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Purification, Classification and Nomenclature of Organic Compounds, Study notes of Organic Chemistry

An overview of the history and modern definition of organic chemistry, as well as the steps involved in the purification and characterization of organic compounds. It discusses various methods of purification, including simple and fractional crystallization, sublimation, distillation, and chromatography. The document also covers qualitative and quantitative analysis, determination of molecular mass and empirical/molecular formulas, and spectroscopic and diffraction methods for determining the structure of organic compounds.

Typology: Study notes

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970 Purification, Classification and Nomenclature of Organic compounds
The word ‘organic’ signifies life. Therefore, all
substances which were obtained directly or indirectly
from living organisms, plants and animals were called
organic compounds and the branch of chemistry which
deals with these compounds was called organic
chemistry.
Modern definition of organic chemistry :
Organic chemistry is a chemistry of hydrocarbons and
their derivatives in which covalently bonded carbon is
an essential constituent.
Berzelius put forward a theory in 1815 known as
vital force theory. According to this theory, "organic
compounds could be prepared only by living organism
under the influence of a mysterious force known as vital
force". Accidental synthesis of urea by Wohler and
synthesis of acetic acid by Kolbe led to the fall of this
theory.
O
NHCNHCNONH ||
224 
COOHCHCHOCH O 3
][
3
Berthelot prepared methane in laboratory and
the most abundant organic compound is cellulose
which is a polymer of glucose. Kekule and Couper
proposed the tetravalency of carbon and wrote the first
structural formula. In 1874, Van't Hoff and Le Bell
suggested a tetrahedron model of carbon.
Purification and Characterisation of organic
compounds
The study of organic compounds starts with the
characterisation of the compound and the
determination of its molecular structure. The procedure
generally employed for this purpose consists of the
following steps :
(1) Purification of organic compounds
(2) Qualitative analysis of organic compounds
(3) Quantitative analysis of organic compounds
(4) Determination of molecular mass of organic
compounds
(5) Calculation of Empirical formula and
Molecular formula of organic compounds
(6) Determination of structure of organic
compounds by spectroscopic and diffraction methods
(1) Purification of organic compounds : A large
number of methods are available for the purification of
substances. The choice of method, however, depends
upon the nature of substance (whether solid or liquid)
and the type of impurities present in it. Following
methods are commonly used for this purpose,
(i) Simple crystallisation
(ii) Fractional crystallisation,
(iii) Sublimation
(iv) Simple distillation
(v) Fractional distillation
(vi) Distillation under reduced pressure
(vii) Steam distillation
(viii) Azeotropic distillation
(ix) Chromatography
(x) Differential extraction
(xi) Chemical methods
(Ammonium
cyanate)
(First organic compound
synthesised in laboratory)
Urea
Acetic acid
(First organic compound
synthesised from its
elements)
Chapter
22
Purification, Classification and Nomenclature of Organic
compounds
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pf5
pf8
pf9
pfa
pfd
pfe
pff
pf12
pf13
pf14
pf15
pf16
pf17
pf18
pf19
pf1a

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The word ‘ organic’ signifies life. Therefore, all

substances which were obtained directly or indirectly

from living organisms, plants and animals were called

organic compounds and the branch of chemistry which

deals with these compounds was called organic

chemistry.

Modern definition of organic chemistry :

Organic chemistry is a chemistry of hydrocarbons and

their derivatives in which covalently bonded carbon is

an essential constituent.

Berzelius put forward a theory in 1815 known as

vital force theory. According to this theory, " organic

compounds could be prepared only by living organism

under the influence of a mysterious force known as vital

force". Accidental synthesis of urea by Wohler and

synthesis of acetic acid by Kolbe led to the fall of this

theory.

O

NHCNO NH C NH

4 ^2   2

CH CHO CH COOH

O    3 

[] 3

Berthelot prepared methane in laboratory and

the most abundant organic compound is cellulose

which is a polymer of glucose. Kekule and Couper

proposed the tetravalency of carbon and wrote the first

structural formula. In 1874, Van't Hoff and Le Bell

suggested a tetrahedron model of carbon.

Purification and Characterisation of organic

compounds

The study of organic compounds starts with the

characterisation of the compound and the

determination of its molecular structure. The procedure

generally employed for this purpose consists of the

following steps :

(1) Purification of organic compounds

(2) Qualitative analysis of organic compounds

(3) Quantitative analysis of organic compounds

(4) Determination of molecular mass of organic

compounds

(5) Calculation of Empirical formula and

Molecular formula of organic compounds

(6) Determination of structure of organic

compounds by spectroscopic and diffraction methods

(1) Purification of organic compounds : A large

number of methods are available for the purification of

substances. The choice of method, however, depends

upon the nature of substance (whether solid or liquid)

and the type of impurities present in it. Following

methods are commonly used for this purpose,

(i) Simple crystallisation

(ii) Fractional crystallisation,

(iii) Sublimation

(iv) Simple distillation

(v) Fractional distillation

(vi) Distillation under reduced pressure

(vii) Steam distillation

(viii) Azeotropic distillation

(ix) Chromatography

(x) Differential extraction

(xi) Chemical methods

(Ammonium

cyanate) (^) (First organic compound

synthesised in laboratory)

Urea

Acetic acid

(First organic compound

synthesised from its

elements)

Acetaldehyde

Chapter

Purification, Classification and Nomenclature of Organic

compounds

(i) Simple crystallisation : This is the most

common method used to purify organic solids. It is

based upon the fact that whenever a crystal is formed,

it tends to leave out the impurities. For crystallisation,

a suitable solvent is one (a) which dissolves more of

the substance at higher temperature than at room

temperature (b) in which impurities are either

insoluble or dissolve to an extent that they remain in

solution (in the mother liquor) upon crystallisation,

(c) which is not highly inflammable and (d) which

does not react chemically with the compound to be

crystallized. The most commonly used solvents for

crystallisation are : water, alcohol, ether, chloroform,

carbon- tetrachloride, acetone, benzene, petroleum

ether etc.

Examples : (a) Sugar having an impurity of

common salt can be crystallized from hot ethanol since

sugar dissolves in hot ethanol but common salt does

not.

(b) A mixture of benzoic acid and naphthalene can

be separated from hot water in which benzoic acid

dissolves but naphthalene does not.

(ii) Fractional crystallisation : The process of

separation of different components of a mixture by

repeated crystallisations is called fractional

crystallisation. The mixture is dissolved in a solvent in

which the two components have different solubilities.

When a hot saturated solution of this mixture is

allowed to cool, the less soluble component crystallises

out first while the more soluble substance remains in

solution. The mother liquor left after crystallisation of

the less soluble component is again concentrated and

then allowed to cool when the crystals of the more

soluble component are obtained. The two components

thus separated are recrystallized from the same or

different solvent to yield both the components of the

mixture in pure form.

Fractional crystallisation can be used to separate

a mixture of KClO 3 (less soluble) and KCl (more

soluble).

(iii) Sublimation : Certain organic solids on

heating directly change from solid to vapour state

without passing through a liquid state, such substances

are called sublimable and this process is called

sublimation.

The sublimation process is used for the separation

of sublimable volatile compounds from non sublimable

impurities. The process is generally used for the

purification of camphor , naphthalene, anthracene,

benzoic acid NH 4 Cl , HgCl 2 , solid SO 2 , Iodine and

salicylic acid etc containing non-volatile impurities.

(iv) Simple distillation : Distillation is the joint

process of vapourisation and condensation. This

method is used for the purification of liquids which boil

without decomposition and contain non-volatile

impurities. This method can also be used for separating

liquids having sufficient difference in their boiling

points. This method can be used to separate a mixture

of

(a) chloroform (b. p. 334 K ) and aniline (b. p. 457

K )

(b) ether (b. p. 308 K ) and toluene (b. p. 384 K )

(v) Fractional distillation : This process is used

to separate a mixture of two or more miscible liquids

which have boiling points close to each other. Since in

this process, the distillate is collected in fractions

under different temperatures, it is known as fractional

distillation. This process is carried out by using

fractionating columns. Fractionating column is a

special type of long glass tube provided with

obstructions to the passage of the vapour upwards and

that of liquid downwards. This method may be used to

separate a mixture of acetone (b. p. 330 K ) and methyl

alcohol (b. p. 338 K ) or a mixture of benzene and

toluene. One of the technological applications of

fractional distillation is to separate different fractions

of crude oil in petroleum industry into various useful

fractions such as gasoline, kerosene oil, diesel oil,

lubricating oil etc.

(vi) Distillation under reduced pressure : This

method is used for the purification of high boiling

liquids and liquids which decompose at or below their

boiling points.

The crude liquid is heated in distillation flask

fitted with a water condenser, receiver and vacuum

pump. As the pressure is reduced, the liquid begins to

boil at a much lower temperature than its normal

boiling point. The vapour is condensed by water

condenser and the pure liquid collects in the receiver.

Glycerol which decomposes at its boiling point

(563 K ) under atmospheric pressure can be distilled

without decomposition at 453 K under 12 mm of Hg.

Similarly, sugarcane juice is concentrated in sugar

industry by evaporation under reduced pressure which

saves a lot of fuel.

(vii) Steam distillation : This method is

applicable for the separation and purification of those

organic compounds (solids or liquids) which (a) are

insoluble in water (b) are volatile in steam (c) possess

a high vapour pressure (10- 15 mm Hg ) at 373 K and (d)

contain non-volatile impurities.

Aniline (b. p. 457 K ) can be purified by steam

distillation since it boils at a temperature of 371.5 K in

Solid Vapour

Heat

Cool

acetone (0.5%) and methanol (3%). Acetic acid can be

separated from this mixture by treating it with milk of

lime when acetic acid forms the calcium salt. The

reaction mixture on distillation gives a mixture of

acetone and methanol (which can be further separated

by fractional distillation into individual components as

mentioned above) while the calcium salt remains as

residue in the flask. The calcium salt is then

decomposed with dil HCl and distilled to afford acetic

acid.

(c) A mixture of 1

o , 2

o and 3

o amines can be

separated using either benzenesulphonyl chloride

( Hinsberg's reagent ) or diethyl oxalate ( Hoffmann's

method ).

(d) Purification of commercial benzene :

Commercial benzene obtained from coal-tar distillation

contains 3-5% thiophene as an impurity which can be

removed by extraction with conc. H 2 SO 4. This

purification is based upon the fact that thiophene

undergoes sulphonation much more easily than

benzene. Thus, when commercial benzene is shaken

with conc. H 2 SO 4 in a separating funnel, thiophene

undergoes sulphonation to form thiophene- 2 - sulphonic

acid which dissolves in conc. H 2 SO 4 while benzene

does not.

  

Roomtemp H 2 SO 4

 H 2 O

After this treatment, the benzene layer is

removed, washed with water to remove unreacted

H 2 SO 4

, dried over anhyd. CaCl 2 and then distilled to

give pure benzene.

(e) Absolute alcohol from rectified spirit : The

rectified spirit (ethanol : H 2 O , 95. 87 : 4. 13 by weight) is

kept over a calculated amount of active quick lime

( CaO ) for few hours and then refluxed. During this

process, water present in rectified spirit combines with

CaO to form Ca ( OH ) 2. When the resulting mixture is

distilled, absolute alcohol distils over leaving behind,

Ca ( OH ) 2.

Drying of Organic Substances. (1) For solids :

Most solids are dried first by pressing them gently

between folds of filter papers. Compounds which

neither decompose on heating nor melt below 100

o C are

dried by keeping them in steam or oven maintained at

o C. Substances, which decompose on heating are

dried by keeping them in a vacuum desiccator

containing a suitable dehydrating agent like fused

CaCl 2 , conc. H 2 SO 4 , P 4 O 10 , solid KOH or NaOH, etc

(desiccant).

(2) For liquids : Organic liquids are generally

dried by keeping them over night in contact with a

dehydrating (desiccating) agent which does not react

chemically with the liquid to be dried. Commonly used

dehydrating agents are quick lime, anhydrous CaCl 2 ,

fused 4

CuSO or CaSO , KOH 4

, metallic sodium or

potassium, etc.

Criteria of purity of organic compounds : The

purity of an organic compound can be ascertained by

determining its some physical constants like m.p., b.p.,

specific gravity, refractive index and viscosity. In usual

practice, sharp m.p. (in case of solids) and boiling point

(in case of liquids) are used as criteria for purity

because their determination is feasible in the

laboratory. A pure organic solid has a definite and sharp

(sudden, rapid and complete) melting point, while an

impure substance has a lower and indefinite melting

point.

(1) Mixed melting point : The melting point of

two thoroughly mixed substances is called mixed

melting point. This can also be used for ascertaining

the purity of a compound.

The substance, whose purity is to be tested, is

mixed with a pure sample of the same compound. The

melting point of the mixture is determined. If the

melting point of the mixture is sharp and comes out to

be the same as that of pure compound, it is sure that

the compound under test is pure. On the other hand, if

the melting point of the mixture is less than the

melting point of the pure compound, the compound in

question is not pure.

(2) Qualitative analysis : (Detection of Elements

The qualitative analysis of an organic compound

involves the detection of all the elements present in it.

Carbon is an essential constituent of an organic

compound whereas hydrogen is nearly always present.

On heating the organic compound with dry cupric

oxide, carbon is oxidized to 2

CO and hydrogen to HO 2

CO 2 is detected by lime water which turns milky while

H 2 O is detected by anhydrous CuSO 4 (white) which

turns it blue. This method is known as copper oxide

test.

C 2 CuO CO 2 2 Cu

Heat     ;

Ca OH CO CaCO H 2 O Milky

2 3 Limewater

HCuO   H 2 OCu

Heat 2^ ;

(Hydrated)

Blue

2 4 2

(Anhydrous)

Colourles

CuSO (^) 4 5 HO CuSO. 5 HO

s

Thiophen

e

(Conc. S )

SO 3 H

Thiophene- 2 - sulphonic

acid

(Dissolves in conc.

)

S

If the substance under investigation is a volatile

liquid or gas, the vapours are passed over heated

copper oxide kept in combustion tube and the gaseous

products are tested as above.

Lassaigne method

This is used to detect nitrogen, halogen and

sulphur. Organic compounds are fused with dry sodium

in a fusion-tube and fused mass after extraction with

HO

2

is boiled and filtered. Filtrate called sodium

extract (S.E.) is used to detect elements (other than C

and H ) and the tests are given in the table.

 Organic compounds being covalents normally do

not have ionisable groups, hence direct test is not

possible.

 Fusion with (^) Na forms soluble salt (like

NaCl , NaCN etc.) which can be easily detected.

 This test fails in case of diazo compounds.

 Sometimes when the amount of nitrogen

present is small, the prussian blue is present in

colloidal form and the solution looks green.

Table : 22.2 Lassaigne method (Detection of elements)

Elemen

t

Sodium Extract (S.E.) Confirmed Test Reaction

Nitroge

n

( S. E .)

NaCN  NaCN

 S.E.+ FeSONaOH 4 , boil

and cool + FeCl conc. HCl 3

Blue or green colour

2 NaCNFeSO 4  Fe ( CN ) 2  Na 2 SO 4

Sodium ferrocyanide

Fe ( CN ) 2  4 NaCN  Na 4 [ Fe ( CN ) 6 ]

Na FeCN FeCl Fe FeCN NaCl

HCl 3 [ ( )] 4 [ ( )] 12

(Prussian blue)

Ferric ferrocyanide

Sulphu

r (S.E.)

2

2 NaS  NaS

 (i) S.E. + sodium nitro

prusside

(ii)S.E+

CH (^) 3 CO 2 H ( CH 3 CO 2 ) 2 Pb

A black ppt.

(i)

(Purple)

4 5 Sodium nitroprusside

Na (^) 2 SNa 2 [ Fe ( CN ) 5 NO ] Na [ Fe ( CN ) NO. S ] or

( Violet)

Sodium thionitroprusside

Na 3 [ Fe ( ONSNa )( CN ) 5 ]

(ii) Na S CHCOO Pb PbS CHCOONa blackppt

CHCOOH 3 .

2 (^3 ) 2 2   3  

Haloge

n

( S. E .)

NaX  NaX

( X = Cl, Br, I )

S.E. HNO 3 (^)  AgNO 3

(i) White ppt soluble in aq

NH 3 confirms Cl.

(ii) Pale yellow ppt

partially soluble in aq. NH 3

confirms Br.

(iii) Yellow ppt insoluble in

aq NH 3 confirms I.

ppt

HNO NaXAgNO   AgX

3 3

soluble

3 32 Whiteppt

AgCl  2 NH ( aq )[ Ag ( NH ) ] Cl

Partially soluble

3 32 Yellowppt.

AgBr  2 NH ( aq )[ Ag ( NH ) ] Br

AgINH 3 ( aq )Insoluble

Nitroge

n and

sulphur

togethe

r

(S.E.)

NaCNS  NaCNS

with excess of Na the

thiocyanate formed

decomposes into

cyanide and sulphide.

NaCNS  2 NaNaCN

Na 2 S

As in test for nitrogen;

instead of green or blue

colour, blood red

colouration confirms

presence of N and S both.

3 NaCNS FeCl [ Fe ( SCN ) or[ Fe ( SCN )] Cl 3 NaCl

(Blood red colour)

Ferric sulphocyanide

Table : 22.3 Other methods for detection of elements

Element Test

Nitrogen Soda lime test : A pinch of an organic compound is heated strongly with soda lime ( NaOHCaO )in a

test tube. If ammonia gas evolves, it indicates nitrogen. 3 3 Acetamide

CH (^) 3 CONH 2 NaOH CHCOONa NH

CaO    .

This test is, however, not reliable since certain compounds like nitro, azo etc do not evolve

NH 3 when heated with soda lime.

Table : 22.4 Quantitative estimation of elements in organic compounds

Element (^) Method and its principle Formula

Carbon and

Hydrogen

Liebig's combustion method : In this method, a

known weight of organic compound is heated

with pure and dry cupric oxide in a steam of pure

and dry oxygen, when carbon is oxidised to

carbon dioxide while hydrogen is oxidised to

water. From the weight of CO 2 and H (^) 2 O , the

percentage of C and H can be calculated.

HO

y O xCO

y C (^) xH y x 2 2 2 4 2

   

  

  

(i) 100 44

12

Weightoforg.compound

Weightof % of

2   

CO C

(ii) 100 18

2

Weightoforg.compound

Weightof % of

2   

HO H

Nitrogen (i) Duma's method : Elemental nitrogen is

converted into molecular nitrogen by a suitable

chemical method and its voiume is changed to

STP data.

C  2 H  3 CuOCO 2  H 2 O  3 Cu

2 N  2 CuON 2  oxide of nitrogen

Oxidesofnitrogen Cu  N 2  CuO

(ii) Kjeldahl's method : Nitrogen in organic

compound is converted into NH 3 by suitable

chemical method which, in turn, is absorbed by

V 1 mL of N 1 H 2 SO 4_._

N ( from organiccompound) conc. H 2 SO 4 ( NH 4 ) 2 SO 4

( NH (^) 4 ) 2 SO 4  2 NaOH  Na 2 SO 4  2 H 2 O  2 NH 3

% of N = 100 22400

28   W

V

Where, V= volume of 2

N in nitrometer (in ml )

at NTP,

W= Weight of substance taken;

W

N V N

  

%of

Note : This method is, however, not applicable

to compounds containing nitrogen in the ring

(e.g. Pyridine, quinoline etc) and compounds

containing nitro and azo (– N = N – ) groups

since nitrogen in these compounds is not

completely converted into ( NH 4 ) 2 SO 4 during

digestion.

Halogens (^) (i) Carius method : The method is based on the

fact that when an organic compound containing

halogen ( Cl, Br, or I ) is heated in a sealed tube

with fuming nitric acid in presence of silver

nitrate, silver halide is formed. From the mass of

silver halide formed, the percentage of the

halogen can be calculated.

100 Massofsubstancetaken

Massof formed

% of   

AgCl Cl

100 Massofsubstancetaken

Massof formed

188

80 % of   

AgBr Br

100 Massofsubstancetaken

Massof formed

235

127 % of   

Agl I

(ii) Schiff's and Piria method : In this method

the accurately weighed organic compound (0.15 –

0.25 g ) is taken in a small platinum crucible with

a mixture of lime and sodium carbonate,

( CaONa 2 CO 3 ). It is now heated strongly and then

cooled and dissolved in dilute nitric acid in a

beaker. The solution is then filtered and the

halide is precipitated with silver nitrate solution.

Halogen is now calculated as in Carius method.

Sulphur Carius method : When an organic compound

containing sulphur is heated with fuming nitric

acid, sulphur is oxidised to sulphuric acid. This is

precipitated as barium sulphate by adding

barium chloride solution. From the amount of

barium sulphate, percentage of sulphur can be

calculated.

2 4

heat SHNO 3 (fuming)   HSO

H SO BaCl BaSO 2 HCl

white ppt

2 4  2  4 

100 Massofsubstancetaken

Massof formed

233

32 % of   

BaSO 4 S

phosphoro

us

Carius method : The organic compound

containing phosphorus is heated with fuming

nitric acid. Phosphorus is oxidised to phosphoric

acid. It is precipitated as magnesium ammonium

100 Massofsubstancetaken

Massof formed

222

62 % ofP  

Mg 2 P 2 O 7

phosphate, MgNH 4 PO 4 , by the addition of

magnesia mixture

( MgSO (^) 4  NH 4 OHNH 4 Cl ). The magnesium

ammonium phosphate is washed, dried and

ignited when it is converted to magnesium

pyrophosphate ( Mg 2 P 2 O 7 ).

MgNH PO MgPO NH HO

heat 2 4 4   227  2 3  2

From the mass of magnesium pyro-phosphate,

the percentage of phosphorus in the compound

can be calculated.

Oxygen (i) The usual method of determining the

percentage of oxygen in an organic compound is

by the method of difference. All the elements

except oxygen present in the organic compound

are estimated and the total of their percentages

subtracted from 100 to get the percentage of

oxygen.

(ii) Aluise's method :. Organic compound

containing oxygen is heated with graphite and

CO formed is quantitatively converted into CO 2

on reaction with I 2 O 5.

Oxygen

Pyrolysis Org. compound 

O C CO

oC 2 2

1100 2  ^ 

5 COI 2 O 5  I 2  5 CO 2

Percentage of oxygen = 100 – (Sum of the

percentages of all other elements)

g

CO

g

O CO

16 44

  2

100 massoforg.compd.

massofCO

44

16 % of

2 O   

(4) Determination of Molecular Mass : The

molecular mass of the organic compounds can be

determined by various methods.

(i) Physical methods for volatile compounds

(a) Victor Meyer's method : Molecular mass of

volatile liquids and solids can be easily determined

from the application of Avogadro hypothesis according

to which the mass of 22.4 litres or 22400 ml of the

vapour of any volatile substance at NTP is equal to the

molecular mass of the substance.

In Victor Meyer's method, a known mass of the

volatile substance is vaporised in a Victor Meyer's tube.

The vapours formed displace an equal volume of air

into a graduated tube. The volume of air collected in

graduated tube is measured under experimental

conditions. This volume is converted to NTP conditions.

Calculations : Mass of the organic substance

Wg

Let the volume of the air displaced be Vml 1

Temperature T 1 K

Pressure (after deducting aqueous tension)

p 1 mm

Let the volume at NTP be V 2 ml

Applying gas equation, 760

1

1 1 2 

T

p V V

V 2 ml of vapours weight at NTP = Wg

 22400 ml of vapour weight at NTP =

M

V

W

2

Alternatemethod : Vapour density of substance

Massof 1 mlofhydrogenatNTP

Massof 1 mlof vapoursatNTP 

or V. D.

  1. 00009

W / V 2  ( Mass of 1 ml of H 2 at

NTP

 0. 00009 g^ or^2 /^22400 )

or V. D.

2 ^0.^00009

V

W

Mol. Mass,

  1. 00009

V

W

M VD

(b) Hofmann's method : The method is applied to

those substances which are not stable at their boiling

points, but which may be volatilised without

decomposition under reduced pressure. A known mass

of the substance is vaporised above a mercury column

in a barometric tube and the volume of the vapour

formed is recorded. It is then reduced to NTP

conditions. The molecular mass of the organic

which will require 1000 ml of a normal alkali solution

for complete neutralisation can be calculated. This

mass of the acid will be its equivalent mass.

Onegramequivalentof alkali

1000 ml 1 N alkalisolu tion One gram equivalent of the

acid

Calculations : Suppose w g of the organic acid

requires V ml N 1 alkali solution for complete

neutralisation.

V ml N 1 alkali solution  wgm acid

So 1000 ml N 1 alkali solution g V N

w 1000

1

 acid

 one gram equivalent acid

Equivalent mass of the acid 1000

1

V N

w

Thus, Molecular mass of the acid = Eq. mass 

basicity

In the case of organic bases, the known mass of

the base is titrated against a standard solution of an

acid. Knowing the volume of the acid solution used, the

mass of the organic base which will require 1000 ml of

a normal acid solution for complete neutralisation can

be calculated. This mass will be the equivalent mass of

the base.

Onegramequivalentof the acid

1000 mlN acid solution One gram equivalent of the

base

Molecular mass of the base (^) Eq. mass acidity

(5) Calculation of Empirical and Molecular

formula

(i) Empirical formula : Empirical formula of a

substance gives the simplest whole number ratio

between the atoms of the various elements present in

one molecule of the substance. For example, empirical

formula of glucose is CH 2 O , i.e. for each carbon atom,

there are two H- atoms and one oxygen atom. Its

molecular formula is however, C 6 H 12 O 6.

Calculation of empirical formula : The steps

involved in the calculation are as follows,

(a) Divide the percentage of each element by its

atomic mass. This gives the relative number of atoms.

(b) Divide the figures obtained in step (i) by the

lowest one. This gives the simplest ratio of the various

elements present.

(c) If the simplest ratio obtained in step (ii) is not

a whole number ratio, then multiply all the figures with

a suitable integer i.e., 2, 3, etc. to make it simplest

whole number ratio.

(d) Write down the symbols of the various

elements side by side with the above numbers at the

lower right corner of each. This gives the empirical or

the simplest formula.

(ii) Molecular formula : Molecular formula of a

substance gives the actual number of atoms present in

one molecule of the substance.

Molecular formula = n Empirical formula

Where, n is a simple integer 1, 2, 3,...... etc. given

by the equation,

Empiricalformulamassof thecompound

Molecularmassof thecompound n

where the molecular mass of the compound is

determined experimentally by any one of the methods

discussed former, empirical formula mass is calculated

by adding the atomic masses of all the atoms present in

the empirical formula.

(iii) Molecular formula of gaseous hydrocarbons

(Eudiometry)

Eudiometry is a direct method for determination

of molecular formula of gaseous hydrocarbons without

determining the percentage composition of various

elements in it and without knowing the molecular

weight of the hydrocarbon. The actual method used

involves the following steps,

(a) A known volume of the gaseous hydrocarbon

is mixed with an excess (known or unknown volume) of

oxygen in the eudiometer tube kept in a trough of

mercury.

(b) The mixture is exploded by passing an electric

spark between the platinum electrodes. As a result,

carbon and hydrogen of the hydrocarbon are oxidised

to CO 2 and HO 2

vapours respectively.

(c) The tube is allowed to cool to room

temperature when water vapours condense to give

liquid water which has a negligible volume as

compared to the volume of water vapours, Thus, the

gaseous mixture left behind in the eudiometer tube

after explosion and cooling consists of only CO 2 and

unused O 2.

(d) Caustic potash or caustic soda solution is then

introduced into the eudiometer tube which absorbs

2

CO completely and only unused 2

O is left behind.

2 NaOHCO 2  Na 2 CO 3  H 2 O

Thus, the decrease in volume on introducing

NaOH or KOH solution gives the volume of 2

CO

formed. Sometimes, the volume of O 2 left unused is

found by introducing pyrogallol and noting the

decrease in volume.

Calculation : From the volume of CO 2 formed and

the total volume of O 2 used, it is possible to calculate

the molecular formula of gaseous hydrocarbon with the

help of the following equation.

1 vol ( / 4 )vol vol / 2 vol

( / 4 ) 2 2 / (^22)

x y x y

C (^) xHy x y O xCO y HO

   

(Negligible volume on

condensation)

From the above equation, it is evident that for one

volume of hydrocarbon,

(a) ( xy / 4 )volume of 2

O is used

(b) x volume of 2

CO is produced

(c) y/ 2 volume of H 2 O vapours is produced which

condense to give liquid H 2 O with negligible volume.

(d) Contraction on explosion and cooling

[( 1  xy / 4 ) x ] 1  y / 4

By equating the experimental values with the

theoretical values from the above combustion equation,

the values of x and y and hence the molecular formula

of the gaseous hydrocarbon can be easily determined.

(6) Determination of structure by spectroscopic

and diffraction methods : The structures of organic

substances are determined by spectroscopic and

diffraction methods.

Classification of organic compounds

Organic compounds have been classified on the

basis of carbon skeleton (structure) or functional

groups or the concept of homology.

(1) Classification based on structure

(i) Acyclic or open-chain compounds : Organic

compounds in which all the carbon atoms are linked to

one another to form open chains (straight or branched)

are called acyclic or open chain compounds. These may

be either saturated or unsaturated. For example,

Butane

CH 3 CH 2 CH 2 CH 3

Isobutane

3 |

3 3

CH

CHCHCH

1 - Butene

3 2 2

CH CHCH  CH

3, 3 - Dimethy l-1-buty ne

3

3

|

|

3

CH

CH

CHCCCH

These compounds are also called as aliphatic

compounds.

(ii) Cyclic or closed-chain compounds : Cyclic

compounds contain at least one ring or closed chain of

atoms. The compounds with only one ring of atoms in

the molecule are known as monocyclic but those with

more than one ring of atoms are termed as polycyclic.

These are further divided into two subgroups.

(a) Homocyclic or carbocyclic : These are the

compounds having a ring or rings of carbon atoms only

in the molecule. The carbocyclic or homocyclic

compounds may again be divided into two types :

Alicyclic compounds : These are the compounds

which contain rings of three or more carbon atoms.

These resemble with aliphatic compounds than

aromatic compounds in many respects. That is why

these are named alicyclic, i.e. , aliphatic cyclic. These

are also termed as polymethylenes. Some of the

examples are,

Aromatic compounds : These compounds consist

of at least one benzene ring, i.e., a six-membered

carbocyclic ring having alternate single and double

bonds. Generally, these compounds have some fragrant

odour and hence, named as aromatic ( Greek word

aroma meaning sweet smell ).

These are also called benzenoid aromatics.

Non-benzenoid aromatics : There are aromatic

compounds, which have structural units different from

benzenoid type and are known as Non-benzenoid

aromatics e.g. Tropolone, azulene etc.

Cyclopropane (^) Cyclobutane Cyclohexane

Naphthalene

(Bicyclic)

Benzene

(Monocyclic)

HO

O

Tropolone

Azulene

Table : 22.

Class Functional group Class Functional group

Olefins/Alkenes (ene) Acid halides (Alkanoyl

halids)

O

C X

||

  (Acylhalide)

Acetylenes/Alkynes

(yne)

CC ^ Amides (Alkanamides) O

C NH

||

  2 (Amide)

Alkyl Halides  F ,  Cl , Br , I (Halo) Acid anhydrides

(Alkanoic anhydrides)

O

C

O

C O

|| ||

   

(Anhydride)

Alcohols (Alkanols) – OH (Hydroxy) Esters (Alkylalkanoates) |

|

||

   C

O

C O (Ester)

Ethers (Alkoxyalkanes) (Alkoxy)

|

|

|

|

 C  O  C 

Cyanides/Nitriles

(Alkanenitrile)

CN (Cyano)

Aldehydes (Alkanals)

O

C H ||

  (Aldehydic) Isocyanides – N C (Isocyano)

Ketones (Alkanones) O

C

||

 ^ (Carbonyl)

Nitro compounds

(Nitroalkanes)

(Nitro)

Carboxylic acid

(Alkanoic acid)

O

C OH

||

^  (Carboxyl)

Amines

(Amino)

(3) Homologous series : A homologous series can

be defined as a group of compounds in which the various

members have similar structural features and similar

chemical properties and the successive members differ in

their molecular formula by 2

CH group.

Characteristics of homologous series

(i) All the members of a series can be represented

by the general formula. For example, the members of

the alcohol family are represented by the formula

C (^) n H 2 n  1 OH where n may have values 1, 2, 3..... etc.

(ii) Two successive members differ in their

formula by  CH 2 group or by 14 atomic mass units

(iii) Different members in a family have common

functional group e.g., the members of the alcohol

family have  OH group as the functional group.

(iv) The members in any particular family have

almost identical chemical properties and their physical

properties such as melting point, boiling point, density,

solubility etc., show a proper gradation with the

increase in the molecular mass.

(v) The members present in a particular series

can be prepared almost by similar methods known as

the general methods of preparation.

(4) Saturated and unsaturated compounds : If,

in an organic compound containing two or more carbon

atoms, there are only single bonds between carbon

atoms, then the compound is said to be saturated, e.g.

ethane, n- propyl alcohol, acetaldehyde etc.

Ethane

|

|

|

|

H

H

C H

H

H

H  C  

n-propy lalcohol

|

|

|

|

|

|

;

H

H

C O H

H

H

C

H

H

HC    

Acetaldehyde

|

| H

H O

H

HCC

On the other hand, if the compound contains at

least one pair of adjacent carbon atoms linked by a

multiple bond, then that compound is said to be

unsaturated , e.g, ethylene, acetylene, vinyl alcohol,

acraldehyde etc.

Acety lene

HCCH

Viny lalcohol

| |

H

C OH

H

H  C  

Acraldehy de

| |

H

C C

H

H  C  

Nomenclature of organic compounds

H

H

C = C

H

H

Ethylen

e

O

H

C = C

N

O

O

N

H

H

Nomenclature means the assignment of names to

organic compounds. There are two main systems of

nomenclature of organic compounds.

(1) Trivial system : This is the oldest system of

naming organic compounds. The trivial name was

generally based on the source, some property or some

other reason. Quite frequently, the names chosen had

Latin or Greek roots. For example,

(i) Acetic acid derives its name from vinegar of

which it is the chief constituent (Latin : acetum =

vinegar).

(ii) Formic acid was named as it was obtained

from red ants. The Greek word for the red ants is

formicus.

(iii) The names oxalic acid ( oxalus ), malic acid

( pyrus malus ), citric acid ( citrus ) have been derived

from botanical sources given in parentheses.

(iv) Urea and uric acid have derived their names

from urine in which both are present.

(v) The liquid obtained by the destructive

distillation of wood was named as wood spirit. Later

on, it was named methyl alcohol (Greek : methu =

spirit; hule = wood).

(vi) Names like glucose (sweet), pentane (five),

hexane (six), etc. were derived from Greek words

describing their properties or structures.

(vii) Methane was named as marsh gas because it

was produced in marshes. It was also named as fire

damp as it formed explosive mixture with air.

Table : 22.6 Common or trivial names of some organic compounds.

Compound Common name Compound Common name

CH 4 Methane CHCl 3 Chloroform

C 2 H 2 Acetylene CHI 3 Iodoform

H 3 CCH 2 CH 2 CH 3 n- Butane CH 3 CN Acetonitrile

( H 3 C ) 2 CHCH 3 Isobutane CH 3 COOH Acetic acid

( H 3 C ) 4 C Neopentane C 6 H 6 Benzene

HCHO Formaldehyde C 6 H 5 CH 3 Toluene

( H 3 C ) 2 CO Acetone C 6 H 5 NH 2 Aniline

CH 3 CH 2 OH Ethyl alcohol C 6 H 5 OH Phenol

CH 3 CONH 2 Acetamide C 6 H 5 OCH 3 Anisole

CH 3 OCH 3 Dimethyl ether C 6 H 5 COCH 3 Acetophenone

( CH 3 CH 2 ) 2 O Diethyl ether C 6 H 5 CONH 2 Benzamide

(2) IUPAC system : In order to rationalise the

system of naming, an International Congress of

Chemists was held in Geneva in 1892. They adopted

certain uniform rules for naming the compounds.

The system of nomenclature was named as Geneva

system. Since then the system of naming has been

improved from time to time by the International Union

of Pure and Applied Chemistry and the new system is

called IUPAC system of naming. This system of

nomenclature was first introduced in 1947 and was

modified from time to time. The most exhaustic rules

for nomenclature were first published in 1979 and later

revised and updated in 1993. The rules discussed in the

present chapter are based on guide books published by

IUPAC in 1979 ( Nomenclature of Organic Chemistry

by J. Rigandy and S.P. Klesney) and 1993 ( A Guide to

IUPAC Nomenclature for Organic Chemistry by R.

Panico, W.H. Powell and J.C. Richer). With the help of

this system, an organic compound having any number

of carbon atoms can be easily named.

IUPAC System of Naming Organic Compounds :

In the IUPAC system, the name of an organic compound

consist of three parts : (i) Word root (ii) Suffix (iii)

Prefix

(i) Word root : The word root denotes the number

of carbon atoms present in the chain.

  • Cl Chloro
  • Br Bromo
  • I Iodo
  • NO Nitroso
  • N = N – Diazo
  • OCH 3 Methoxy
  • OC 2 H 5 Ethoxy
  • NO 2 Nitro
  • NH 2 Amino
  • OH Hydroxo

Thus, a complete IUPAC name of an organic

compound may be represented as:

Prefix + word root + Primary suffix + Secondary

suffix

Word root : Pent (five C – C – C – C – C )

Primary suffix : ene (double bond at C – 2)

Secondary suffix : oic acid (– COOH group)

Prifix : Bromo (– Br group at C – 4)

IUPAC name : Bromo + pent + ene + oic acid or 4-

Bromopent - 2 - en- 1 - oic acid

Classification of carbon atoms in organic

compounds

The carbon atoms in an alkane molecule may be

classified into four types as primary (

o ), secondary

o ), tertiary (

o ) and quaternary (

o ). The carbon

atoms in an organic compound containing functional

group can be designated as , , , .

3

1

|

3

3 1

3

1

3

1

|

|

4 2

2 3

1

CH

CH CH

CH

CH

CH CH C

o

o o

o

o

o o o    

CH  CH  CH  CH  OH

   

3 2 2 2

CH CH CHO

CH

CH  CH   

   

2 2

3

|

3

Alkyl groups

These are univalent groups or radicals obtained

by the removal of one hydrogen atom from a molecule

of a paraffin. The symbol ' R ' is often used to represent

an alkyl group.

( ) ( )

(Alkane) 2 2 2 1 (Alkylgroup)

 

   

 

R H R

C H CnHn

H n n

Alkyl groups are named by dropping-ane from the

name of corresponding paraffin and adding the ending–

yl.

Parent saturated

hydrocarbon

Name of

the alkyl

group

Structure

Methane Methyl CH 3 –

Ethane Ethyl CH 3 – CH 2 –

Propane n- Propyl CH 3 – CH 2 – CH 2 –

Butane n - Butyl CH 3 – CH 2 – CH 2 – CH 2

Alkyl groups derived from saturated hydrocarbons

having three or more carbon atoms exist in isomeric

forms.

Similarly, removal of different H atoms in

pentane gives the following radicals :

nPentyl

CH CHCHCH CH

3 2 2 2 2

Isopenty l

3

3 2 2 |

CH

CH CHCH CH ;

Neopenty l

3

3

3 2

|

CH

CH

CH CCH  ;

secPenty l

3 2 2 3 |

CHCHCHCH CH ;

Penty l

3

3 2 3

CH

|

|

tert

CHCCH CH

Table : 22.11 Unsaturated groups or radicals

Group Common

name

IUPAC name

CH 2  CH  Vinyl Ethenyl

2 1

CH 2 CH CH

Allyl 2 - Propenyl

CH  CH  CH 

1

3

  • 1 - Propenyl

HCC  Acetylide Ethynyl

1

2

2

HC C CH

Propargyl 2 - Propynyl

General rules for naming organic compounds

In the common system, all the isomeric alkanes

(having same molecular formula) have the same parent

name. The names of various isomers are distinguished

by prefixes. The prefix indicates the type of branching

in the molecule. For example,

Prefix (^) Pri.

suffix

OH

O

CH CH C

Br

CHCH    

3 2 1 ||

|

5 4

3

Sec.

suffix

Functional

group

Butane

3 2 2 3

n

CHCHCHCH

CH 3 CH 2 CH 2 CH 2  n Butyl

CH  s ec .Butyl 3

3 2

CH

CHCH

(1) Prefix n- ( normal ) is used for those alkanes in

which all the carbon atoms form a continuous chain

with no branching.

Butane

3 2 2 3 n

CHCHCH CH^ ;

Pentane

3 2 2 2 3 n

CHCHCHCH CH

(2) Prefix iso is used for those alkanes in which

one methyl group is attached to the next-to-end carbon

atom (second last) of the continuous chain.

Isobutane

3

|

3 3

CH

CHCHCH

Isopentane

3

|

3 2 3

CH

CH  CH  CHCH

Isohexane

3

|

3 2 2 3

CH

CHCHCHCHCH

(3) Prefix neo is used for those alkanes which

have two methyl groups attached to the second last

carbon atom of the continuous chain.

Neopentane

3

3 |

|

3 3

CH

CH

CHCCH

Neohexane

3

3 |

|

3 2 3

CH

CH

CH  C  CH  CH

IUPAC system of nomenclature of complex

compounds

The naming of any organic compound depends on

the name of normal parent hydrocarbon from which it

has been derived. IUPAC system has framed a set of

rules for various types of organic compounds.

(1) Rules for Naming complex aliphatic

compounds when no functional group is present

(saturated hydrocarbon or paraffins or Alkanes)

(i) Longest chain rule : The first step in naming

an organic compound is to select the longest continuous

chain of carbon atoms which may or may not be

horizontal (straight). This continuous chain is called

parent chain or main chain and other carbon chains

attached to it are known as side chains (substituents).

Examples :

3

3 |

|

2 3

3

|

3 2 2

CH

CH

C CH CH

CH

CHCHCHCH   

If two different chains of equal length are

possible, the chain with maximum number of side

chains or alkyl groups is selected.

(ii) Position of the substituent : Number of the

carbon atoms in the parent chain as 1, 2, 3,....... etc.

starting from the end which gives lower number to the

carbon atoms carrying the substituents. For examples,

A(Correct)

5 4 3 2 | 1

C C C C C

X

B(Wrong)

1 2 3 4 | 5

C C C C C

X

The number that indicates the position of the

substituent or side chain is called locant.

2 Methy lpentane

3

|

3

2 1

2

3

2

4

3

5

CH

CH CH CH CH CH

3 Ethy lhexane

2 2 3

4 | 5 6

3

3

2

2

3

1

CH CH CH

CH CH CH CH CH

(iii) Lowest set of locants : When two or more

substituents are present, then end of the parent chain

which gives the lowest set of the locants is preferred

for numbering.

This rule is called lowest set of locants. This

means that when two or more different sets of locants

are possible, that set of locants which when compared

term by term with other sets, each in order of

increasing magnitude, has the lowest term at the first

point of difference. This rule is used irrespective of the

nature of the substituent. For example,

Set oflocants:2,3, 5 (Correct)

3

|

3

2 1

3

|

3

2

4

3

|

6 5

3

CH

CH CH

CH

CH CH

CH

H C  CH    

Set oflocants:2,4, 5 (Wrong)

3

|

3

5 6

3

|

4

2

3

3

|

1 2

3

CH

CH CH

CH

CH CH

CH

H C  CH    

The correct set of locants is 2, 3, 5 and not 2, 4, 5.

The first set is lower than the second set because at the

first difference 3 is less than 4. (Note that first locant

is same in both sets 2; 2 and the first difference is with

the second locant 3, 4. We can compare term by term as

2 - 2, 3 - 4 (first difference), 5 - 5. Only first point of

difference is considered for preference. Similarly for

the compounds,

3

|

3

2 1

2

3

3

|

2

4

2

5

2

7 6

3

|

8

2

9

3

10

CH

CH CH CH

CH

CH CH CH CH

CH

C HCHCH       

Set of locants : 2, 7, 8 (Correct)

3

|

3

9 10

2

8

3

|

2

7

2

6

2

4 5

3

|

3

2

2

3

1

CH

CH CH CH

CH

CH CH CH CH

CH

C HCHCH       

Set of locants : 3, 4, 9 (Wrong)

First set of locants 2, 7, 8 is lower than second set

3, 4, 9 because at the first point of difference 2 is lower

than 3.

Lowest sum rule : It may be noted that earlier,

the numbering of the parent chain containing two or

more substituents was done in such a way that sum of

Substituen

ts

Substituen

ts

Parent

chain

3

3

3

2

| (^12)

3

3

|

3

2 3

2

| 1

3

1

2

2

2

3

2

| (^54)

2

6

2

7

2

8

3

9

CH

CH C CH CH

CH

CH CH CH

CH CH CH CH C CH CH CH CH

  

 

       

The substituent dimethyl is cited first because it

is alphabetized under d. Similarly,

3

3

10

2

9

2

| 7 8

2 5 3

2

6

2

3 | 4 | 5

2

2

3

1

CH

CH CH CH CH

CH CH

C HCHCHCCHCH    

When the names of two or more complex

substituents are composed of identical words, priority

for citation is given to the substituent which has lowest

locant at the first cited point of difference within the

complex substituent. For example,

3

1

2

2

2

3

2

| (^54)

2

7 6

2

8

2

9

2

10

2

11

3

12

C HCHCHCHCHCHCHCHCHCHCHC H

  • 5(1-methyl butyl)- 7 - (2-methyl butyl) dodecane

The substituent (1-methylbutyl) is written first

because it has lower locant than the substituent (2-

methylbutyl).

When the same complex substituent (substituted

in the same way) occurs more than once, it is indicated

by the multiplying prefix bis (for two), tris (for three),

tetra kis (for four) etc.

3

|

3

2 1

2

3

2

| 5 4

2

6

2

7

2

8

2

9

3

10

CH

C HCHCHCHCHCCHCHCHCH

(viii) Cyclic hydrocarbons : These compounds

contain carbon chain skeletons which are closed to

form rings. The saturated hydrocarbons with ring of

carbon atoms in the molecule are called cycloalkanes.

These have the general formula Cn H 2 n.

The cyclic compound is named by prefixing cyclo

to the name of the corresponding straight chain alkane.

For example,

If side chains are present, then the rules given in

the previous section are applied. For example,

When more than one side chains are present, the

numbering is done beginning with one side chain so

that the next side chain gets the lower possible

number. For example,

When a single ring system is attached to a single

chain with a greater number of carbon atoms or when

more than one ring system is attached to a single chain,

then it is named as cycloalkylalkanes. For example,

Complex

substituent

( 2-methylpropyl) 5 - (1, 1-Dimethylpropyl) – 5 - (2-methylpropyl)

nonane

Complex

substituent

( 1, 1-

dimethylpropyl)

2 - (^3) methylbutyl

|

3

4

2

2 3

2

| 1

CH

C HCHCHCH

3

4

2

3

2

2

3 | 1

CH CH C H

CH

C H   

1 -

methylbutyl

Complex

substituent

( 1, 1-

4 - (1, 1-Dimethylpropyl) – 3 - ethyl dimethylpropyl)-4, 7-

dimethyldecane

3

3

| 2

| 1

3 CH

CH

H CC

1,

dimethylpropyl

5, 5-Bis (1, 1-dimethylpropyl)- 2 -

methyldecane

1,

dimethylpropyl

3

3

| 1

2

2

3

3

CH

C H  CH  C  CH

3

|

3

1 |

2

2

3

3

CH

C H  CH  C  CH

C 3 H 6 ,

Cyclopropane

C 4 H 8 , Cyclobutane

C 5 H 10 ,

Cyclopentane

C 6 H 12 ,

Cyclohexane

CH 3

Methylcyclohexa

ne

C 2 H 5

Ethylcyclopentan

e

 CH 2  CH 2  CH 2 

CH 3

1, 3 - Bis (2-methylcyclopropyl) CH 3

propane

Cyclohexyl

cyclohexane

CH 2 CH 3

2

3

CH 3 CH 3

1

3 - Ethyl-1, 1-dimethylcyclohexane

(Not 1 - Ethyl-3,3-

dimethylcyclohexane)

CH 3

CH 3

1.3-

Dimethylcyclobutane

1 2

(^43)

CH 2 CH 2 CH 3

6 2

3

4

5

CH 3

1 - Methyl- 3 - propylcyclohexane

(Not 5 - Methyl- 1 - propyl

cycloalkane)

1

3

4

2

3

2

2

2

1

C HCHCHC H

1 - Cyclopropyl butane

 CH 2  CH 2  CH 2  CH 2  CH 3

1 -

Cyclobutylpentane

In case of substituted cycloalkenes, the double

bond is given the lowest possible number and

numbering is done in such a way that the substituents

get the lowest number.

According to the IUPAC system of Nomenclature,

certain trivial or semi- systematic names may be used

for unsubstituted radicals. For example, the following

names may be used,

( CH (^) 3 ) 2 CH ^ Isopropyl

3

|

3 2

CH

CHCHCHSec - Butyl

( CH 3 ) 2 CHCH 2  CH 2  Isopentyl

3

3 |

|

3 2

CH

CH

CHCHC

tert - Pentyl

( CH 3 ) 2 CHCH 2  Isobutyl

( CH (^) 3 ) 3 C  tert-Butyl

( CH 3 ) 3 CCH 2  Neopentyl

( CH (^) 3 ) 2 CHCH 2  CH 2  CH  Isohexyl

However, when these are substituted, these

names cannot be used as such. For example,

3 |

2

3 , 3 Diethy l 5 isopropy l 4 methy loctane

2 3

|

3

1

2

| 3 2

3

|

4

3

|

3

|

5

2

6

2

7

3

8

CH

CH

CH CH

C CH CH

CH

CH

CH

CH CH

CH CH CH CH

   

   

  

It may be noted that while writing the

substituent's name in alphabetical order, the prefixes

iso - and neo- are considered to be part of the

fundamental name. However, the prefixes sec - and tert-

are not considered to be the part of the fundamental

name.

(2) Rules for IUPAC names of polyfunctional

organic compounds

Organic compounds which contain two or more

functional groups are called polyfunctional compounds.

Their IUPAC names are obtained as follows,

(i) Principal functional group : If the organic

compound contains two or more functional groups, one

of the functional groups is selected as the principal

functional group while all the remaining functional

groups (also called the secondary functional groups)

are treated as substituents. The following order of

preference is used while selecting the principal

functional group.

Sulphonic acids > carboxylic acids > anhydrides >

esters > acid chlorides > acid amides > nitriles >

aldehydes > ketones > thiols > alcohols >alkenes >

alkynes.

All the remaining functional groups such as halo

(fluoro, chloro, bromo, iodo), nitroso (– NO ), – nitro (–

NO 2 ), amino (– NH 2 ) and alkoxy (– OR ) are treated as

substituents.

Table : 22.

Order of

preference

Preflx Suffix (ending)

  • SO 3 H Sulpho Sulphonic acid
  • COOH Carboxy – oic acid
  • COOR Alkoxy carbonyl Alkyl alkanoate
  • COX Haloformyl Oyl halide
  • CONH 2 Carbamoyl – amide
  • CN Cyano – nitrile
  • CHO Formyl – al

C = O Keto – one

  • OH Hydroxy – ol
  • NH 2 Amine – amine

C = C – – ene

  • CC – – – yne
  • O – Epoxy –
  • X Halo –
  • NO 2 Nitro –

(ii) Selecting the principal chain : Select the

longest continuous chain of carbon atoms containing

the principal functional group and maximum number of

secondary functional groups and multiple bonds, if any.

(iii) Numbering the principal chain : Number the

principal chain in such a way that the principal

functional group gets the lowest possible number

followed by double bond and triple bond and the

substituents, i.e.

Principal functional group > double bond > triple

bond > substituents

(iv) Alphabetical order : Identify the prefixes and

the positional numbers (also called locants) for the

secondary functional groups and other substituents and

place them in alphabetical order before the word root.

CH 3

CH 3

3

2

1

2, 3 - Dimethylcyclopent-

1ene

CH 3

2

1

3 - Methylcyclohex- 1 - ene