enthalpy (b) entropy (c) volume (d) refractive index

2. Attempt the following (Any Six) : [12]

(i) Define the terms : (2)

(a) Intrinsic energy

(b) Heat of reactions at constant pressure and volume.

(ii) Distinguish between the rate of a reaction and the rate constant. (2)

(iii) Explain :

(a) Normality (b) Molarity (2)

(iv) With the help of catalytic hydrolysis of methyl acetate, explain the pseudo first order reaction. (2)

(v) Explain the types of thermal energy. (2)

CHEMISTRY

Time : 3 Hrs. M.M. : 70

GENERAL INSTRUCTIONS(i) All questions are compulsory.

(ii) Internal choice is given to Q. 2,3 and 4.

(iii) Use of calculator is not allowed.

(iv) Section I and Section II has weightage of 35 marks each.

SECTION - I

Page 3 of 14

(vi) Write down the reactions taking place during discharging of lead accumulator. (2)

(vii) Draw a neat and labelled diagram for standard Hydrogen gas electrode. (2)

(viii) Give important ores of zinc metal. (2)

3. Attempt the following (Any Three) [9]

(i) What are fuel cells ? Give construction and working of 2 2H O− fuel cell. (3)

(ii) The heat of formation of carbondioxide and carbon monoxide are 1396.6kJmol−− and 110.5kJ− per

mole respectivley at constant pressure and at 296 K. Calculate the heat of combustion of carbon

monoxide at constant volume and at 298K, if R = 1 18.314JK mol− − . (3)

(iii) How is 2SO an air pollutant ? (3)

(iv) Calculate the osmatic pressure at 288 K of a solution of urea ( )2 2NH CONH containing 33 10−× kg

of it into half litre of the solution. (3)

4. (A) (i) How will you determine the molar mass of a non-volatile substances by Landsberger and Walker

method. (4)

(ii) Write the cell reaction and find E.M.F of the following cell at 298 K. (3)

Z naq

Z n + +

1a =

a qC u + +

1a =( )s

C u

0 00.76 , 0.34Zn CuE V E V= = −

OR

(B) (i) Define the rate- law . (1)

(ii) The decy constant of a radio element is 24.62 10−× per hour. What fraction of it will remain after 40

hours. (3)

(iii) Calculate the enthalpy change for the formation of 5.4kg of water form the following reaction (data):

( ) ( ) ( )22 2

1

2g g l

H O H O+ → , 1284.5H kJmol

−∆ = (3)

SECTION - II5. Select and write the most appropriate answer from the given alternatives for each subquestion :

[7]

(i) Dettol consists of : (1)

(a) Cresol and ethanol (b) Xylenol and terphenol

(c) Chloroxylenol and terpeneol (d) Xylene and terpeneol

(ii) In isomeric alcohols, the correct order of boiling point is : (1)

(a) 2 3 1° > ° > ° (b) 3 2 1° > ° > ° (c) 1 2 3° > ° > ° (d) 2 1 3° > ° > °(iii) Which of the following is an oil ? (1)

(a) Triolein (b) Tripalmitin (c) Tristearin (d) all of them

(iv) The correct order of increasing reactivity of H X− towards alkenes is : (1)

(a) HF HBr HI HCl< < < (b) HF HBr HCl HI< < <

(c) HF HI HBr HCl< < < (d) HF HCl HBr HI< < <

Page 4 of 14

(v) The compound which is metamer of diethyl ether is : (1)

(a) 1-methoxy propane (b) 1-buttanol

(c) 2-methoxy propane (d) both (a) and (c)

(vi) Natural silk is a : (1)

(a) polyester (b) polyamide (c) polyacids (d) polysaccharide

(vii) Acetaldol is the condensation product of : (1)

(a) 2 molecules of ethanal (b) 2 molecules of propanone

(c) ethanal and methanal (d) ethanal and propanone

6. Attempt the following (Any Six) : [12]

(i) What are monosaccharides and disaccharides ? (2)

(ii) Give the prepartion of Nylon-6 (2)

(iii) What is the action of hot HI on (1) Diethyl ether (2) Ethyl methyl ether ? (2)

(iv) What are Lanthanides and Actinides ? (2)

(v) What is the action of following reagents on phenol ?

(a) Bromine water (b) Conc. 3HNO ? (2)

(vi) Alcohols have higher boiling point than the corresponding alkanes. Explain ? (2)

(vii) How will you convert (2)

(a) Nitromethane to methanamine

(b) Methyl cyanide to ethylamine

(viii) How will you obtain aceticanhydride from acetic acid ? (2)

7. Attempt the following (Any Three) [9]

(i) How will you prepare acetaldehyde and acetone from geminal dihalides ? (3)

Explain silver mirror test.

(ii) Explain optical activity of 2-chlorobutane. (3)

(iii) What is the action of potassium dichromate in presence of dil. 2 4H SO on (3)

(a) Propan-1-ol (b) Propan-2-ol

Write the structure of 3-chloro-2, 2 dimethyl hexan-1-ol

(iv) Give points of distinction between oils and fats ? Define the term conjugated proteins. (3)

8. (A) (i) Write a note on aldol consdensation. (3)

(ii) What are esters ? How will you prepare ethyl acetate from (a) silver acetate (b) acetic anhydride ? How

is acetamide prepared from acetic acid ? (4)

OR

(B) (i) Define Artificial fibre. Give any two uses of terylene.

How will you prepare.

(1) Ethanamine from ethyl bromide.

(2) Iso-propylamine from an oxime. (4)

(ii) How is chloroform prepared from ethanol ? (3)

SOLU

TIONS

Page 6 of 14

SOLUTIONS OF PAPER - 1SECTION - I

1. i. (b) ii. (b) iii. (d) iv. (a) v. (a)

vi (b) vii. (d)

2. (i) Internal or Intrinsic energy : In may be defined

as a certain fixed amount of energy, which is

stored in a system (substance). It is the sum of

all kinds of enegies present in a substance . It is

denoted as ‘E’

Heat of reaction at constant pressure : Heat

of reaction at a constant pressure and at a given

temperature is defined as the enthalpy change

of a system, when number of moles as indicated

by the chemical equation have completely

reacted at that constant pressure it is donoted

as H∆ heat of reaction at constant volume : Heat

of reaction at a constant volume and at a given

temperature is defined as the enthalpy change

of a system, when number of moles as indicated

by the chemical equation have completely

reacted at that constant volume.It is denoted as

E∆ .

(ii) Distinction : The difference between the rate of

a reaction and its rate constant can be

summarized as follow :

Rate of reaction Rate constant

1 Rate of a reaction is the

speed by which the

reactants are converted

into the products.

Rate constant

becomes the rate of

the reaction, if the

concentration of the

reactants is unity

2

It is measured as the

rate of decrease in the

oncentration of the

products with time

It is the

proportionally

constant in the rate

law.

3

Rate of a reaction

depends upon the initial

concentration of the

reactants.

Rate constant is

independent of the

initial concentration

of the reactants

4

The unit of the rate of

reaction

is always mol ( or

The unit of rate

constant depends

upon the order of the

1 1dm s− −

(iiii) (a) Normality : The normality of a solution may be

defined as the number of gram equivalents of the

solute dissolved in one litre ( 3dm ) of the solution.

Thus Normality (N) =

3( )

Number of gram equivalents of solute

Volume of solution in litre dm

or Normality (N) =

31000

.( )

Number of gram equivalents of solute

Volume of solution in ml cm×

The unit of normality is gram equivalent per litre

( )3dm i.e. g equiv 3dm−(b) Molarity : Molarity of a solution may be

defined as the number of moles of the solute

dissolved per ( )3dm of the solution. It is denoted

by ‘M’ .

Thus Molarity (M) =

3( )

Number of moles of solute

Volume of solution in litres dm or

Molarity(M) =

31000

( )

Number of moles of solute

Volume of solution in ml cm×

The unit of molarity is mol 3dm− . A solution

having its molarity one is called molar solution.

Similarly deci molar, centimolar etc. Solutions

can also be obtained. The molarity of a solution

changes with temperature

Thus

Number of moles of solute (a) =

Mass of solute in gram

Molar mass of the solute

(iv) Definition : A chemical reaction of higher order

can be converted into the first order by taking the

other reactant in a large excess. Such a reaction

is called the pseudo first order reaction.

Explanation : Consider the hydrolysis of methyl

acetate ( )3 3CH COOCH in presence of an

acidic medium.

3 3 2C H C O O C H H O+

(M ethyl acetate) (Excess)( )H HCl

Hydrolysis

+

→

3 3C H C O O H C H O H+

(Acetic acid) (M ethanol)

In the above hydrolysis reaction the molecularity

of the reaction in two (involving two reactants

methyl acetate and water) and the order of the

reaction appears to be 2. The rate law equation

for the above reaction can be written as

Rate (r) = [ ][ ]3 2k CH COOH H O

However, the above reaction is generally carried

out by taking water is large excess over methyl

acetate.

Thus the concentration of water 2H O remains

virtually constant.

Thus the rate of the above reaction depends upon

only the concentration of methyl acetate and the

Page 7 of 14

order of the reaction becomes one. The rate law

equation for this reaction can be written as,

Rate (r) = [ ]3 3CH COOCH

(where [ ]2k H O = constant)

Therefore the hydrolysis of methyl acetate taking

place in a large excess of water is called the

pseudo first order reaction or pseudo

unimolecular reaction.

(v) Types of thermal (kinetic) energies are : (1)

Translational energy (2) Rotational energy (3)

Vibrational energy

Translational energy : It may be defined as

the energy produced by a system by virtue of

translational motion in a straight line path by the

molecules along the three co-ordinate axes. It is

a kind of kinetic energy.It is written is as .tranE

(2) Rotational energy : It is defined as the energy

produced by a system by virtue of rotational motion

of its molecules about the axis perpendicular to

the line joining their nuclei. It is a kind of kinetic

energy. It is written as .rotE

(3) Vibrational energy : It is defined as the

energy produced by a system by virture of

vibrational motion of the atoms of its molecules

with respect to each other along their line of

centres. It is a kind of kinetic energy.

It is written as .vibE

(vi) Reactions taking place during discharging of lead

accumulator : (1) When the cell is in operation,

Pb gets dissolved at the negative electrode to

form 2Pb+ ions. These 2Pb

+ ions combine with

2

4SO −ions (obtained from dissociation of

2 4H SO ) and form insoluble salt 4PbSO .Thus

( ) ( ) ( )2

2 4 42 4 2

aq aq aqH SO H SO

+ −→ +And the reaction at the negative electrode would

be, ( )2

2s

Pb Pb e+ −→ + ........... (oxidation)

( )2 2

4 4 sPb SO PbSO

+ −+ →

Net reaction : ( ) ( ) ( )2

4 42

s aq sPb SO PbSO e

− −+ → +... (oxi.)

(2) At the positive electrode lead dioxide ( 2PbO )

is reduced to 4PbSO by accepting electrons.

Thus ( ) ( ) ( )2

2 44 2

s aq aqPbO H SO e

+ − −+ + + →

( ) ( )242

sPbSO H O l+ ...... (reduction)

The total reaction in discharging of the cell can

be written as :

( ) ( ) ( )22 42

s s aqPb PbO H SO+ + →

( ) ( )242 2

sPbSO H O l+

(vii)

Glass jacket

Pure and dry H gas at 1 atm pressure2Glass tube

Mercury

Hydrogen bubblesPlatinised platinumplate

1M H Cl solution

Platinum wire

Fig : Standard hydrogen gas electrode

(viii) Ores of zinc : (1) The chief ore of zinc is zinc

sulphide, ZnS known as zinc blende (sphalerite),

which is found in Belgium, America, Burma and

Canada.

(2) Other important ore of zinc is zinc carbonate,

3ZnCO which is known as calamine

(smithsonite)

(3) Less important ores are (i) Wilemite 2 4Zn SiO

(ii) Franklinite 2 3,ZnO Fe O (iii) Zincite ZnO (iv)

Zincspinel 2 3,ZnO Al O

3. (i) Fuel cells are the votaic cells which convert the

energy produced from the combustion of some

fuels like 2H and 2O gases directly into electrical

energy. In these cells the different reactants are

supplied to the electrodes continuously. The most

common example is the hydrogen oxygen fuel

cell. The comversion of chemical energy of the

fuel into electrical energy involves the combustion

of fuel to liberate heat. This heat energy is utilized

to generate steam for spinning the turbines, which

are connected to electrical generator.

Hydrogen -Oxygen fuel cell :

Construction : It is a voltaic cell in which 2H and

2O gases are used. In this type of a cell hydrogen

and oxygen gases are bubbled into a concentrated

aqueous solutions of sodium hydroxide or

potassium hydroxide through the porous carbon

electrodes (which contain a small amount of

catalyst, like ,Pt Ag etc.) hydrogen is fed into

the anode compartment where it is oxidized. The

oxygen is fed into cathode compartment where

it is reduced. The diffusion rates of these gases

are carefully regulated to get the maximum

Page 8 of 14

efficiency. This cell is also called Bacon-cell after

the name of its inventor (Fig,.)

+ CathodeAnode

H O(g)2

H gas2 O gas2

Electrolyte(concentratedalkali so lution)

Porous cabonelectrodes im pregnatedw ith catalyst

Fig : Fuel cell

Working (reactions) : The anode and cathaode

reactions taking place in the cell can be written

as follows

Anode reaction :

( ) ( ) ( )2 22 4 4 4aq

H g OH H O l e− −+ → +

Cathode reaction :

( ) ( )2 22 4 4 aqO g H O l e OH− −+ + →

Overall or net reaction :

( ) ( ) ( )2 2 22 2H g O g H O l+ →The cell runs continuously as long as the

reactants are fed into it. These fuel cells are more

efficient than conventionally used methods of

generating electricity on a large scale by burning

hydrogen and carbon fuels. These fuel cells

convert the energy of the fuel directly into

electricity. The voltage of this cell is 1.2 V

(ii) Given (1) T=298 K (2) R= 1 18.314JK mol− −

(3) Heat of formation of CO at constant pressure

= 1396.6kJmol−= −

(4) Heat of formation of 2CO at constant pressure

= 1110.5kJmol −= − , E∆ = ?

The required chemical equaiton for the formation

of ( )2 gCO and ( )g

CO cab e written as follows :

(1) ( ) ( ) ( )2 2s g gC O CO+ → ;

1396.6H kJmol

−∆ = −

(2) ( ) ( ) ( )2

1

2s s g

C O CO+ → ;

1110.5H kJmol−∆ = −

The required thermo chemical equation for the

combustion of ( )2 gCO can be written as follows

(3) ( ) ( ) ( )2 2

1

2g g g

CO O CO+ → ; H∆ = ?

Equation (3) can be obtained by subtracting

equation (2) from equation (1)

Hence ( ) ( ) ( )2 2

1

2g g g

O CO CO→ −

[ 396.6] [ 110.5]H∆ = − − −

∴ ( ) ( ) ( )2 2

1

2g g g

CO O CO+ → ;

286.1H kJ∆ = −∴ Heat of combustion at constant pressure

( H∆ )= 286.1kJ− (iii) Advantages of fuel cells over ordinary

batteries : The important advantages of fuel cells

over ordinary batteries are :

(1) High efficiency : The cells convert the

chemical energy of a fuel directly into electricity

and therefore these cells are more efficient than

the conventional methods of generating electricity

on a large scale. Nearly 60-70% efficiency has

been obtained in this type of cell.

(2) Continuous source of energy : In this cell,

there is no electrode material to be replaced as

in case of ordinary battery. In this the fuel can be

fed continuously to produce power and that is

the reason. 2 2H O− fuel cells have been used

in space crafts.

(3) Pollution free working : In these cell, no

objectionable byproducts are produced and

therefore they do not cause any pollution

problems.

(iv) We have V RTπ = (Formula)

Molar mass of urea ( )2 2NH CONH

( ) ( ) ( ) ( )14 2 1 4 12 1 16 1= × + × + × + × =

160g mol

−

= 3 160 10 kg mol

− −×

Wt. of solute = 33 10 kg

−×

Volume of solution = 30.5dm

R= 1 10.082lit atmK mol− −− , 288 , ?T k π= =

Now 33 10 kg

−× of urea is present in 30.5dm

∴ 360 10−× kg (1 mol of it will be present in

33

3

0.5 60 1010

3 10dm

−

−

× ×= =

× , ∴ 3

10V dm=

Now 10 0.082 288π × = ×

or0.082 288

2.36210

atmπ×

= =

Osmotic pressure = 2.362 atm

Page 9 of 14

4. (A) (i) Determination of molecular mass by Landsberger

and Walker method :

Principle : To determine the molecular mass of

a non-volatile solute, a known mass of the solute

is dissolveding known mass of a suitable boiling

solvent. The elevation in boiling point is

determined. If the b

K value of the solvent is known

then the malar mass of the solure can be

calculated.

Description os the apparataus : (1) The

apparatus consists of a hard glass boiling tube

(graduated tube) with a hole, almost at its top

portion, containing a bulb (Fig)

Dropp ingfannel

Two holedstopper

Roundbottomflask

Tripedstand Barner

Pureboilingsolven t

Bent tube

Porousrosehead

To condenser

Solvent

Outer vessel (jacket)

Boiling tube

Hole

Cock

Two holedstopper

Sensitivetherm om eter

Fig : Landsberger and Walker method

(2) The boiling tube is corked with a two holed

rubbeer stopper. A sensitive thermometer is

inserted in one of the holes of the rubber stopper.

In the other hole a bent glass tube (delivery tube)

if fitted whose one end is in the form of a perforated

rose head. This end remains immersed in the

solvent of the boiling tube. The other end of the

bent tube is almost in the neck of a round bottom

glass flask containing the pure solvent.

(3) The flask is corked with a two holed stopper.

In the other hole of the cork a dropping or thistle

funnel is fitted.There is an outer vessel (jacket)

which covers the boiling tube and protects it from

damage. The jacket is connected to a condenser,

which helps to condense the escaping vapour.

(4) The flask is heated with the help of a heating

arrangement.

Procedure : (1) A small quanitity (5-10 mL) of

the pure solvent is placed in the boiling tube. The

solvent is added in the round bottom flask with a

dropping funnel and then it is heated to get the

vapours of the solvent.

(2) The vapour of the solvent is then passed into

the boiling tube ad get condensed. The solvent

in the boiling tube starts boiling. When the

temperature of the solvent becomes steady, it is

noted down ( correct boiling point of the solvent)

(3) Now, the passage of vapours of the solvent

from the flask is stopped and the apparaturs is

cooled at the room temperature.

(4) A weighed quantity of the non-volatile solute

(say w gram) is then added in the boiling tube

from the top of it.

(5) The solute dissolves in the the solvent and

the boiling point of the solution (temperature

becomes constant) so obtained is noted down

as explained before. This gives the boiling point

of the solution.

(6) The difference in temperatures is the elevation

in the boiling point (b

T∆ )

(7) After the second observation the flow of solvent

vapours from the flask is stopped. The volume of

the solvent in the boiling tube is measured (the

volume increases continuously due to

condensation of vapour)

(8) Knowing the density the mass of the solvent

present in the solution can be calculated.

(9) The volume occupied by the solute is not

considered (as it is negligible)

Observations :

(1) Weight of non-volatile solute = 2W g

(2) Weight of the solvent = 1W g [if, volume and

density of the solvent is known its weight

( )V d× can be determined)

(3) Let the molecular mass of solute = M

Calculations : Molality of the solution (m)

= 1000Number of moles of solute

Weight of solvent in grams×

∴m =2

1

1000w

M w

××

(Since the number of moles of solute = 2

w

M)

But b

T∆ =b

K m×

∴2

1

1000b

b

K wT

M w

× ×∆ =

×[

bK =molal boiling point elevation constant)

∴2

1

1000b

b

K wM

T w

× ×=

∆ ×

∴ Molar mass (M) =2

1

1000b

b

K w

T w

× ×∆ ×

Result : It the volues of mass of sulute and the

solvent molal boiling elevation constant of solvent

and the elevation in boiling point is known the

Page 10 of 14

molar mass (molecular weight) of the solute can

be determined.

(ii) Reaction at anode :

( )2

2aqsZn Zn e

+ −→ + (oxidation)

Reaction at cathode :

( )2

2aq sCu e Cu

+ −+ → (reduction)

∴ Net cell reaction

( ) ( )2 2

aq aqs sZn Cu Cu Zn

+ ++ → +

∴ 0 0 0

( ) ( )cell oxi anode oxi cathodeE E E= −

( )0.76 0.34= − − 1.10V=

∴ E.M.F. of the cell 1.10V=5. (B) (i) Rate law : The rate law or rate equation may be

defined as the mathematical expression, which

denotes the experimentally observed rate of a

reaction in terms of the concentrations of the

reacting species which influenece the rate of the

reaction.

(ii) (1) 24.62 10λ −= × per hour

(2) 40t = hours (3) 1

0

?N

N=

Q

010

1

2.303log

N

t Nλ

=

.

∴0

10

1

log2.303

N t

N

λ =

2

4.62 10 40 1.848

2.303 2.303

−× ×= = 0.8024=

∴0

1

N

N= A.L. ( )0.8024 6.345=

∴ 1

0

10.1576

6.345

N

N= = ,∴ Fraction left =

0.1576

(iii) 3

. 5.4

. . 18 10

Wt of water gn

Mol Wt of water kg−= =×

3

2[ 2 16 18 18 10 ]H O g kg−= + = = ×

5.4300

0.018moles=

For 1 mole of 2H O , heat of formation

284.5kJ= −∴ for 300 moles of water, heat of formation

284.5 300kJ= − × , 85350kJ= −Heat evolved, 48.535 10H kJ∆ = − ×

SECTION - II5. i. (c) ii. (c) iii. (a) iv. (d)

v. (d) vi. (b) vii. (a)

6. (i) (1) Mono saccharides : These are the basic

(simple and small) units of carbohydrates and

cannot be further hydrolysed into smaller units

or molecules. They have genral formula

( )2 nCH O where n = 3 to 7.,

e.g. Glucose 6 12 6C H O

(2) Disaccharideds : These are the

carbohydrates which can be further hydrolysed.

On hydrolysis they yield two molecules of

monosaccharides.

e.g. Sucrose, Maltose, Lactose ( 12 22 11C H O )

(ii) Preparation of Nylon-6 : The starting materail

for prepartion of Nylon-6 is ∈-caprolactum. When

∈-caprolactrum is heated to 533 K in an inert -

atmosphere, polymer of nylon 6 is obtined. This

polymer on spinning is converted into the fibers

of nylon-6.

Reaction :

2H

2H

(E- Capro lactam)

( )

2

533

/

K

Polymersiationinert atm N

∆→

( )2 5C N C H- - -

HO

Nylon-6 polym er

(iii) (1) Hot HI on diethyl ether :

(D iethyl ether)

2 5 2 5 2C H O C H H I- - +373

Hot

K→

(E thyl iod ide)

2 5 22C H I H O+

(2) Hot HI on ethyl methyl ether :

(E thylmethyl ether)

2 5 32C H O C H H I- - +

373

Hot

K→

(M ethyl iodide) ( E thyl iodide)

3 2 5 2C H I C H I H O+ +

(iv) Lanthanides : It is a series of the elements which

are placed according to their increasing atomic

Page 11 of 14

numbers in the periodic table. It involves the filling

of 4 f -orbitals following lanthanum (La, 57) and

ending to letetium (Lu 71)

Actinides : It is a series of the elements which

are placed according to their increasing atomic

number in the periodic table. It involves the filling

of 5 f -orbitals, following actinium (Ac 89) and

ending to lawrencium (Lw 103)

(v) Action of phenol :

(a) With Bromine water :

O H

(Phenol)

( )23Br water+ →

O H

(2,4,6 - Trib romophenol)

B r

B rB r

(b) With conc. 3HNO :

O H

(C arbolic acid )

2 4

.Conc

H SO→

O H

(P icric acid)2N O

2N O2O N

(vi) Alcohols possess OH− as the functional group.

* The electro negative of oxygen is 3.5 and that

of hydrogen is 2.1. Defference in their electro

negativity is 1.4.

* Due to this large difference in their electro

negativities the OH− group in alcohols becomes

polar in nature forming hydrogen bondings. Thus

alcohol molecules posses hydrogen bondings

and all the molecules get associated to each

other.

* Due to these hydrogen bondings the separation

of alcohol molecules from one another during

boiling requires more energy than for the

separation of their corresponding alkane

molecules.

* In alkanes no such hydrogen bondings are

present and hence less energy is required to

separate their molecules.

Therefore alcohols have higher boiling points than

their corresponding alkanes.

(vii) (a) Nitromethane to methanamine :

Reduction of nitromethane by tin and HCl

[ ]3 26C H N O H- +

(N itro m ethane)

Reduction

Sn HCl+→3 2 22C H N H H O- +

(M e thy la m in e )

(b) Methyl cyanide to ethylamine :Reduciton of methyl cyanide by sodium and

alcohol.

34[ ]C H C N H- = +

(M e th y l cyan ide )

/

Reduciton

Na alcohol→ 3 2 2C H C H N H- -

(E thy lam ine )

(viii) Anhydride formation : When ethanoic acid is

treated with some dehydrating agent like

2 5P O anhydride (acetic anhydride) is formed.

(Ace tic acid)

3C H C O H- +

O

3C H C O H- +

O

2 5P O+ Dehydration

∆→

2O H O+3C H C O-

3C H C O-

(Acetic anhydride)

7. (i) Preparation acetaldehyde and acetone from geminal

dihalides :

(1) Acetaldehyde : When ethylidenedichloride

or 1,1-dichloroethane is treated with aqueous

KOH finally acetaldehyde is formed .

3C H C H C l-

(E thy lidene d ich lo ride ) or

3C H C C l- -

H

C

K O H

K O H

Acueous

∆→ 2KCl +

Page 12 of 14

3C H C O H- -

H

(Unstable)

O H

→ 2 3H O C H C O+ - -

H

(Acetaldehyde)

(2) Acetone : When 2, 2-dichloro propane is

treated with aqueous KOH finally acetone is

formed.

3C H C C l- -

3C H

C l

K O H

K O H

(2,2 d ichloropropane)

Aqueous

∆→ 2KCl +

3C H C O H- -

3C H

(Unstable)

O H

→ 2 2H O C H C O- - =

3C H

(Acetone )

Silver Mirror test : Tollen’s reagent is anammonical silver nitrate solution

( )3 2Ag NH OH . When an aldehyde is heated

with Tollen’s reagent, a silver mirror is formed

on the walls of the test tube or a greyish black

precipitate of colloidal silver is obtained. Aldehyde

is oxidised to the corresponding carboxylic

acid.e.g.

( )3 3 22C H C O A g N H O H- = +

H

(Aceta ldehyde)(To llen 's reagent)

∆→

3 3 24 2C H C O O H N H H O A g- + - +

(ii) Optical activity of 2-chlorobutane : The

general formula of 2-chlorobutane can be written

as follows:

3 2 3CH CHCl CH CH− − − or

3 2 5C H C C H- -

H

C l(1) Presence of asymmetric carbon atom : In

2-chlorobutane, there is only one asymmetric

carbon atom containing four different groups.

According to the formula 2na =(a= number of isomers, n= number of asymmetric

carbon atoms). Since it has only one asymmetric

carbon atom, 12 2a = = isomers of 2-

chlorobutane are possible.

(2) Non-superimposable object and mirrorimage : The two isomers of 2-chlorobutane are

non-identical, non-superimposable objects and

mirror images of each other. Thus they represent

the two optically active isomers of 2-chlorobutane

namely, Dextro-rotatory and Laevo-rotatory.

C

3C H

C l

H2 5C H

C

3C HC l

H2 5C H

(I) (II)

(I)d or (+) 2-chlofrobutane

or

H C C l- -

3C H

(I) (II)

(Ii)l o r (-) 2-chlo frobu tane

2 5C H

C l C H- -

3C H

2 5C H

(3) If structure (I) represents the (d)-isomeric form

of 2-chlorobutane (rotates the plane of plane

polarised light towards right) strucutre (II)

represents the (l) isomeric form (rotates the plane

of plane polarised light towards left) of the same

or vice-versa. They are called enantiomers.

(4) dl mixture and its optical inactivity : When

equal amounts (equimolar quantities) of the two

isomer (I) and (II) are mixed, the resulting mixture

is found to be optically inactive because of the

external compensation of optical rotation. This

form is known as racemic form and represented

as ( )+− or (dl) 2-chlorobutane.

(iii) Action 2 2 7 2 4K Cr O dil H SO+ :

(a) On Propan-1-ol :

3 2 2[ ]C H C H C H O H O- - - +

(1-Propanol)( )

2 4dil H SO

Oxidation→

3 2 2C H C H C H O H O- - -

(Propiona ldehyde)

3 2 [ ]CH CH CHO O− − + Oxidation→

3 2C H C H C O O H- -

(Propanoic acid)

Page 13 of 14

(b) On Propan-2-ol :

3 3[ ]C H C H O H C H O- - +

(2 -P ropan o l)

( )2 2 7 2 4K Cr O dil H SO

Oxidation

+→

3 3 2C H C C H H O- - +

O

(Acetone)

3 3 4[ ]C H C C H O- - +

ODestructive

Oxidation→

3 2 2C H C O O H C O H O- + +

(A ce tic ac id )

Structure :

3 2 2 2H C C H C H C H C C H O H- - - - - -

3C H

3C HC l

(3-chloro-2,2-dimethyl hexan-1-ol)

(iv) Conjugated proteins : Conjugated proteins are

thoseproteins which yield α − amino acids along

with some non-proteinous compounds (known as

prosthetic groups). The non-proteinous part can

be separated from the proteins by careful

hydrolysis.

Oil Fats

1 Oils are liquid at ordinary

(room) temperature

Fats are soilds at ordinary

(room) temperature

2 Oils contain a large

proportion of unsaturated

fatty acids

Fats contain a large

proportion of saturated fatty

acids

3 Oils have lower melting

and boiling points than fats

Fats have higher melting

and boiling points than oils

4 Oils can be hydrogenated

to get fats

Fats are soild and cannot be

hydrogenated

5Oils can be assimilated in

the stomach

Fats can be assimilated with

difficulty in the stomach

6 Due to irregular carbon-

Carbon bonds in the chain

oils do not have close

packing of molecules

Due to regular carbon-

Carbon bonds in the chain,

fats have close packing of

molecules

For example : Nucleo proteins, glycoproteins,

phosphoproteins and chronoproteins etc.

8. (A)(i) Aldol condensation : It is the reaction of those

aldehydes and certain ketones (acetone)

which contain an α hydrogen atom such

as acetaldehyde etc.

When such aldehydes are treated with a dilute

base like NaOH (10%) or aqueous 2 3Na CO or

dilute acid, then the two molecules combine to

form β -hydroxy aldehyde.

The β -hydroxy aldehyde contains two functional

groups, aldehydic (-ald), CHO− and alcoholic

(-ol) OH− and hence it is called an aldol. This

reaction is called aldol condenstion.

An aldol is formed when α -carbon of one

molecule gets attached to the carbonyl carbon

( )C O- of the other molecule by migration of

α -hydrogen of one molecule to the carbonyl

oxygen of the other molecule forming a hydroxyl

( )OH− group.

3 2C H C O H H C C O- = + - =

H Hα

2(Acetadehde)

(10%)

300

dil NaOH

K→

3 2C H C H C C O- - - =

H Hα

(Acetaldol)

O H

An aldol loses a water molecule on heating with

a trace of a mineral acid to form an α , β -

unsaturated aldehyde.

3C H C C C O- - - =

H

(Acetaldol)

O H

H H

H

( )

2

mineral acid

( )dehydrationH O

∆

−

→

3C H C C C O- = - =

H

( Unsaturated aldehyde)α,β

H H

αβ

or(2-Butenal or Cro tona ldehyde)

(ii) Esters : Esters are the derivatives of carboxylic

acids in which ‘H’ atom of the carboxylic group

( )COOH−is replaced by an alkyl group (R)

(C arboxy lic ac id )

R C O O H-H

R

−+→

(E s te r)

R C O O R- -

Preparation of ethylacetate :

(a) From 3CH COOAg : When silver acetate,

3CH COOAg is heated with ethyl bromide, ethyl

Page 14 of 14

acetate is obtained.

(Ag acetate )

3 2 5C H C O O A g C H B r+

Alcohol

∆→

(E thyl acetate)

3 2 5C H C O O C H A g B r+

(b) From acetic anhydride : When acetic

anhydride is heated with ethanol in presence of

a catalyst pyridine, ethyl acetate is obtained.

3C H C O

(E thanol)3C H C O

2 5O C H O H+

(Acetic anhydride)

Pyridine

∆→

(E thyl aceta te)

3 3 2 5C H C O O H C H C O O C H+

Amide formation : When ethanoic acid is

heated with ammonia, ammonium acetate is

formed which gives rise to acetamide on heating.

(Ace tic acid)3 3

C H C O O H N H+ ∆→(Amm onium acetate)

3 4C H C O O N H-

3 4CH COONH− Dehydration

∆→

(Acetam ide)

2 3 2H O C H C O N H+ -

8.(B) (i) Artificial fibre : Artificial fibre may be defined

as material in the form of a thin filament, which

can be woven or spun into a fabric and it is

obtained artifically.

e.g. Terylene fibres, nylon fibres etc.

Uses of terylene fiberes : (1) It is widely used

in the manufacture of textile.

(2) These fibers are blended with cotton and wool

to get terycot, terywool etc.

(1) Preparation of ethanamine from 2 5C H Br :

3 2C H C H B r- -

(E thyl brom ide)

3NH

alcoholic→ 3 2 2C H C H N H H B r- - -

(E thyl am ine)

(2) Iso propylamine from acetoxime :

34[ ]C H C N O H H- = - +

(Acetoxime)

3C H

/

Reduction

Na Hg Water−→

3 2 2C H C H N H H O- - +

(Iso propyl am ine)

3C H

(ii) Preparation of chloroform : A paste of

bleaching powder is placed in a round bottom

flask and ethyl alcohol is added into it slowly by

a dropping funnel. The flask is fitted with a water

condenser. It is heated on a sand bath for about

3-4 hours. Chloroform distils over at its boiling

point 334 K and collected in a receiver. It is dried

by fused 2CaCl and then purified by redistillation.

Reactions taking place during the formation of

chloroform can be written as follows:

(1) Decompostion of bleaching powder :

When bleaching powder is mixed with water it

decomposes to form calcium hydroxide (lime

water) and chlorine.

2 2C a O C l H O+

(B leaching powder)

∆→ C a

(Calcium hydroxide)

O H

O H2Cl+

(2) Oxidation of ethyl alcohol :

(E thyl a lchol)

3 2 2H C C H O H C l- - +

Oxidation→

(A ce ta ldehyde)

3 2H C C H O H C l- +

Ethyl alcohol is oxidised to acetaldehyde in

presence of chlorine.

(3) Chlorination of acetaldehyde :

(A ce ta ldeh yde )

3 23H C C H O C l- + Chlorination→

(Chloral)

33C l C C H O H C l- +

Acetaldehyde gets chlorinated to form chloral

(trichloroacetaldehyde) in presence of excess of

chlorine.

(4) Hydrolysis of chloral :

C a

(Chloral)

3O H C l C C H O+ -

3O H C l C C H O+ -

(Calciumhydroxide)

Hydrolysis

∆→

32CHCl + C a

(Chloroform )

H C O O

H C O O

or (Calcium form ate)

2( )H C O O C a

Finally calcium hydroxide hydrolyses chloral to

form chloroform.