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

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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 ¾¾^D ® NH2 – C – NH2 ClCH2 COOH ¾¾^Zⁿ ^/¾^H¾^C^l ® CH3 – COOH

(Ammonium cyanate)

Urea

(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

Heat

Solid

Cool

Vapour

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 (100^oC) and ethanol (78.1^oC) differ by 21.9^oC, 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 78.13^oC. 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
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, a-amino acids and
		inorganic compounds)

Note :®R_f 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

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.

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 1^o, 2^o and 3^o 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

(Conc.)

¾¾Roo¾m t¾em¾p ®

SO3 H

H 2 O

Thiophene

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 100^oC are dried by keeping them in steam or oven maintained at 110^oC. 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

CO2

and hydrogen to

H 2 O .

CO2 is

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

H 2 O

is detected by anhydrous

CuSO4

(white) which turns it blue.

C+ 2CuO ¾¾^He¾^at ® CO2 + 2Cu ; Ca(OH)₂ + CO₂ ¾¾® CaCO₃ + H₂O

Lime water Milky

H 2 + CuO ¾¾^He¾^at ® H 2 O + Cu ; CuSO4 + 5H 2 O ¾¾® CuSO4 .5H 2 O

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

Sulphur

Na + C + N ¾¾^D ® NaCN

(S.E.)

2Na + S ¾¾^D ® Na₂S

(S.E.)

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 ¾¾^H¾^C^l ® 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 ¾¾^C^H¾3CO¾^O¾^H ® 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. CH3 CONH 2 + NaOH ¾¾^C^a¾^O ® CH3 COONa + NH 3 . This test

Acetamide

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

Na2 SO4 + 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

H 2 O , 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

Nitrogen

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 ¾¾^D ® xCO2 + y H 2 O

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 ¾¾^D ®(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

Sulphur

phosphorous

Oxygen

(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) ¾¾^h^e¾^at ® H 2 SO4

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 ) .

2MgNH 4 PO4 ¾¾^h^e¾^at ® Mg 2 P2 O7 + 2NH 3 + H 2 O

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 ¾^C ® 2CO

5CO + I 2 O5 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

N H SO

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 ´

233

wt.of BaSO₄

wt. of organic compound

´ 100

= 32 ´

233

0.35

0.2595

´ 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 =

wt. of H ₂ O
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

0.2

´ 0.12 ´ 100 = 6.666

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

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 ´

188

Mass of AgBr formed Mass of substance taken

´ 100 =

188

´ 0.12 ´ 100 = 34 (Approx.)

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 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)

0.50

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

0.00009

( 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

Where,

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

Where,

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 ¾¾^h^e¾^a^t ®

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 = 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

B2 H 2 PtCl6 ¾¾^h^e¾^a^t ® 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 + 410 = w = Mass of platinum salt ;

2E = éw ´ 195 – 410ù

; E = 1 éw ´ 195 – 410ù

195

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–m–l 1–N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–m–l 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

H 2 O

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

CO2

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 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

H 2 O

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 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

Mass of 45ml. of H ₂ at NTP

(Density of H 2 = 0.089 gL^–¹ = 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

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

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

(c)

C2 H 2 OCl

(d)

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]

(a)

C3 H6 O3

CH 2 O

C2 H6 O2

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 H6 O3

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 2 O2

(c)

C3 H6 O2

(d)

CH 2 O

Solution: (a) Calculation of empirical formula

Element	Percentage	At. mass	*Relative number* of *atoms*	*Simplest ratio of* *atoms*
Carbon	34.62	12	34.62 = 2.88 12	2.88 = 1´ 3 = 3 2.88
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 16	3.84 = 1.33 ´ 3 = 4 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 º 0.1075g

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

acid

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

(a)

C6 H6

(b)

C2 H6

(c)

C3 H 4

(d)

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 = P1 V1 ´ 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 =

2.8

0.6825

´ 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

(a)

C2 H 2

(b)

C2 H 4

(c)

C3 H6

(d)

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 CH2CH2CH3

Butane

CH3

CH₃ – CH – CH₃

Isobutane

CH3 CH2CH = CH2

1-Butene

CH3

CH₃ – C – C º CH

CH3

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

Pyrrolidine

Piperidine

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

Furan

Pyrrole

Thiophene

Organic compounds

Pyridine

Quinoline

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

– C – X (Acylhalide)

– C – NH₂ (Amide)

O O

|| ||

– C – O – C –

(Anhydride)

|| |

– C – O – C – (Ester)

–C º N (Cyano)

– N C (Isocyano)

O (Nitro)

–N O ¯

Acetylenes/Alkynes

Amides

Alkyl Halides

Acid anhydrides

Alcohols

Esters

Ethers

Cyanides/Nitriles

Aldehydes

Isocyanides

Ketones

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 – C – C

| | | | | |

H H H H H H H

Ethane

n-propyl alcohol

Acetaldehyde

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

Acetylene

H – C = C – OH

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)

–ane

–ene

–yne

Alkane

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
acids
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

CH₃CH₂OH

CH₃CH₂CN

Eth

Prop

an (e)*

ane

nitrile

Ethanol

Propanenitrile

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,

CH4

CH3CH3

CH3CH2CH3

: 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 H₃ – C H – C H₂ – C H₂ –

Prefix

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)

CH₃ – CH– CH = CH– C– OH

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 (1^o), secondary (2^o),

tertiary (3^o) and quaternary (4^o). 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 C – ³o CH –¹o CH

d g b a

Functional group

3 2

1o CH

CH₃ – CH – CH₂ – CH₂ – CHO

CH3

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 ¾¾^–¾^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

hydrocarbon

Name of the alkyl group

Structure

Parent saturated

hydrocarbon

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

Propane

CH 3

CH –

;

Isopropyl

CH3CH2CH2CH3

n – Butane

CH3 CH 2

CH –

sec.Butyl

CH3

CH₃ – CH – CH₃

Isobutane

CH3

CH₃ – CH – CH ₂ –

CH3

CH₃ – CH– CH₃ –

Isobutyl

Tertiary

CH 3

butyl (t – Butyl)

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

CH3

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

n–Pentyle

CH3

Isopentyl

CH3

sec -Pentyl

tert -Pentyl

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 º C –	Acetylide	Ethynyl
2 1 CH₂ = C H – C H₂ –	Allyl	2-Propenyl		2 1 HC º C– CH₂ –	Propargyl	2-Propynyl
1 CH₃ – CH = C H –	–	1-Propenyl

Different classes of organic compounds

	3 2 1	But But Eth Prop But But	-ene -ene -yne -yne -yne -yne	– – – – – –	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 Alkynes	2 1 C H º C H 3 2 1
(CnH2n-2)	C H₃ – C º C H
	4 3 2 1
	C H₃ C H₂ – C º C H
	4 3 2 1
	C H₃ – C º C– C 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 1 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 1 H C OOH 2 1 C H₃ C OOH 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	Meth	-ane	-ol	Methanol
(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	-ane -ane	-al -al	Methanal Ethanal
		Prop	-ane	-al	Propanal
		But	-ane	-al	Butanal
6. Ketones (Cn H 2n O)		Prop But	-ane -ane	-one -one	Propanone Butanone
		Pent	-ane	-one	2-Pantanone
		Pent	-ane	-one	3-Pentanone
7. Carboxylic acid (Mono)		Meth	-ane	-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 -ane	-oyl chloride -oyl chloride	Ethanoyl chloride Propanoyl chloride
		But	-ane	-oyl chloride	Butanoyl chloride
9. Acid amides		Eth	-ane	Amide	Ethanamide
(RCONH₂)		Prop	-ane	Amide	Propanamide
10. Esters	4 3 2 1 C H3 C H2 C H2 C ONH2 1 H C OOCH₃	But Meth	-ane -ane	Amide -oate	Butanamide Methyl methanoate

(RCOOR¢) 2 1

Eth

-ane

-oate

Ethyl ethanoate

C H3 C OOC2H5

2 1 Eth

-ane

-oate

Methyl ethanoate

C H3 C OOCH3

3 2 1

Prop

-ane

-oate

Ethyl propanoate

Anhydrides

(RCO)₂ O

Amines

C H3 – C H2 C OOC2H5

(CH₃ – CO)₂O

(C2H5 – CO)2 O

C H3 NH2

Eh Prop

Meth

-ane

-oic anhydride

amine

Ethanoic anhydride Propanoic anhydride

Methanamine

(R – NH2) 2 1

Eth

-ane

amine

Ethanamine

Primary

C H3 C H2 NH2

3 2 1

Prop

-ane

amine

Propanamine

C H3 C H2 C H2 NH2

3 2 1

Prop

-ane

amine

Propanamine

C H3 C HNH2 C H3

Cyanides or ² ¹

Eth

-ane

nitrile

Ethane nitrile

nitriles (R–

C H3 C N

CN)

3 2 1

Prop

-ane

nitrile

Propane nitrile

C H3 C H2 C N

4 3 2 1

But

-ane

nitrile

Butane nitrile

Alkyl halides

(CnH2n+1 X)

C H3 C H2 C H2 C N

C H3Cl

Meth

-ane

Chloro

Chloromethane

2 1 Eth

-ane

Chloro

Chloroethane

C H3 C H2Cl

3 2 1

Prop

-ane

Bromo

Bromopropane

C H3 C H2 C H2Br

3 2 1

Prop

-ane

Bromo

Bromopropane

Ethers

C H₃ C HBr C H₃

C H3OCH3

Meth

-ane

Methoxy

Methoxymethane

(R– O – R) 2 1

Eth

-ane

Methoxy

Methoxyethane

C H3 C H2OCH3

2 1 Eth

-ane

Ethoxy

Ethoxyethane

Nitro compounds

C H3 C H2OC2H5

C H3 NO2

Meth

-ane

Nitro

Nitromethane

(R – NO ) ² ¹

Eth

-ane

Nitro

Nitroethane

2 C H3 C H2 NO2

Prop

-ane

Nitro

Nitropropane

3 2 1

C H3 C H2 C H2 NO2

Prop

-ane

Nitro

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

CH3CH2CH2CH3

n-Butane

CH3 CH2CH2CH2CH3

n-Pentane

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

Isobutane

CH3 – CH – CH2CH3

CH3

Isopentane

CH3 – CHCH2CH2CH3

CH3

Isohexane

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

CH3

CH3 – C – CH3

CH3

Neopentane

CH3

CH3 – C – CH2 – CH3

CH3

Neohexane

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 :

Substituents

CH3

Parent chain

2 1

CH2 –CH3

3| 4

CH3

CH₃ – C H – C H – CH₂ – 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.

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

Substituent

1 2 3

4| 5

C– C– C– C– C

A (Correct)

C– C– C– C– C

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

2-Methylpentane

CH3

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

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

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

8 7 6

5 4 3 2 1

CH3 – CH2 – CH – CH – CH2 – CH2 – CH2 – CH2 – CH – CH3

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

CH3

1 2 3

CH3

4 5

6 7 8

CH3

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

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 ; CH₃ – CH – CH – CH₂ – CH₂ – CH₃

2| 1 | 6 7

CH2 –CH3 CH2 –CH3

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

CH3

1 2| 3

CH3

4| 5

CH3

5 4| 3

CH3

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 – CH – CH – CH2 – CH2 – CH2 – CH2 – CH – 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 H 2 – C H – C H – C H 2 – C H 2 – C H 2 – C H 2 – C H – C H3

CH3

Set of locants = 3, 4, 9

CH3

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,

CH3

5 4 3 |2 1

C H₃ – C H – C H₂ – C– C H₃

CH3

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

CH3

5 4 3| 2 1 1 2| 3|

C H₃ – C H ₂ – C– C H – C H₃

CH3

Ethyl-2, 3-dimethylpentane

C H3 – C –

CH3

C– CH 2 – CH3

4 |

CH3

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 H– C H – C H2 – C H3

C H3 – C H 2 – C

C – C H ₂ – C H₃

CH3

C2H5

CH3

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

CH3

|1 2 3

CH3 – C– C H 2 – C H3

6 |5 4 3

( 1, 1-dimethylpropyl)

2 1

|1 2 3

CH₂–CH–CH₃

CH3

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

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

Complex substituent ( 2-methylpropyl)

Tags: Chapter 13 Purification, classification and nomenclature of organic compound 1

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Lassaigne method

Other methods for detection of elements

Quantitative estimation of elements in organic compounds

(iii) Chemical methods

(5) Calculation of Empirical and Molecular formula

(1) Classification based on structure

Characteristics of homologous series

Common or trivial names of some organic compounds.

Unsaturated groups or radicals

(1) Rules for Naming complex aliphatic compounds when no functional group is present (saturated hydrocarbon or paraffins or Alkanes)

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