NCERT Solutions for Class 11 Chemistry Chapter 8 Organic Chemistry – Some Basic Principles and Techniques

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Chemistry
Chapter : 12
Chapter Name : Organic Chemistry - some basic principles and techniques

Q12.1 What are hybridisation states of each carbon atom in the following compounds ?
CH = C = O,
2

CH CH = CH ,
3 2

(CH ) CO,3 2

CH = CHCN,
2

C 6 H6

1 2

Answer. (i) C H =C= 0 2

C-1 is sp hybridised 2

C-2 is sp hybridised

(ii) CH − CH = CH
3 2

C-1 is sp hybridised 3

C-2 is sp hybridised 2

C-3 is sp hybridised 2

(iii) (CH ) CO 3 2

C-1 is sp hybridised 3

C-2 is sp hybridised 2

C-3 is sp hybridised 3

(iv) CH = CHCN 2

C-1 is sp hybridised 2

C-2 is sp hybridised 2

C-3 is sp hybridised

(v) C H 6 6

All carbon atoms in benzene are sp hybridised. 2

Page : 370 , Block Name : Exercise

Q12.2 Indicate the σ and π bonds in the following molecules :
C H ,
6 6

C H
6 ,Q12

CH Cl ,
2 2

CH2 = C = CH2 ,
CH3 NO2 ,
HCONHCH3

Answer. (i) C H e 6

There are six C-C sigma bonds , six C-H bonds , and three C = C resonating bonds in the
given compound.

(ii) C H
6 12

There are six C-C sigma bonds , twelve C-H bonds sigma in the given compound.

(iii) CH Cl ,
2 2

There are two C-H sigma bonds , two C-Cl bonds sigma in the given compound.

(iv) CH 2 = C = CH2 ,

There are two C-C sigma bonds , four C-H bonds and two C=C π bonds in the given compound.

(v) CH NO ,
3 2

There are three C-H sigma bonds , one C-N sigma bond and one N-O sigma bond and one N=O
π bond in the given compound.

(vi) HCONHCH 3

There are two C-N sigma bonds , four C-H sigma bond , N-H sigma bond and one C=O π bond
in the given compound.

Page : 370 , Block Name : Exercise

Q12.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4- one.

Answer. The bond line formulate of the given compounds are:

Page : 370 , Block Name : Exercise

Q12.4 Give the IUPAC names of the following compounds :

Answer.

Page : 370 , Block Name : Exercise

Q12.5 Which of the following represents the correct IUPAC name for the compounds
concerned ?
(a) 2,2-Dimethylpentane or 2-Dimethylpentane
(b) 2,4,7- Trimethyloctane or 2,5,7-Trimethyloctane
(c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
(d) But-3-yn-1-ol or But-4-ol-1-yne.

Answer. (a) The pre x di in tine IUPAC narne Indicates that two identical substituent groups
are present in the parent chain. Since two methyl groups are present in the of the C-2 of the
parent chain of the given compound, the correct IUPAC name of the given compound is 2,2-
dimethylpentane.
(b) locant number 2, 4, 7 is lower than 2 , 5, 7 hence ,the IUPAC name of the given compound
is 2,4,7-trimethyloctane.
(c) It the substituents are present in the equivalent position ot the parent chain, then the
lower number is given to the one that comes rst in the name according to the Alphabetical
order. Hence, the correct IUPAC name of the given compound is 2-chIcro -4-methylpentane.
(d) Two functional groups — alcoholic and alkyne -are present in the given compound, The
principal functional group is the alcoholic group. Hence, the parent chain will be suf xed with
ol.The alkyne group is present in of the C-3 of the present chain . hence,the correct IUPAC
name of the given compound is but But-3-yn-1-ol.

Page : 370 , Block Name : Exercise

Q12.6 Draw formulas for the rst ve members of each homologous series beginning with the
following compounds.
(a) H − COOH
(b) CH COCH
3 3

(c) H − CH = CH 2

Answer. The rst ve members of homogeneous series beginning with the given compounds
are shown as follows:
(a)
H − COOH - methanoic acid

CH − COOH -ethanoic acid
3

CH − CH − COOH -propanoic acid
3 2

CH − CH − CH − COOH -Butanoic acid
3 2 2

CH − CH − CH − CH − COOH -pentanoic acid
3 2 2 2

(b)
CH COCH - propanone
3 3

CH COCH CH -butanone
3 2 3

CH COCH CH CH -pentan-2-one
3 2 2 3

CH COCH CH CH CH -Hexan-2-one
3 2 2 2 3

CH COCH CH CH CH CH - Heptan-2-one
3 2 2 2 2 3

(c)
H − CH = CH -Ethene
2

CH − CH = CH - propene
3 2

CH − CH − CH = CH - 1-Butene
3 2 2

CH − CH − CH − CH = CH - 1-pentene
3 2 2 2

CH − CH − CH − CH − CH = CH - 1- Hexene
3 2 2 2 2

Page : 370 , Block Name : Exercise

Q12.7 Give condensed and bond line structural formulas and identify the functional group(s)
present, if any, for :
(a) 2,2,4-Trimethylpentane
(b) 2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial

Answer. (a) 2,2,4-trimethylpentane
condensed formula:
(CH3 ) CHCH2 C(CH3 )
2 3

Bond line formula:

(b) 2-hydroxy-1,2,3-propanetricarboxylic acid
condensed formula:
(COOH)CH2 C(OH)(COOH)CH2 (COOH)

The functional groups present in the given compound are carboxylic acid (-COOH) and
alcoholic
(-OH) groups.

(c) Hexanedial
condensed formula:
(CHO)(CH2 ) (CHO)
4

The functional group present in the given compound is aldehyde ( -CHO).

Page : 370 , Block Name : Exercise

Q12.8 Identify the functional groups in the following compounds.

Answer. The functional group present in the given compounds are:
(a) Aldehyde ( -CHO)
Hydroxyl (-OH)
Methoxy (-OMe)
C=C double bond

(b) Amino −N H 2

Ketone (C=O),
Diethylamine (N(C H ) )
2 s 2

(c) Nitro −NO 2

C=C double bond

Page : 370 , Block Name : Exercise

Q12.9 Which of the two: O NCH CH O or CH CH O is expected to be more stable and
− −
2 2 2 3 2

why ?

Answer. NO group is an electron-withdrawing group. Hence, it shows -1 effect. By
2

withdrawing the electrons toward it, the decreases the negative charge on the compound,
thereby stabilising it. On the other hand, ethyl group is an electron-releasing group.
Hence, the ethyl group shows +I effect. This increases the negative charge on the compound,
thereby destabilising it Hence, O NCH CH O is expected to be more stable than
2 2 2
−

−
CH3 CH2 O

Page : 370 , Block Name : Exercise

Q12.10 Explain why alkyl groups act as electron donors when attached to a π system.

Answer. when an alkyl group is attached to a Π system, it acts as an electron-donor group by
the process of hyperconjugation. To understand this concept better, let us take the example of
propene.

In hyperconjugation, the sigma electrons of the C—H bond of an alkyl group are delocalised.
This group is directly attached to an atom of an unsaturated system. The delocalisation occurs
because of a partial overlap of a sp —s sigma bond orbital with an empty p orbital of the Π
3

bond of an adjacent carbon atom.
The process of hyperconjugation in propene is shown as follows:

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the Π
electrons, making the molecule more stable.

Page : 370 , Block Name : Exercise

Q12.11 Draw the resonance structures for the following compounds. Show the electron shift
using curved-arrow notation.
(a) C H OH
6 5

(b) C H NO
6 5 2

(c) CH CH = CHCHO
3

(d) C H − CHO
6 5

(e) C H − CH
6
˙
5 2

(f) CH CH = CHCH
3 2

Answer. (a) the structure of C H OH is :
6 5

The resonating structures of phenol are represented as:

(b) the structure of C H NO is :
6 5 2

The resonating structures Of nitro benzene are represented as:

(c) the structure of CH CH = CHCHO is :
3

The resonating structures of the given compound are represented as:

(d) the structure of C H
6 5 − CHO is :

The resonating structures of benzaldehyde are represented as:

(e) C H − CH
6
˙
5 2

The resonating structures of the given compound are represented as:

(f) CH CH = CHCH
3 2

The resonating structures of the given compound are represented as:

Page : 370 , Block Name : Exercise

Q12.12 What are electrophiles and nucleophiles ? Explain with examples.

Answer. An electrophile is a reagent that t*es away an electron pair. In other words, an
electron-seeking reagent is called m electrophile Electrophiles are electron- (E ) +

Electrophiles are electron-de cient md can receive m electron pair.
Carbocations CH CH and neutral molecules having functional groups such as carbonyl
∗
3 2

group ( C=O ) are examples of electrophiles.
A nucleophile is a reagent that brings an electron pair. [n other words, a nucleus-sed«ing
reagent is called a nucleophile (Nu:).
For example:OH ,NC such as H O − −
2

And ammonia also act as nucleophiles because of the presence of a lone pair

Page : 370 , Block Name : Exercise

Q12.13 Identify the reagents shown in bold in the following equations as nucleophiles or
electrophiles:
(a) CH COOH + HO → CH COO + H O
3
−
3
−
2
~
(b) CH COCH + CN → (CH ) C(CN)(OH)
3 3 a 2
+

(c) C H
6 6 + CH3 CO→ C6 H5 COCH3

Answer. Electrophiles Ye electron-de cient species and cm receive an electron pair. On the
other hand, nucleophiles are electron-rich species and cm donate their electrons.
(a) CH COOH + HO → CH COO + H O
3
−
3
−
2

Here, OH acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus- seeking
−

species.

(b) CH COCH + CN ⟶ (CH ) C(CN) + (OH)acts as a nucleophile as it is electron-
3 3 3 2

rich species. i.e., it is a nucleus- seeking species.

+

(c) C H
6 6 + CH3 CO→ C6 H5 COCH3
+

Here , CH 3 CO acts as an electrophile as it is an electron-de cient species.

Page : 370 , Block Name : Exercise

Q12.14 Classify the following reactions in one of the reaction type studied in this unit.
(a) CH CH Br + HS → CH CH SH + Br
− −
3 2 3 2

(b) (CH ) C = CH + HCl → (CH ) ClC − CH
3 2 2 3 2 3

(c) CH CH Br + HO → CH = CH + H O + Br
3 2
−
2 2 2
−

(d) (CH ) C − CH OH + HBr → (CH ) CBrCH CH CH + H O
3 3 2 3 2 2 2 3 2

Answer. (a) It is an example of substitution reaction as in this reaction the bromine group in
bromoethane is substituted by the -SH group.
(b) It is an example of addition reaction as in this reaction two reactant molecules combine to

form a single product.
(c) It is an example of elimination reaction as in this reaction hydrogen and bromine are
removed from bromoethane to give ethene,
(d) In this reaction, substitution takes place, followed by a rearrangement of atoms and groups
of atoms.

Page : 370 , Block Name : Exercise

Q12.15 What is the relationship between the members of following pairs of structures ? Are
they structural or geometrical isomers or resonance contributors ?

Answer. (a) Compounds having the same molecular formula but with different structures are
called structural isomers. The given compounds have the same molecular formula but they
differ in the position of the functional group (ketone group).

In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II,
ketone group is at the C-2 of the parent chain (hexane chain). Hence, the given pair represents
structural isomers.

(b) Compounds having the same molecular formula, the sarne constitution, end the sequence
of covalent bonds, but with different relative position of their atoms in space are called
geometrical isomers

In structures I and II, the relative position of Deuterium (D) and hydrogen (H) in space are
different. Hence, the given pairs represent geometrical isomers.

(c) The given structures are canonical structures or contributing structures. They are
hypothetical and individually do not represent any real molecule. Hence, the given pair

represents resonance structures, called resonance isomers.

Page : 371 , Block Name : Exercise

Q12.16 For the following bond cleavages, use curved-arrows to show the electron ow and
classify each as homolysis or heterolysis. Identify reactive intermediate produced as free
radical, carbocation and carbanion.

Answer. (a) The bond cleavage using curved-arrows to show the electron ow of the given
reaction can be represented as :

It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with
the bonded atom. The reaction intermediate formed is a free radical.

(b) The bond cleavage using curved-arrows to show the electron ow of the given
reaction can be represented as :

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared
pair of electrons remains with the carbon of propanone. The reaction intermediate formed is
carbanion.

(c) The bond cleavage using curved-arrows to show the electron ow of the given reaction can
be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the
shared pair of electrons remains with the bromine ion. The reaction intermediate formed
is a carbocation,

(d) The bond cleavage using curved-arrows to show the electron ow of the given
reaction can be represented as

It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of

electrons remains with one of the fragments. The intermediate formed is a carbocation.

Page : 371 , Block Name : Exercise

Q12.17 Explain the terms Inductive and Electromeric effects. Which electron displacement
effect explains the following correct orders of acidity of the carboxylic acids?
(a) Cl CCOOH > Cl CHCOOHClCH COOH
3 2 2

(b)CH CH COOH > (CH ) CHCOOH > (CH ) C ⋅ COOH
3 2 3 2 3 3

Answer. Inductive effect
The permanent displacement of sigma (o) electrons along a saturated chain, Rhenever an
electron withdrawing or electron donating group is present, is called inductive effect
Inductive effect could be + I effect or - I effect. When an atom or group attracts electrons
towards itself more strongly than hydrogen, it is said to possess - effect. For
Example ,

When an atom or group attracts electrons towards itself less strongly than hydrogen, it is said
to possess + I effect. For example,

Electrometric effect
It involves the complete transfer of the shared pair of n electrons to either of the two atoms
linked by multiple bonds in the presence of an attacking agent, For example,

Electrometric effect could be + E effect or -E effect.
+ E effect: When the electrons are transferred towards the attacking reagent
- E effect: When the electrons are transferred away from the attacking reagent

(a) Cl CCOOH > Cl CHCOOHClCH COOH
3 2 2

The order of acidity can be explained on the basis of Inductive effect ( -I effect). As the
number of chlorine atoms increases, the -I effect increases. With the increase in -I effect, the
acid strength also increases accordingly .

(b)CH CH COOH > (CH ) CHCOOH > (CH ) C ⋅ COOH
3 2 3 2 3 3

The order of acidity can be explained on the basis of inductive effect( I effect). As the number
of alkyl groups increases, the + I effect also increases. With the increase in +I effect, the acid
strength also increases accordingly.

Page : 371 , Block Name : Exercise

Q12.18 Give a brief description of the principles of the following techniques taking an example
in each case.
(a) Crystallisation
(b) Distillation
(c) Chromatography

Answer. (a) Crystallisation
Crystallisation is one of the most commonly used techniques for the puri cation of solid
organic compounds. Principle: It is based on the difference in the solubilities of the compound
and the impurities in a given solvent, The impure compound gets dissolved in the solvent in
which it is sparingly soluble at room temperature, but appreciably soluble at higher
temperature. The solution is concentrated to obtain a nearly saturated solution. On cooling
the solution, the pure compound crystallises out and is removed by ltration.
For example, pure aspirin is obtained by recrystallising crude aspirin. Approximately 2 4 g of
crude aspirin is dissolved in about 20 mL of ethyl alcohol. The solution is heated (if necessary)
to ensure complete dissolution, The solution is then left undisturbed until some crystals start
to separate out The crystals are then ltered and dried.

(b) Distillation
This method is used to separate volatile liquids from non-volatile impurities or a mixture of
those liquids that have a suf cient difference in their boiling points. principle: It is based on
the fact that liquids having different boiling points vapourise at different temperatures, The
vapours are then cooled and the liquids so formed are collected separately .
For example, a mixture of chloroform (b.p = 334 K) and aniline (b.p = 457 K) can be separated
by the method of distillation. The mixture is taken in a round bottom ask tted with a
condenser It is then heated Chloroform, being more volatile, vaporizes rst and passes into
the condenser. In the condenser, the vapours condense and chloroform trickles down. In the
round bottom ask, aniline is left behind.

(c) Chromatography
It is one of the most useful methods for the separation and puri cation of organic compounds.

principle: It is based on the difference in movement of individual components of a mixture

through the stationary phase under the in uence of mobile phase.
For example, a mixture of red and blue ink can be separated by chromatography. A drop of the
mixture is placed on the chromatogram. The component of the ink, which is less adsorbed on
the chromatogram, moves with the mobile phase while the less adsorbed component remains
almost stationary.

Page : 371 , Block Name : Exercise

Q12.19 Describe the method, which can be used to separate two compounds with different
solubilities in a solvent S.

Answer. Fractional crystallisation is the method used for separating two compounds with
different solubilities in a solvent S. The process of fractional crystallisation is carried out in
four steps.
(a) Preparation of the solution: The powdered mixture is taken in a ask and the solvent is
added to it slowly end stirred simultaneously. The solvent is added till the solute is just
dissolved in the solvent. This saturated solution is then heated.
(b) Filtration of the solution: The hot saturated solution is then ltered through a lter paper
in a China dish
(c) Fractional crystallisation: The solution in the China dish is now allowed to cool. The less
soluble compound crystallises rst, while the more soluble compound remains in the solution,
After separating these crystals from the mother liquor, the latter is concentrated once again.
The hot solution is allowed to cool and consequently, the crystals of the more soluble
compound are obtained.
(d) Isolation and drying: These crystals are separated from the mother liquor by ltration.
Finally, the crystals are dried.

Page : 371 , Block Name : Exercise

Q12.20 What is the difference between distillation, distillation under reduced pressure and
steam distillation ?

Answer. The differences among distillation, distillation under reduced pressure, and steam
distillation are given in the following table.

Page : 371 , Block Name : Exercise

Q12.21 Discuss the chemistry of Lassaigne’s test.

Answer. Lassaigne's test
This test is employed to detect the presence of nitrogen, sulphur, halogens, and phosphorus in
an organic compound These elements are present in the covalent form in an organic
compound. These are converted into the ionic form by fusing the compound with sodium
metal.
Δ

Na + C + N ⟶ NaCN

2Na + S ⟶ Na2 S

Δ

Na + X ⟶ NaX

The cyanide, sulph de, and halide of sodium formed are extracted from the fused mass by
boiling it in distilled water. The extract so obtained is called Lassaigne's extract. This
Lassaigne's extract is then tested for the presence of nitrogen, sulphur, halogens, and
phosphorous.
Test for nitrogen

Chemistry of the test
In the Lassaigne's test for nitrogen in an organic compound. the sodium fusion extract is
boiled With iron (II) sulphate and then acidi ed With sulphuric acid, In the process. sodium
cyanide rst reacts With ron (II) sulphate and forms sodium hexacyanoferrate (II). Then. on
heating With sulphuric acid. some iron (II) gets oxidised to form iron (Ill) hexacyanoferrate
(II). Which is Prussian blue in colour. The chemical equations involved in the reaction can be
represented as
− 2+ +
6CN + Fc ⟶ [Fe(CN)6 ]

-> Fe [Fe(CN) ] xH O
4− 3+
3[Fe(CN)6 ] + 4Fe 4 6 3 2

(b) Test of sulphur
(i) Lassaigne's extract + lead acetate ->Black precipitate
Chemistry ot the test
In the Lassaigne's test for sulphur in an organic compound, the sodium fusion extract is
acidi ed with acetic acid and then lead acetate is added to it. The precipitation of lead
sulphide, which is black in colour, indicates the presence of sulphur in the compound.
2− 2+
S + Pb ⟶ PbS

(ii) Lassaigne's extract + sodium nitroprusside-> violet colour
Chemistry of the test
The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet colour
also indicates the presence of sulphur in the compound,
(violet)
2− 2− −4
S + [Fe(CN) NO] s⟶ [Fe(CN) NOS] s

It in an organic compound, both nitrogen and sulphur are present, then instead of NaCN.
formation of NaSCN takes place.
Na + C + S +S -> NaSCN
This NaSCN (sodium thiocyanate) gives a blood red colour. Prussian colour is not formed due
to the absence of tree cyanide ions.
3∗ − 2+
Fe + SCN ⟶ [Fe(SCN)]

(Blood red)
(c) Test for halogens

Chemistry of the test
In the Lassaigne's test for halogens in an organic compound. the sodium fusion extract is
acidi ed with nitric acid and then treated with silver nitrate.
− +
X + Ag ⟶ AgX

If nitrogen and sulphur both are present in the organic compound. then the Lassaigne's
extract is boiled to expel nitrogen and sulphur. which would otherwise interfere in the test for
halogens.

Page : 372 , Block Name : Exercise

Q12.22 Differentiate between the principle of estimation of nitrogen in an organic compound
by (i) Dumas method and (ii) Kjeldahl’s method.

Answer. In Dumas method. a known quantity of nitrogen containing organic compound is
heated strongly with excess of copper oxide in an atmosphere of carbon dioxide to produce
tree nitrogen in addition to carbon dioxide and water. The chemical equation involved in the
process can be represented as
CxHyNz + (2x + y/2)cuO ⟶ xCO2 + y/2H2 O + z/2N2 + (2x + y/2)Cu

The traces of nitrogen oxides can also be produced in the reaction, which can be reduced to
dinitrogen by passing the gaseous mixture over a heated copper gauge, The dinitrogen
produced is collected over an aqueous solution of potassium hydroxide. The volume of
nitrogen produced is then measured at room temperature and atmospheric pressure, On the
other hand, in Kjeldahl's method, a known quantity of nitrogen containing organic compound
is heated with concentrated sulphuric acid, The nitrogen present in the compound is
quantitatively converted into ammonium sulphate. It is then distilled with excess of sodium
hydroxide. The ammonia evolved during this process is passed into a known volume of H SO 2 4

, The chemical equations involved in the process are

Organic compound ->(NH ) SO4 2 4

(NH4 ) SO4 + 2NaOH ⟶ Na2 SO4 + 2NH3 + 2H2 O
2

2NH3 + H2 SO4 (⟶ (NH4 ) SO4
2

The acid that is left unused is estimated by volumetric analysis (titrating it against a standard
alkali) and the amount of ammonia produced can be determined. Thus. the percentage of

nitrogen in the compound can be estimated. This method cannot be applied to the
compounds. in which nitrogen is present in a ring structure, and also not applicable to
compounds containing nitro and azo groups.

Page : 372 , Block Name : Exercise

Q12.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an
organic compound.

Answer. Estimation of halogens
Halogens are estimated by the Carius method. In this method, a known quantity of organic
compound is heated with fuming nitric acid in the presence of silver nitrate, contained in a
hard glass tube called the Carius tube. taken in a furnace. Carbon and hydrogen that are
present in the compound are oxidized to form CO and H O respectively and the halogen
2 2

present in the compound is converted to the torm ot AgX. This AgX is then ltered, washed,
dried, and weighed.
Let the mass of organic compound be m g.
Mass ot AgX formed = m g 1

1 mol of AgX contains 1 mol of X .
1 mol of AgX contains 1 mol of X.
Therefore ,
Mass of halogen in m g of AgX =
Atomic mass of X×m1 g

1
Molecular mass of AgX

Thus ,% of halogen will be =
Atomic mass of X×m1 ×100

Molecular mass of AgX×m

In this method. a known quantity of organic compound is heated with either fuming nitric
acid or sodium peroxide in a hard glass tube called the Carius tube. Sulphur. present in the
compound. is oxidized to form sulphuric acid. On addition of excess of barium chloride to it.
the precipitation of barium sulphate takes place. This precipitate is then ltered, washed,
dried, and weighed.
Let the mass of organic compound be m g.
Mass of BaSO formed m g
4 1

1 mol of BaSO 233 g BaSO =32 g of sulphur.
4 4

Therefore ,m
32×m1
1 g of BaSO 4 contains
233

Thus , % of sulphur =
32×m1 ×100

233×m

Estimation of phosphorus
In this method, a known quantity of organic compound is heated with fuming nitric acid.
Phosphorus. present in the compound. is oxidized to form phosphoric acid. By adding
ammonia and ammonium molybdate to the solution. phosphorus can be precipitated as
ammonium phosphomolybdate.
Phosphorus can also be estimated by precipitating it as MgNH PO by adding magnesia
4 4

mixture, which on ignition yields Mg P O . 2 2 7

Let the mass of organic compound be m g.
Mass of ammonium phosphomolybdate formed m g 1

Molar mass of ammonium phosphomolybdate =1877 g
Thus the percentage of phosphorus =
31×m1 ×100
%
1877×m

If P is estimated as Mg P O .
2 2 7

Then the percentage of phosphorus =
62×m1 ×100
%
222×m

Page : 372 , Block Name : Exercise

Q12.24 Explain the principle of paper chromatography.

Answer. In paper chromatography. chromatography paper is used. This paper contains water
trapped in it. which acts as the stationary phase. On the base of this chromatography paper,
the solution of the mixture is spotted. The paper strip is then suspended in a suitable solvent.
which acts as the mobile phase. This solvent rises up the chromatography paper by capillary
action and in the procedure. it ows over the spot. The components are selectively retained on
the paper (according to their differing partition in these two phases). The spots of different
components travel with the mobile phase to different heights. The paper so obtained (shown
in the given gure) is known as a chromatogram.

Page : 372 , Block Name : Exercise

Q12.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing
halogens?

Answer. While testing the Lassaigne's extract for the presence of halogens. it is rst boiled
with dilute nitric acid. This is done to decompose NaCN to HCN and Na S to H S and to expel
2 2

these gases. That is. it any nitrogen and sulphur are present in the form of NaCN andNa S .
2

then they are removed. The chemical equations involved in the reaction are represented as
NaCN + HNO3 ⟶ NaNO3 + HCN

Na2 S + 2HNO3 ⟶ 2NaNO3 + H2 S

Page : 372 , Block Name : Exercise

Q12.26 Explain the reason for the fusion of an organic compound with metallic sodium for
testing nitrogen, sulphur and halogens.

Answer. Nitrogen. sulphur. and halogens are covalently bonded in organic compounds. For
their detection. they have to be rst converted to ionic form. This is done by fusing the organic

compound with sodium metaL This is called "Lassaigne's test". The chemical equations
involved in the test are
Na + C + N ⟶ NaCN

Na + S + C + N ⟶ NaSCN

2Na + S ⟶ Na2 S

Na + X ⟶ NaX

( X = Cl,Br,I )
Carbon, nitrogen, sulphur, and halogen come from organic compounds.

Page : 372 , Block Name : Exercise

Q12.27 Name a suitable technique of separation of the components from a mixture of calcium
sulphate and camphor.

Answer. The process of sublimation is used to separate a mixture of camphor and calcium
sulphate, In this process. the sublimable compound changes from solid to vapour state
without passing through the liquid state. Camphor is a sublimable compound and calcium
sulphate is a non- sublimable solid. Hence. on heating. camphor will sublime while calcium
sulphate will be left behind.

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Q12.28 Explain, why an organic liquid evaporates at a temperature below its boiling point in
its steam distillation ?

Answer. In steam distillation. the organic liquid starts to boil when the sum of vapour pressure
due to the organic liquid (pl) and the vapour pressure due to water (p2) becomes equal to
atmospheric pressure (p). that is. p = p1 + p1 Since p1 p2. organic liquid will vapourise at a
lower temperature than its boiling point.

Page : 372 , Block Name : Exercise

Q12.29 Will CCl give white precipitate of AgCl on heating it with silver nitrate? Give reason
4

for your answer.

Answer. CCl will not give the white precipitate of AgCl on heating it with silver nitrate. This
4

is because the chlorine atoms are covalently bonded to carbon in CCl . To obtain the
4

precipitate. it should be present in ionic form and for this. it is necessary to prepare the
Lassaigne's extract of CCl .
4

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Q12.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved
during the estimation of carbon present in an organic compound?

Answer. Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence.
carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as
2KOH + CO2 ⟶ K2 CO3 + H2 O

Thus. the mass of the U-tube containing KOH increases. This increase in the mass ot U-tube
gives the mass of CO produced. From its mass. the percentage of carbon in the organic
2

compound can be estimated.

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Q12.31 Why is it necessary to use acetic acid and not sulphuric acid for acidi cation of sodium
extract for testing sulphur by lead acetate test?

Answer. Although the addition of sulphuric acid will precipitate lead sulphate. the addition of
acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to
common ion effect. Hence. it is necessary to use acetic acid for acidi cation of sodium extract
for testing sulphur by lead acetate test.

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Q12.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being
oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this
substance is subjected to complete combustion.

Answer. Percentage of carbon in organic compound = 69 %
That is. 100 g of organic compound contains 69 g of carbon 69 g of carbon.
:.0.2 g of organic compound will contain = 69×0.2

100
= 0.138g of C
Molecular mass of carbon dioxide. CO 44 g
2

That is. Q12 gm of carbon is contained in 44 g of CO . 2

Therefore. 0.138 g of carbon will be contained in = 44×0.138

Q12
=0.506g of CO 2

Thus. 0.506 g of CO will be produced on complete combustion of 0.2 g of organic
2

compound.Percentage of hydrogen in organic compound is 4.8 .
i.e.. 100 g of organic compound contains 4.8 g of hydrogen.
Therefore. 0.2 g of organic compound will contain = 0.0096g of H
4.8×0.2

100

It is known that molecular mass Of water ( H O) is 18 g.
2

Thus. 2 g of hydrogen is contained in 18 g of water.
0.0096 g of hydrogen will be contained in 18×0.0096

2
= 0.0864g of water.Thus , 0.0864 g of water
will be produced on complete combustion of 0.2 g of the organic compound.

Page : 372 , Block Name : Exercise

Q12.33 A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s
method. The ammonia evolved was absorbed in 50 ml of 0.5 M H SO . The residual acid
2 4

required 60 mL of 0.5 M solution of NaOH for neutralisation.

Find the percentage composition of nitrogen in the compound.

Answer. Given that. total mass of organic compound = 0.50 g
60 mL of 0.5 M solution of NaOH was required by residual acid for neutralisation.
60 mL of 0.5 M NaOH solution = mL of 0.5M 60

2
= 30 mL of 0.5 M H SO
H2 SO4 2 4

:.Acid consumed in absorption of evolved ammonia is (50 - 30) mL =20 mL
Again. 20 mL of 0.5 M H SO = 40 mL of 0.5 MNH
2 4 3

Also. since 1000 mL of 1 MNH contains 14 g of nitrogen.
3

40 mL of 0.5 M NH will contain =
14×40
3 × 0.5
1000

= 0.28 g of N
Therefore. percentage of nitrogen in 0.50 g of organic compound = = 56%
0.28
× 100
0.50

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Q12.34 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius
estimation. Calculate the percentage of chlorine present in the compound.

Answer. Given that.
Mass of organic compound is 0.3780 g.
Mass of AgCl formed = 0.5740 g
1 mol of AgCl contains 1 mol of Cl.
Thus. mass of chlorine in 0.5740 g of AgCl
= 35.5×0.5740

143.32

= 0.1421 g
Percentage of chlorine = = 37.59 %
0.1421
× 100
0.3780

Hence, the percentage of chlorine present in the given organic chloro compound is 37.59%

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Q12.35 In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur
compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the
given compound.

Answer. Total mass of organic compound = 0.468 g [Given)
Mass of barium sulphate formed = 0.668 g (Given]
1 mol of BaSO = 233 g of BaSO = 32 g of sulphur
4 4

Thus. 0.668 g of BaSO contains
4 = 0.0917 g of sulphur
32×0.668

233

Therefore. percentage of sulphur = = 19.59 %
0.0197
× 100
0.468

Hence. the percentage of sulphur in the given compound is 19.59 %.

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Q12.36 In the organic compound CH 2 = CH − CH2 − CH2 − C = CH , the pair of

hybridised orbitals involved in the formation of: C 2 − C3 bond is:
(a) sp − sp 2

(b) sp − sp 3

(c) sp − sp
2 3

(d) sp − sp
3 3

Answer. CH = CH − CH − CH − C ≡ CH
2 2 2

In the given organic compound. the carbon atoms numbered as 1. 2. 3. 4. 5. and 6 are sp,
sp, sp sp , sp . andsp hybridized respectively.
3 3 2 2

Thus. the pair of hybridized orbitals involved in the formation of C2-C3 bond is sp - sp 3

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Q12.37 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour
is obtained due to the formation of:
(a) Na [Fe(CN) ]
4 6

(b) Fe [Fe(CN) ]
4 6 3

(c) Fe [Fe(CN) ]
2 6

(d) Fe [Fe(CN) ]
3 6 4

Answer. In the Lassaigne's test for nitrogen in an organic compound. the sodium fusion
extract is boiled with iron (II) sulphate and then acidi ed with sulphuric acid. In the process.
sodium cyanide rst reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II).
Then. on heating with sulphuric acid. some iron (II) gets oxidised to form iron (Ill)
hexacyanoferrate (II). which is Prussian blue in colour The chemical equations involved in the
reaction can be represented as
xH2 O
− 2+ 4− 4− 3+
6CN + Fe ⟶ [Fe(CN)6 ] 3[Fe(CN)6 ] + 4Fe ⟶ Fe4 [Fe(CN)6 ] ⋅ xH2 O
3

Prussian blue
Hence , the Prussian blue colour is due to the formation of Fe [Fe(CN) ] 4 6 3

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Q12.38 Which of the following carbocation is most stable ?
⋆

(a) (CH ) C⋅ C H
3 3 2

+

(b) (CH ) 3 3
C

(c) CH CH CH
3 2 2

+

(d) CH 3 C HCH2 CH3

+

Answer. (CH ) 3 3
C is a tertiary carbocation. A tertiary carbocation is the most stable
carbocation due to the electron releasing effect of three methyl groups. An increased + I effect
by three methyl groups stabilizes the positive charge on the carbocation.

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Q12.39 The best and latest technique for isolation, puri cation and separation of organic
compounds is:
(a) Crystallisation
(b) Distillation
(c) Sublimation
(d) Chromatography

Answer. Chromatography is the most useful and the latest technique of separation and
puri cation of organic compounds. It was rst used to separate a mixture of coloured
substances.

Page : 372 , Block Name : Exercise

Q12.40 The reaction: CH CH I + KOH(aq) → CH CH OH + KI is classi ed as :
3 2 3 2

(a) electrophilic substitution
(b) nucleophilic substitution
(c) elimination
(d) addition

Answer. CH CH I + KOH
3 2 (aq)
⟶ CH3 CH2 OH + KI

It is an example of nucleophilic substitution reaction. The hydroxyl group of KOH (OH–) with
a lone pair of itself acts as a nucleophile and substitutes iodide ion in CH CH I to form
3 2

ethanol.

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