Chapter 13 Purification, classification and nomenclature of organic compound 1 by TEACHING CARE Online coaching and tuition classes

Chapter 13 Purification, classification and nomenclature of organic compound 1





The word ‘organic’ signifies life. Therefore, all substances which were obtained directly or indirectly from living organisms, plants and animals were called organic compounds and the branch of chemistry which deals with these compounds was called organic chemistry.

Modern definition of organic chemistry : Organic chemistry is a chemistry of hydrocarbons and their derivatives in which covalently bonded carbon is an essential constituent.

Berzelius put forward a theory in 1815 known as vital force theory. According to this theory, “organic compounds could be prepared only by living organism under the influence of a mysterious force known as vital force”. Accidental synthesis of urea by Wohler and synthesis of acetic acid by Kolbe led to the fall of this theory.



NH4CNO ¾¾D ® NH2  – CNH2                                            ClCHCOOH ¾¾Zn /¾H¾Cl ® CH3  – COOH


(Ammonium cyanate)


(First organic compound synthesised in laboratory)

(Chloroacetic acid)

Acetic acid

(First organic compound synthesised from its elements)


Berthelot prepared methane in laboratory and the most abundant organic compound is cellulose which is

a polymer of glucose. Kekule and Couper proposed the tetravalency of carbon and wrote the first structural formula. In 1874, Van’t Hoff and Le Bel suggested that four bonds in the most carbon compounds are directed towards the corners of a tetrahedron.

The study of organic compounds starts with the characterisation of the compound and the determination of its molecular structure. The procedure generally employed for this purpose consists of the following steps :

  • Purification of organic compounds
  • Qualitative analysis of organic compounds
  • Quantitative analysis of organic compounds
  • Determination of molecular mass of organic compounds
  • Calculation of Empirical formula and Molecular formula of organic compounds
  • Determination of structure of organic compounds by spectroscopic and diffraction methods
  • Purification of organic compounds : A large number of methods are available for the purification of The choice of method, however, depends upon the nature of substance (whether solid or liquid) and the type of impurities present in it. Following methods are commonly used for this purpose,
    • Simple crystallisation (ii) Fractional crystallisation (iii) Sublimation (iv) Simple distillation

(v) Fractional distillation (vi) Distillation under reduced pressure (vii) Steam distillation

(viii) Azeotropic distillation (ix) Chromatography (x) Differential extraction (xi) Chemical methods

  • Simple crystallization : This is the most common method used to purify organic solids. It is based upon the fact that whenever a crystal is formed, it tends to leave out the For crystallization, a suitable solvent is selected (a) which dissolves more of the substance at higher temperature than at room temperature (b) in which impurities are either insoluble or dissolve to an extent that they remain in solution (in the mother liquor) upon crystallization, (c) which is not highly inflammable and (d) which does not react chemically with the compound to be crystallized. The most commonly used solvents for crystallization are : water, alcohol, ether, chloroform, carbon- tetrachloride, acetone, benzene, petroleum ether etc.




Examples : (a) Sugar having an impurity of common salt can be crystallized from hot ethanol since sugar dissolves in hot ethanol but common salt does not.

(b) A mixture of benzoic acid and naphthalene can be separated from hot water in which benzoic acid dissolves but naphthalene does not.

Note :®Sometimes crystallization can be induced by adding a few crystals of the pure substance to the concentrated solution. This is called seeding.

  • Fractional crystallization : The process of separation of different components of a mixture by repeated crystallizations is called fractional crystallization. The mixture is dissolved in a solvent in which the two components have different solubilities. When a hot saturated solution of this mixture is allowed to cool, the less soluble component crystallises out first while the more soluble substance remains in solution (mother liquor). The mother liquor left after crystallization of the less soluble component is again concentrated and then allowed to cool when the crystals of the more soluble component are The two components thus separated are recrystallized from the same or different solvent to yield both the components of the mixture in pure form.


Fractional crystallization can be used to separate a mixture of

KClO3 (less soluble) and KCl (more soluble).


  • Sublimation : Certain organic solids on heating directly change from solid to vapour state without passing through a liquid state, such substances are called sublimable and this process is called







The sublimation process is used for the separation of sublimable volatile compounds from non sublimable impurities. The process is generally used for the purification of camphor, naphthalene, anthracene, benzoic acid NH4Cl, HgCl2 , solid SO2 , Iodine and salicylic acid etc containing non-volatile impurities.

  • Simple distillation : Distillation is the joint process of vaporisation and This method is used for the purification of liquids which boil without decomposition and contain non-volatile impurities. This method can also be used for separating liquids having sufficient difference in their boiling points. This method can be used to separate a mixture of
  • chloroform (b. 334 K) and aniline (b. p. 457 K)
  • ether (b. 308 K) and toluene (b. p. 384 K)
  • benzene (b. 353 K) and aniline (b. p. 457 K)

Note :®Although the b.p. of H2O (100oC) and ethanol (78.1oC) differ by 21.9oC, they can not be separated by simple distillation since they form an azeotrope.

  • Fractional distillation : This process is used to separate a mixture of two or more miscible liquids which have boiling points close to each Since in this process, the distillate is collected in fractions under different temperatures, it is known as fractional distillation. This process is carried out by using fractionating columns. Fractionating column is a special type of long glass tube provided with obstructions to the passage of the vapour upwards and that of liquid downwards. This method may be used to separate a mixture of acetone (b. p. 330 K) and methyl alcohol (b. p. 338 K) or a mixture of benzene and toluene. One of the technological applications of fractional distillation is to separate different fractions of crude oil in petroleum industry into various useful fractions such as gasoline, kerosene oil, diesel oil, lubricating oil etc.




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

  • Distillation under reduced pressure (vacuum distillation) : In this process, a liquid is distilled much below its boiling 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.

  • 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

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.

  • 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 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.

  • 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 This method was described by Tswett in 1906.
  • 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 The stationary phase is either a solid or a liquid while the moving phase may be a liquid or a gas.
  • 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
Thin-layer chromatography Liquid/Solid Qualitative           analysis         (identification          and
    characterization of organic compounds)
High               performance              liquid Liquid/Solid Qualitative and quantitative analysis
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,  a-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 ( Rf value). This gives the relative adsorption of each component of the mixture.

R Distance moved by the substance from the base line

f          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 Rf



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.

  • 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

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.

  • 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 For example,


  • 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.


  • 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 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.

  • A mixture of 1o, 2o and 3o amines can be separated using either benzenesulphonyl chloride (Hinsberg’s reagent) or diethyl oxalate (Hoffmann’s method).
  • Purification of commercial benzene : Commercial benzene obtained from coal-tar distillation contains 3-


5% thiophene as an impurity which can be removed by extraction with conc.

H 2 SO4 . This purification is based


upon the fact that thiophene undergoes sulphonation much more easily than benzene. Thus, when commercial


benzene is shaken with conc.

H 2 SO4

in a separating funnel, thiophene undergoes sulphonation to form thiophene-


  • sulphonic acid which dissolves in

H 2 SO4

while benzene does not.


  • H 2 SO4


¾¾Roo¾m  t¾em¾p ®


  • H 2 O



Thiophene-2-sulphonic acid (Dissolves in conc. H2SO4 )


After this treatment, the benzene layer is removed, washed with water to remove unreacted

over anhyd. CaCl 2 and then distilled to give pure benzene.

H 2SO4 , dried


  • 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 alcohol distils over leaving behind, Ca(OH)2 .

Ca(OH)2 . When the resulting mixture is distilled, absolute


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

P4 O10 , solid KOH or NaOH, etc.

CaCl2 , conc.

H2SO4  ,


  • 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 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.

  • 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


and hydrogen to

HO .

CO2    is


detected by lime water which turns milky while This method is known as copper oxide test.


is detected by anhydrous


(white) which turns it blue.


C+ 2CuO ¾¾He¾at ® CO2  + 2Cu  ;  Ca(OH)2 + CO2 ¾¾® CaCO3 + H2O

Lime water                                     Milky

H 2  + CuO ¾¾He¾at ® HO + Cu  ;  CuSO4 + 5HO ¾¾® CuSO4 .5HO


Colouriess (Anhydrous)

Blue (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


  • Fusion with Na forms soluble salt (like

NaCl, NaCN

etc.) which can be easily detected.


  • This test fails in case of diazo
  • Sometimes when the amount of nitrogen present is small, the prussian blue is present in colloidal form and the solution looks

Lassaigne method (Detection of elements)


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





Na + C + N ¾¾D ® NaCN







2Na + S ¾¾D ® Na2S


S.E.+ FeSO4 + NaOH , boil and cool + FeCl3 + conc.HCl Blue or green colour


(i)      S.E. + sodium nitro prusside

(ii)   S.E+

 CH3CO2 H + (CH3CO2 )2 Pb

A black ppt.

 2NaCN + FeSO4 ¾¾® Fe(CN)2 + Na2SO4

 Fe(CN)2 + 4 NaCN ¾¾® Na4 [Fe(CN)6 ]

3Na4 [Fe(CN)6 ] + 4 FeCl3  ¾¾H¾Cl ® Fe4 [Fe(CN)6 ]3 + 12NaCl

Prussian blue


(i)        Na2S + Na2[Fe(CN)5 NO] ¾¾® Na4 [Fe(CN)5 NO.S]

deep violet


(ii)     Na2S + (CH3COO)2 Pb ¾¾CH¾3CO¾O¾H ® PbS ¯+ 2CH3COONa

black ppt.




Other methods for detection of elements

Element                                                                                 Test

Nitrogen              Soda lime test : A pinch of an organic compound is heated strongly with soda lime (NaOH + CaO) in a test tube.  If  ammonia  gas  evolves, it  indicates  nitrogen.  CHCONH 2 + NaOH ¾¾Ca¾O ® CHCOONa + NH.  This  test


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

Sulphur                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.

Na2 CO3  + S + 3O ¾¾® Na2 SO4  + CO2 .   The   fused   mass   is


NaSO4 + BaCl 2 ¾¾® BaSO4 + 2NaCl

(White ppt.)

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.



Halogens             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 + Cl2 ¾¾®

2NaCl + Br2

tums CS 2 layer orange

; 2NaI + Cl2 ¾¾®

2NaCl + I 2 turns CS 2 layerviolet


Phosphorus        Phosphorus is detected by fusing the organic compound with sodium peroxide when phosphorus is converted into sodium phosphate.   2P + 5 Na2 O2 ¾¾® 2Na3 PO4 + 2Na2 O .    The fused mass is extracted with


 HO , 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.



  • 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, 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 Hydrogen










and Liebig’s  combustion  method  :  In  this  method,  a            (i) % of C =           Weight of CO2          ´ 12 ´ 100

known weight of organic compound is heated with                            Weight of org. compound     44

pure and dry cupric oxide in a steam of pure and                                          Weight of H O              2

dry  oxygen,  when  carbon  is  oxidised  to  carbon         (ii) % of H =                               2            ´       ´ 100

dioxide while hydrogen is oxidised to water. From                             Weight of org. compound     18

the weight of CO2 and H 2 O , the  percentage  of  C and H can be calculated.

Cx Hy  + æ x +  y öO2  ¾¾D ® xCO2  + y HO


ç       4 ÷                                2

è          ø


(i) Duma’s method : Elemental nitrogen is converted         Oxides of nitrogen + Cu ¾¾® N 2 + CuO

into  molecular  nitrogen    by    suitable  chemical

method and its voiume is changed to STP data.                   % of N    28 ´ V ´ 100

22400 W

C H N + CuO ¾¾® xCO + y H O + z N + (Cu)

x       y       z                                                 2         2     2             2     2                                    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    NH       by   suitable                                 W


chemical  method  which,  in  turn,  is  absorbed  by       Note  :  This  method  is,  however,  not  applicable  to

 V1mL  of  N1 H2 SO4 .                                                                       compounds containing nitrogen in the ring (e.g. Pyridine, quinoline etc) and compounds containing

 N(from organic compound) + conc. H2SO4 ¾¾D ®(NH4 )2 SO4                       nitro  and  azo  (–  N  =  N  –)  groups  since  nitrogen  in

(NH  )  SO   + 2NaOH ¾¾® Na  SO   + 2H  O + 2NH                                                  these  compounds  is  not  completely  converted  into

4 2      4                              2     4         2              3                         (NH   ) SO    during digestion.

4 2          4

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

143.5     Mass of substance taken

% of Br = 80 ´ Mass of AgBr formed ´ 100

188     Mass of substance taken

% 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.

Carius   method   :   When   an   organic   compound     % of S   32 ´ Mass of BaSO4 formed (W1 ) ´ 100

containing sulphur is heated with fuming nitric                      233  Mass of substance taken (W)

acid, sulphur is  oxidised to  sulphuric acid.  This is precipitated as barium sulphate by adding barium chloride solution. From the amount of barium sulphate, percentage of sulphur can be calculated.

 S + HNO3 (fuming) ¾¾he¾at ® HSO4

 H 2 SO4 + BaCl 2 ¾¾® BaSO4  + 2HCl

Carius method : The organic compound containing phosphorus is heated with fuming nitric acid. Phosphorus is oxidised to phosphoric acid. It is precipitated as magnesium ammonium

phosphate,     MgNH4 PO4 ,    by    the    addition    of

% of P = 62 ´ Mass of Mg 2 P2O7 formed (W1 ) ´ 100

222       Mass of substance taken (W)















Percentage of oxygen = 100 – (Sum of the percentages of all other elements)






     O º CO º CO2

16 g             44 g

% of O 16 ´       mass of CO 2       ´ 100

44    mass of org. compd.

magnesia mixture

(MgSO4 + NH4 OH + NH4 Cl ). The magnesium ammonium phosphate is washed, dried and ignited when it is converted to magnesium pyrophosphate (Mg 2 P2 O7 ) .

2MgNHPO4  ¾¾he¾at ® MgPO7  + 2NH 3  + HO

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 in an organic compound is by the method of difference. All the elements except oxygen present in the organic compound are estimated and the total of their percentages subtracted from 100 to get the percentage of oxygen.

(ii)              Aluise’s method :. Organic compound containing oxygen is heated with  graphite  and CO formed is quantitatively converted into  CO2 on reaction with I2O5.

Org. compound ¾¾Py¾ro¾lys¾is ® Oxygen


 O2  + 2C ¾¾110¾0 ¾C ® 2CO

5CO + IO5 ¾¾® I 2 + 5CO2



Example : 1       The percentage of

N 2 in urea is about                                                                              [KCET (Med.) 2001]


(a) 18.05                      (b) 28.29                      (c) 46.66                      (d) 85.56

Solution: (c)      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%



Example : 2        58 ml. of


are used to neutralize ammonia given by 1g of organic compound. Percentage of


5     2           4

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

Mass of substance

= 1.4 ´ 1 ´ 58 = 16.2

1 ´ 5


Example : 3 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 BaSO4

  1. wt. of organic compound

´ 100

= 32 ´





´ 100

= 18.52%


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 =

  1. wt. of H 2 O
  2. wt. of organic compound

´ 2 ´ 100


= 0.18 ´ 2

0.2     18

´ 100 = 10


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



% H =   2



´ 0.12 ´ 100 = 6.666



\% 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

22400 ´ weight of compound

28 ´ 31.1 ´ 100 = 15.55%

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 Mass of substance taken

´ 100 =




´ 0.12 ´ 100 = 34 (Approx.)



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 2 SO4


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





\ vol. of 1N H 2 SO4 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.


  • Determination of Molecular Mass : The molecular mass of the organic compounds can be determined by various
    • Physical methods for volatile compounds
  • 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 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 = T1 K

Pressure (after deducting aqueous tension) = p1mm

Let the volume at NTP be = V2 ml

Applying gas equation,


V   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 )


or V. D. =             W            ; Mol. Mass, M = 2 ´ V.D. =             2W         

V2 ´ 0.00009                                               V2 ´ 0.00009

  • 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 The molecular mass of the organic substance can be calculated by the application of following relationship,


Mol. Mass =

Mass of the substance volume of the vapours at NTP

´ 22400


  • 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 :

  • Elevation in boiling point : Mass = 1000 Kb ´w

W ´ DT



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

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

Mass of the solvent


  • Depression in freezing point : M Mass = 1000 Kf´ w

W ´ DT



Kf = Molal depression constant of the solvent, w = Mass of the compound, W = Mass of the solvent


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

(iii)    Chemical methods

  • Silver salt method for acids : It is based on the fact that silver salt of an organic acid on heating gives residue of metallic


RCOOAg ¾¾he¾at ®

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 saltMass 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 = éw ´ 108 – 107ù

x                                êë x                                                                             úû

If n be the basicity of the acid, then Mol. Mass of the acid = éw ´ 108 – 107ù ´ n

êë x                                                                            úû


  • 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

BHPtCl6  ¾¾he¾at ® Pt

Molecular mass of the salt = Mass of platinum salt


B2 H 2 PtCl6 .


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 + 410wMass of platinum salt ;

2E = éw ´ 195 – 410ù

; E = 1 éw ´ 195 – 410ù



x           Mass of platinum

êë x

ûú            2 êë x                                      úû


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

where n is the acidity of the base.

  • 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 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–ml 1Nalk–a–li s–olu–ti–on

One gram equivalent of alkali

º One gram equivalent of the acid


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

V ml N1 alkali solution º w g acid


So 1000 ml N1 alkali solution º   w         ´ 1000 g

V ´ N1

Equivalent mass of the acid º   w      ´ 1000

V ´ N1

acid º one gram equivalent acid


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–ml N–ac–id–so–lut–ion  One gram equivalent of the acid

º One gram equivalent of the base


Molecular mass of the base = Eq. mass ´ acidity

(5)  Calculation of Empirical and Molecular formula

  • 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 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 H12 O6 .

Calculation of empirical formula : The steps involved in the calculation are as follows,

  • Divide the percentage of each element by its atomic This gives the relative number of atoms.
  • Divide the figures obtained in step (i) by the lowest This gives the simplest ratio of the various elements present.
  • If the simplest ratio obtained in step (ii) is not a whole number ratio, then multiply all the figures with a suitable integer e., 2, 3, etc. to make it simplest whole number ratio.
  • Write down the symbols of the various elements side by side with the above numbers at the lower right corner of This gives the empirical or the simplest formula.




  • Molecular formula : Molecular formula of a substance gives the actual number of atoms present in one molecule of the

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.

  • 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 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
  • The mixture is exploded by passing an electric spark between the platinum As a result, carbon


and hydrogen of the hydrocarbon are oxidised to CO2 and


vapours respectively.


  • 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 .


  • Caustic potash or caustic soda solution is then introduced into the eudiometer tube which absorbs completely and only unused O2 is left 2NaOH + CO2 ¾¾® Na2CO3 + H2O



Thus, the decrease in volume on introducing NaOH or KOH solution gives the volume of




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 Hy + (x + y / 4)O2 ¾¾® xCO2 + y / 2H 2 O


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,

  • (x + y / 4) volume of O2 is used
  • x volume of CO2 is produced


  • y/2 volume of H 2O

vapours is produced which condense to give liquid


with negligible volume.


  • 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

éM = mol. wt. of compound            ù


Solution: (c)

M = W ´ 22400 ; M = 0.2 ´ 22400 = 80

êW = Mass of organic compoundú


V2                       56

ê                                                            ú

êëV2  = Volume of air at STP           úû


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 gL1 )                                                                                                [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

Mass of 45ml. of H 2 at NTP

(Density of H 2 = 0.089 gL1 = 0.000089 g / ml.)

=           0.24

45 ´ 0.000089

= 59.93


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 = éw ´ 108 – 107ù ´ n

êë x                                     úû

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

\Mol. wt. of acid = é 0.228 ´ 108 – 107ù ´ 2 = [152 – 107] ´ 2 = 90

êë 0.162                          úû

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),


1 éw

ù      1 é 100                   ù      1

1                  æw = mass of platinum salt ö



E  =  2 êë x  ´ 195 – 410úû  =  2 êë 32.5 ´ 195 – 410úû  =  2 [600 – 410] =  2 ´ 190 = 95 ç 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


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


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


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]


  • CHClO (b)

CH 2 ClO


C2 H 2 OCl


ClCH 2 O


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 2 O and molecular weight 90.The molecular formula of it is (C = 12, H = 1 and O= 16)                                                            [CPMT 2000; MP PET 2002]



C3 HO3


C2 HO2

C2 H 2 O




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


n = Molecular weight Empirical weight

= 90 = 3



\Molecular formula = Empirical formula ´ n

= CH 2 O ´ 3 = C3 HO3


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]



C3 H 4 O4


C2 H 2 O2


C3 HO2


CH 2 O


Solution: (a)                                                      Calculation of empirical formula

Element Percentage At. mass Relative number

of atoms

Simplest ratio of


Carbon 34.62 12 34.62 = 2.88


2.88 = 1´ 3 = 3


Hydrogen 3.84 1 3.84 = 3.84 3.84 = 1.33 ´ 3 = 4
      1 2.88
Oxygen 61.54 (by difference) 16 61.54 = 3.84


3.84 = 1.33 ´ 3 = 4


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 º 0.1075g

acid ; 20 ´ 0.1, ml 1 N NaOH º 0.1075g




So 1000 ml 1 N NaOH º

0.1075 ´ 1000g acid º 53.75g

20 ´ 0.1

acid ; Eq. mass of the acid = 53.75


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


n = Mol. mass Emp. mass

= 107.50 » 1 ;       \Molecular formula = C

104.0                                                 3

H 4 O4


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]



C6 H6


C2 H6


C3 H 4


C7 H8


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

Calculation of empirical formula

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,


V  PV1 ´ T2


= 1 ´ 1 ´ 273 = 0.6825 litre



2       T1        P2

400 ´ 1


0.6825 litre of the gas weigh = 2.8 g






22.4 litre of the gas weigh =




´ 22.4 = 91.89 » 92


n = Mol. mass Emp. mass

= 92 = 1



Molecular formula = Empirical formula = C 7 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



C2 H 2


C2 H 4


C3 H6


C4 H8


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 Hy + (x + y / 4)O2 ¾¾® xCO2          +

y / 2H 2 O


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

  • Determination of structure by spectroscopic and diffraction methods : With the advancement of scientific developments, new techniques have been designed to determine the structures of
    • 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 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.
    • Diffraction methods : The diffraction methods help to determine the complete three dimensional structure of the molecules including bond length, bond angle The commonly used diffraction methods are: X- ray diffraction, neutron diffraction, electron diffraction etc.

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

  • 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 These may be either saturated or unsaturated. For example,












CH3 CH2CH  = CH2







3, 3-Dimethyl-1-butyne


These compounds are also called as aliphatic compounds.

  • Cyclic or closed-chain compounds : Cyclic compounds contain at least one ring or closed chain of 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.
  • Homocyclic or carbocyclic : These are the compounds having a ring or rings of carbon atoms only in the 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)

These are also called benzenoid aromatics.

Naphthalene (Bicyclic)



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.



Tropolone                                                  Azulene

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




Oxirane or Epoxyethane

Tetrahydrofuran (THF)

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,









Organic compounds





Acyclic or open chain compounds                                                                            Cyclic or closed chain compounds

Homocyclic or carbocyclic compounds                                                                             Heterocyclic compounds




Alicyclic compounds                        Aromatic compounds                               Alicyclic heterocyclic compounds

Aromatic heterocyclic compounds


Benzenoid aromatics                    Non-benzenoid aromatics


  • 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 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


C º C



F, – Cl, – Br, – I (Halo)



–OH (Hydroxy)



|                   |

C O C – (Alkoxy)

|                   |


C H (Aldehydic)





–  C (ketonic)

Acid halides O


CX (Acylhalide)



–  C NH2 (Amide)

O           O

||                  ||

–  C O C



||                   |

–  C O C (Ester)



C º N (Cyano)


– N     C (Isocyano)


O (Nitro)

N     O        ¯






Alkyl Halides


Acid anhydrides









Aldehydes Isocyanides



Nitro compounds




  • 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

  • All the members of a series can be represented by the general For example, the members of the


alcohol family are represented by the formula Cn H2n +1OH

where n may have values 1, 2, 3……. etc.


  • Two successive members differ in their formula by – CH 2 group or by 14 atomic mass units (12 + 2 ´ 1).
  • Different members in a family have common functional group g., the members of the alcohol family

have – OH group as the functional group.

  • 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
  • The members present in a particular series can be prepared almost by similar methods known as the general methods of
    • 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, g. ethane, n-propyl alcohol, acetaldehyde etc.

H      H                                  H   H     H                                      H                O

|       |                                     |       |       |                                         |


H C C H;

H C C C O H;



|       |                                     |        |       |                                         |

H      H                                   H     H     H                                      H                H



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.


H     H

H                           H                                                                         |       |

H      H

|       |                O


C = C

H       Ethylene    H

;      H C º C H



Vinyl alcohol

H C = C C

Acraldehyde           H


Note :®The double bond between carbon and oxygen atoms is not a sign of unsaturation as in acetaldehyde or acetone.

Nomenclature means the assignment of names to organic compounds. There are two main systems of nomenclature of organic compounds.

  • Trivial system : This is the oldest system of naming organic 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,
  • Acetic acid derives its name from vinegar of which it is the chief constituent (Latin : acetum = vinegar).
  • Formic acid was named as it was obtained from red The Greek word for the red ants is formicus.




  • The names oxalic acid (oxalus), malic acid (pyrus malus), citric acid (citrus) have been derived from botanical sources given in
  • Urea and uric acid have derived their names from urine in which both are
  • The liquid obtained by the destructive distillation of wood was named as wood Later on, it was named methyl alcohol (Greek : methu = spirit; hule = wood).
  • Names like glucose (sweet), pentane (five), hexane (six), were derived from Greek words describing their properties or structures.
  • Methane was named as marsh gas because it was produced in 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
CH4 Methane CHCl3 Chloroform
C2H2 Acetylene CHI3 Iodoform
H3CCH2CH2CH3 n-Butane CH3CN Acetonitrile
(H3C)2CHCH3 Isobutane CH3COOH Acetic acid
(H3C)4C Neopentane C6H6 Benzene
HCHO Formaldehyde C6H5CH3 Toluene
(H3C)2CO Acetone C6H5NH2 Aniline
CH3CH2OH Ethyl alcohol C6H5OH Phenol
CH3CONH2 Acetamide C6H5OCH3 Anisole
CH3OCH3 Dimethyl ether C6H5COCH3 Acetophenone
(CH3CH2)2O Diethyl ether C6H5CONH2 Benzamide
  • IUPAC system : In order to rationalise the system of naming , an International Congress of Chemists was held in Geneva in 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

  • Word root : The word root denotes the number of carbon atoms present in the


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
  • Suffix : The word root is linked to the suffix which may be primary or secondary or
  • Primary suffix : A primary suffix is added to the word root to indicate whether the carbon chain is saturated or
Type of carbon chain Primary suffix General name
Saturated (C – C) Unsaturated (C = C)

Unsaturated (C º C)





Alkene 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.

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

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




an (e)*






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

  • Prefix : There are many groups which are not regarded as functional groups in the IUPAC name of the 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 :
  • 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. e. alkane – ane + yl = alkyl.





For example,




:     Methane               becomes

:     Ethane                 becomes

:     Propane               becomes

CH3    – CH3CH2 – CH3CH2CH2 –

:     Methyl

:     Ethyl

:     Propyl etc.


  • 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 The other functional groups are regarded as substituents and are indicated by prefixes.
Substituent Prefix Substituent Prefix Substituent Prefix
–F Fluoro – NO Nitroso – NO2 Nitro
–Cl Chloro – N = N – Diazo – NH2 Amino
–Br Bromo –OCH3 Methoxy –OH Hydroxo
–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 :



3           2             1

Functional group

Word root : But


C H3C H C H2C H2



Primary suffix : – ane Secondary suffix : –ol Prefix : Chloro


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

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




5            4          3


2         1||

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

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




Sec. suffix

Secondary suffix : oic acid (– COOH group)

Prifix : Bromo (– Br group at C – 4)


Prefix                          Pri. suffix

IUPAC name : Bromo + pent + ene + oic acid or 4-Bromopent -2-en-1-oic acid

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 a, b, g, d.


1o CH

d                   g                    b                 a


|  3                                                        CH3 – CH2 – CH2 – CH2 – OH


1o CH

2o    CH   4o    C3o  CH1o    CH

d                   g                  b                 a

Functional group


3                 2



1o CH




1o CH






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  ¾¾¾H ® 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 alkyl group Structure Parent saturated


Name of the alkyl group Structure
Methane Methyl CH3 Propane n-Propyl CH3 – CH2 – CH2
Ethane Ethyl CH3 –  CH2

Butane n-Butyl CH3 – CH2 – CH2 – CH2

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


CH3CH2CH2 – n – Propyl                                               CH3 CH2CH2CH2  – n – Butyl


CH 3 CH 2 CH 3


CH 3






n – Butane

CH3 CH 2
























CH 3




butyl (t – Butyl)


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






CH3CH 2CH 2CH 2CH 2 – ; CH3CHCH2CH2  – ; CH3CCH 2 – ; CH3CHCH 2CH 2CH3 ;  CH3CCH 2CH3









sec -Pentyl


tert -Pentyl



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 º C Acetylide Ethynyl
2              1

 CH2 = C H C H2

Allyl 2-Propenyl 2        1

 HC º CCH2

Propargyl 2-Propynyl

 CH3CH = C H


Different classes of organic compounds



  3             2               1   But But



Eth Prop But














1-   Butene 2-Butene



Ethyne Propyne 1-Butyne

2-   Butyne

  C H3 C H = C H2
  4             3             2               1  
  C H3 C H2 C H = C H2
  4             3                2          1  
  C H3 C H = C H C H3
3.          Acetylenes or


2                1

C H º C H

3                 2         1

 (CnH2n-2) C H3C º C H
4             3                 2         1
  C H3 C H2C º C H
  4                  3         2       1
  C H3C º CC H3
4.          Monohydric Alcohols 1

 C H3OH  

2             1

 C H3 C H2OH

3             2            1

 C H3 C H2 C H2OH

3             2                   1

 C H3 C HOH C H3

4            3             2            1

C H3 C H2 C H2 C H2OH

4             3             2                   1

C H3 C H2 C HOH C H3


 H C HO                               

2            1

 C H3 C HO

3             2             1

 C H3 C H2 C HO

4             3             2            1

 C H3 C H2 C H2 C HO

3             2         1

 C H3 C OC H3

4             3            2         1

 C H3 C H2 C OC H3

5             4            3             2         1

C H3 C H2 C H2 C O C H3

5             4            3          2            1

C H3 C H2 C O C H2 C H3


 H C OOH                            

2            1


3             2            1

 C H3 C H2 C OOH

4             3             2            1

 C H3 C H2 C H2 C OOH

2            1

 C H3 C OCl

3             2            1

 C H3 C H2 C OCl

4             3             2            1

 C H3 C H2 C H2 C OCl

2            1

 C H3 C ONH2

3             2            1

 C H3 C H2 C ONH2









 (CnH2n+1OH) Eth -ane -ol Ethanol
  Prop -ane -ol 1-Propanol
  Prop -ane -ol 2-Propanol
  But -ane -ol 1-Butanol
  But -ane -ol 2-Butanol
5.          Aldehydes

(Cn H 2n O)

Meth Eth  






Methanal Ethanal
  Prop -ane -al Propanal
  But -ane -al Butanal
6.          Ketones

(Cn H 2n O)

Prop But  






Propanone Butanone
  Pent -ane -one 2-Pantanone
  Pent -ane -one 3-Pentanone
7.          Carboxylic acid              (Mono)  





-oic acid


Methanoic acid

 (CnH2nO2) Eth -ane -oic acid Ethanoic acid
  Prop -ane -oic acid Propanoic acid
  But -ane -oic acid Butanoic acid
8.          Acid

Chlorides (RCOCl)

Eth Prop -ane


-oyl chloride

-oyl chloride

Ethanoyl chloride Propanoyl chloride
  But -ane -oyl chloride Butanoyl chloride
9.          Acid     amides  








 (RCONH2) Prop -ane Amide Propanamide


10.         Esters

4             3             2            1

 C H3 C H2 C H2 C ONH2














Methyl methanoate





(RCOOR¢)                2             1




Ethyl ethanoate




2            1                                                                  Eth



Methyl ethanoate




3                  2            1




Ethyl propanoate



  1. Anhydrides

(RCO)2 O



  1. Amines

C H3 – C H2 C OOC2H5


(C2H5 – CO)2 O




 C H3 NH2


Eh Prop









-oic anhydride

-oic anhydride


Ethanoic anhydride Propanoic anhydride




(R NH2)                 2             1







 C H3 C H2 NH2


3             2            1




  • Propanamine


C H3 C H2 C H2 NH2


3             2                     1




  • Propanamine


C H3 C HNH2 C H3


  1. Cyanides or      2        1




Ethane nitrile


nitriles        (R–

 C H3 C N



3             2            1




Propane nitrile


 C H3 C H2 C N


4             3            2            1




Butane nitrile


  1. Alkyl  halides

(CnH2n+1 X)

 C H3 C H2 C H2 C N  


 C H3Cl






2            1                                                                  Eth





 C H3 C H2Cl


3             2            1




  • Bromopropane


 C H3 C H2 C H2Br  


3             2                1




  • Bromopropane



  1. Ethers

 C H3 C HBr C H3









(R– O – R)               2            1






 C H3 C H2OCH3


2             1                                                                 Eth





  1. Nitro compounds

 C H3 C H2OC2H5


 C H3 NO2







(R – NO )                2        1






2                       C HC HNO2




  • Nitropropane


3            2            1


C H3 C H2 C H2 NO2




  • Nitropropane


3             2                     1

C H3 C HNO2 C H3

In the common system, all the isomeric alkanes (having same molecular formula) have the same parent name. The names of various isomers are distinguished by prefixes. The prefix indicates the type of branching in the molecule. For example,

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







  • 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





CH3  – CHCH3




CH3  – CH CH2CH3









  • Prefix neo is used for those alkanes which have two methyl groups attached to the second last carbon atom of the continuous




CH3 – C CH3






CH3 – C CH2  – CH3







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)

  • 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 :












Parent chain


2           1

CH2 –CH3

3|          4









(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.

  • Position of the substituent :Number of the carbon atoms in the parent chain as 1, 2, 3,…

starting from the end which gives lower number to the carbon atoms carrying the substituents. For examples,





5      4      3


2|    1



1      2      3


4|    5



A (Correct)


B (Wrong)


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

5              4              3              2            1                                              1              2              3

C H3  – C H2 – C H2 – C H C H3                                            C H3 – C H2 – C H C H C H3

|                                                                                       4|            5         6





CH2 –CH2 –CH3

3 – Ethylhexane




  • 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

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,


6      5           4

3           2            1

1      2           3

4           5            6


HCC H C H 2  – C H C H C H3                           HCC H C H 2 – C H C H C H3















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

8           7            6

5              4             3              2            1


CH3 – CH2 – CHCHCH2 – CH2 – CH2 – CH2 – CHCH3

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




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)









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                                                                                   CH2 –CH3

3            4|             5               6                  7                                5            4|            3               2                  1

CH3  – C H C H   C H2 –  C H2 – CH3 ;            CH3CHCHCH2CH2CH3

2|        1                                                                                       | 6           7

CH2 –CH3                                                                               CH2 –CH3

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



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

8           7            6              5             4

3             2            1


CH3 – CH2 – CHCHCH2 – CH2 – CH2 – CH2 – CHCH3









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 H 2  – C H C H C H 2 – C H 2 – C H 2  – C H 2 – C H C H3






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.

  • Presence of more than one same substituent : If the same substituent or side chain occurs more than once, the prefixes di, tri, tetra…………….. , are attached to the names of the substituents. For example,


5              4            3             |2    1








2, 2, 4 – Trimethylpentane

  • 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


CH 2 CH3




5              4              3|    2            1                                                           1              2|          3|


C H3C H 2CC H C H3



  • Ethyl-2, 3-dimethylpentane

C H3 – C     



CCH 2  – CH3

4 |



5 |           6

CH 2 – C H3

3-Ethyl-2, 2, 3-trimethylhexane


  • 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

C2H5      CH3


6              5              4           3            2              1

1              2             3|

4|      5              6


C H3  – C H2 – C HC H C H2 – C H3

C H3 – C H 2  – C

  • C C H 2C H3











3- Ethyl – 4 – methyl hexane                                                                                  3, 3-Diethyl-4, 4-dimethyl hexane

  • Naming the complex substituents (or substituted substituents) : If the substituent on the parent chain is complex (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              5            6              7              8              9

C H3 – C H 2 – C H 2 – C H 2 – C H C H 2 – C H 2 – C H 2 – C H3



Complex substituent



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              5           6

|7          8

9             10            11


CH3 – CH2 – CH2 – CH2 – CH CH2 – C H CH2 – CH2 – CH2 – CH3



Complex substituent (1, 2-dimethylpropyl)



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).






9              8              7

Complex substituent

|1    2              3

CH3 – CC H 2  – C H3

6              |5    4              3



( 1, 1-dimethylpropyl)

2              1


|1          2         3





C H3 – C H 2 – C H 2 – C H 2 – CC H 2 – C H 2 – C H 2 – C H3






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


Complex substituent ( 2-methylpropyl)

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