chapter 12 – organic chemistry some basic principles...

Chapter 12 – Organic Chemistry Some Basic Principles and Techniques Class XI Chemistry

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Question 12.1:

What are hybridisation states of each carbon atom in the following compounds?

CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6

Answer

(i)

C–1 is sp2 hybridised.

C–2 is sp hybridised.

(ii)




(iii)

C–1 and C–3 are sp3 hybridised.


(iv)



C–3 is sp hybridised.

(v) C6H6

All the 6 carbon atoms in benzene are sp2 hybridised.

Question 12.2:

Indicate the σ and π bonds in the following molecules:

C6H6, C6H12, CH2Cl2, CH2 = C = CH2, CH3NO2, HCONHCH3

Answer

(i) C6H6


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There are six C–C sigma ( ) bonds, six C–H sigma ( ) bonds, and three C=C pi (

) resonating bonds in the given compound.

(ii) C6H12

There are six C–C sigma ( ) bonds and twelve C–H sigma ( ) bonds in the given

compound.

(iii) CH2Cl2

There two C–H sigma ( ) bonds and two C–Cl sigma ( ) bonds in the given

compound.

(iv) CH2 = C = CH2


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There are two C–C sigma ( ) bonds, four C–H sigma ( ) bonds, and two C=C pi

( ) bonds in the given compound.

(v) CH3NO2

There are three C–H sigma ( ) bonds, one C–N sigma ( ) bond, one N–O sigma

( ) bond, and one N=O pi ( ) bond in the given compound.

(vi) HCONHCH3

There are two C–N sigma ( ) bonds, four C–H sigma ( ) bonds, one N–H sigma

bond, and one C=O pi ( ) bond in the given compound.

Question 12.3:

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

Answer

The bond line formulae of the given compounds are:

(a) Isopropyl alcohol


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(b) 2, 3–dimethyl butanal

(c) Heptan–4–one

Question 12.4:

Give the IUPAC names of the following compounds:

(a)

(b)

(c)

(d)


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(e)

(f) Cl2CHCH2OH

Answer

(a)

3–phenyl propane

(b)

2–methyl–1–cyanobutane

(c)

2, 5–dimethyl heptane

(d)

3–bromo–3–chloroheptane

(e)

3–chloropropanal

(f) Cl2CHCH2OH

1, 1–dichloro–2–ethanol


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Question 12.5:

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 prefix di in the IUPAC name indicates that two identical substituent groups are

present in the parent chain. Since two methyl groups are present in the C–2 of the

parent chain of the given compound, the correct IPUAC 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) If the substituents are present in the equivalent position of the parent chain, then

the lower number is given to the one that comes first in the name according to the

alphabetical order. Hence, the correct IUPAC name of the given compound is 2–chloro–

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

suffixed with ol. The alkyne group is present in the C–3 of the parent chain. Hence, the

correct IUPAC name of the given compound is But–3–yn–1–ol.

Question 12.6:

Draw formulas for the first five members of each homologous series beginning with the

following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2

Answer

The first five members of each homologous series beginning with the given compounds

are shown as follows:

(a)

H–COOH : Methanoic acid

CH3–COOH : Ethanoic acid


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CH3–CH2–COOH : Propanoic acid

CH3–CH2–CH2–COOH : Butanoic acid

CH3–CH2–CH2–CH2–COOH : Pentanoic acid

(b)

CH3COCH3 : Propanone

CH3COCH2CH3 : Butanone

CH3COCH2CH2CH3 : Pentan-2-one

CH3COCH2CH2CH2CH3 : Hexan-2-one

CH3COCH2CH2CH2CH2CH3 : Heptan-2-one

(c)

H–CH=CH2 : Ethene

CH3–CH=CH2 : Propene

CH3–CH2–CH=CH2 : 1-Butene

CH3–CH2–CH2–CH=CH2 : 1-Pentene

CH3–CH2–CH2–CH2–CH=CH2 : 1-Hexene

Question 12.7:

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)2CHCH2C (CH3)3

Bond line formula:

(b) 2–hydroxy–1, 2, 3–propanetricarboxylic acid

Condensed Formula:

(COOH)CH2C(OH) (COOH)CH2(COOH)


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Bond line formula:

The functional groups present in the given compound are carboxylic acid (–COOH) and

alcoholic (–OH) groups.

(c) Hexanedial

Condensed Formula:

(CHO) (CH2)4 (CHO)

Bond line Formula:

The functional group present in the given compound is aldehyde

(–CHO).

Question 12.8:

Identify the functional groups in the following compounds

(a)

(b)

(c)


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Answer

The functional groups present in the given compounds are:

(a) Aldehyde (–CHO),

Hydroxyl (–OH),

Methoxy (–OMe),

C=C double bond

(b) Amino (–NH2),

Ketone (C = O),

Diethylamine (N(C2H5)2 )

(c) Nitro (–NO2),

C=C double bond

Question 12.9:

Which of the two: O2NCH2CH2O– or CH3CH2O

– is expected to be more stable and why?

Answer

NO2 group is an electron-withdrawing group. Hence, it shows –I effect. By withdrawing

the electrons toward it, the NO2 group 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, O2NCH2CH2O– is expected to be more stable

than CH3CH2O–.

Question 12.10:

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.


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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 sp3 –s sigma bond orbital with an

empty p orbital of the π 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.


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Question 12.11:

Draw the resonance structures for the following compounds. Show the electron shift

using curved-arrow notation.

(a) C6H5OH (b) C6H5NO2 (c) CH3CH = CH – CHO

(d) C6H5CHO (e) (f)

Answer

(a) The structure of C6H5OH is:

The resonating structures of phenol are represented as:

(b) The structure of C6H5NO2 is:

The resonating structures of nitro benzene are represented as:

(c) CH3CH = CH – CHO

The resonating structures of the given compound are represented as:


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(d) The structure of C6H5CHO is:

The resonating structures of benzaldehyde are represented as:

(e) C6H5CH2⊕

The resonating structures of the given compound are:

(f) CH3 CH = CH CH2⊕

The resonating structures of the given compound are:

Question 12.12:

What are electrophiles and nucleophiles? Explain with examples.

Answer


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An electrophile is a reagent that takes away an electron pair. In other words, an

electron-seeking reagent is called an electrophile (E+). Electrophiles are electron-

deficient and can receive an electron pair.

Carbocations and neutral molecules having functional groups such as carbonyl

group ( ) are examples of electrophiles.

A nulceophile is a reagent that brings an electron pair. In other words, a nucleus-seeking

reagent is called a nulceophile (Nu:).

For example: OH–, NC–, carbanions (R3C–), etc.

Neutral molecules such as H2Ö and ammonia also act as nulceophiles because of the

presence of a lone pair.

Question 12.13:

Identify the reagents shown in bold in the following equations as nucleophiles or

electrophiles:

Answer

Electrophiles are electron-deficient species and can receive an electron pair. On the other

hand, nucleophiles are electron-rich species and can donate their electrons.

Here, HO– acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-

seeking species.

Here, –CN acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-

seeking species.


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Here, acts as an electrophile as it is an electron-deficient species.

Question 12.14:

Classify the following reactions in one of the reaction type studied in this unit.

(a) CH3CH2Br + HS– → CH3CH2SH + Br–

(b) (CH3)2 C = CH2 + HCl → (CH3)2 ClC–CH3

(c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br–

(d) (CH3)3 C – CH2 OH + HBr → (CH3)2 CBrCH2CH3 + H2O

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.

Question 12.15:

What is the relationship between the members of following pairs of structures? Are they

structural or geometrical isomers or resonance contributors?

(a)

(b)

(c)


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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 same constitution, and 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.

Question 12.16:

For the following bond cleavages, use curved-arrows to show the electron flow and

classify each as homolysis or heterolysis. Identify reactive intermediate produced as free

radical, carbocation and carbanion.

(a)

(b)


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(c)

(d)

Answer

(a) The bond cleavage using curved-arrows to show the electron flow 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 flow of the given


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 flow of the given


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.


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(d) The bond cleavage using curved-arrows to show the electron flow of the given


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.

Question 12.17:

Explain the terms Inductive and Electromeric effects. Which electron displacement effect

explains the following correct orders of acidity of the carboxylic acids?

(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH

(b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH

Answer

Inductive effect

The permanent displacement of sigma (σ) electrons along a saturated chain, whenever

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 – I 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 π 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


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– E effect: When the electrons are transferred away from the attacking reagent

(a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH

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) CH3CH2COOH > (CH3)2 CHCOOH > (CH3)3 C.COOH

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.

Question 12.18:

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 purification of solid

organic compounds.

Principle: It is based on the difference in the solubilites 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 filtration.


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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 filtered 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 sufficient 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 flask

fitted with a condenser. It is then heated. Chloroform, being more volatile, vaporizes

first and passes into the condenser. In the condenser, the vapours condense and

chloroform trickles down. In the round bottom flask, aniline is left behind.

(c) Chromatography

It is one of the most useful methods for the separation and purification of organic

compounds.

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

mixture through the stationary phase under the influence 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.

Question 12.19:

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.


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(a) Preparation of the solution: The powdered mixture is taken in a flask and the

solvent is added to it slowly and 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 filtered through a

filter paper in a China dish.

(c) Fractional crystallisation: The solution in the China dish is now allowed to cool.

The less soluble compound crystallises first, 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

filtration. Finally, the crystals are dried.

Question 12.20:

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.

Distillation Distillation under

reduced pressure

Steam distillation

1. It is used for the

purification of compounds

that are associated with

non-volatile impurities or

those liquids, which do not

decompose on boiling. In

other words, distillation is

used to separate volatile

liquids from non-volatile

This method is used to

purify a liquid that tends

to decompose on boiling.

Under the conditions of

reduced pressure, the

liquid will boil at a low

temperature than its

boiling point and will,

therefore, not

It is used to purify an

organic compound, which

is steam volatile and

immiscible in water. On

passing steam, the

compound gets heated up

and the steam gets

condensed to water. After

some time, the mixture of


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impurities or a mixture of

those liquids that have

sufficient difference in

boiling points.

decompose. water and liquid starts to

boil and passes through

the condenser. This

condensed mixture of

water and liquid is then

separated by using a

separating funnel.

2. Mixture of petrol and

kerosene is separated by

this method.

Glycerol is purified by

this method. It boils with

decomposition at a

temperature of 593 K. At

a reduced pressure, it

boils at 453 K without

decomposition.

A mixture of water and

aniline is separated by

steam distillation.

Question 12.21:

Discuss the chemistry of Lassaigne’s test.

Answer

Lassaigne’s test

This test is employed to detect the presence of nitrogen, sulphur, halogens, and

phosphorous 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.

The cyanide, sulphide, 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


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Lassaigne’s extract is then tested for the presence of nitrogen, sulphur, halogens, and

phosphorous.

(a) 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 acidified with sulphuric acid. In the process,

sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate

(II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III)

hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved

in the reaction can be represented as

(b) Test for sulphur


In the Lassaigne’s test for sulphur in an organic compound, the sodium fusion extract is

acidified 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.



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The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet

colour also indicates the presence of sulphur in the compound.

If in an organic compound, both nitrogen and sulphur are present, then instead of NaCN,

formation of NaSCN takes place.

Na + C + N + S → NaSCN

This NaSCN (sodium thiocyanate) gives a blood red colour. Prussian colour is not formed

due to the absence of free cyanide ions.

(c) Test for halogens


In the Lassaigne’s test for halogens in an organic compound, the sodium fusion extract is

acidified with nitric acid and then treated with silver nitrate.

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.


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Question 12.22:

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 free

nitrogen in addition to carbon dioxide and water. The chemical equation involved in the

process can be represented as

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 H2SO4. The chemical equations involved in the process are

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.

Question 12.23:

Discuss the principle of estimation of halogens, sulphur and phosphorus present in an

organic compound.


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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 CO2 and H2O

respectively and the halogen present in the compound is converted to the form of AgX.

This AgX is then filtered, washed, dried, and weighed.

Let the mass of organic compound be m g.

Mass of AgX formed = m1 g

1 mol of Agx contains 1 mol of X.

Therefore,

Mass of halogen in m1 g of AgX

Estimation of Sulphur

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 filtered, washed, dried, and weighed.


Mass of BaSO4 formed = m1 g

1 mol of BaSO4 = 233 g BaSO4 = 32 g of Sulphur

Therefore, m1 g of BaSO4 contains

Estimation of phosphorus


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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 MgNH4PO4 by adding magnesia

mixture, which on ignition yields Mg2P2O7.


Mass of ammonium phosphomolybdate formed = m1 g

Molar mass of ammonium phosphomolybdate = 1877 g

If P is estimated as Mg2P2O7,

Question 12.24:

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 flows 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 figure) is known

as a chromatogram.


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Question 12.25:

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 first boiled with

dilute nitric acid. This is done to decompose NaCN to HCN and Na2S to H2S and to expel

these gases. That is, if any nitrogen and sulphur are present in the form of NaCN and

Na2S, then they are removed. The chemical equations involved in the reaction are

represented as

Question 12.26:

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 first 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


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Carbon, nitrogen, sulphur, and halogen come from organic compounds.

Question 12.27:

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.

Question 12.28:

Explain, why an organic liquid vaporises 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 (p1) and the vapour pressure due to water (p2) becomes equal

to atmospheric pressure (p), that is, p = p1 + p2

Since p1 < p2, organic liquid will vapourise at a lower temperature than its boiling point.

Question 12.29:

Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for

your answer.

Answer


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CCl4 will not give the white precipitate of AgCl on heating it with silver nitrate. This is

because the chlorine atoms are covalently bonded to carbon in CCl4. To obtain the

precipitate, it should be present in ionic form and for this, it is necessary to prepare the

Lassaigne’s extract of CCl4.

Question 12.30:

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

Thus, the mass of the U-tube containing KOH increases. This increase in the mass of U-

tube gives the mass of CO2 produced. From its mass, the percentage of carbon in the

organic compound can be estimated.

Question 12.31:

Why is it necessary to use acetic acid and not sulphuric acid for acidification 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 acidification of

sodium extract for testing sulphur by lead acetate test.

Question 12.32:

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


of 34




Percentage of carbon in organic compound = 69 %

That is, 100 g of organic compound contains 69 g of carbon.

∴0.2 g of organic compound will contain

Molecular mass of carbon dioxide, CO2 = 44 g

That is, 12 g of carbon is contained in 44 g of CO2.

Therefore, 0.138 g of carbon will be contained in = 0.506 g of CO2

Thus, 0.506 g of CO2 will be produced on complete combustion of 0.2 g of organic

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

It is known that molecular mass of water (H2O) is 18 g.

Thus, 2 g of hydrogen is contained in 18 g of water.

∴0.0096 g of hydrogen will be contained in of water

Thus, 0.0864 g of water will be produced on complete combustion of 0.2 g of the organic

compound.

Question 12.33:

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 H2SO4. The residual acid 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 H2SO4 = 30 mL of 0.5 M H2SO4


of 34




∴Acid consumed in absorption of evolved ammonia is (50–30) mL = 20 mL

Again, 20 mL of 0.5 MH2SO4 = 40 mL of 0.5 MNH3

Also, since 1000 mL of 1 MNH3 contains 14 g of nitrogen,

∴ 40 mL of 0.5 M NH3 will contain = 0.28 g of N

Therefore, percentage of nitrogen in 0.50 g of organic compound = 56 %

Question 12.34:

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

∴ Percentage of chlorine

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

.

Question 12.35:

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]


of 34




Mass of barium sulphate formed = 0.668 g [Given]

1 mol of BaSO4 = 233 g of BaSO4 = 32 g of sulphur

Thus, 0.668 g of BaSO4 contains of sulphur = 0.0917 g of sulphur

Therefore, percentage of sulphur = 19.59 %

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

Question 12.36:

In the organic compound CH2=CH–CH2–CH2–C≡CH, the pair of hydridised orbitals

involved in the formation of: C2 – C3 bond is:

(a) sp – sp2(b) sp – sp3(c) sp2 – sp3 (d) sp3– sp3

Answer

In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are

sp, sp, sp3, sp3, sp2, and sp2 hybridized respectively. Thus, the pair of hybridized orbitals

involved in the formation of C2-C3 bond is sp – sp3.

Question 12.37:

In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is

obtained due to the formation of:

(a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)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 acidified with sulphuric acid. In the process,

sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate

(II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III)

hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved

in the reaction can be represented as


of 34




Hence, the Prussian blue colour is due to the formation of Fe4[Fe(CN)6]3.

Question 12.38:

Which of the following carbocation is most stable?

(a) (b) (c)

(d)

Answer

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.

Question 12.39:

The best and latest technique for isolation, purification 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

purification of organic compounds. It was first used to separate a mixture of coloured

substances.

Question 12.40:

The reaction:

is classified as :

(a) electrophilic substitution (b) nucleophilic substitution


of 34




(c) elimination (d) addition

Answer

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 CH3CH2I to

form ethanol.

[XI – Chemistry] 94

UNIT-12

IUPAC NOMENCLATURE

(1) Longest Chain Rule :

Longest chain in molecule is selected.

1 2 3 4 5 6 7 8 9C–C–C–C–C–C–C–C–C

C 7C 8C

(2) Lowest Sum Rule :

The numbering is done is such a way that the branched carbon atoms of the

parent chain get the lowest possible number or locant.

1 2 3 4 5 6 7 8 9C–C–C–C–C–C–C–C–C

C C CCase (I)

9 8 7 6 5 4 3 2 1C–C–C–C–C–C–C–C–C

C C CCase (II)

In case (I), the sum of locants (2 + 6) is lower than the sum of locants (4 + 8)

in case (II). Hence, the numbering done in case (I) is “right.”

(3) The names of side chains and the substituents are then prefixed to the name

of parent alkane and position of substituents is indicated by appropriate

numbers. The numbers are separated from the prefixes by hyphens (–).

(4) Alphabetical arrangement of side chains/substituents :

Thus, the name for compound shown above is : 6–ethyl–2–methyl nonane.

Note that there is no break between methy and nonane, i.e., the name is

always written as a single word.

(5) We write di, tri, tetra, penta for two, three, four, five identical subtituents

and so on. The numbers (locants) are separated by commas. Note that di, tri,

95 [XI – Chemistry]

tetra, penta, hexa etc do not determine the alphabetical order of the

subtituents.

CH –CH–CH –CH–CH–CH –CH3 2 2 3

2 4 6

CH3 CH3 CH2 CH3

6 - Ethyl -2, 4 - dimethyl heptane

(6) (a) If two substituents are found in equivalent positions, the lower number

(locant) is given to the one that comes first in alphabetical order.

CH –CH –CH–CH–CH –CH3 2 32

CH3 C H52

6 5 4 3 2 1

1 2 3 4 5 6

3- Ethyl - 4 - methyl hexane and not 4- ethyl - 3- methyl hexane

(b) The name of complex chains (substitued substituents) are enclosed in

parentheses and numbering of C-atoms is carried such that the C-atom of

substituted substituent attached to main chain is given number 1.

5

CH3CH

2CH

2CH

2CHCH

2CH

2CH

2CH

2CH

3

1 2 3

H3C–CH–CH

2CH

3

5-(1-Methylpropyl) decane, in place of 5-Isobutyldecane

(c) Composite or complex groups are represented by prefixing bis for 2, tris

for 3, tetrakis for 4, pentakis for 5 identical composite groups and so on,

and entire group is enclosed in parentheses (brackets). For example,

CH2Cl

|

Cl–CH2–C–CH

2–Cl

|

CH2Cl

1, 3-Dichloro-2, 2-bis (chloromethyl) propane and IUPAC name of DDT is

Cl— —CH— —Cl

|

CCl3

-1, 1,1-trichloroethane 2, 2-bis (4-Chlorophenyl)


(d) Name of a substituted substituent is considered to begin with the first

letter of its complete name.

F F

| |

H3C–H

2C–HC–CH CH

2–CH

3

| |

CH3–[CH

2]4–CH

2–CH–CH

2–CH–[CH

2]3–CH

3

7-(1, 2-Diflurobutyl)-5-ethyltridecane

IUPAC Nomenclature for alkenes and alkynes :

(a) Multiple bond is preferred for numbering and represented by lowest

number of C-atoms.

H2C =

CH

CH

2CH

3: But-1-ene and not 1-Butene

H3C

C

C

CH

2CH

3: Pent-2-yne and not 2-pentyne

(b) When two or more similar multiple bonds are present, suffix ‘a’ to the

root word.

CH2 =

CH

CH =

CH

2H

2C =

CH

CH =

CH

CH =

CH

2

Buta-1, 3-diene Hexa-1, 3, 5-triene

(c) When double and triple bonds are present on the same position from

the ends in an organic compound, then double bond is preferred for

naming and the compound is named as alkenyne, e.g,

CH2 =

CH

CH

CH : But-1-en-3-yne

(d) When double and triple bonds are present on different positions, lowest

sum rule will be followed and the compound is named as alkenyne, e.g.,

HC C

CH

2CH =

CHCH

2CH

2CH

3Oct-4-en-1-yne

Functional groups : The groups which determine the properties and

functions of an organic compound are known as functional groups.

When a functional group is present, a secondary suffix for functional group

is added by replacing ending ‘e’ of the parent name if the name of functional

groups starts from vowels y i.e., a, e, i, o and y.

CHARACTERISTICS (FUNCTIONAL) GROUPS

Suffixes and prefixes for some important characteristic groups in IUPAC

nomenclature (in alphabetical order). (C) designates a carbon atom included

in the name of the parent hydride and does not belong to a group designated by

a suffix or a prefix.

Class Formula Prefix Suffix

Acid halides CO-halogen halodarbonyl- carbonyl halide

(C)O-halogen -oyl halide


Alcohols, Phenols OH hydroxy- -ol

Aldehydes CHO formyl- -carbaldehyde

(C)HO oxo- -al

Amides CO-NH

2carbamoyl- -carboxamide

(C)O-NH

2 -amide

Amines NH

2amino- -amine

Carboxylates COO carboxylato -carboxylate

(C)OO -oate

Carboxylic acids COOH carboxy- -carboxylic acid

(C)OOH -oic acid

Ethers OR (R)-oxy-

Ester (of carboxylic acids) COOR (R)oxycarbonyl- (R).....Carboxylate

(C)OOR (R)......oate

Ketones > (C)

O oxo- -one

Nitriles C

N cyano- -carbonitrile

C

N -nitrile

Salts (of carboxylic acids) COOM+ (cation).....

carboxylate

Sulphonic acids SO

2–OH sulpho -sulphonic acid

Important Notes :

• When the name of suffix starts with a consonent (except a, e, i, o, u and y),

ending ‘e’ of the parent hydride is retained.

CH3CH

2OH : ethanol

HOCH2CH

2OH : ethane-1, 2-diol

Here ‘e’ is retained since first letter of diol, i.e., ‘d’ is a consonent.

• When a functional group and a side chain/substituent is present, then

functional group is preferred for naming and numbering.

OH

|

BrCH2CH

2CH

2CHCH

3: 5-bromopentan-2-ol

• When two or more functional groups are present, the order of preference is :

COOH

SO

3H > (

CO)

2O >

COOR' >

COX >

CONH

2 >

CN

> CHO >–

O||C

NH

2 > C

C > C

C


• When two or more functional groups are present, name of higher order

functional group is suffixed and the rest are prefixed. Examples are :

HOCH3CH

2CH

2COOH : 4-hydroxybutanoic acid

3 2

O||

CH C CH CH CH CHO : 5-Oxohex-2-enol

CH3COCH

2COOC

2H

5 : Ethyl 3-oxobutanoate

: 5-Hydroxycyclopent-2-enecarbaldehyde

: 2-Hydroxypropane-1, 2, 3-tricarbonitrile

: 3-Formylpropane-1, 2, 3-tricarboxylic acid

: N, 2-Dimethylcyclopentanecarboxamide

: Ethyl 2-methylcyclopentanecarboxylate

CHO

OH

OH

NC–CH –C–CH –CN

CN

2 2

HOOC–CH–C–CH–COOH

CHO

COOH

CONHCH3

CH3

COOC H2 5

CH3


: 2-Nitropropane-1, 2, 3-tricarbonyl chloride

•2

O O O O O|| || || || ||C OH, C H, C NH , C X, C OR and CN are chain terminators.

• In cyclic compound, the word ‘cyclo’ is prefixed and lowest sum rule is

applied for prefixing the side chains/substituents.

Cyclopropane Cyclobutene

Cyclopentanone 3-Chloro-1, 1-dimethylcyclohexane

• In aromatic compounds, the following trival names have been retained by

IUPAC system.

Benzene Toluene

Benzyl Benzyl alcohol

Benzyl chloride Benzaldehyde

NO2

COCl–CH –C–CH –COCl2 2

COCl

O

Cl

CH3

CH2– CH2OH

CH2Cl CHO


Benzoic acid o-/p-/m-xylenes

Acetophenone Benzophenone

Phenol Aniline

Benzonitrile Anisole

• When a chain of two or more carbon atoms having a functional group or a

multiple bond is attached with the benzene ring, then ring is taken as

substituent and is names as phenyl group, e.g.,

2-Phenylethan-1-ol 3-Phenylpropanal

When the benzene ring is further substituted, then the name of the substituted

phenyl group is enclosed in parentheses for example :

2-(3-Bromophenyl) ethan-1-ol 3-(4-Bromophenyl) propanal

COOH

CH3

CH3

COCH3

C

O

OH NH2

CN OCH3

CH2CH2OH CH2CH2CHO

CH2CH2OH

Br

CH2CH2CHO

Br


Purification and Characterization of Organic Compounds

(1) Lassaigne’s test for nitrogen : Lassiagne’s extract is heated with FeSO4

solution in presence of alkali, the solution is cooled and acidified with

dil. H2SO

4. If a green or blue colouration is obtained, it confirms the

presence of N in the organic compound. The chemisty of the test is :

From organic compound

4 2 4 42 2 6

Sod. ferrocyanide

3+

4 46 6 3

Ferric ferrocyanide (Prussian Blue)

Na C N NaCN

2NaCN FeSO Fe CN Na SO ; Fe CN +4NaCN Na Fe CN

3Na Fe CN 4Fe Fe Fe CN

+12Na

This test is very delicate and is given by all compounds containing C and N.

As such NH2NH

2, NH

4Cl, NaNO

3 etc. do not respond to this test since they do

not contain carbon.

Formation of blood red colour indicates the presence of both N and S.

From organic compound Sod. thiocyanate or Sod. sulphocyanide

+ 23+ +

Ferric thiocyanate (blood red colour)

Na C N S NaSCN

Fe N a SC N Fe SCN Na

(2) Detection of sulphur : If S is present, during fusion with Na metal, Na2S is

formed which may be tested as follows :

(i) With sodium nitroprusside, violet colouration is produced.

2 2 5 4 5

Sodium nitroprusside (Violet colour)

Na S Na Fe(CN) NO Na Fe(CN) NOS

(ii) With lead acetate, black ppt. of PbS is formed.

2 3 32 (Black ppt.)Na S CH COO Pb PbS 2CH COONa

(3) Detection of halogens :

(a) Lassaigne’s test : When the organic compounds is fused with Na metal,

the halogens combine with Na to form sodium halides.

The presence of these halides is tested with AgNO3 solution.

(i) A white ppt. soluble in NH4OH and insoluble in dil. HNO

3 indicates

chlorine.

Na X NaX


(i) NaCl + AgNO3

AgCl + NaNO3

AgCl + 2NH2OH [Ag(NH

3)2]Cl + 2H

2O

(ii) NaBr + AgNO3

AgBr + NaNO3

(iii) NaI + AgNO3 AgI + NaNO

3

(ii) A pale yellow ppt. partially soluble in ammonia indicates bromine.

(iii) A yellow ppt. insoluble in ammonia indicates iodine.

If the organic compound also contain N or S, the sodium extract is first

boiled with dil. HNO3 to decompose any cyanides or sulphides, otherwise these

will form ppt. with AgNO3 solution.

(4) Detection of phosphorus : Phosphorus is detected by fusing the organic

compound with sodium peroxide when phosphorus is converted into sodium

phosphate.

Fuse2 2 3 4 25Na O 2P 2Na PO 2Na O

3 4 3 3 4 3Na PO 3HNO H PO 3NaNO

4 4 33

3 4 4 4 324 3 2

Ammonium phosphomolybedate

NH PO .12MoOH PO 12 NH MoO 21HNO

21NH NO 12H O.

The fused mass is extracted with H2O and then boiled with conc. HNO

3

and then ammonium molybdate is added. Appearance of yellow ppt. or

colouration due to the formation of ammonium phosphomolybdate indicates

the presence of phosphorus.

(5) Estimation of Carbon and Hydrogen : Liebig’s method : A known mass

of the organic compound is heated strongly with excess of dry copper oxide

in a current of dry air or oxygen (free from CO2) when carbon present in the

organic compound is oxidised to CO2 and hydrogen to H

2O.

2 2C 2CuO CO 2Cu; 2H CuO H O Cu

Percentage of carbon 2Mass of CO formed12100

44 Wt. of substance taken

Perentage of hydrogen 2Mass of H O formed2100


(6) Estimation of Nitrogen :

(a) Dumas method :


2 2

2

2

C 2CuO CO 2Cu; 2H CuO H O Cu

N CuO N Small amounts of oxides of nitrogen

Oxides of nitrogen Cu CuO N

Percentage of nitrogen 3

2Vol. of N (in cm ) at STP28100


(b) KJELDAHL’s Method :

2NaOH

2 4 4 2 4 2 4 3 2Organic compound H SO (NH ) SO Na SO 2NH 2H O

3 2 4 4 2 42NH H SO (NH ) SO

Percentage of nitrogen

1.4 Molarity of the acid Vol. of acid used Basicity of the acid

Mass of substance taken

(7) Estimation of Halogens : Carius method :

Percentage of chlorine 35.5 Mass of AgCl formed

100143.5 Mass of substance taken

Percentage of bromine 80 Mass of AgBr formed

100188 Mass of substance taken

Percentage of iodine 127 Mass of AgI formed


Percentage of sulphur 4Mass of BaSO formed32100


(8) Estimation of Phosphorus : A known mass of the organic compound is

heated with fuming HNO3 in a Carius tube when P of the organic compound

is oxidized to H3PO

4. Phosphoric acid thus formed is precipitated as

magnesium ammonium phosphate by adding magnesia mixture (a solution

containing MgCl2, NH

4Cl and NH

4OH).

Percentage of phosphorus 2 2 7Mass of Mg P O formed62



ORGANIC CHEMISTRY

1. Write IUPAC names of the following :

(a) (b) (c)

(d) (e) (f)

2. Write IUPAC names of the following organic compounds :

(a) H2C CH CH

2NH

2(b)

3 2

O||

CH C CH COOH

(c) 2 2HO CH CH CH OH|OH

(d)

(e) (f) 3 3

3

CH O C H CH|

CH

(g) 2

Br|

HOOC CH C CH CHO (h) CH2–COOH

| CH–CHO | CH

2–COOH

(i) COOH (j) HO CH

2 CH

CH CH

2CN

|

COOH

3. How many and bonds are present in the following organic compounds :

(a) (b) H2C CH CH CH C CH

(c) N C CH CH C N (d)

OH

O–

O

OH

CH3

OH

NO2CHO

Cl Br

CN

O

Br HBr CHO


(e) 2 2

O||

H C CH C CH CH

4. Draw the structures of the following organic compounds :

(a) 2,2-dimethylhexane (b) 4-methylpentanal

(c) 2-methylpropan-2-ol (d) cyclohexylcarbaldehyde

(e) pent-3-enoic acid (f) 4-nitrobenzoic acid

(g) ethylproponoate (h) ethoxybenzene

(i) 4-hydroxybutanenitrile (j) 4-phenylbut-1-ene

5. Mention the states of hybridisation of each carbon atom in the following

compounds :

(a) C H2 C O

(b) CH3 C CH

(c) H3C

C = O H

3C

(d) C6H

6

(e) H2C CH CH

2 C CH

6. Define the following terms with suitable examples :

(a) Nucleophile (b) Electrophile

(c) Free radicals (d) Carbocation

(e) Carbanion

7. Identify the reagents shown in bold in the following equations as nucleophiles

or electrophiles :

(a) 3 3 2CH COOH CH COO H O

H O

(b) –

3 3 3 3CH COCH CH C CN O

C N

(c) 6 5 6 5 3C H C H COCH H

3CH CO

(d) 3 2 3 2CH CH Cl CH CH C CH Cl

HC C

(e) 6 5 6 5 2C H C H NO H

2N O


8. Which of the following carbocation is most stable :

(a)

(b) 3 23CH CCH

(c) 3 3CH C

(d) 3 2 2CH CH C H

(e) 3 2 3H C C H CH CH

Arrange the above carbocations in the decreasing order of their stability.

9. Explain the following technique of separation of organic compounds with

suitable examples :

(a) Crystallization

(b) Distillation

(c) Sublimation

(d) Chromatography

(e) Differential extraction

10. Discuss the chemistry of Lassaigne’s test. Explain the reason for the fusion

of an organic compound with metallic sodium for testing nitrogen, sulphur

and halogens.

11. Name a suitable technique of separation of the components from a mixture of :

(a) calcium sulphate and camphor

(b) water and aniline

(c) a mixture of dyes

(d) copper sulphate and sodium chloride

(e) a mixture of amino acids

(f) glycerol and spentlye in soap industry.

12. Explain the terms inductive effect and electromeric effect. Arrange the

following compounds in the increasing order of their acidic and basic strength :

C+


(a) ClCH2COOH, Cl

3CCOOH and Cl

2CHCOOH

(b) CH3CH

2COOH, (CH

3)2CHCOOH and (CH

3)3CCOOH

(c) CH3NH

2, CH

3NHCH

3, (CH

3)3N, C

6H

5NHCH

3

13. Write the isomerism among the following compounds :

CH3

CH3

| |

(a) CH3CH

2CH

2CH

2CH

3 , CH

3–CH–CH

2CH

3 H

3–C–CH

3

|

OH CH3

|

(b) CH3CH

2CH

2OH and CH

3CH–CH

3

O ||

(c) CH3CH

2CHO and H

3C–C–CH

3

(b) CH3–O–CH

2CH

2CH

2CH

3 and CH

3CH

2OCH

2CH

3

O OH

|| |

(e) CH3–C–CH

2COOC

2H

5 and CH

3–C = CHCOOC

2H

5

14. An organic compound was found to contain C = 39.13%, H = 8.64% and

remaining is oxygen. Calculate the empirical formula of the compound.

[Ans. C3H

8O

3]

15. An organic compound gave the following results on analysis :

C = 17.39%, H = 1.45%, Br = 57.97%, O = 23.19%. Molar mass of the

com pound w as found to be 276 g m ol1. Calculate its molecular formula.

[Ans. C4H

4Br

2O

4]

16. 0.185 g of an organic compound when treated with concentrated nitric acid

and then silver nitrate in Carius method gave 0.320 g of silver bromide.

Calculate the percentage of bromine in the compound. [Ans. 73.6%]

17. A sample of 0.50 g of an organic compound was treated to Kjeldahl’s method.

The ammonia evolved was absorbed in 50 mL of 0.5M H3SO

4. The residual

acid required 60mL of 0.5M NaOH for neutralisation. Find the percentage

composition of nitrogen in the compound. [Ans. 56%]

18. A welding fuel gas contain carbon and hydrogen only. Burning a small sample

of it in oxygen, gives 3.38 carbondioxide, 0.690 g water and no other


products. A volume of 10.0 L (measured at STP) of this welding gas found

to weigh 11.6 g. Calculate :

(a) Empirical formula

(b) Molar mass of the gas

(c) Molecular formula

19. An organic compound on analysis gave following data

(i) 0.25g of compound on complete combustion gave 0.37g CO2 and 0.2g

of water.

(ii) 0.25g of compound on analysis by Duma’s method gave 32 ml of

nitrogen at N.T.P.

Calculate the parcentage of C, H and N in compound.

chapter 12 – organic chemistry some basic principles...

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