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Published by rajusingh79, 2019-08-09 08:38:11

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

Free Flip-Book Chemistry Class 11th by Study Innovations. 534 Pages

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Note :Fractional distillation can not be used to separate a mixture of liquids which form azeotropes, i.e.,

constant boiling mixtures.

(vi) Distillation under reduced pressure (vacuum distillation) : In this process, a liquid is distilled much
below its boiling point. It is of special advantage for the purification of liquids which decompose near 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 Hg pressure. Similarly, sugarcane juice is concentrated in sugar industry by
evaporation under reduced pressure.

(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 presence
of steam. Other compounds which can be purified by steam distillation are: nitrobenzene, bromobenzene,
o-nitrophenol, salicylaldehyde, o-hydroxyacetophenone, essential oils, turpentine oil etc.

(viii) Azeotropic distillation : Azeotropic mixture is a mixture having constant boiling point. The most
familiar example is a mixture of ethanol and water in the ratio of 95.87 : 4.13 (a ratio present in rectified spirit). It
boils at 78.13oC. The constituents of an azeotropic mixture can't be separated by fractional distillation. Hence a
special type of distillation (azeotropic distillation) is used for separating the constituents of an azeotropic mixture.

In this method a third compound is used in distillation. The process is based on the fact that dehydrating
agents like C6 H6, CCl4 , diethyl ether etc. depress the partial pressure of one of the original components. As a

result, the boiling point of that component is raised sufficiently and thus the other component will distil over.

Dehydrating agents having low boiling point (e.g. C6 H6, CCl4, ether) depress the partial pressure of alcohol

more than that of water; on the other hand, dehydrating agents having high boiling point (glycerol, glycol) depress
the partial pressure of water more than that of alcohol.

(ix) Chromatography : This is a modern method used for the separation of mixtures into its components,
purification of compounds and also to test the purity of compounds. The name chromatography is based on the
Greek word chroma meaning colour and graphy for writing because the method was first used for the separation of
coloured substances found in plants. This method was described by Tswett in 1906.

(a) Principle of chromatography : The technique of chromatography is based on the difference in the rates at
which the components of a mixture move through a porous medium (called stationary phase) under the influence of
some solvent or gas (called moving phase). Thus, this technique consists of two phases- one of these is a stationary
phase of large surface area while the second is a moving phase which is allowed to move slowly over the stationary
phase. The stationary phase is either a solid or a liquid while the moving phase may be a liquid or a gas.

(b) Types of chromatography : Depending upon the nature of the stationary and the mobile phases, the
different types of chromatographic techniques commonly used are,


Type of Chromatography Mobile/Stationary Phase Uses

Adsorption or column Liquid/Solid Large scale separations

chromatography

Thin-layer chromatography Liquid/Solid Qualitative analysis (identification and
characterization of organic compounds)

High performance liquid Liquid/Solid Qualitative and quantitative analysis

chromatography

Gas-liquid chromatography (GLC) Gas/Liquid Qualitative and quantitative analysis

Paper or Partition chromatography Liquid/Liquid Qualitative and quantitative analysis of polar
organic compounds (sugars, α-amino acids and
inorganic compounds)

Note :Rf value (retention factor) : The movement of the substances relative to the solvent is expressed in

terms its retention factor ( R f value). This gives the relative adsorption of each component of the mixture.

Rf = Distance moved by the substance from the base line
Distance moved by the solvent from the base line

It is constant for a given substance (component) under a given set of conditions. Therefore, it is possible to
identify the various components by determining their R f values.

It is also possible to estimate the components quantitatively by measuring the intensity of colours developed
by them on reacting with suitable reagents.

It is determined in Thin layer chromatography (TLC) and Ascending paper chromatography.

(x) Differential extraction : This method is used for the separation of an organic compound (solid or
liquid) from its aqueous solution by shaking with a suitable solvent (e.g. ether, benzene, chloroform, carbon
tetrachloride etc.) in a separating funnel. The solvent selected should be immiscible with water but should dissolve
the organic compound to an appreciable extent.

It is important to note that extraction is more efficient (i.e., more complete) when a given volume of the
extracting solvent is used in several installments than if all the volume is used in one installment.

This method is normally applied to nonvolatile compounds. For example, benzoic acid can be extracted from
its water solution using benzene.

(xi) Chemical methods : Besides these physical methods, a number of chemical methods have also been
used to separate a mixture of organic compounds. These methods are based upon the distinguishing chemical
properties of one class of organic compounds from the others. For example,

(a) Phenols can be separated from carboxylic acids on treatment with an aqueous solution of NaHCO3 .
Since acids dissolve in NaHCO3 solution evolving CO2 but phenols usually do not react.

(b) Destructive distillation of wood gives pyroligneous acid which contains acetic acid (10%), 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 1o, 2o and 3o 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 2SO4 . 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 2SO4 in a separating funnel, thiophene undergoes sulphonation to form thiophene-
2-sulphonic acid which dissolves in conc. H 2SO4 while benzene does not.

+ H 2SO4 Room temp → + H2O
S (Conc.)
S SO3 H
Thiophene Thiophene-2-sulphonic acid

(Dissolves in conc. H2SO4 )

After this treatment, the benzene layer is removed, washed with water to remove unreacted H 2SO4 , dried

over anhyd. CaCl2 and then distilled to give pure benzene.

(e) Absolute alcohol from rectified spirit : The rectified spirit (ethanol : H2O,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 100oC are dried by
keeping them in steam or oven maintained at 110oC. Substances, which decompose on heating are dried by
keeping them in a vacuum desiccator containing a suitable dehydrating agent like fused CaCl2 , conc. H2SO4 ,

P4 O10 , solid KOH or NaOH, etc.

(2) For liquids : Organic liquids are generally dried by keeping them over night in contact with a dehydrating
(desicating) agent which does not react chemically with the liquid to be dried. Commonly used dehydrating agents
are quick lime, anhydrous CaCl2 , fused CuSO4 or CaSO4 , KOH , 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.

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.

Note :Boiling point is not as reliable a test of purity as is the melting point for the solids. There are many

liquids which are miscible with other liquids and mixtures have fixed boiling points (azeotrope). Thus
other physical properties are being used for deciding the purity.

(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 seldom absent. On heating

the organic compound with dry cupric oxide when carbon is oxidized to CO2 and hydrogen to H 2O . CO2 is
detected by lime water which turns milky while H 2O is detected by anhydrous CuSO4 (white) which turns it blue.
This method is known as copper oxide test.

C+ 2CuO Heat → CO2 + 2Cu ; Ca(OH)2 + CO2 → CaCO3 + H2O
Lime water Milky

H 2 + CuO Heat → H 2O + Cu ; CuSO4 + 5H 2O → CuSO4 .5H 2O
Colouriess Blue
(Anhydrous) (Hydrated)

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 is fused with dry sodium in a fusion-
tube and fused mass after extraction with H 2O 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 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.

Lassaigne method (Detection of elements)

Element Sodium Extract (S.E.) Confirmed Test Reaction

Nitroge Na + C + N ∆ → NaCN S.E.+ FeSO4 + NaOH , boil 2NaCN + FeSO4 → Fe(CN)2 + Na2SO4
n (S.E.) and cool + FeCl3 + conc.HCl
Blue or green colour
Fe(CN)2 + 4 NaCN → Na4[Fe(CN)6]
3Na4[Fe(CN)6] + 4FeCl3 HCl → Fe4[Fe(CN)6]3 + 12NaCl

Prussian blue

Sulphur 2Na + S ∆ → Na2S (i) S.E. + sodium nitro (i) Na2S + Na2[Fe(CN)5 NO] → Na4[Fe(CN)5 NO.S]
(S.E.) prusside deep violet
(ii)S.E+
(ii) Na2S + (CH3COO)2 Pb CH3COOH → PbS ↓+ 2CH3COONa
CH3CO2H + (CH3CO2)2 Pb
black ppt.
A black ppt.


Haloge Na + Cl ∆ → NaCl S.E. +HNO3 + AgNO3 NaX + AgNO3 HNO3 → AgX ↓
n ppt
(S.E.) (i) White ppt soluble in aq
Nitroge NH3 confirms Cl AgCl + 2NH3(aq) → [Ag(NH3)2]Cl
n and Na + C + N + S ∆ → NaCNS White ppt soluble
sulphur (ii) Yellow ppt partially
togethe (S.E.) soluble in aq. NH3 confirms NaCNS + FeCl3 → [Fe(CNS)]Cl2 + NaCl
r Br .
(iii) Yellow ppt insoluble in blood red colour
aq NH3 confirms I

As in test for nitrogen;
instead of green or blue
colour, blood red colouration
confirms presence of N and
S both.

Other methods for detection of elements

Element Test
Nitrogen
Sulphur Soda lime test : A pinch of an organic compound is heated strongly with soda lime (NaOH + CaO) in a test
Halogens
tube. If ammonia gas evolves, it indicates nitrogen. CH 3CONH 2 + NaOH CaO → CH 3COONa + NH 3 . This test
Phosphorus Acetamide

is, however, not reliable since certain compounds like nitro, azo etc do not evolve NH3 when heated with
soda lime.

Oxidation test : Sulphur can also be tested by oxidation test. The organic compound is fused with fusion
mixture (a mixture of sodium carbonate and potassium nitrate). The sulphur, if present in the organic

compound, is oxidised to sodium sulphate. Na2CO3 + S + 3O → Na2SO4 + CO2 . The fused mass is

dissolved in water and the solution is acidified with hydrochloric acid. Barium chloride solution is then
added. The formation of a white precipitate indicates the presence of sulphur.

Na2SO4 + BaCl 2 → BaSO4 + 2NaCl .
(White ppt.)

Beilstein's test (copper wire test) : A clean copper wire is heated in the Bunsen flame till it does not impart
any green colour to the flame. The heated end is dipped in the organic compound and heated again. The
appearance of a green or bluish green flame due to the formation of volatile cupric halides indicates the
presence of some halogen in the organic compound. Though this test is very sensitive yet it does not
confirm the presence of halogens in an organic compound since certain organic compounds like urea,
thiourea, pyridine, organic acids etc. Which do not contain halogens give this test due to the formation of
volatile cupric cyanide. It does not tell as to which halogen is present.
Special test for bromine and iodine (layer test) : Boil a portion of the Lassaigne's extract with nitric acid.
Add a few drops of CS2 and then add chlorine water slowly with constant shaking.

An orange colouration in CS2 layer confirms the presence of bromine where as a violet colouration in the

layer confirms the presence of iodine. 2NaBr + Cl 2 → 2NaCl + Br2 ; 2NaI + Cl 2 → 2NaCl + I 2
tums CS2 layer orange turns CS2 layerviolet

Phosphorus is detected by fusing the organic compound with sodium peroxide when phosphorus is

converted into sodium phosphate. 2P + 5Na2O2 → 2Na3 PO4 + 2Na2O . The fused mass is extracted with
H 2O , boiled with conc. HNO3 and then ammonium molybdate is added. Appearance of yellow ppt. or
colouration due to the formation of ammonium phosphomolybdate indicates the presence of phosphorus.

Na3 PO4 + 3HNO3 ∆ → H 3 PO4 + 3NaNO3


H 3 PO4 + 12(NH 4 )2 MoO4 + 21HNO3 →(NH 4 )3 PO4 .12MoO3 + 21NH 4 NO3 + 12H 2O
Amm. molybdate Amm. Phosphomolybdate
(yellow ppt.)

Oxygen There is no satisfactory qualitative method for the detection of oxygen. However, its presence can be
inferred indirectly.

(i) If the organic compound is heated alone in a dry test tube in presence of nitrogen, the formation of water
drops on cooler parts of the tube may indicate the presence of oxygen.

(ii) The presence of oxygen can be inferred by testing the presence of functional groups known to contain
oxygen, e.g., hydroxyl (–OH), aldehydic (–CHO), carboxyl (–COOH), etc.

(3) Quantitative analysis (Estimation of Elements) : After qualitative analysis of elements, the next step in
the determination of molecular formula of an organic compound is the estimation of various elements by mass, i.e.
finding the percentage composition of the substance by mass. The various methods commonly employed for the
estimation of principal elements are discussed in table.

Quantitative estimation of elements in organic compounds

Element Method and its principle Formula

Carbon and Liebig's combustion method : In this method, a (i) % of C = Weight of CO2 × 12 × 100
Hydrogen known weight of organic compound is heated with Weight of org. compound 44
pure and dry cupric oxide in a steam of pure and
dry oxygen, when carbon is oxidised to carbon (ii) % of H = Weight of H 2O 2 × 100
dioxide while hydrogen is oxidised to water. From Weight of org. compound × 18
the weight of CO2 and H 2O , the percentage of C

and H can be calculated.

Cx Hy +  x + y O2 ∆ → xCO2 + y H2O
 4 2

Nitrogen (i) Duma's method : Elemental nitrogen is converted Oxides of nitrogen + Cu → N 2 + CuO
Halogens into molecular nitrogen by suitable chemical
method and its voiume is changed to STP data. % of N = 28 × V × 100
22400 W
y z
Cx Hy Nz + CuO → xCO2 + 2 H2O + 2 N2 + (Cu) Where, V= volume of N 2 in nitrometer (in ml) at
NTP, W= Weight of substance taken.

(ii) Kjeldahl's method : Nitrogen in organic % of N = 1.4 × N1 × V1
compound is converted into NH3 by suitable W
chemical method which, in turn, is absorbed by
V1mL of N1H 2SO4 . Note : This method is, however, not applicable to
compounds containing nitrogen in the ring (e.g.
N(from organic compound) + conc. H 2SO4 ∆ →(NH4 )2 SO4 Pyridine, quinoline etc) and compounds containing
nitro and azo (– N = N –) groups since nitrogen in
(NH4 )2 SO4 + 2NaOH → Na2SO4 + 2H 2O + 2NH3 these compounds is not completely converted into
(NH 4 )2 SO4 during digestion.

(i) Carius method : The method is based on the fact % of Cl = 35.5 × Mass of AgCl formed × 100
that when an organic compound containing halogen 143.5 Mass of substance taken
(Cl, Br, or I) is heated in a sealed tube with fuming
nitric acid in presence of silver nitrate, silver halide % of Br = 80 × Mass of AgBr formed × 100
is formed. From the mass of silver halide formed, 188 Mass of substance taken
the percentage of the halogen can be calculated.
% of I = 127 × Mass of Agl formed × 100
235 Mass of substance taken


(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, (CaO + Na2CO3 ) . 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 % of S= 32 × Mass of BaSO4 formed (W1 ) × 100
phosphorous containing sulphur is heated with fuming nitric 233 Mass of substance taken (W)
acid, sulphur is oxidised to sulphuric acid. This
Oxygen is precipitated as barium sulphate by adding
barium chloride solution. From the amount of
barium sulphate, percentage of sulphur can be
calculated.

S + HNO3 (fuming) heat → H 2SO4

H 2SO4 + BaCl2 → BaSO4 + 2HCl

Carius method : The organic compound % of P = 62 × Mass of Mg 2 P2 O7 formed (W1 ) × 100
containing phosphorus is heated with fuming 222 Mass of substance taken (W)
nitric acid. Phosphorus is oxidised to phosphoric
acid. It is precipitated as magnesium ammonium
phosphate, MgNH4 PO4 , by the addition of

magnesia mixture

(MgSO4 + NH 4 OH + NH 4 Cl ). The magnesium

ammonium phosphate is washed, dried and
ignited when it is converted to magnesium
pyrophosphate (Mg2 P2O7 ) .

2MgNH 4 PO4 heat → Mg 2 P2O7 + 2NH 3 + H 2O

From the mass of magnesium pyro-phosphate,
the percentage of phosphorus in the compound
can be calculated.

(i) The usual method of determining the Percentage of oxygen = 100 – (Sum of the percentages
percentage of oxygen in an organic compound is of all other elements)
by the method of difference. All the elements
except oxygen present in the organic compound O≡ CO ≡ C44Og2
are estimated and the total of their percentages 16 g
subtracted from 100 to get the percentage of
oxygen. % of O = 16 × mass of CO 2 × 100
44 mass of org. compd.
(ii) Aluise's method :. Organic compound
containing oxygen is heated with graphite and
CO formed is quantitatively converted into CO2
on reaction with I2O5.

Org. compound Pyrolysis → Oxygen

O2 + 2C 1100oC → 2CO

5CO + I 2O5 → I 2 + 5CO2

Example : 1 The percentage of N 2 in urea is about [KCET (Med.) 2001]
Solution: (c)
(a) 18.05 (b) 28.29 (c) 46.66 (d) 85.56

Urea (NH 2CONH 2 ) has molecular weight 60 and weight of nitrogen is 28


In 60gm. of urea nitrogen present = 28gm.

In 100gm. of urea nitrogen present = 28 × 100 = 46.66%
60

Example : 2 58 ml. of N H 2 SO4 are used to neutralize ammonia given by 1g of organic compound. Percentage of
5

nitrogen in the compound is

(a) 34.3 (b) 82.7 (c) 16.2 (d) 21.6

Solution: (c) % of N = 1.4 × Normality of acid × Volume of acid = 1.4 × 1 × 58 = 16.2
Example : 3 Mass of substance 1× 5

0.2595g of an organic substance in a quantitative analysis yielded 0.35g of the barium sulphate. The

percentage of sulphur in the substance is [CPMT 2000; AFMC 2001]

(a) 18.52g (b) 182.2g (c) 17.5g (d) 175.2g

Solution: (a) % of S = 32 × wt. of wt.of BaSO4 × 100 = 32 × 0.35 × 100 = 18.52%
233 organic compound 233 0.2595

Example : 4 0.2g of an organic compound on complete combustion produces 0.18g of water, then the percentage of

hydrogen in it is [Pb. CET 1988]

(a) 5 (b) 10 (c) 15 (d) 20

Solution: (b) % of H= wt. of wt. of H 2O × 2 × 100 = 0.18 × 2 × 100 = 10
organic compound 18 0.2 18

Example : 5 If 0.2g of an organic compound containing carbon, hydrogen and oxygen on combustion, yielded 0.147g

carbon dioxide and 0.12g water. What will be the content of oxygen in the substance [AFMC 1998]

(a) 73.29% (b) 78.45% (c) 83.23% (d) 89.50%

Solution: (a) %C = 12 × 0.147 × 100 = 20.045
44 0.2

%H = 2 × 0.12 × 100 = 6.666
18 0.2

∴% O = (100 − 20.045 − 6.666) = 73.289 = 73.29 (approx)

Example : 6 0.25g of an organic compound gave 31.1ml., of N 2 by Duma's method. Calculate the % of N in this

compound. [MP PET 1997]

(a) 16 (b) 18.32 (c) 45.46 (d) 15.55

Solution: (d) % N (By Duma's method) = 28 × Volume of N 2 at NTP × 100 = 28 × 31.1 × 100 = 15.55%
22400 × weight of compound 22400 × 0.25

Example : 7 0.15g of an organic compound gave 0.12g of AgBr by Carius method. The percentage of bromine in the

compound is

(a) 20 (b) 10 (c) 30 (d) 34

Solution: (d) % of Br = 80 × Mass of AgBr formed × 100 = 80 × 0.12 × 100 = 34 (Approx.)
188 Mass of substance taken 188 0.15

Example : 8 0.50g of an organic compound was Kjeldahlised and the NH 3 evolved was absorbed in a certain volume of
1N H 2SO4 . The residual acid required 60 Cm3 of N/2 NaOH. If the percentage of N is 56 then the volume of
1N H 2SO4 taken was

(a) 30ml. (b) 40ml. (c) 50ml. (d) 60ml.

Solution: (c) Let the vol. of 1NH2SO4 taken = V of ml.
Now 60ml. of N / 2 NaOH = 30 ml. of 1N NaOH = 30ml. of 1N H 2SO4

Thus, vol. of acid left unused = 30 ml. of 1N H 2SO4


∴ vol. of 1N H 2SO4 used = (V – 30) ml.

Now % of N is given by the relation, % of N = 1.4 × N1 ×V or 56 = 1.4 × 1 × (V − 30) or V = 50 ml.
W 0.50

(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 22400ml 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 = W g

Let the volume of the air displaced be = V1 ml ; Temperature = T1K
Pressure (after deducting aqueous tension) = p1mm

Let the volume at NTP be = V2 ml
Applying gas equation,

V2 = p1 × V1 × 273
T1 760

22400 ml of vapours weight at NTP = M (mol. mass); V2 ml of vapours weight at NTP = Wg

22400 ml of vapour weight at NTP = W × 22400 = M
V2

or Vapour density of substance = Mass of 1 ml of vapours at NTP
Mass of 1 ml of hydrogen at NTP

or V. D. = W / V2 ( Mass of 1 ml of H 2 at NTP = 0.00009 g or 2 / 22400 )
0.00009

or V. D. = V2 W ; Mol. Mass, M = 2× V .D. = V2 2W
× 0.00009 × 0.00009

(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 substance can be calculated by the application
of following relationship,

Mol. Mass = Mass of the substance × 22400
volume of the vapours at NTP

(ii) Physical methods for Non-volatile substances : The molecular mass of a non-volatile organic
compound can be determined by noting either the elevation in boiling point of the solvent (Ebullioscopic method)
or the depression in freezing point of the solvent (Cryoscopic method) produced by dissolving a definite mass of the


substance in a known mass of the solvent. The molecular mass of the compound can be calculated from the
following mathematical relationships :

(a) Elevation in boiling point : Mol. Mass = 1000 Kb × w
W × ∆T

Where, Kb = Molal elevation constant of the solvent, w = Mass of the compound, W = Mass of the solvent

∆T = Elevation in boiling point of the solvent (determined experimentally)

(b) Depression in freezing point : Mol. Mass = 1000 K f × w
W × ∆T

Where, K f = Molal depression constant of the solvent, w = Mass of the compound, W = Mass of the solvent

∆T = Depression in freezing point of the solvent (determined experimentally)

(iii) Chemical methods
(a) Silver salt method for acids : It is based on the fact that silver salt of an organic acid on heating gives
residue of metallic silver.

RCOOAg heat → Ag
Silver salt Silver (residue)

From the mass of silver salt taken and the mass of the silver residue obtained, the equivalent mass of the silver
salt can be calculated.

Equivalent mass of silver salt = Mass of silver salt
Equivalent mass of silver Mass of silver

Knowing the equivalent mass of silver salt, the equivalent mass of the acid can be obtained. The molecular
mass of an acid can be determined with the help of the following relationship,

Mol. mass of the acid = Equivalent mass of the acid × basicity
Calculations : (i) Mass of silver salt taken = wg (ii) Mass of metallic silver = x g

Eq. mass of silver salt = w ; Eq. mass of silver salt = w × 108
Eq. mass of silver x x

Let the equivalent mass of the acid be E. In the preparation of silver salt, a hydrogen atom of the carboxylic
group is replaced by a silver atom.

Thus, Equivalent mass of silver salt = E − 1 + 108 = E + 107

Thus, E + 107 = w × 108 or E = w × 108 − 107
x  x

If n be the basicity of the acid, then Mol. Mass of the acid = w × 108 − 107 × n
 x

(b) Platinichloride method for bases : Organic bases combine with chloroplatinic acid , H 2 PtCl6 to form
insoluble platinichlorides, which, on ignition, leave a residue of metallic platinum. Knowing the mass of platinum

salt and the mass of metallic platinum, the molecular mass of the platinum salt can be determined. Let B represents

one molecule of the base. If the base is mono-acidic, the formula of the salt will be B2 H 2 PtCl6 .

B2 H 2 PtCl6 heat → Pt

Molecular mass of the salt Mass of platinum salt
Atomic mass of platinum = Mass of platinum


Let E be the equivalent mass of the base.

Molecular mass of the salt = 2E + 2 + 195 + 213 = 2E + 410

So 2E + 410 = w = Mass of platinum salt ; 2E =  w × 195 − 410 ; E = 1 w × 195 − 410
195 x Mass of platinum  x 2  x

Mol. mass of the base = Eq. mass × acidity = E × n

where n is the acidity of the base.

(c) Volumetric method for acids and bases : Molecular mass of an acid can be determined by dissolving a

known mass of the acid in water and titrating the solution against a standard solution of an alkali using

phenolphthalein as indicator. Knowing the volume of alkali solution used, the mass of the acid, 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.

1000ml 1Nalkali solution ≡ One gram equivalent of the acid

One gram equivalent of alkali

Calculations : Suppose w g of the organic acid requires V ml N1 alkali solution for complete neutralisation.
V ml N1 alkali solution ≡ w g acid

So 1000 ml N1 alkali solution ≡ w × 1000 g acid ≡ one gram equivalent acid
V × N1

Equivalent mass of the acid ≡ V w × 1000
× N1

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.

1000ml Nacidsolution ≡ One gram equivalent of the base

One gram equivalent of the acid

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 2O , i.e. for each carbon atom, there are two H-atoms and one oxygen atom. Its molecular formula is
however, C6 H12O6 .

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,

n = Molecular mass of the compound
Empirical formula mass of the compound

where as 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 CO2 and H 2O 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 CO2 and unused O2 .

(d) Caustic potash or caustic soda solution is then introduced into the eudiometer tube which absorbs CO2

completely and only unused O2 is left behind. 2NaOH + CO2 → Na2CO3 + H2O

Thus, the decrease in volume on introducing NaOH or KOH solution gives the volume of CO2 formed.
Sometimes, the volume of O2 left unused is found by introducing pyrogallol and noting the decrease in volume.

Calculation : From the volume of CO2 formed and the total volume of O2 used, it is possible to calculate the
molecular formula of gaseous hydrocarbon with the help of the following equation.

Cx H y + (x + y / 4)O2 → xCO2 + y / 2H 2O

1 vol (x + y / 4) vol x vol y / 2 vol

(Negligible volume on condensation)
From the above equation, it is evident that for one volume of hydrocarbon,
(a) (x + y / 4) volume of O2 is used
(b) x volume of CO2 is produced
(c) y/2 volume of H 2O vapours is produced which condense to give liquid H 2O with negligible volume.
(d) Contraction on explosion and cooling = [(1 + x + y / 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.

Example : 9 In victor mayer's method 0.2 gm of an organic substance displaced 56 ml. of air at STP the molecular weight

of the compound [Kerala (Med.) 2003]


(a) 56 (b) 112 (c) 80 (d) 28

W 0.2 M = mol. wt. of compound 
V2 56 W = Mass of organic compound
Solution: (c) M = × 22400 ; M = × 22400 = 80

V2 = Volume of air at STP 

Example : 10 0.24 g of a volatile liquid on vaporization gives 45ml. of vapours at NTP. What will be the vapour density of

the substance. (Density of H 2 = 0.089 gL−1 ) [CBSE PMT 1996]

(a) 95.39 (b) 39.95 (c) 99.53 (d) 59.93

Solution: (d) Vapour density = Mass of 45ml. of vapours at NTP = 0.24 = 59.93
Mass of 45ml. of H 2 at NTP 45 × 0.000089

(Density of H 2 = 0.089 gL−1 = 0.000089 g / ml.)

Example : 11 If 0.228 g of silver salt of dibasic acid gave a residue of 0.162 g of silver on ignition then molecular weight of

the acid is [AIIMS 2000]

(a) 70 (b) 80 (c) 90 (d) 100

Solution: (c) Molecular weight of the acid = w × 108 − 107 × n
 x

Where, w = wt. of silver salt, x = wt. of metallic silver, n = basicity of acid

∴Mol. wt. of acid =  0.228 × 108 − 107 × 2 = [152 − 107] × 2 = 90
 0.162

Example : 12 Platinum salt of an organic base contains 32.5% platinum. Hence, equivalent weight of the base is

(a) 95 (b) 190 (c) 600 (d) 300

Solution: (a) Equivalent weight of base (E),

E = 1 w × 195 − 410 = 1  100 × 195 − 410 = 1 [600 − 410] = 1 × 190 = 95  w = mass of platinum salt 
2  x 2  32.5 2 2 x = mass of platinum

Example : 13 Insulin contains 3.4% sulphur. The minimum molecular weight of insulin is [MP PET 1993]

(a) 350 (b) 470 (c) 560 (d) 940

Solution: (d) Minimum mass of sulphur = wt. of its one atom = 32

 3.4 gms of sulphur present in 100 gms.; ∴ 32 gms of sulphur present in = 100 × 32 = 940
3.4

Example : 14 Haemoglobin contains 0.33% of iron by weight . The molecular weight of haemoglobin is approximately

67200. The number of iron atoms (At. wt. (Fe) = 56) present in one molecule of haemoglobin are

(a) 6 (b) 1 (c) 4 (d) 2

Solution: (c) Total wt. of Fe in haemoglobin = 0.33 × 67200 = 221.76 ; At. wt. of Fe = 56
100

Total no. of Fe atoms per molecule of haemoglobin = 221.76 = 3.96 ; Thus no. of Fe atoms =4
56

Example : 15 If a compound on analysis was found to contain C = 18.5%, H = 1.55%, Cl = 55.04% and O= 24.81%,

then its empirical formula is [AIIMS 1998]

(a) CHClO (b) CH 2ClO (c) C2 H 2OCl (d) ClCH 2O

Solution: (a) C:H : Cl : O = 18.5 : 1.55 : 55.04 : 24.81 =1:1:1:1 ; ∴ Empirical formula = CHClO
12 1 35.5 16

Example : 16 An organic compound has been found to possess the empirical formula CH 2O and molecular weight 90.The

molecular formula of it is (C = 12, H = 1 and O= 16) [CPMT 2000; MP PET 2002]

(a) C3 H 6O3 (b) CH 2O (c) C2 H 6O2 (d) C2 H 2O


Solution: (a) Molecular weight = 90 ; Empirical weight of CH 2O = 12 + 2 + 16 = 30

n = Molecular weight = 90 = 3
Empirical weight 30

∴Molecular formula = Empirical formula × n = CH 2O × 3 = C3 H6O3

Example : 17 A dibasic organic acid gave the following results : C = 34.62%, H = 3.84%. 0.1075g of this acid consumes 20

ml of 0.1N NaOH for complete neutralisation. Find out the molecular formula of the acid. [Roorkee 1979]

(a) C3 H 4 O4 (b) C2 H 2O2 (c) C3 H 6O2 (d) CH 2O

Solution: (a) Calculation of empirical formula

Element Percentage At. mass Relative number Simplest ratio of
of atoms atoms

Carbon 34.62 12 34.62 = 2.88 2.88 = 1×3 = 3
12 2.88

Hydrogen 3.84 1 3.84 = 3.84 3.84 = 1.33 × 3 = 4
1 2.88

Oxygen 61.54 (by 16 61.54 = 3.84 3.84 = 1.33 × 3 = 4
difference) 16 2.88

Empirical formula of the acid = C3 H 4 O4 ; Empirical formula mass = (3 × 12) + (4 × 1) + (4 × 16) = 104
Calculation of molecular mass : 20 ml 0.1 N NaOH ≡ 0.1075g acid ; 20 × 0.1, ml 1 N NaOH ≡ 0.1075g acid

So 1000 ml 1 N NaOH ≡ 0.1075 × 1000g acid ≡ 53.75g acid ; Eq. mass of the acid = 53.75
20 × 0.1

Mol. Mass of the acid = Eq. mass × basicity = 53.75 × 2 = 107.50

n = Mol. mass = 107.50 ≈ 1; ∴ Molecular formula = C3 H 4 O4
Emp. mass 104.0

Example : 18 A hydrocarbon contains 10.5g of carbon for each one gram of hydrogen. The mass of 1 litre of hydrocarbon

vapours at 127oC and 1 atmospheric pressure is 2.8 g. Find out the molecular formula. [IIT 1980]

(a) C6 H 6 (b) C2 H 6 (c) C3 H 4 (d) C7 H 8

Solution: (d) Carbon : Hydrogen :: 10.5 : 1

Calculation of empirical formula

Element Percentage At. mass Relative number of Simplest ratio
atoms

Carbon 10.5 × 100 = 91.3 12 91.3 = 7.6 7.6 = 1×7 = 7
115 12 7.6

Hydrogen 1 × 100 = 8.7 1 8.7 = 8.7 8.7 = 1.14 × 7 = 8
11.5 1 7.6

Empirical formula = C7 H8

Empirical formula mass = (12 × 7) + (1 × 8) = 92

Calculation of molecular mass,

Experimental conditions : V1 = 1 litre, P1 = 1 atm, T1 = 127 + 273 = 400 K

NTP conditions : V2 = ? , P2 = 1 atm, T2 = 273 K

Applying gas equation,

V2 = P1 V1 × T2 = 1 × 1 × 273 = 0.6825 litre
T1 P2 400 × 1

0.6825 litre of the gas weigh = 2.8 g


22.4 litre of the gas weigh = 2.8 × 22.4 = 91.89 ≈ 92
0.6825

n = Mol. mass = 92 = 1
Emp. mass 92

Molecular formula = Empirical formula = C7 H8

Example : 19 10 ml of a gaseous hydrocarbon were mixed with 100 ml of oxygen and exploded in a eudiometer tube. The
volume of the residual gases was 95 ml of which 20 ml was absorbed by caustic soda and the remainder by
pyrogallol. The molecular formula of the gas is

(a) C2 H 2 (b) C2 H 4 (c) C3 H 6 (d) C4 H 8

Solution: (a) Vol. after explosion and cooling i.e., vol. of CO2 formed + unsolved O2 = 95 ml

Vol. after introducing KOH solution, i.e., vol. of CO2 formed = 20 ml

Vol. absorbed by alkaline pyrogallol, i.e, vol. of unsured O2 = 95 − 20 = 75 ml

∴ Vol. of O2 used = 100 − 75 = 25 ml.

Applying the combustion equation,

Cx H y + (x + y / 4)O2 → xCO2 + y / 2H 2O

10 ml 10 (x + y / 4)ml 10x ml negligible volume

Equating theoretical and experimental values,
10x = 20 and 10 (x+y/4) = 25
∴x = 2 and 10 (2+y/4) = 25

or y = 2 ; Thus the hydrocarbon is C2 H 2

(6) Determination of structure by spectroscopic and diffraction methods : With the advancement of
scientific developments, new techniques have been designed to determine the structures of compounds.

(i) Spectroscopic methods : All spectroscopic methods involve either the absorption of radiation or the
emission of radiation. The common types of spectroscopic methods used these days are ultra-violet (U. V.), infra-
red (I. R.), nuclear magnetic resonance (N. M. R.), mass spectroscopy etc. Mass spectroscopy determines the
molecular mass of the compounds. In fact, it is the best available method to determine the molecular mass. The
other spectroscopic methods detect the presence of functional groups present in the molecule.

(ii) Diffraction methods : The diffraction methods help to determine the complete three dimensional
structure of the molecules including bond length, bond angle etc. The commonly used diffraction methods are: X-
ray diffraction, neutron diffraction, electron diffraction etc.

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,


CH3CH 2CH2CH3 C| H3 CH3CH2CH = CH2 CH3
Butane CH3 − CH − CH3 1-Butene |

Isobutane CH3 − C − C ≡ CH
|
CH3
3, 3-Dimethyl-1-butyne

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,

Cyclopropane Cyclobutane Cyclohexane

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).

Benzene (Monocyclic) Naphthalene (Bicyclic)

These are also called benzenoid aromatics.

Note :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.

O
HO

Tropolone Azulene

(b) Heterocyclic compounds : Cyclic compounds containing one or more hetero atoms (e.g. O, N, S etc.) in

the ring are called heterocyclic compounds. These are of two types :

Alicyclic heterocyclic compounds : Heterocyclic compounds which resemble aliphatic compounds in

their properties are called Alicyclic heterocyclic compounds. For example,

O O O N N
H H
Oxirane or Epoxyethane Tetrahydrofuran (THF) O
Pyrrolidine Piperidine
1, 4-Dioxane


Aromatic heterocyclic compounds : Heterocyclic compounds which resemble benzene and other
aromatic compounds in most of their properties are called Aromatic heterocyclic compounds. For example,

O N S N QuinolineN
H
Furan Thiophene Pyridine
Pyrrole
Organic compounds

Acyclic or open chain compounds Cyclic or closed chain compounds

Homocyclic or carbocyclic compounds Heterocyclic compounds

Alicyclic compounds Aromatic compounds Alicyclic heterocyclic Aromatic heterocyclic
compounds compounds

Benzenoid aromatics Non-benzenoid aromatics

(2) Classification based on functional groups : A functional group is an atom or group of atoms in a
molecule that gives the molecule its characteristic chemical properties. Double and triple bonds are also considered
as functional groups.

All compounds with the same functional group belong to the same class. Various classes of compounds
having some of the common functional groups are listed in the table.

Class Functional group Class Functional group
Olefins/Alkenes C=C Acid halides
−C ≡ C − Amides O
Acetylenes/Alkynes Acid anhydrides
−F, − Cl, − Br, − I (Halo) ||
Alkyl Halides Esters
− C − X (Acylhalide)
Alcohols –OH (Hydroxy) O

||

− C − NH2 (Amide)

OO

|| ||

− C − O− C −
(Anhydride)

O

|| |

− C − O − C − (Ester)
|

Ethers || Cyanides/Nitriles −C ≡ N (Cyano)

− C − O − C − (Alkoxy)

||

Aldehydes −C − H (Aldehydic) Isocyanides – N C (Isocyano)
||
O

Ketones O Nitro compounds O (Nitro)
–N O ↓
||

− C − (ketonic)


Carboxylic acid O Amines –N H
H
|| (Amino)

− C − OH (Carboxyl)

(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 CH2 .

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 CnH2n+1OH 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 (12 + 2 × 1) .

(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.

HH HH H HO|
|| | ||
H − C − C − H; H − C − C− C − O − H; H − C − C
|| | | | |
H
HH H H H H
Acetaldehyde
Ethane n-propyl alcohol

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.

Note :H H HH HH O
H || || H
C=C ; H−C≡C−H
H Ethylene Acetylene H − C = C − OH H −C =C −C
Vinyl alcohol Acraldehyde

The double bond between carbon and oxygen atoms is not a sign of unsaturation as in

acetaldehyde or acetone.

Nomenclature of organic compounds

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.

Common or trivial names of some organic compounds.

Compound Common name Compound Common name
Methane Chloroform
CH4 Acetylene CHCl3 Iodoform
C2H2 n-Butane CHI3 Acetonitrile
H3CCH2CH2CH3 Isobutane CH3CN Acetic acid
(H3C)2CHCH3 Neopentane CH3COOH Benzene
(H3C)4C Formaldehyde C6H6 Toluene
HCHO Acetone C6H5CH3 Aniline
Ethyl alcohol C6H5NH2 Phenol
(H3C)2CO Acetamide C6H5OH Anisole
CH3CH2OH Dimethyl ether C6H5OCH3 Acetophenone
CH3CONH2 Diethyl ether C6H5COCH3 Benzamide
CH3OCH3 C6H5CONH2
(CH3CH2)2O

(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.

Chain length Word root Chain length Word root

C1 Meth- C11 Undec-

C2 Eth- C12 Dodec-

C3 Prop- C13 Tridec-

C4 But- C14 Tetradec-


C5 Pent- C15 Pentadec-

C6 Hex- C16 Hexadec-

C7 Hept- C17 Heptadec-

C8 Oct- C18 Octadec-

C9 Non- C19 Nonadec-

C10 Dec- C20 Eicos

(ii) Suffix : The word root is linked to the suffix which may be primary or secondary or both.

(a) Primary suffix : A primary suffix is added to the word root to indicate whether the carbon chain is
saturated or unsaturated.

Type of carbon chain Primary suffix General name

Saturated (C – C) –ane Alkane

Unsaturated (C = C) –ene Alkene

Unsaturated (C ≡ C) –yne Alkyne

If the parent chain contains two, three or more double or triple bonds, then the numerical prefixes such as di
(for two), tri (for three), tetra (for four), etc. are added to the primary suffix.

Note :It may be noted that extra 'a' is added to the word root if the primary suffix to be added begins with

a consonant (other than a, e, i, o, u). For example, for two double bonds, suffix is diene and if it is to
be added to word root but (for 4C atoms), it becomes butadiene.

(b) Secondary suffix : A secondary suffix is then added to the word root after the primary suffix to indicate the
functional group present in the organic compound.

Class of org. Functional group Secondary suffix Class of org. Functional group Secondary suffix
compound compound
–OH –ol –COCl –oyl chloride
Alcohols –CHO –al Acid chlorides – CONH2 –amide
Aldehydes >C = O –one Acid amides –nitrile
Ketones –COOH –oic acid Nitriles – C≡ N –amine
Amines – NH2
Carboxylic thiol
acids –COOR alkyl.... oate Thiol –SH
Esters

It may be noted that while adding the secondary suffix to the primary suffix, the terminal 'e' of the primary
suffix (i.e. ane, ene and yne) is droped if the secondary suffix begins with a vowel but is retained if the secondary
suffix begins with a consonant. For example

Organic compound Word root Primary suffix Secondary suffix IUPAC name

CH3CH2OH Eth an (e)* ol Ethanol
CH3CH2CN Prop ane nitrile Propanenitrile

The terminal 'e' from the primary suffix has been dropped because the secondary suffix i.e. 'ol' begins with a

vowel 'o'.

(iii) Prefix : There are many groups which are not regarded as functional groups in the IUPAC name of the

compound. These are regarded as substituents or side chains. These are represented as prefixes and are

placed before the word root while naming a particular compound. These may be :

(a) Alkyl groups : These groups contain one hydrogen atom less than the alkane. These are named by

substituting the suffix ane of the name of the corresponding alkane by yl. i.e. alkane – ane + yl = alkyl.


For example,

CH 4 : Methane becomes CH 3 − : Methyl

CH3CH3 : Ethane becomes CH3CH2 − : Ethyl

CH3CH2CH3 : Propane becomes CH3CH2CH2 − : Propyl etc.

(b) Functional groups not regarded as principal functional groups : If a compound contains more than one

functional group, then one of the functional group is regarded as principal functional group and is treated as

secondary suffix. The other functional groups are regarded as substituents and are indicated by prefixes.

Substituent Prefix Substituent Prefix Substituent Prefix
Nitro
–F Fluoro – NO Nitroso – NO2 Amino
Hydroxo
–Cl Chloro –N=N– Diazo – NH2

–Br Bromo –OCH3 Methoxy –OH

–I Iodo –OC2H5 Ethoxy

Thus, a complete IUPAC name of an organic compound may be represented as:

Prefix + word root + Primary suffix + Secondary suffix

For example : Functional group

4 32 1 OH Word root : But
C H3 CH C H2 C H2 Primary suffix : – ane
− | − − − Secondary suffix : –ol
Prefix : Chloro
Prefix Cl

IUPAC name : Chloro+but+ane+ol; 3-Chloro butan-1-ol

(Number 1 and 3 represent the positions of suffix and prefix)

5 O Word root : Pent (five C – C – C – C – C)
43 2 1|| Primary suffix : ene (double bond at C – 2)
CH3 − CH − CH CH − C− OH Secondary suffix : oic acid (– COOH group)
| = Prifix : Bromo (– Br group at C – 4)

Prefix Br Sec. suffix

Pri. suffix
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 (1o), secondary (2o),
tertiary (3o) and quaternary (4o). The carbon atoms in an organic compound containing functional group can be

designated as α, β, γ, δ.

1o CH3 δγ βα
| CH3 − CH2 − CH2 − CH2 − OH
Functional group
1o CH3 −2o CH2 −4o C − 3o CH −1o CH3 δ γ βα
|| CH3 − CH CH2 − CH2 − CHO
1o CH3 1o CH3 | −

CH3

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) CnH2n+2 −H → CnH2n+1 (Alkyl group)

(R − H) (R−)


Alkyl groups are named by dropping-ane from the name of corresponding paraffin and adding the ending–yl.

Parent saturated Name of the Structure Parent saturated Name of the Structure
hydrocarbon alkyl group hydrocarbon alkyl group

Methane Methyl CH3 – Propane n-Propyl CH3 – CH2 – CH2 –
Ethane Ethyl Butane n-Butyl CH3 – CH2 – CH2 – CH2 –
CH3 – CH2


Alkyl groups derived from saturated hydrocarbons having three or more carbon atoms exist in isomeric forms.

CH 3 CH 2CH 3 CH3CH2CH2 − n − Propyl ; CH3CH 2CH 2CH 2 − n − Butyl
Propane CH 3
CH 3 CH − Isopropyl CH3CH2CH2CH3 CH 3CH 2 CH − sec.Butyl
n − Butane CH 3

|CH3 Isobutyl
CH 3 − CH − CH 2 − Tertiary butyl (t - Butyl)
|CH3
CH 3 − CH − CH 3 CH3 |CH3 CH3
CH −
Isobutane − | −

Similarly, removal of different H atoms in pentane gives the following radicals :

CH 3 C|H3
|
CH 3CH 2CH 2CH 2CH 2 − ; CH3|CHCH2CH2 − ; CH 3CCH 2 ; CH CHCH CH CH ; CH CCH CH
n− Pentyle | − 3 | 2 2 3 3 | 2 3

CH 3 CH 3 sec −Pentyl tert −Pentyl
Isopentyl
Neopentyl

Note :The prefix sec-or tert-before the name of the group indicate that the H-atom was removed from a

secondary or tertiary carbon atom respectively.

Unsaturated groups or radicals

Group Common name IUPAC name Group Common name IUPAC name
CH2 = CH − vinyl Ethenyl HC ≡ C − Acetylide Ethynyl
Allyl 2-Propenyl
21 21 Propargyl 2-Propynyl
– 1-Propenyl
CH2 = C H − C H2 − HC ≡ C− CH2 −

1

CH3 − CH = C H −

Different classes of organic compounds

S. No. Homologous Structural formula Root Primary Secondary IUPAC Name
1. series word suffix suffix
1 Meth -ane – Methane
2. Paraffins or Eth -ane – Ethane
Alkanes C H4 Prop -ane – Propane
But -ane – Butane
(Cn H 2n+ 2 ) 21
Eth -ene – Ethene
Olefins or C H3 C H3 Prop -ene – Propene
Alkenes
321
(CnH2n)
C H3 C H2 C H3

4321

C H3 C H2 C H2 C H3

21

C H2 = C H2


32 1 But -ene – 1-Butene
C H3 C H = C H2 But -ene – 2-Butene

432 1 Eth
C H3 C H2 C H = C H2 Prop
But
43 21 But
C H3 C H = C H C H3
Meth
3. Acetylenes or 21 Eth -yne – Ethyne
Alkynes Prop -yne – Propyne
CH ≡CH Prop -yne – 1-Butyne
But -yne – 2-Butyne
3 21 But
(Cn H 2n− 2 ) C H3 − C ≡ C H -ane -ol Methanol
Meth -ane -ol Ethanol
43 21 Eth -ane -ol 1-Propanol
C H3 C H2 − C ≡ C H Prop -ane -ol 2-Propanol
But -ane -ol 1-Butanol
4 3 21 -ane -ol 2-Butanol
Prop
C H3 − C ≡ C− C H3 But -ane
Pent -ane
4. Monohydric 1 Pent -ane
Alcohols -ane
C H3OH Meth
Eth -ane
(CnH2n+1OH) 21 Prop -ane
But -ane
C H3 C H2OH -ane
Eth
321 Prop -ane
But -ane
C H3 C H2 C H2OH -ane
Eth -ane
32 1 Prop
C H3 C HOH C H3 But -ane
-ane
4321 Meth -ane

C H3 C H2 C H2 C H2OH -ane
-ane
432 1 -ane
C H3 C H2 C HOH C H3
-ane
5. Aldehydes 1 -al Methanal
(Cn H 2nO) -al Ethanal
H C HO -al Propanal
-al Butanal
21
-one Propanone
C H3 C HO -one Butanone
-one 2-Pantanone
321 -one 3-Pentanone

C H3 C H2 C HO

4321

C H3 C H2 C H2 C HO

6. Ketones 3 21
(Cn H 2nO)
C H3 C O C H3

4 3 21

C H3 C H2 C O C H3

5 4 3 21

C H3 C H2 C H2 C O C H3

5 4 32 1

C H3 C H2 C O C H2 C H3

7. Carboxylic 1 -oic acid Methanoic acid
acid (Mono) -oic acid Ethanoic acid
H C OOH -oic acid Propanoic acid
(CnH2nO2) -oic acid Butanoic acid
21
C H3 C OOH

321

C H3 C H2 C OOH

4321

C H3 C H2 C H2 C OOH

8. Acid 21 -oyl chloride Ethanoyl chloride
Chlorides -oyl chloride Propanoyl chloride
(RCOCl) C H3 C OCl -oyl chloride Butanoyl chloride

321

C H3 C H2 C OCl

4321

C H3 C H2 C H2 C OCl

9. Acid amides 21 Amide Ethanamide
(RCONH2) Amide Propanamide
C H3 C ONH2 Amide Butanamide

321
C H3 C H2 C ONH2

4321

C H3 C H2 C H2 C ONH2

10. Esters 1 -oate Methyl methanoate

H C OOCH3


(RCOOR′) 21 Eth -ane -oate Ethyl ethanoate
Eth -ane -oate Methyl ethanoate
11. Anhydrides C H3 C OOC2H5 Prop -ane -oate Ethyl propanoate
(RCO)2 O
21 Eh -ane -oic Ethanoic anhydride
Prop -ane anhydride Propanoic
C H3 C OOCH3 anhydride
-ane -oic
3 21 -ane anhydride Methanamine
-ane Ethanamine
C H3 − C H2 C OOC2H5 -ane amine 1-Propanamine
2-Propanamine
(CH3 − CO)2O -ane amine
(C2H5 − CO)2O -ane
-ane amine
12. Amines 1 Meth
(R − NH2) Eth -ane amine
Primary C H3 NH2 Prop -ane
Prop -ane
21 -ane
Eth
C H3 C H2NH2 Prop -ane
But -ane
321 -ane
Meth
C H3 C H2 C H2NH2 Eth -ane
Prop -ane
32 1 Prop -ane
C H3 C HNH2 C H3 -ane
Meth
13. Cyanides or 21 Eth nitrile Ethane nitrile
nitriles (R– Eth nitrile Propane nitrile
C H3 C N nitrile Butane nitrile
Meth
CN) 321 Eth Chloro Chloromethane
Prop Chloro Chloroethane
C H3 C H2 C N Prop Bromo 1-Bromopropane
Bromo 2-Bromopropane
4321

C H3 C H2 C H2 C N

14. Alkyl halides 1
(Cn H 2n+1 X )
C H3Cl

21
C H3 C H2Cl

321

C H3 C H2 C H2Br

32 1
C H3 C HBr C H3

15. Ethers 1 Methoxy Methoxymethane
(R– O – R) Methoxy Methoxyethane
C H3OCH3 Ethoxy Ethoxyethane

21 Nitro Nitromethane
Nitro Nitroethane
C H3 C H2OCH3 Nitro 1-Nitropropane
Nitro 2-Nitropropane
21

C H3 C H2OC2H5

16. Nitro 1
compounds
C H3 NO2
(R – NO2)
21

C H3 C H2NO2

321

C H3 C H2 C H2NO2

32 1
C H3 C HNO2 C H3

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,

(1) Prefix n-(normal) is used for those alkanes in which all the carbon atoms form a continuous chain with no
branching.

CH3CH2CH2CH3 CH3CH 2CH 2CH2CH3
n−Butane n−Pentane

(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.


CH3 − CHCH3 CH3 − CH − CH2CH3 CH3 − CHCH2CH2CH3
| | |
CH3 CH3 CH3
Isobutane Isopentane Isohexane

(3) Prefix neo is used for those alkanes which have two methyl groups attached to the second last carbon

atom of the continuous chain.

CH3 C| H3
CH3 − C − CH2 − CH3
|
|
CH3 − C − CH3
| CH3
CH3
Neopentane Neohexane

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 :

Substituents

CH3
|
CH 3 − CH − CH 2 − CH 2 − C − CH 2 − CH 3
|| Parent chain

CH3 CH3

Substituents 21
56
1 2C| H33 C H2 −C H3 C H2 −C H3

4| 3| 4
CH3 − C − CH2 − C − CH3 CH3 − C H − C H − CH2 − CH3
|| 5|
CH3 CH3 CH2
(Longest chain consists of six carbon atoms) 6| 7
CH2−CH3
(Longest chain consists of seven carbon atoms)

Note :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,

Substituent

X X
5 4 3 2| 1 1 2 3 4| 5

C− C− C− C− C C− C− C− C− C
A (Correct) B (Wrong)

The number that indicates the position of the substituent or side chain is called locant.

5 4 3 21 123

C H3 − C H2 − C H2 − C H − C H3 C H3 − C H2 − C H − C H − C H3
| 4| 5 6
CH3 CH2 −CH2 −CH3
2− Methylpentane 3 − Ethylhexane


(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,

65 4 32 1 12 3 45 6
H3 C− C H − C H2 − C H− C H − C H3 H3 C− C H − C H2 − C H− C H − C H3
| || | ||
CH3 CH3 CH3 CH3 CH3 CH3
Set of locants : 2, 3, 5 (Correct) Set of locants : 2, 4, 5 (Wrong)

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,

10 9 87 6 5 43 21
CH3 − CH2 − CH −CH − CH2 − CH2 − CH2 − CH2 − CH − CH3 Set of locants : 2, 7, 8 (Correct)

|| |
CH3 CH3 CH3

1 2 3 4 5 6 7 8 9 10
C H3 − C H2 − C H −C H − C H2 − C H2 − C H2 − C H2 − C H − C H3
Set of locants : 3, 4, 9 (Wrong)

|| |
CH3 CH3 CH3

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 the locants is the lowest. This rule is called lowest sum rule.

For example, the carbon chain of alkanes given below should be numbered as indicated in structures A and not

according to structure B.

CH2−CH3 4|CH2 −CH3
3 4| 5 6 7 5 3 2 1
CH3 − C H − C H − C H2 − C H2 − CH3 ; CH3 − C H − C H − C H2 − C H2 − CH3
2| 1 |6 7
CH2−CH3 CH2−CH3

(A) correct Sum of locants =3+4=7 (B) wrong Sum of locants =4+5=9

CH3 CH3 CH3 CH3
1 2| 3 4| 5 5 4| 3 2| 1
C H3 − C− C H − C H − C H3 C H3 − C− C H − C H − C H3
|| ||
CH3 CH2−CH3 CH3−CH2 CH3

(A) Correct Sum of locants =2+2+3+4=11 (B) Wrong Sum of locants =2+3+4+4=13

Note :According to latest IUPAC system of nomenclature, the lowest set of locants is preferred even if it

violates the lowest sum rule. For example,

10 9 87 6 54 3 21
CH3 − CH2 − CH − CH − CH2 − CH2− CH2 − CH2 − CH − CH3
|| |
CH3 CH3 CH3

Structure (A) Set of locants = 2, 7, 8
Sum of locants = 2 + 7 + 8 =17


1 2 3 4 5 6 7 8 9 10
C H3 − C H2 − C H − C H − C H2 − C H2− C H2 − C H2 − C H − C H3
|| |
CH3 CH3 CH3
Set of locants = 3, 4, 9
Structure (B) Sum of locants = 3 + 4 + 9 =16

This compound is numbered as 2, 7, 8 and not as 3, 4, 9 in accordance with latest lowest set of locants rule,
even though it violates lowest sum rule.

(iv) Presence of more than one same substituent : If the same substituent or side chain occurs more

than once, the prefixes di, tri, tetra ..........etc., are attached to the names of the substituents. For example,

CH3
5 43 |2 1
C H3 − C H − C H2 − C− C H3
||
CH3 CH3
2, 2, 4- Trimethylpentane

(v) Naming different substituents : If two or more different substituents or side chains are present in the

molecule, they are named in the alphabetical order along with their appropriate positions.

CH 2CH 3 CH3 CH3
5 4 3| 2 1 1 2| 3|
C H3 − C H2 − C− C H − C H3 C H 3 − C − C− CH 2 − CH 3
| |
CH3 4|
3-Ethyl-2, 3-dimethylpentane CH3
CH 3

5| 6

CH 2 − C H 3

3-Ethyl-2, 2, 3-trimethylhexane

(vi) Naming different substituents at equivalent position : In case, there are different alkyl substituents
at equivalent positions, then numbering of the parent chain is done in such a way that the alkyl group which comes
first in the alphabetical order gets the lower number.

6 5 43 2 1 C2H5 CH3
C H3 − C H2 − C H− C H − C H2 − C H3
|| 1 2 3| 4| 5 6
CH3 C2H5
3-Ethyl- 4 - methyl hexane C H3 − C H2 − C − C− C H2 − C H3
||
C2H5 CH3
3, 3-Diethyl-4, 4-dimethyl hexane

(vii) Naming the complex substituents (or substituted substituents) : If the substituent on the parent chain
is complex (i.e. it is branched) it is named as substituted alkyl group by numbering the carbon atom of this group
attached to the parent chain as 1. The name of such substituent is given in brackets in order to avoid confusion with
the numbering of the parent chain. For example,

1 2 3 4 56 7 8 9
C H3 − C H2 − C H2 − C H2 − C H − C H2 − C H2 − C H2 − C H3
1|
CH −CH3
2|
CH −CH3 Complex substituent

3|
CH3

5-(1, 2-Dimethylpropyl) nonane

The name of the complex substituent is always written in brackets.
While deciding the alphabetical order of the various substituents, the name of the complex substituent is
considered to begin with the first letter of the complete name. It may be remembered that in case of simple


substituents, however, the multiplying prefixes are not considered. The names of simple substituents are first
alphabetized and then multiplying prefixes are inserted. For example,

CH2 −CH3
1 2 3 4 56 |7 8 9 10 11
C H3 − C H2 − C H2 − C H2 − C H − C H2 − C H − C H2 − C H2 − C H2 − C H3
1|
CH −CH3
2|
Complex substituent
CH −CH3 (1, 2-dimethylpropyl)
3|
CH3

5-(1, 2-Dimethylpropyl)-7-ethyl undecane

It may be noted that dimethyl propyl (a complex substituent) is alphabetized under d and not under m.
Therefore, it is cited before ethyl (e).

CH3 Complex substituent
|1 2 3 ( 1, 1-dimethylpropyl)
CH 3 − C− C H 2 − C H 3
9 8 7 6 |5 4 3 2 1
C H3 − C H2 − C H2 − C H2 − C− C H2 − C H2 − C H2 − C H3
|1 2 3
C H2−C H−C H3
| Complex substituent
CH3 ( 2-methylpropyl)

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


Purification, nomenclature and classification of organic compounds

The substituent dimethyl is cited first because it is alphabetized under d. Similarly,

C2H5 CH3 CH3
1 2 3| 4| 5 6 |7 8 9 10
CH3 − CH2 − CH − C− CH2− CH2− C H − CH2 − CH2 − CH3

|1
H C C CH
3 − − 3 Complex substituent
|2 ( 1, 1-dimethylpropyl)

CH 3

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

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,
CH 3
|1 2 3 4 1-methylbutyl

C H − C H2 − C H2 − C H3

12 11 10 9 8 76 |5 4 3 2 1
C H3 − C H2 − C H2 − C H2 − C H2 − C H − C H2 − C H − C H2 − C H2 − C H2 − C H3

|1 2 3 4
C H2 − C H − C H2 − C H3
| 2-methylbutyl
CH3

– 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.

CH3 1,1 dimethylpropyl

3 2 |1 CH
C H C H C− 4
10 9 8 7 3 − 6 2 − 3 3 21
|5
C H3 − C H2 − C H2 − C H2 − C H2 − C− C H2 − C H2 − C H − C H3
|
3 2 1| CH3

C H3 − C H2 − C− C H3
|
CH3 1,1 dimethylpropyl

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

(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 H2n .

The cyclic compound is named by prefixing cyclo to the name of the corresponding straight chain alkane.
For example,

C3H6, Cyclopropane C4H8, Cyclobutane C5H10, Cyclopentane C6H12, Cyclohexane

If side chains are present, then the rules given in the previous section are applied. For example,


CH3 C2H5

Methylcyclohexane Ethylcyclopentane

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,

CH3 2 CH3 CH3 CH3
1 1 1
2
4 3 CH3 62 CH2CH2CH3 3
53
1.3-Dimethylcyclobutane CH2CH3
4

1-Methyl-3-propylcyclohexane 3-Ethyl-1, 1-dimethylcyclohexane
(Not 1-Ethyl-3,3-dimethylcyclohexane)
(Not 5-Methyl-1-propyl cycloalkane)

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 a named as cycloalkylalkanes. For example,

1234 − CH 2 − CH 2 − CH 2 − CH 2 − CH 3

− C H2 − C H2 − C H2 − C H3 1-Cyclobutylpentane

1-Cyclopropyl butane

CH3

− CH 2 − CH 2 − CH 2 −

In case of 1, 3-Bis (2-methylcyclopropyl) ptrhopeandeoublCe Hbo3 nd is given the Cyclohexyl cyclohexane and numbering is

substituted cycloalkenes, lowest possible number

done in such a way that the substituents get the lowest number.

CH3 CH3

2 3 CH3
2

11

3-Methylcyclohex-1-ene 2, 3-Dimethylcyclopent-1ene

Note :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,

(CH3)2CH − Isopropyl (CH3)2CH − CH2 − Isobutyl

CH 3 − CH 2 −CH − Sec- Butyl (CH3)3C − tert-Butyl

| Isopentyl
CH3 tert-Pentyl

(CH3)2CH − CH2 − CH2 − (CH3)3C − CH2 − Neopentyl
(CH3)2CH − CH2 − CH2 − CH − Isohexyl
CH3

|

CH3 − CH2 − C −

|

CH3


However, when these are substituted, these names cannot be used as such. For example,

CH(CH3)2
1 2 3 |4 5 6 7 8 9 10
C H3 − C H2 − C H2 − C H − C H − C H2 − C H2 − C H2 − C H2 − C H3
|
CH3 −CH −CH2 −CH3
5−sec −Butyl−4−isopropyldecane

C|H3 CH3
|
CH2 CH − CH3
8 7 65 4 |3 2 1 1 2 |3 4 56 7 8
H3 H2 CH C− C H2 − H3 ; H3 H2 C H − C H2 − C H − C H2 H2 H3
C − C − C H2 − C H− − | C C − C − −C − C
| |
CH3 −CH CH2 −CH3 1|
CH3 Cl − C − CH3

| 2|
CH3 CH3
3,3−Diethyl −5−isopropyl −4 −methyloctane 5 −(1− chloro-1- methylethyl)- 3- isopropyloctane

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 (–NO2),
amino (–NH2) and alkoxy (–OR) are treated as substituents.

The decreasing order of seniority among the principal groups

Order of preference Preflx Suffix (ending) Order of preference Preflx Suffix (ending)
– SO3H Sulphonic acid Hydroxy – ol
– COOH Sulpho – oic acid – OH Amine – amine
– COOR Carboxy Alkyl alkanoate – NH2 – – ene
Alkoxy C= C
– COX carbonyl – yne
– CONH2 Haloformyl Oyl halide –C≡C– – –
–C≡N Carbamoyl – amide –O– Epoxy –
– CHO Cyano – nitrile –X Halo –
>C=O Formyl – al – NO2 Nitro
Keto – one

(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.

OH Substituent

1 Principal functional group

C OOH

4-Hydroxy pentanoic acid

5 4| 3 2
C H3 − C H − C H2 − C H2 −

Some other examples are :

O O
5 4|| 3 2 1 4 3|| 2
C H3 − C − C H 2 C H 2 C OOH ; 5 43 21 ; 1
C H3 C H C H2 C H − C H3 C H3 − C − C H2 − C HO
(−COOH is principal group) − | − − | (−CHO is principal group)
4-oxopentanoic acid) 3 - Oxobutanal
NH2 OH

(−OH is principal group)
4- Aminopentan- 2-ol

O
2||
1 H3 − C − 3 H2 − 4 H2 − OH ; 321 3 21
C H3 C H C OOH
C C C NC − C H2 − C H2 − C OOH ; − | −
(−COOH is principal group)
(−CO is principal group) 3-cyanopropanoic acid
4 -Hydroxybutane- 2-one
NH2

(−COOH is principal group)
2-Aminopropanoic acid

4 3C|H3 2 1 ; 4 3 21 ; 4 3 21 ; 4 32 1
C− C H3 CH− C||− C OOH C H C H2 C|H C OOH C H3 C C H C OOH
C H3 − | C− C H3 − | O | 2 − − F − − | = −
|| CH3
OH O OH COOC2H5

(−CO is principal group) (−COOH is principal group) (−COOH is principal group) 3-Carbethoxybut-2-en-1-oic acid
3-Hydroxy-3-methylbutan- 2-one 3-Methyl- 2-oxo butanoic acid 2-Fluoro-4-hydroxybutanioc acid or 3-Carbethoxy-2-butenoic acid

6 5 C4|H3 3 O
2|| 1
21 5 43 54 3 21
H C ≡ C− C H − C HOH − C H 2 C OCl ; C H2 = C H − C H2 − C − C H3 ; OH C− C H 2 − C H = C H − C HO
3-Hydroxy-4-methylhex-5-yn-1-oyl chloride Pent-4-en- 2-one Pent-2-ene-1, 5-dial
or 3-Hydroxy-4-methylhex-5-ynoyl chloride

CH 3
2|
6 5 4 3 1 43 21
C H3 − C H2 − C H = C H − C H C HO ; HOO C− C H = C H − C OOH
2-Methylhex -3-en-1-al
But-2-ene-1, 4-dioic acid (or But-2-enedioic acid)

(3) Polyfunctional compounds containing more than two like functional groups : According to
latest convention (1993 recommendations for IUPAC nomenclature), if an unbranched carbon chain is directly
linked to more than two like functional groups, the organic compound is named as a derivative of the parent alkane

which does not include carbon atoms of the functional groups. For example,

CN COOH
3 2| 1 3| 2 1
NC − C H 2 − C H − C H 2 − CN 54
HOOC − C H 2 C H 2 − C H − C H 2 C H 2 − COOH
Propane-1, 2, 3-tricarbonitrile Pentane-1, 3, 5-tricarboxylic acid
(formerly 3-cyanopentane-1, 5-dinitrile) (formerly 4-carboxyheptane-1, 7-dioic acid)

Following the above rule, citric acid may be named as,


COOH
1 2| 3
HOOC C H2 C− C H2 COOH 2-Hydroxypropane-1, 2, 3-tricarboxylic acid (formerly
− − | − 3-carboxy-3-hydroxypentane-1, 5-dioic acid )

OH

Note :If none of the choices in a multiple choice question follows 1993 recommendations, the choice
corresponding to earlier system of nomenclature as given in parentheses should be taken as correct.
If, however, all the three like groups are not directly linked to the unbranched carbon chain, the

carbon atoms of the two like groups are included in the parent chain while the third which forms the
side chain is considered as a substituent group. For example,

65 4 3C|H2CN2 (substituent group) 65 4 C3|H2CO2OH (substituent group)
1 1
N CCH2 CH2 − C H − CH2 C N HOO C C H 2 C H 2 − C H − C H 2 C OOH
3-(Cyanomethyl) hexane-1, 6-dinitrile 3-(Carboxymethyl) hexane-1, 6-dioic acid

Bond-line Notation of organic compounds

Sometimes, the bonds between carbon atoms are represented by lines. For example, n-hexane has a
continuous chain of six carbon atoms which may be represented as,
`

CH 3 − CH 2 − CH 2 − CH 2 − CH 2 − CH 3

n − Hexane

In this notation, the carbon atoms are represented by line ends and intersections. It is assumed that the
required number of hydrogen atoms are present wherever they are necessary to satisfy the tetravalency of carbon. A
single line represents a single bond (C – C), two parallel lines represent a double bond (C = C) and three parallel
lines represent a triple bond (C ≡ C). For example,

1 3-Ethyl-1, 1-dimethylcyclohexane 3
2 (Not 1-ethyl-3, 3-dimethylcyclohexane) 2
1
3 24
1-Methyl-3-propylcyclohexane 13 3-Methyl cyeclohex-1-ene

42 1 Buta-1, 3-diene 13
53 2

Pent-1-ene 1,3-Dicyclohexyl propane

13 5 2 4 24
2 4 1 3
3
5 1
2-Methylbuta-1, 3-diene (a line at
1-Cyclobutyl pentane Penta-1, 4-diene position 2 corresponds to CH3 group )
31
2 1 2 OH
OH 2 4

3 1 35

3-Cyclohexyl butan-2-ol 1, 3-Dimethyl cyclohex-1-ene 4-Hydroxy-4-methyl pentan-2-one


2 3 O
3 4 2 OH 1

1 5 26
6 35
4
5 3, 3, 5-trimethylhex-1-en-2-ol 4

3-Ethyl penta-1, 3-diene Br 5, 6-Dimethylcyclohex-2-en-1-one
O
2 46
42 1 35

31 Hexa-1, 3, 5-triene

1-Bromo-3-methyl pentan-2-one

Nomenclature of Bicyclic compounds

Many hydrocarbons and their derivatives contain two fused or bridged rings. The carbon atoms common to
both rings are called bridge head atoms and each bond or chain of carbon atoms connecting both the bridge head
atoms is called as bridge. The bridge may contain 0, 1, 2.... etc. carbon atoms. For example,

Bridge head atom Bridge head atom

CH Zero carbon bridge CH One carbon bridge

One carbon CH2 CH2 One carbon bridge Two C|H 2 CH  Two
bridge carbon CH 2 |  carbon
bridge CH2 2 bridge

CH CH 2 

Bridge head atom CH

Bridge head atom

These bicyclic compounds are named by attaching the prefix 'bicyclo' to the name of the hydrocarbon having
the same total number of carbon atoms as in the two rings. The number of carbon atoms in each of the three
bridges connecting the two bridge head carbon atoms is indicated by arabic numerals, i.e., 0, 1, 2.....etc. These
arabic numerals are arranged in descending order; separated from one another by full stops and then enclosed in
square brackets. The complete IUPAC name of the hydrocarbon is then obtained by placing these square brackets
containing the arabic numerals between the prefix bicyclo and the name of alkane. For example,

Bicyclo [2, 2, 1] heptane Bicyclo [3, 1, 1] heptane Bicyclo [2, 2, 2] octane Bicyclo [4, 4, 0] decane Bicyclo [1, 1, 0] butane
(also called norbornane) (also called decalin)

If a substituent is present, the bicyclic ring system is numbered. The numbering begins with one of the bridge
head atoms, proceeds first along the longest bridge to the second bridge head atom, continues along the next
longest bridge to the first bridge head atom and is finally completed along the shortest path. For example,

1 Cl 321 9 2
27 8 CH3 1 63
36 8 46
5 5 74
4 5
7
2, 6, 6-Trimethylbicyclo [3, 1, 1] hept-2-ene
8-Chlorobicyclo [3, 2, 1] octane 8-Methylbicyclo [4, 3, 0] nonane


Nomenclature of spiro compounds

Compounds in which one carbon atom is common to two different rings are called spiro compounds. The

IUPAC name for a spiro compound begins with the word spiro followed by square brackets containing the number
of carbon atoms, in ascending order, in each ring connected to the common carbon atom and then by the name of
the parent hydrocarbon corresponding to the total number of the carbon atoms in the two rings. The position of
substituents are indicated by numbers ; the numbering beginning with the carbon atom adjacent to the common
carbon and proceeding first around the smaller ring and then around the larger ring and finally ending on the

common carbon atom. For example,

56 1 8 9 1 2 Cl
2 7 510 4 3
4 7
3 6

Spiro [2.4] heptane 2-Chlorospiro [3.5] decane

Nomenclature of special compounds

Unbranched assemblies consisting of two or more identical hydrocarbon units joined by a single bond are
named by placing a suitable numerical prefix such as bi for two, ter for three, quater for four, quinque for five etc.
before the name of the repititive hydrocarbon unit. Starting from either end, the carbon atoms of each repititive
hydrocarbon unit are numbered with unprimed and primed arabic numerals such as 1, 2, 3...., 1', 2', 3' ....., 1'', 2'',
3''..... etc. The points of attachment of the repititive hydrocarbon units are indicated by placing the appropriate
locants before the name. For example,

2 2' 34 4'' 3''
1 1'
21 1' 2' 1'' 2''
3 3'

1, 1' -Bicyclopropane

4' 3'

1, 1', 2', 1'' -Tercyclobutane

As an exception, unbranched assemblies consisting of benzene rings are named by using appropriate prefix
with the name phenyl instead of benzene. For example,

32 2' 3' 2'' 3''

4 1 4' 4''
1' 1''

56 6' 5' 6'' 5''

1, 1', 4', 1'' -Terphenyl

Note :If two atoms/groups of same preference occupy identical positions from either end of the parent

chain, the lower number must be given to the atom/group whose prefix comes first in the
alphabetical order. For example,

4321 54 32 1
C H3 −CH C− CH − CH3
Cl C H 2 C H 2 C H 2 C H 2 Br − |
1-Bromo-4-chlorobutane | ||
OCH2CH3
OCH3 O

2-Ethoxy-4-methoxypentan-3-one

When two or more prefixes consist of identical words, the priority for citation is given to that group
which contains the lowest locant at the first point of difference. For example,

3


Cl

22 3

1− 1 H2 − 2 H2 −1 4 Cl

C C

1-(2-chlorolphenyl)-2-(4-chlorophenyl) ethane

If a compound contains a benzene ring coupled to an alicyclic ring, it is named as a derivative of
benzene, i.e. compound having lowest state of hydrogenation. For example,

56 23

4 1 1 4 NO2

3 2 CH3

1-(2-Methylcyclohexyl)-4-nitro benzene

In the common system of nomenclature, prefixes iso and neo are used only for compounds containing
an isopropyl group, (CH3 )2 CH and a tert-butyl group, (CH3 )3 C respectively at the end of the carbon
chain.

Nomenclature of simple aromatic compounds

Aromatic compounds are those which contain one or more benzene rings in them. An aromatic compound
has two main parts : (1) Nucleus, (2) Side chain
(1) Nucleus :The benzene ring represented by regular hexagon of six carbon atoms with three double bonds
in the alternate positions is referred to as nucleus. The ring may be represented by any of the following ways,

CH

HC CH

HC CH (ii) (iii) (iv)
CH
(i)

(2) Side chain : The alkyl or any other aliphatic group containing at least one carbon atom attached to the
nucleus is called side chain. These are formed by replacing one or more hydrogen atoms in the ring by alkyl
radicals i.e., R (R may be − CH 3 , − C2 H5 , − C3 H7 etc.)

Ring or nucleus

R Side chain

If one atom of hydrogen of benzene molecule is replaced by another atom or group of atoms, the derivative
formed is called monovalent substituted derivative. It can exist only in one form because all the six hydrogens

of benzene represent equivalent positions. For example, C6 H5 X , where X is a monovalent group.
When two hydrogen atoms of benzene are replaced by two monovalent atoms or group of atoms, the

resulting product can have three different forms. These forms are distinguished by giving the numbers. The position
occupied by one of the substituent is given as 1 and the other position is numbered in a clockwise direction.

(i) Ortho (or 1, 2-) : The compound is said to be ortho (or 1, 2-) if the two substituents are on the adjacent
carbon atoms.

(ii) Meta (or 1, 3-) : The compound is said to be meta or (1, 3-) if the two substituents are on alternate carbon
atoms.


(iii) Para (or 1, 4-) : The compound is said to be para or (1, 4-) if the two substituents are on diagonally
situated carbon atoms.

Ortho, meta and para are generally represented as o-, m- and p- respectively as shown below,

X X (1, 3-) Meta X
1X
1 1
2 (1, 2-) Ortho 2
2
(1, 4-) para

3

3X 4

X

Aryl group : The radicals obtained by removal of one or more hydrogen atoms of the aromatic hydrocarbon
molecules are known as aryl radicals or aryl groups. For example,

CH2– CH– –C–

Phenyl Benzyl Benzal Benzo
(From nucleus) From side chains

Nomenclature of different aromatic compounds : The names of few simple aromatic compounds are
given below :

Hydrocarbons CH3 CH3 CH3
CH3
CH2CH3

CH3

CH3 CH3 H3C CH3

Benzene Toluene Ethylbenzene 1, 3-Dimethylbenzene 1, CH3 1, 3, 5-Trimethyl
(m-Xylene) benzene (Mesitylene)
4-Dimethylbenzene
(p-Xylene)
1, 2-Dimethyl benzene
(o-xylene)

The aromatic hydrocarbons may also contain two or more benzene rings condensed together.

Naphthalene Anthracene Phenanthrene

Halogen derivatives Side chain substituted CHCl2
CH2Cl
Nuclear substituted Cl CCl3
Cl CH3 Cl

Cl

Chlorobenzene 2-Chlorotoluene ,1,2-Dichlorobenzene Phenyl chloromethane Phenyl dichloromethane Phenyl trichloromethane
(o-Chlorotoluene) (o-Dichlorobenzene)
(Benzyl chloride) (Benzal chloride) (Benzo chloride)


Hydroxy derivatives OH
OH CH3 OH
CH2OH OH OH
OH

Phenol 2-Methyl phenol Benzyl alcohol 1, 2-Dihydroxybenzene OH OH
o-Cresol (Catechol)
Ethers 1, 3-Dihydroxybenzene 1, 4-dihydroxybenzene
OCH2CH3 (Resorcinol)

OCH3 O

Methoxybenzene Ethoxybenzene Phenoxybenzene
(Anisole) (Phenetole)

Aldehydes and ketones (Nuclear substituted) CO

CHO COCH3

Benzaldehyde Methyl phenyl ketone Diphenyl ketone
(Acetophenone) (Benzophenone)
Carboxylic acids (Nuclear substituted)
COOH
COOH COOH
COOH CH3 COOH COOH

COOH OH

Benzoic acid o-Toluic acid 2-Hydroxy benzoic acid Phthalic acid Isophthalic COOH COOH
(o-salicylic acid)
acid Terephthalic acid

Acid derivatives O COOC2H5
OO OC CO

C Cl C NH2 COOCH3 COOC2H5 COOC6H5

Benzoyl chloride Benzamide Methyl benzoate Benzoic anhydride Ethyl benzoate Phenyl benzoate Br

Ethyl 4-bromobenzoate


Amino derivatives (Nuclear substituted) Sulphonic acids Nitro derivatives
SO3H
NH2 CH3 CH2NH2 NO2 NO2
NH2

NO2

Aniline (Aminobenzene) 2-Amino toluene Benzyl amine Benzene sulphonic acid Nitrobenzene 1, 3-Dinitrobenzene
or (m-Dinitrobenzene)

o-Toluidine

Some tips for nomenclature of aromatic compounds : For IUPAC nomenclature of substituted
benzene compounds, the substituent is placed as prefix to the word benzene. It may be noted that common names
of many substituted benzene compounds are still universally used. Some important tips for nomenclature of organic
compounds are given below,

(i) When the benzene ring is named as substituent on the other molecule, it is named as phenyl group. It is
treated in the nomenclature just like the name of an alkyl group. It is abbreviated as Ph. For example,

CH2CH2CHO Ph 45
or Ph CH 2CH 2CHO 3-Phenylpropanal
|3 2 1 C2 H5

C H = C H C OOH 3| 2 1
3-Phenylprop-2-enoic acid
C6 H5 − C − C H − C H3
||
C6H5 OH

1 2 34

CH 2 − C ≡ C − CH 3 or Ph C H 2 − C ≡ C− C H3 Phenylbut-2-yne

(ii) Disubstituted, trisubstituted or tetrasubstituted benzenes are named by using the numbers for the positions
of the substituents.
CH3 Cl NO2
1 Cl
1 1
2 2

3 2 53
1, 2-Dichlorobenzene NO2 4 NO2
CH3
1, 3, 5-Trinitrobenzene
1, 3-Dimethyl benzene

(iii) If different groups are attached to the benzene ring, then the following rules are kept in mind,

(a) The principal group is fixed as number 1.

(b) The numbering of the chain is done in any direction (clockwise or anticlockwise) which gives lower

number to the substituent.

(c) The substituents are written in alphabetical order. For example,

OH NH2 NH2 6 COOH
1 Cl 1 CH3 1 2 Br
4 5
2 35 2 3
HO
2-Chlorophenol 26 3 4 NO2
1
CH3 2-Bromo-5-hydroxy-
COOH 3-nitrobenzoic acid
2, 3-Dimethyl aniline
4-Aminobenzoic acid
(–COOH is principal group)


CHO 1 Cl NH2 NO2
6 1 2 CH3
1 42 NO2 12
2 53
O2N 3 4 3
3
1-Chloro, 2, 4, dinitrobenzene C2H5 4 Cl
4 CH3
4-Ethyl-2-methylaniline
OCH3 3-Chloro-4-methyl nitrobenzene

4-Methoxy benzaldehyde

OCH3 OH COH
1 2 Cl
12 12
3
3 3
4
4 CH3 4 OMe
CH3 CH3 OH

2-Chloro-4-methyl anisole 3, 4-Dimethylphenol 4-Hydroxy-3-methoxy
benzaldehyde


General organic chemistry

The chemical properties of an element depend on the electronic configuration of the outershell. Carbon has four
electrons in its outershell.

According to the ground state electronic configuration of carbon, carbon is divalent. Tetravalency of carbon can
be explained by promoting one 2s - electron to a 2pz orbital. Some energy must be supplied to the system in order to

effect this promotion. This promotion requires energy about 96 kcal/mol, but this energy is more than regained by the
concurrent farmation of chemical bonds.

6 C → ↑↓ ↑↓ ↑↑ ↑↓ ↑ ↑↑↑

1s2 2s2 2px 2p′y 2p′z 1s2 2s1 2px′ 2py′ 2pz′

Carbon in ground state Carbon in excited state

The four valencies of carbon atom are similar and they are 109o28′
symmetrically arranged around the carbon atom. According to Le Bel and 1.09Å C
Van’t Hoff the four valencies of carbon do not lie in one plane. They are
directed towards the corners of a regular tetrahedron with carbon atom at the
centre and the angle between any two valencies is 109o28′.

Bonding in organic compounds.

The organic compounds are carbon compounds consisting of one or more carbon atoms. Carbon must form
only covalent bonds, i.e., it should share its valency electrons with other atoms.

According to the modern concept, a covalent bond is formed between two atoms if there is an overlapping of
an atomic orbital of one atom with an atomic orbital of another atom. The overlapping is possible by two ways,

(1) End to end overlapping : This type of overlapping is possible between s − s, s − px and px − px atomic
orbitals. The molecular bond formed is termed as sigma (σ ) bond.

(2) Sidewise or parallel or lateral overlapping : Such overlapping is possible between p − p atomic

orbitals. The molecular bond formed is termed as pi (π ) bond.

σ -Bond π -Bond

Formed by End to End overlap of AO’s. Formed by lateral overlap of p -orbitals.

Has cylindrical charge symmetry about bond axis. Has maximum charge density in the cross-
Has free rotation sectional plane of the orbitals.

No free rotation, i.e., frozen rotation

Low energy Higher energy

Only one σ bond can exist between two atoms, One or two π bonds can exist between two atoms.
Sigma bonds are directional. Thus the geometry of the
molecule depends on the σ bonds. π bonds are not directional. Geometry of the
molecule not depends on π bond.
Area of overlapping is higher hence bond is stronger.
Area of overlapping is small hence bond is weaker.

σ bond can have independent existance. π bond always exist along with a σ bond and π
bond is formed after the formation of σ bond.

Hybridisation in Organic Compounds.

(1) The process of mixing atomic orbitals to form a set of new equivalent orbitals is termed as hybridisation.
There are three types of hybridisation encountered in carbon atom. These are,

(i) sp 3 hybridisation (involved in saturated organic compounds containing only single covalent bonds),

(ii) sp 2 hybridisation (involved in organic compounds having carbon linked by double bonds) and


(iii) sp hybridisation (involved in organic compounds having carbon linked by a triple bonds).

Type of Number of Number of unused Bond Bond Geometry % s-character
hybridisation orbitals used p-orbitals angle
Nil Four -σ Tetrahedral 25 or 1/4
sp3 1s and 3p One Three -σ 109.5° Trigonal 33.33 or 1/3
sp2 1s and 2p One -π 120°
Two -σ 50 or 1/2
sp 1s and 1p Two Two -π 180° Linear

(2) Determination of hybridisation at different carbon atoms : It can be done by two methods,
(i) First method : In this method hybridisation can be know by the number of π − bonds present on that
particular atom.

Number of π– bond/s 0 1 2
Type of hybridisation sp3 sp2 sp

O
||
Examples : (i) CH3 − CH = CH − C − CH3 (ii) CH 2 = C = CH 2
↓ ↓ ↓
↓ sp2 sp2 ↓ ↓ ↓ sp ↓
sp3 sp2 sp3 sp 2 sp 2

(iii) CH 3 − CH = CH − CH 2 − C ≡ N (iv) HC ≡ C − CH = CH 2
↓ ↓ ↓↓ ↓ ↓↓
↓ sp 2 sp 2 ↓ sp sp sp sp sp 2 ↓
sp 3 sp 3 sp 2

Note :  In diamond carbon is sp3 hybridised and in graphite carbon is sp 2 hybridised.

This method can not be used for those atoms of the molecule which have positive charge, negative
charge or odd electron.

(ii) Second method : (Electron pair method)
ep = bp + lp; where ep = electron pair present in hybrid orbitals , bp = bond pair present in hybrid orbitals
Number of bp = Number of atoms attached to the central atom of the species

First atom Central atom Central atom
Second atom HH
12 2 1
C=C H
H− C ≡ C− H
HH ↓ 3 bp = 3
bp = 2

bp = 3 Third atom

Number of lp’s can be determined as follows,
(a) If carbon has π - bond/s or positive charge or odd electron, than lp on carbon will be zero.
(b) If carbon has negative charge, then lp will be equal to one.
Number of electron pairs (ep) tells us the type of hybridisation as follows,


ep 2 3456
Type of hybridisation sp
sp2 sp 3 sp 3d sp 3d 2

Example :

⊕ Θ • (iv) CH ≡ CΘ (v) Θ
bp=↓1
(i) CH 2 = CH (ii) CH 2 = CH (iii) CH 2 = C − CH 3 lp = 1 CH 3 − CH − CH 3
↓ ↓ | ↓
CH 3 ep = 2, sp
bp = 2 bp = 2 bp = 3 bp=3
lp = 0 lp = 1 lp = 0 lp = 1
ep = 4, sp 3
ep = 2, sp ep = 3, sp 2 ep = 3, sp 2

(3) Applications of hybridisation
(i) Size of the hybrid orbitals : Since s - orbitals are closer to nucleus than p - orbitals, it is reasonable to

expect that greater the s character of an orbital the smaller it is. Thus the decreasing order of the size of the three
hybrid orbitals is opposite to that of the decreasing order of s orbital character in the three hybrid orbitals.

sp3 > sp2 > sp

(ii) Electronegativity of different orbitals

(a) Electronegativity of s-orbital is maximum.

(b) Electronegativity of hybrid orbital ∝ % s-character in hybrid orbitals

Orbital sp sp 2 sp 3
s-cha%racste-rcihnadraecctreerasingorde5r 0and electro3n3.e3g3ativity in d2e5creasingorde→r

Thus sp-hybrid carbon is always electronegative in character and sp3 - hybrid carbon is electropositive in
character. sp 2 -hybrid carbon can behave as electropositive (in carbocation) as well as electronegative (in
carbanion) in character.

⊕ ⊕

CH 3 − CH 2 CH2 = CH
sp 2 sp

Electropositive carbon Electronegative carbon
having positive charge

(c) Electronegativity of different hybrid and unhybrid orbitals in decreasing order is as follows

s >sp >sp2> sp3 >p→
% s - character in decreasing order and electronegativity in decreasing order.

(iii) Bond length variation in hydrocarbons

% s orbital character ∝ C − C 1 length ∝ C − H 1 length
bond bond


Bond type (C – H) Bond length Bond type (C – C) Bond length
sp3 − s (alkanes) 1.112Å sp3 − sp3 (alkanes) 1.54 Å
sp 2 − s (alkenes) 1.103Å sp 2 − sp 2 (alkenes) 1.34Å
sp − s (alkynes) 1.08Å 1.20Å
sp − sp (alkynes)

Note :  C–C bond length in benzene lies between single and double bond due to resonance. (1.40Å).

(iv) Bond strength in hydrocarbons : The shorter the bond, the greater the compression between atomic
nuclei and hence greater the strength of that bond is. Thus the bond formed by sp hybridised carbon is strongest

(i.e., it has maximum bond energy) while that formed by sp3 hybridised carbon is the weakest (i.e., it has minimum

bond energy). This is evident by the bond energies of the various types of C − H and C − C bonds.

Bond type (C – H) Bond energy Bond type (C – C) Bond energy
(kcal/mole) (kcal/mole)

sp3 − s (in alkanes) 104 sp3 − sp3 (in alkanes) 80 – 90

sp 2 − s (in alkenes) 106 sp 2 − sp 2 (in alkenes) 122 – 164

sp − s (in alkynes) 121 sp − sp (in alkynes) 123 – 199

(v) Acidity of hydrocarbons
(a) Hydrogen present on electronegative carbon is acidic in character.
(b) Acidity of hydrogen is directly proportional to the electronegativity of atom on which hydrogen is present.

Thus

H −O−HNH3 CH≡CH →
Electronegativity of atom in decreasing order
Acidity of compounds in deceasing order

(c) Acidity of hydrocarbon ∝ % s-character

CH ≡ CH CH 2 = CH 2 CH 3 − CH 3
25
% s-character 50 33.33 50

pKa 25 44

s- character in decreasing order and acidity in decreasing order

Note :  Acidity ∝ Ka and Acidity ∝ 1 (pKa = − log Ka)
pKa

Order of acidic nature of alkynes is, HC ≡ CH > HC ≡ C − CH3
The relative acidic character follows the order;
H2O > ROH > HC ≡ CH > NH3 > CH2 ≡ CH2 > CH3 − CH3
Obviously, the basic character of their conjugate bases follows the reverse order, i.e.,

CH 3CH  > CH2 = CH  > NH  > HC ≡ C > RO  > HO
2 2


Dipole moment of Organic Compounds.

(1) Due to differences in electronegativity polarity developes between two adjacent atoms in the molecule
(i.e., in a bond). The degree of polarity of a bond is called dipole moment. Dipole moment is represented by µ and
its unit is Debye (D).

µ = e×l

Where, e = magnitued of separated charge in e.s.u., l = internuclear distance between two atoms i.e., bond
length in cm.

The dipole moment is denoted by arrow head pointing towards the positive to the negative end (↦).
(2) Dipole moment of the compound does not depend only on the polarity of the bond but also depends on
the shape of the molecule.Dipole moment of symmetrical compound is always zero, ( µ = 0 ). Symmetrical
compounds are those compounds which fulfil following two conditions,

(i) Central atom is bonded with the same atoms or groups.Examples : H 2 , BF3 , CSySm2m, CetHric2al =moCleHcu2le,sCH ≡ CH

(ii) Central atom should have no lone pair of electrons.

.. ..

Examples : CCSylm4 ,mCeHtric4a,lBmHole3c,uCleOs 2 UnsyHmm2 eOtr,icalHm2olSecules

Note :  Compounds which have regular tetrahedral structure has no dipole moment.

(3) µ ∝ electronegativity of central atom or surrounding atoms present on the central atom of the molecule.

CHF3ElectroCnHegCatliv3ity in decCreHasBinrg3orderCHI3 →

µ is also in decreasing order

NHE3lectronegPaHtiv3ity of central aAtosmHis3indecreaSsibnHg 3 →

order µ is also in decreasing order

Note :  Decreasing order of dipole moment in CH3Cl, CH 2Cl2, CHCl3 and CCl4 is

CH3Cl > CH2Cl2 > CHCl3 > CCl4

µ = 1.86 D 1.62 D 1.03 0
Alkynes has larger dipole moment because the electronegativity of sp − C is more than that of

sp2 − C .

(4) µ cis > µ trans in geometrical isomers.

(5) Dipole moment of the trans derivative of the compound (a)(b)C = C(a)(b) will only be zero if both a and

b will be in the form of atoms.

HX HH

Example : C=C C=C ( X = Cr, Br, I or F )

XH X X

µ=0 µ≠0

If both will not be atoms then µ trans may or may not be zero.


If group have non-linear moments, then the dipole moment of the trans isomer will not be zero. If group have
linear moments, then the dipole moment of the trans isomer will be zero.

H3C CH 3 H CH 3
C=C
Example : C=C

HH H3C H

µ =0.33D µ =0 D

(6) Dipole moment of disubstituted benzene
(i) When both groups X and Y are electron donating or both groups are electron with drawing

XX

Y or Y

Then, µ = µ 2 + µ 2 + 2µ1 µ 2 cosθ
1 2

Where, µ1 = dipole moment of bond C − X , µ 2 = dipole moment of bond C − Y , θ = angle between X and Y.

If value of θ will be more, then cosθ will be less. Hence, dipole moment will be as,

o−derivative µ>inmd-edcereriavsaintigveord>er p-derivative →

(ii) When one group is electron with drawing and the other group is electron donating then,

µ= µ12 + µ 2 − 2µ1 µ 2 cosθ
2

Hence, dipole moment is as follows,

o−derivative µ>in mde-cdreeraivsiantgivoerde>rp-derivative →

Steric effect.

On account of the presence of bulkier groups at the reaction centre, they cause mechanical interference and
with the result that the attacking reagent finds it difficult to reach the reaction site and thus slows down the reaction.
This phenomenon is called steric hinderance or steric effect.

(1) Tertiary alkyl halides having bulky groups form tertiary carbocation readily when hydrolysed because of
the presence of the three bulky groups on the carbon having halogen.
Steric strain is released
CH Steric strain around this carbon (less strained species)
|3 (More strained species) ⊕

H 3C − C| − Cl → H 3C − C − CH 3

CH ↑ 3
3 CH

(2) Primary alkyl halide having quaternary β -carbon does not form transition state because of the steric strain

around α -carbon by the β -carbon. To release the strain it converts into carbocation.

|CH3 Bulky group

CH 3 −C CH 2 − Cl
|
CH
3 Strained carbon due to
bulky group present

around this carbon.

(3) Steric strain inhibits the resonance. This phenomenon is known as steric inhibitions of resonance.


Intermolecular forces.

In an ionic compound, the structural units are ions. These ions are held together by very powerful electrostatic
forces are known as inter ionic forces. On the other hand, in non-ionic (covalent) compounds, the structural units
are molecules. These molecules are held together by very weak forces are known as intermolecular forces or
secondary forces. Secondary forces are of the following types,

(1) Dipole – dipole forces. (2) Van der Waal’s forces. (3) Hydrogen bond.

(1) Dipole – Dipole Forces : These forces exist between polar molecules which have permanent dipoles.
The interactions of the permanent dipole in different molecules are called dipole-dipole forces (DF).

Magnitude of DF depends on the dipole moment (µ) of the bond of the compound and intermolecular distance (d),

DF ∝ µ ∝ 1 (i.e. these forces are effective only over short distance)
d4

Example CH 3 − Cl, CH 3 − Br, CH 3 − I
d in increasing order, µindecreasing orderandDF alsoindecreasing order →

(2) Vander Waal’s forces : These forces exist between non-polar molecules. The intermolecular electrostatic
attractions between nuclei of one molecule and electrons of the other molecule are called Vander Waal’s forces

(VF). Magnitude of VF depends on the number of electrons (e − ) and protons (p) in the molecule as well as on the

intermolecular distance (d),

VF ∝ number of e − and p; ∝ 1 / d7 ; ∝ MW; ∝ Surface area of the molecule and

∝ Symmetry of the molecule (symmetry of molecule decreases intermolecular distance (d)).

(3) Hydrogen bonding : An electrostatic attractive force between the covalently bonded hydrogen atom of
one molecule and an electronegative atom (such as F, O, N ) of the other molecule is known as hydrogen

bonding.

Examples of H-bonding in between the two molecules of same or different compounds are

H – F --- H– F H – O --- H – O HH RR RH H O–H
|| || || ||
Hydrogen fluoride HH H – N --- H – N O – H --- O – H ||
(associated) Water || O – H --- O – H H – N --- H
HH Alcohol
(associated) Ammonia (associated) Alcohol and Water |a
(associated)
Ha

Ammonia and water

Nature and Importance of Hydrogen bonding

(i) Hydrogen bond is merely an electrostatic force rather than a chemical bond.
(ii) Hydrogen bond never involves more than two atoms.
(iii) Bond energy of hydrogen bond is in the range of 3 to 10 kcal/mol or 10 to 40 kJ/mol, i.e., about 1/10th
the energy of a covalent bond.
(iv) With the increase of electronegativity of the atom to which hydrogen is covalently linked, the strength of
the hydrogen bond increases.
(v) All the three atoms in X – H- - - X lie in a straight line.
(vi) The bond length of hydrogen bond is of the order of 250 to 275 pm.
The relative order of these intermolecular forces is,
Hydrogen bonding > dipole-dipole forces > Vander waal’s forces.


Mechanism of organic reactions.

When a chemical reaction takes place between two or more chemical species, new products are formed. This
change is represented by a chemical equation. In a chemical equation, reactants are written on the left hand side
while the products are written on the right hand side. The two are separated by an arrow (→). The reactants
normally consists of two species,

(1) Substrate : The species, which is attacked by some other chemical species, is called a substrate.
(2) Reagent : The species, which attacks the substrate in order to get the major product, is called a reagent.
Thus, Substrate + Reagent → Products.

Normally, a substance and a chemical reagent form a highly energetic species, called activated complex,
before it changes into the product. In certain cases, a relatively energetically more stable species than the activated
complex may also be formed. It is called reaction intermediate. Thus, a chemical reaction, in general, may follow
either of the following two paths,

Path I : Substrate + Reagent → Activated complex → Products

Path II : Substrate + Reagent → Activated complex → Intermediate → Products.

The detailed step by step description of a chemical reaction is called mechanism of a reaction which is only a
hypothesis. If the reaction mechanism involves the breaking and making of bonds simultaneously without the
formation of any intermediate, it is called concerted mechanism. On the other hand, if the reaction mechanism
involve the formation of intermediates before the formation of products, it is called non-concerted mechanism.

Activated Complex Activated Complex
Reactive intermediate

Enthalpy
Enthalpy
Reactants Reactants

Products Products
Progress of reaction
Progress of reaction
Non-concerted mechanism (formation of
Concerted mechanism (formation of activated complex and reactive intermediate).
activated complex only).

Enthalpy curves for concerted and Non-concerted mechanisms

To understand clearly the mechanism of various organic reactions, it is essential to have knowledge about the
following concepts;

• Electronic displacements in covalent bonds,

• Cleavage (fission or breaking) of covalent bonds,

• Nature of attacking reagents.

Electronic displacement in covalent bonds.

It is observed that most of the attacking reagents always possess either a positive or a negative charge,
therefore for a reaction to take place on the covalent bond the latter must possess oppositely charged centres. This is
made possible by displacement (partial or complete) of the bonding electrons. The electronic displacement in turn
may be due to certain effects, some of which are permanent and others are temporary. The former effects are
permanently operating in the molecule and are known as polarisation effects, while the latter are brought into
play by the attacking reagent and as soon as the attacking reagent is removed, the electronic displacement
disappears; such effects are known as the polarisability effects.


Electronic displacement

Polarisation effect (permanent) Polarisability effect (temporary)

Inductive Mesomeric Hyperconjugative Inductomeric Electromeric
effect effect effect effect effect

Inductive effect or Transmission effect

(1) When an electron with drawing (X) or electron-releasing (Y) group is attached to a carbon chain, polarity
is induced on the carbon atom and on the substituent attached to it. This permanent polarity is due to electron
displacement due to difference in electronegativities. This is called inductive effect or simply as I – effect.

C − C − C − C Non polar

Cδδδδ + Cδδδ + Cδδ + Cδ + Xδ− Cδδδδ − Cδδδ − Cδδ − Cδ − Yδ+

(2) Important features of Inductive effect
(i) It is a permanent effect in the molecule or ion.

(ii) It operates through σ bonds.

(iii) It is generally observed in saturated compounds.

(iv) The shared pair of electrons although permanently shifted towards more electronegative atom, yet
remains in the same valence shell.

(v) As a result of electron shifting, the more electronegative end acquires partial negative charge and the other
acquires partial positive charge.

(vi) The inductive effect is not confined to the polarization of one bond but is transmitted along a chain of
carbon atoms through σ bonds. However, the effect is insignificant beyond second carbon in the chain.

(vii) Inductive effect brings changes in physical properties such as dipole moment, solubility, etc. It affects the
rates of the reaction.

(viii) Carbon-hydrogen bond is taken as a standard of inductive effect. Zero effect is assumed for this bond.
Atoms or groups which have a greater electron withdrawing capacity than hydrogen are said to have–I effect
whereas atoms or groups which have a greater electron releasing power are said to have +I effect.

N⊕ H3 > NO2 > CN > SO3 H > CHO > CO > COOH > COCl > COOR > CONH2 > F > Cl > Br > I > OH > OR > NH2 > C6 H5 > H

–I power of groups in decreasing order with respect to the reference H

ter. alkyl > sec. alkyl > pri. alkyl > CH3 > H

+ I power in decreasing order with respect to the reference H

+ I power ∝ number of carbon’s in same type of alkyl groups

CH3 − CH2 − CH2 − CH2 − > CH3 − CH2 − CH2 − > CH3 − CH2 −

+ I power in decreasing order in same type of alkyl groups


(3) Application of Inductive effect

(i) Magnitude of positive or negative charge : Magnitude of +ve charge on cations and magnitude of
–ve charge on anions can be compared by + I or – I groups present in it.

Magnitude of +ve charge ∝ + I power 1 the group Magnitude of –ve charge ∝ − 1 the group
of I power of

Magnitude of +ve charge ∝ –I power of the group Magnitude of –ve charge ∝ + I power of the group

(ii) Reactivity of alkyl halide : + I effect of methyl group enhances – I effect of the halogen atom by
repelling the electron towards tertiary carbon atom.

CH3 CH3

H3C C X > H 3C CH X > CH 3 CH 2 X > CH 3 X

CH3

Tertiary > Secondary > Primary > Methyl

(iii) Dipole moment : As the inductive effect increases, the dipole moment increases.

CH3 I CH3 Br CH3 Cl

1.64 D 1.79 D 1.83 D

Inductive effect increases

(iv) Relative strength of the acids (Acidic nature of − COOH )

(a) An acid may be defined as a species that has the tendency to loose proton. Furthermore, the strength of
an acid depends on the tendency to release proton when the acid is dissolved in water.

OO
|| ||
R − C− OH ⇌ R − C− O + H ⊕
(b) The relative strength of acids are measured in their ionisation constants ( Ka or pKa values).

HA ⇌ H⊕ + A ; Ka = [H ⊕ ][A ] ; pKa = − log Ka
[HA]
Acid

Greater the value of Ka or lower the value of pKa stronger will be the acid.

(c) Any group or atom showing +I effect decreases the acid strength as it increases the negative charge on
the carboxylate ion which holds the hydrogen firmly. Alkyl groups have + I effect.

Thus, acidic nature is, HCOOH > CH3COOH > C2H5COOH > C3 H7COOH > C4 H9COOH
(+ Inductive effect increases, so acid strength decreases)

Formic acid, having no alkyl group, is the most acidic among these acids.

(d) The group or atom having – I effect increases the strength as it decreases the negative charge on the
carboxylate ion. Greater is the number of such atoms or groups (having – I effect), greater is the acid strength.

Thus, acidic nature is, CCl3COOH > CHCl2COOH > CH 2ClCOOH > CH 3COOH
Trichloro Dichloro Monochloro Acetic acid
acetic acid acetic acid acetic acid

(– Inductive effect increases, so acid strength increases)


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