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13. ORGANIC CHEMISTRY I) SOME BASIC PRINCIPLES AND TECHNIQUES

Synopsis : Introduction and classification :

1. Based on their source, chemical compounds are classified into 3 types by Lemery:

i) Mineral source ii) Vegetable source iii) Animal source Vital force theory : • Lavoiser found that the compounds from vegetable and animal sources contain similar

composition. • Then Berzelius classified the above 3 types into 2 types and he coined the new terms Organic and

Inorganic. i) Organic compounds: These are present in living beings (plants, animals etc.) ii) Inorganic compounds: These are present in minerals

• But now based on the structure and chemical behaviour, the compounds are classified as i) Organic compounds ii) Inorganic compounds

• Organic compounds are carbon compounds and the study of chemistry of carbon compounds is called organic chemistry.

• Organic compounds contain Carbon, Hydrogen, essentially and Oxygen, Nitrogen, Sulphur, Phosphorus, Halogen etc frequently.

• Father of organic chemistry is Wohler who synthesized the first organic compound in laboratory from inorganic substances.

NH4Cl+KCNO ⎯⎯⎯ →⎯−KCl

cyanateammonium4CNONH ⎯→⎯Δ NH2–

O||C –NH2

• With the synthesis of urea vital force theory of Berzelius was discarded, according to which organic compounds cannot be synthesized without any vital force.

• Kolbe synthesised CH3COOH from its elements. • Berthelot synthesised CH4

• Carbon is tetravalent as it contains four unpaired electrons in its excited state configuration. The tetravalency of carbon was given by Vanthoff and Lebel who were awarded the first nobel prize in chemistry.

• Ground state configuration of 'C' : oz

1y

1x

226 p2p2p2s2s1C →

Excited state configuration of 'C' : 1z

1y

1x

126 p2p2p2s2s1C →

Carbon alone forms about 10 millions of organic compounds where as the remaining elements together could form just about 50000 compounds.

Largest number of compounds formed by carbon because of its marked features. 1) Highest catenation 2) Tetravalency 3) Ease of formation of multiple bonds

• Natural sources of organic compounds are coal, petroleum, natural gas, animals and plants.

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Classification of organic compounds : Classification based on carbon chain : Carbon compounds are classified into various types based on the nature of functional

groups:

Name Formula Functional

group Name of the functional

group

1) Alkanes (paraffins) R – H →C–C← Single bond 2) Alkenes (olefins) R–CH =CH2 > C = C< Double bond 3) Alkynes (acetylenes) R – C ≡ CH –C ≡ C – Triple bond 4) Alkyl halides (Haloalkanes)

R – X –X Halogen

5) Alcohols (Alkanols)

R – OH –OH Hydroxy

6) Ethers R –O–R – O – Ether 7) Amines (Amino alkanes) R – NH2 – NH2 Amino 8) Aldehydes (Alkanals)

R – CHO – CHO Aldehyde

9) Ketones (Alkanones) R –CO– R > C = O Keto 10) Carboxylic acids(Alkanonic acids)

R – COOH – COOH Carboxyl

11) Esters (Alkyl alkanoates

R – COOR – COOR Ester

12) Amides R – CONH2 –CONH2 Amide 13) Cyanides R – CN – CN Cyanide 14) Nitro compounds R – NO2 – NO2 Nitro 15) Sulphonic acids R – SO3H –SO3

H Sulphonic Acid

• Open chain compounds: Carbon atoms are linked to one another to form straight chains or branched ones but not rings.

• Open chain compounds are also called aliphatic compounds. Ex: Alkanes, alkenes, alkynes and their derivatives.

• Cyclic compounds: Carbon atoms are linked to one another to form ring. • Based on the number of rings, they may be monocyclic or polycyclic.

Homocyclic: Ring is formed by carbon atoms only. Ex: Benzene, Phenol, Toluene, Cyclopropane etc.

Saturated Ex. Alkanes

Open chain (Aliphatic or Acylic)

Closed chain (Ring or cylic )

Unsaturated Ex. Alkenes, Alkynes

Homocylic (Carbocyclic )

Heterocyclic Ex. Furan, Pyrrole, Pyridine, Thiophene Alicyclic

Ex. Cycloalkanes Aromatic

Ex. Benzene

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Heterocyclic: Along with carbons some hetero atom like, N, O, S, P, etc is involved in forming the ring.

Alicyclic: These are cyclic compounds but resemble open chain compounds in properties. Ex.: Cycloalkanes, cycloalkenes, cycloalkynes Aromatic: The ring compounds which resemble benzene in structure and properties are called aromatic compounds.

• These may be homocyclic or heterocyclic. Ex.:

Bonding in carbon compounds :

• In all its compounds, carbon undergoes only 3 types of hybridisation i.e. sp, sp2 and sp3.

• The tetravalency of carbon is possible in its excited state configuration, as it contains four unpaired electrons

• The energy required for the promotion of electron from 2s orbital to 2p orbital is 501.6 kJ/mol. • sp3 hybridisation is found in alkanes. sp3 hybrid carbon forms four single bonds or four sigma

bonds. The ratio of s – character to p – character in each sp3 hybrid orbital is 1 : 4. sp3 hybrid carbon forms C – C and C – H bonds. Shape of the molecule is tetrahedral and bond angle is 109°.

• sp2 hybridisation is found in alkenes. sp2 hybrid carbon forms 3 sigma bonds and 1pi bond or one double bond and two sigma bonds. The ratio of s – character to p – character in sp2 hybrid orbital is 1 : 2. Shape of molecule is trigonal planar and bond angle is 120°.

• sp hybridisation is found in alkynes and cumulative dienes. SP hybrid carbons forms 2σ and 2π bonds i.e. 1 triple and 1 single or 2 double bonds.

• The ratio of s character to p character in sp hybrid orbital is 1 : 1. • Shape of molecule is linear and bond angle is 180°.

• • • •

Furan O • •

ThiophenS

N

Pyridene Pyrrole

N H

• •

Benzene

OH

Phenol Napthalene

• •

Furan O • •

• •

Thiophene S • •

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3. STRUCTURAL FORMULAE (OR) CARBON SKELETON DIAGRAMS 1. In bond line diagrams the atoms other than C & H are only shown. Hydrogen atoms if they are

bonded to Hetero atoms like N & O are shown. 2. 3 2 2 3CH CH CH CH Butane 3 2 2 2 2CH CH CH CH NH Butanamine 3. Cyclo propane Cyclo propane 4 Cyclo butane 5. Chloro cyclo pentane 6.

7.

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8.

4.METHODS OF PURIFICATION OF ORGANIC COMPOUNDS Methods of purification of organic Compounds:

Organic compounds obtained either from natural sources or synthetic processes in the laboratories are contaminated with impurities. These are purified by different methods.

I. Methods of purification of solids: 1.Crystallisation:- • This method is useful to purify the solid organic compounds. • The principle involved in this method is that the compound should be insoluble at low

temperature but soluble at higher temperatures in the given solvent. Impurities are either insoluble or soluble and go into filtrate.

• The insoluble impurities are removed by filteration in hot condition. • The crystals of the compound are seperated by filtering under reduced pressure using Buckner

funnel. • Repeated crystallisation is required and the coloured impurities are removed by adsorbing them

with activated charcoal. 2. Sublimation : - • It is used to purify activated solid organic compounds which undergoes sublimation. • If the compound is sublimating the impurity should not sublimate. • If the compound to be purified has high vapour pressure below its melting point and sublimates

readily on heating and the impurities don’t sublimate. • The pure compound is seperated by scratching the watch glass. • If the sublimating substances have a low vapour pressure or decompose on heating before

sublimation, then the sublimation is carried out under low pressure. II. Methods of purification of liquids:- 1.Simple distillation:- The vapourisation of a liquid by heating and subsequent condensation of

vapours by cooling is known as distillation. • This process is useful for purification of liquids contaminated with non volatile impurities. • The liquids that have boiling point difference greater than 400 C can be purified by this method. 2.Fractional distillation:- • This process is useful for the purification of liquids having boiling point difference less than 400C. • In this process, liquid with high b.p. is collected in the condenser and liquid with low b.p. is

collected in the receiver. 3.Distillation under reduced pressure:- • This method is useful to purify liquids that have very high boiling points and those which

decompose at or below their boiling points. • At reduced pressure the boiling point of a liquid is also reduced. Hence its decomposition is prevented.

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4. Steam distillation:- • The liquids insoluble in water possess high boiling point and steam volatile and the impurities are

not steam volatile are purified by this method. • Principle in steam distillation is - sum of the vapour pressure of organic liquid (P1) and that of

water (P2) is equal to atmospheric pressure (P) and mixture boils. • The water layer and the organic liquid layer are seperated using separating funnel. Ex:- Aniline is purified by this method from Aniline - water mixture Relative masses obtained in steam distillation process is calculated by a formula =

0

0B

A A A A A

B B B B

W n M p MW n M p M

= = ; Where A BW and W are the masses of A and B

A BM and M are the molecular masses of A and B 0 0

A BP and P are the vapour pressures of A and B At 098.5 C water and aniline have vapour pressures 717 torr and 43 torr. In steam distallation at

098.5 C , the relative masses obtained are 2 717 18 3.2343 93

H O

Aniline

WW

×= =

×

5. Solvent extraction( differential extraction) :- • In this method, organic compound is separated from its aqueous solution by using its solubility

which is different with organic solvent and water. 6. Chromatography:- • It was discoverd by Tswett in (1906). He separated chlorophyll and Xanthophyll and other

compounds by using adsorbent 3CaCO . It is classified into three types based on the physical states of stationary phase and mobile phase. • Chromatography involves the three steps a) Adsorption and retention of a mixture of substances on the stationary phase and separation of

adsorbed substances by the mobile phase to different distance on the stationary phase. b) Recovery of the substances separated by a continuous flow of the mobile phase (known as

elution) c) Qualitative and quantitative analysis of the eluted substances

S.No. Chromatography Process Stationary Phase Mobile Phase 1. Column chromatography (Adsorption) Solid Liquid

2. Liquid – liquid partition chromatography Liquid Liquid

3. Paper chromatography Liquid Liquid

4. Thin layer chromatography (TLC) Liquid (or) solid Liquid

5. Gas – liquid chromatography (GLC) Liquid Gas

6. Gas – solid chromatography (GSC) Solid Gas

7. Ionic change chromatography Solid Liquid

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• Important chromatography techniques are 1) Adsorption chromatography : a) Column chromatography, b) Thin layer chromatography 2) Partition chromatography : a) Paper chromatography a) Column Chromatography • The principle used in this method is differential adsorption. • In the column chromatography the components of a mixture are separated by a column of

adsorbent (stationary phase) packed in a glass tube. • The mixture to be adsorbed on the adsorbent is placed at the top of the stationary phase. • A suitable eluant, either a single solvent or a mixture of solvents is allowed to flow down the

column slowly. • The most readily adsorbed substances are retained near the top and others come down

accordingly to various distances. b)Thin layer chromatography (TLC): • This also based on the adsorption differences. • The glass plate which is coated with adsorbent (Eg : Silica gel, alumina) as a thin layer (0.2mm

thick) is called TLC plate or chromoplate. • The plate is then kept in a closed jar containing the eluant. The relative adsorption of a component of the mixture is expressed in terms of RETARDATION

FACTOR (Rf) value.

fDistance moved by thesubstance from base line (X)R =Distance moved by thesolvent from base line (Y)

• The colourless compounds which fluorescence are detected with ultraviolet light. • Spots of compounds are even detected by allowing them to adsorb iodine, when they give brown

spots. • Some times an appropriate reagent is sprayed, as ninhydrin solution to detect aminoacids. c) Partition Chromatography • This is based on continuous differential partitioning of components of a mixture between the

stationary phase and the mobile phase. • In paper chromatography, a special paper called chromatography paper contains water trapped in

it which acts as the stationary phase. • The chromatography paper spotted with the solution of the mixture at the base is suspended in a

suitable solvent or a mixture of solvents. This solvent (s) acts as the mobile phase. • The solvent rises up the paper by capillary action and moves over the spot. • The paper selectively retains different components as per their differing partition in mobile and

stationary phases and is known as chromatogram. • The spots of the separated coloured compounds are detected.

5. QUALITATIVE ANALYSIS OF ORGANIC COMPOUNDS

Qualitative analysis of organic compounds 1) Detection of carbon and hydrogen • Carbon and hydrogen are detected with - CuO (Cupric Oxide) • If carbon is present, its forms CO2 gas. The CO2 gas turns lime water milky + → + ↑2C 2CuO 2Cu CO ( ) + → ↓2 32

Ca OH CO CaCO

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If hydrogen is present in the compound it forms H2O vapour. H2O vapours when passed over white anhydrous CuSO4 turns it blue by forming CuSO45H2O.

+ → + 22H CuO Cu H O

( ) ( )+ →4 2 4 2

white blueCuSO 5H O CuSO 5H O

2) Detection of halogen, nitrogen and sulphur ( Lassaigne’s test or sodium extract test) • In Lassaigne’s test, the compound is heated strongly in an ignition tube to fuse with sodium metal.

Here nitrogen present in the organic compound is converted as .CN–

+ + →Na C N NaCN from organic compound sodium cyanide • If sulphur is present in the organic compound is converted as 2S − + ⎯⎯→ 22Na S Na S sodium sulphide If halogens are present in the organic compound are converted as X − 22Na X 2NaX(X Cl, Br, I)+ → = • The above obtained fused mass is extracted with water by plunging the red hot ignition tube in

distilled water and the contents are boiled for 10 minutes and filtered. The filtrate is called sodium extract.

i) Test for nitrogen: • Take a portion of the sodium extract, if it is not alkaline add NaOH solution to it and then freshly

prepared ferrous sulphate solution. • Now add 2 or 3 drops of 3FeCl solution, cool, acidify with conc HCl. • A prussian blue or green precipitate (or) coloration is observed if nitrogen is present. ( )4 2 42

2FeSO NaOH Fe OH Na SO+ → +

( ) ( )42 66 2NaCN Fe OH Na Fe CN NaOH⎡ ⎤+ → +⎣ ⎦

sodium ferrocyanide ( ) ( )4 3 46 6 3

3 4 12Na Fe CN FeCl Fe Fe CN NaCl⎡ ⎤ ⎡ ⎤+ → +⎣ ⎦ ⎣ ⎦

Ferric ferrocyanide (prussian blue) ii) Test for sulphur : • To a portion of the Na extract add freshly prepared sodium nitroprusside solution. • A deep violet colouration takes place.

( ) ( )2 42

5 5S Fe CN NO Fe CN NOS

− −− ⎡ ⎤ ⎡ ⎤+ →⎣ ⎦ ⎣ ⎦ nitroprusside violet • If N,S both are present gives thiocyanate, Na C N S NaSCN+ + + → thiocyanate

• To this if 3FeCl solution is added, ( )33

6Fe SCN Fe SCN

−+ − ⎡ ⎤+ → ⎣ ⎦ .

• It gives only blood red colour. • No Blood red colour or prussian blue colour indicates that there are no N or S. If sodium fusion is carried out with excess of ‘Na’ the thiocyanate decomposes to yield CN– and S2–. 22Na SCN Na NaCN Na S+ → +

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• Sodium fusion extract is acidified with acetic acid and then lead acetate solution is added 2 2Pb S PbS− −+ → black ppt iii) Tests for halogens: The sodium fusion extract is acidified with nitric acid. Now it is treated

with 3AgNO solution. Ag X AgX+ −+ → • If white precipitate is formed that is soluble 4NH OH in solution, the halide is Cl− . Hence

chlorine is present. • If pale yellow precipitate, sparingly soluble 4NH OH in solution is formed it is bromide ( )Br−

and the halogen is bromine. • If an yellow precipitate, almost insoluble in 4NH OH solution is formed it is ( )I − iodide and the

halogen is iodide. c) Detection of phosphorus: • The compound is heated with an oxidising agent say sodium peroxide. The ‘P’ in the organic

compound is oxidised to 34PO − . The solution is boiled with 3HNO and treated with ammonium

molybdate. • A canary yellow precipitate formation indicates the presence of phosphorous in the compound. . 3 4 3 3 4 33 3Na PO HNO H PO NaNO+ → + ( ) ( )3 4 4 4 3 4 4 3 22 3

12 21 .12 12H PO NH MoO HNO NH PO MoO H O+ + → + ammonium phosphomolybdate d) Detection of oxygen: • There is no direct test for oxygen. • The organic compound when heated in pure nitrogen atmosphere if water droplets are formed on

the walls of the test tube oxygen is detected. • Otherwise after determining the % composition of all elements in the compounds if it does not

come to 100%, the remaining is of oxygen. • Even by detecting functional groups like , , ,OH CHO COOH NO− etc oxygen is detected.

6. QUANTITATIVE ORGANIC ANALYSIS

Quantitative Organic Analysis 1. Estimation of Carbon and hydrogen:- (Leibig method) • A known weight of the organic substance is taken and completely burnt in excess of air &

Copper (II) oxide. Carbon changes to CO2 and hydrogen changes to H2O as follows.

⎛ ⎞ ⎛ ⎞+ + → +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x y 2 2 2y yC H x O xCO H O4 2

• The CO2 and H2O obtained are passed through weighed U tubes Containing anhydrous CaCl2 and caustic potash respectively. The increased weight of these two tubes give the weight of H2O formed and weight of CO2 formed.

= × ×2Wt.of CO formed12% of C 10044 Wt. of Organic compound

2Wt.of H Oformed2% of H= × ×10018 Wt. of Organic compound

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• After calculating the percentage composition all C and H in the compound if their sum is not 100, then the difference is assumed as oxygen.

2. Estimation of Nitrogen: • There are two methods to estimate nitrogen in the given compound. They are i. Duma’s method ii. Kjeldahl’s method i. Duma’s method:- • In this method a known weight of organic compound is heated strongly with dry curpic oxide.

Carbon and hydrogen get oxidised to carbondioxide and water vapour. Nitrogen if present is converted to N2 is collected over KOH solution and its volume is determined at S.T.P

⎛ ⎞ ⎛ ⎞+ + → + + + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

x y z 2 2 2y y z yC H N 2x CuO xCO H O N 2x Cu2 2 2 2

• = × ×28 Volume of Nitrogen atS.T.P% of N 100

22400 Weight of organic compound

ii. Kjeldahl’s method:- • In this method the given organic compound is treated with Conc.H2SO4 in the presence of small

amounts of 4CuSO to convert nitrogen into ammonium sulphate. Then ammonium sulphate is treated with excess of NaOH to liberate NH3 gas.

• The ammonia evolved is neutralised with excess of Conc H2SO4. which is relatively more in amount than that is required to neutralise NH3 gas. Now, the excess of acid is titrated with standard alkali solution. From this the amount of H2SO4 used to neutralise NH3 formed is calculated and from that

percentage of nitrogen is calculated. ( )+ →2 4 4 42

Oraganic compounds H SO NH SO ( ) + → + +4 4 2 4 2 32

NH SO 2NaOH Na SO 2H O 2NH ( )+ →3 2 4 4 42

2NH H SO NH SO

• × ×=

1.4 N V% of NWt.of organiccompound

• Where N = Normality of acid V = Volume of acid (ml) neutralised by ammonia

3. Estimation of halogens (Carius method) • In this method a known weight of organic compound is heated with fuming nitric acid in the

presence of Ag NO3 in a hard glass tube called Carius tube. • Carbon and hydrogen of the compound are oxidised to CO2 and H2O. Halogen forms Silver

halide (AgX). It is filtered, washed dried & weighed.

• ( ) = × ×At.wtof X Wt.of AgX% of Halogen X 100

Mol.wtof AgX Wtof Oraganic compound

• = × ×35.5 Wt.of AgCl formed% of Cl 100

143.5 Wtof Oraganic compound

• = × ×80 Wt.of AgBr formed% of Br 100

188 Wtof Oraganic compound

• 127 Wt.of Ag I formed% of I 100235 Wt of Oraganiccompounds

= × ×

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4. Estimation of sulphur:- • A known weight of organic compound containing sulphur is heated with Na2O2 or fuming HNO3

in a carius tube. The sulphur in it is oxidised to H2SO4. The acid is precipitated as BaSO4 by adding excess of BaCl2 solution. The precipitate is filtered, washed, dried and weighed.

• 4Wt. of BaSO32% of S 100233 Wt.of Oraganiccompound

= × ×

5. Estimation of Phosphorus: • To estimate phosphorus in the given organic compound, a known mass of organic compound is

heated with fuming nitric acid in a Carius tube. Phosphorus is oxidised to phosphoric acid. The acid is precipitated as ammonium phosphomolybdate ( )4 43

NH PO .12MoO3 by adding ammonia and ammonium molybdate solutions.

• Sometimes, the acid is precipitated by adding magnesia mixture. • Magnesia mixture is obtained by dissolving 100.0g. of MgCl2. 6 H2O and 100.0g.of NH4Cl in

water, then adding 50.0 ml of conc.NH4OH and diluting the solution to 1000 ml. • A Precipitate of magnesium ammonium phosphate ( 4 4Mg NH PO ) is formed which on ignition

gives magnesium pyrophosphate (Mg2P2O7).

• 2 2 7Wt. of Mg P O2 31% of P 100222 Wt.of Oraganic compound×

= × × (or)

• ( )4 4 33Wt. of NH PO .12MoO31% of P 1001877 Wt.of Oraganic compound

= × ×

6. Estimation of Oxygen: • The percentage of oxygen in an organic compound is found indirectly by the difference between

100 and total percentage of other elements. • In the direct method, a known mass of organic compound is decomposed by heating in a stream

of nitrogen gas. The mixture of gaseous product containing oxygen is passed over red hot coke to convert all oxygen in those oxides to CO. Then the mixture is passed through warm iodine pentoxide ( I2 O5 ) to convert CO to CO2 and liberating iodine.

• Compound 2 2

HeatN O⎯⎯⎯→ + other gaseous products ;

137322 2KC O CO+ ⎯⎯⎯→ ; 2 5 2 25 5I O CO I CO+ → +

Homologous series : • The series of organic compounds having a common difference of – CH2 between any two

successive members is called homologous series. • The classification and study of the members of the homologous series is called homology and the

members of the series are called homologues. Characteristic features of homologous series :

i) There is a common difference of – CH2 between two successive members. There is a common difference of 14 in molecular weight between two successive members.

ii) They possess similar chemical properties. iii) There is regular gradation in their physical properties iv) They can be prepared by similar methods. v) They can be represented by a general molecular formula.

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Ex. Alkanes – CnH2n + 2

Alkenes – CnH2n

Alkynes – Cn H2n – 2 Alkyl halides – CnH2n+1 “X” Alcohols and ethers – CnH2n+2O Aldehydes and ketones – CnH2nO Carboxylic acids and esters – CnH2nO2

Types of Carbons and Hydrogens : • Primary Carbon (1° – carbon) : It is bonded to just one another carbon or to no other carbon. • Secondary carbon (2° – carbon) : It is bonded to two other carbons. • Tertiary carbon (3° – carbon) – It is bonded to three other carbons. • Quaternary carbon (4° carbon) : It is bonded to four other carbons. Types of hydrogens :

• Primary hydrogen : Hydrogen attached to primary carbon. • Secondary hydrogen : Hydrogen attached to secondary carbon. • Tertiary hydrogen : Hydrogen attached to tertiary carbon.

Ex.

1

31 2 3 1

3 2 3

3311

CH|

CH CH CH C CH||

CHCH

°

° ° ° °

°°

− − − −

In the above structure, five primary carbons, one secondary carbon, one tertiary carbon and one quaternary carbon are present. Similarly, the number of primary hydrogens – 15, secondary hydrogens – 2, tertiary hydrogen – 1, and there is no quaternary hydrogen.

IUPAC nomenclature: IUPAC name of compound consists of

i) Root word ii) Suffix iii) Prefix • Root word gives the number of carbons in parent skeleton • Suffix gives the nature of the functional group • Prefix gives the nature of the substituent. Side chains: i) Alkyl: It contains one hydrogen less than that of alkane. Ex.

CH3 – Methyl C2H5 – Ethyl CH3 – CH2 – CH2 – n – propyl

3

3

CH CH|CH

− − iso propyl

CH3 – CH2 – CH2 – CH2 – n–Butyl 3 2

3

CH CH CH|CH

− − − Iso butyl

32|

3 CHCHCHCH −−− Sec – butyl

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3

3 3

CH|

CH C CH|

− − ter – butyl

CH3 – CH2 – CH2 – CH2 – CH2 – n – pentyl (Amyl) 3 2 2

3

CH CH CH CH|CH

− − − − Iso pentyl (Iso amyl)

3 2 5

3

|CH C C H

|CH

− − Ter-Pentyl

3

3 3

3

CH|

CH C CH|

CH

− − Neo – Pentyl

Alkenyl : It contains one hydrogen less than that of Alkene. Ex : CH2 = CH – Ethenyl ( Vinyl) CH3 – CH = CH – 1–Propenyl CH2 = CH – CH2 – Allyl CH3– CH2 – CH = CH – 1–Butenyl Alkynyl: It contains one hydrogen less than that of alkyne

Ex: HC ≡ C – ethynyl H3C– C ≡ C – 1– propynyl H3C – CH2 – C ≡ C – 1– butynyl • When all the carbons are present in a straight chain, the alkane is called normal alkane. • When all the carbons are not present in straight chain, it may be called iso alkane. Ex: CH3 – CH2 – CH2 – CH3 n – Butane

3 3

3

CH CH CH|CH

− − Iso butane

Root word : No. of carbons: Root word 1 Meth 2 Eth 3 Prop 4 But 5 Pent 6 Hex 7 Hept 8 Oct 9 Non 10 Dec

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1. Longest chain rule : Select the continuous chain of carbons having maximum number of carbon atoms.

2. Lowest sum rule: Gives the lowest possible numbers to the substituents and functional groups. 3. Longest chain rule can be violated to include the double bonds triple bonds, functional groups. 4. If two or more functional groups are present, senior functional group is given suffix and junior

functional group is given prefix. 5. If two or more different substituents are present at various positions, consider the lowest sum

irrespective of the nature of substituents. 6. If two similar substituents are present at identical positions from opposite ends, then follow

alphabetical order to give lowest number. 7. If two or more carbon chains having the same number of carbons are present then the chain having

more number of branches is selected as parent chain. 8. If the compounds contact more than one functional group the principal group form the suffix while

the other functional group is considered as the substitutent. 9. The preference order is COOH > Acid derivatives > CHO > CN > C = O > – OH > NH2 > – O –

> C = C > C ≡ C. Name Formula Suffix Prefix

Carboxylic acid – COOH oic acid carboxy Acid chloride – COCl oyl chloride Chloro formlye Acid amide – CONH2 Amide Carbomyl Ester – COOR ate Alkoxy carbonyl Aldehyde – CHO al Aldo or formyl Cyanide – CN nitrile Cyano

Ketone – OC|

= one Keto or oxo

Alcohol – OH ol Hydroxy Amine – NH2 amine Amino Ether – O – – Alkoxy

Alkene −=−||CC ene ene

Alkyne −≡− CC yne yne

Substituents: Which are given only prefixes: – X (Cl, Br, I) halo – ONO nitrite – NO2 nitro – NO nitroso – OR Alkoxy – R Alkyl

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EXERCISES : Structure IUPAC name

1.

3

3 3

3 3

CH|

CH CH C CH| |CH CH

− − − 2, 2, 3 – trimethyl butane

2.

CH3 CH CH2 CH3

CH

CH3CH3

2, 3 – dimethyl pentane

3. 3

3 2 5

CH CH CH CH| |CH C H

− − − 2, 3 – dimethyl pentane

4. 3 2 2 3

3 2 5

CH CH CH CH CH CH| |CH C H

− − − − − 3–ethyl – 4 – methyl hexane

5.

CH3 CHCH2 CH3

CH

CH3CH3

CH2 CH2 CH2

4 (1– methyl ethyl ) heptane

6.

CH3 CHCH2 CH3

CH

CH2 CH2 CH2

CH

CH3

CH3

CH3

CH3

2, 3, 6 – trimethyl – 4 − propyl heptane

7.

CH3 CHCH2 C2H5

CH

CH2 CH2 CH2

C

CH3

CH3

CH3

CH2

CH3

5 – (1, 2, 2 – trimethyl propyl) nonane

8. CH3 CH

CH = CH2

CH2 CH3

3 – methyl pent – 1 – ene

9. CH2 = CH – CH = CH2 But – 1, 3 – diene

10. CH3 CHCH2

CH

CH2 CH2

CH2

CH3CH2

3 – propyl hex – 1 – ene

11. CH ≡ C – CH = CH – CH3 Pent – 3 – ene – 1 – yne 12. CH ≡ C – CH = CH2 But – 1 – en – 3 – yne

13. 3 3CH CH CH CH

| |Cl Br

− − − 2 – Bromo – 3 – chloro butane

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14. 3 2 2 3

2

CH CH CH CH CH CH CH CH| | |NO Cl Br

− − − − − − − 4 – bromo – 3 – chloro – 2 – nitro octane

15. CH3 – O – C2H5 Methoxy ethane 16. C2H5 – O – C2H5 Ethoxy ethane

17. 3 3

3

CH O CH CH|CH

− − − 2 – Methoxy propane

18. 3 2 3

2 5

CH CH CH CH CH CH|OC H

− = − − − 4 – Ethoxy hex – 2 – ene

19. 3 3CH CH CH

|OH

− − propan – 2 – ol

20. 3 3CH CH CH CH CH

|OH

− − = − pent – 3 – en – 2 – ol

21.

3

3

3

CH|

CH C OH|

CH

− − 2 – methyl propan – 2 – ol

22. CH3 CH CH2 CH3

OCH3

CH

OH 2 – Methoxy pentan – 3 – ol

23.

O||

H C|H

− Methanal

24. CH3 – CHO Ethanal

25. 3

3

CH CH CHO|CH

− − 2 – Methyl propanal

26. 3

3

CHOH| |

H C CH CH CHO− − − 3 – Hydroxy – 2– Methyl Butanal

27. 2CH CH CH CHO

|OH

− − − 2 – Hydroxy – But – 3 – enal

28. 3 3CH C CH

||O

− − Propanone

29. 3

3

O||

H C C CH CHO|CH

− − − 3 – keto – 2– Methyl Butanal

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17

30.

O||

H C|

OH

− Methanoic acid

31. 3

3

CH|

H C CH COOH− − 2– Methyl propanoic acid

32. 3 2

3

CHO|

H C CH CH CH COOH|CH

− − − − 3 – Aldo −2 − Methyl pentanoic acid

33. COOH|COOH

Ethan − dioic acid

34. 3H C OCH||O

− − Methyl methanoate

35. CH3 –COOC2 H5 Ethyl ethanoate

36. 3 3CH CH C O CH| ||Cl O

− − − − Methyl – 2– chloro propanoate

37. CH3 – CH2 – NH2 Ethanamine

38. 3 3

2

CH CH CH|NH

− − Propan – 2− amine

39. 3 3

2

CH CH CH CH| |OH NH

− − − 3 – Amino – Butan – 2– ol

40. CH3 – NH – CH3 N – methyl amino methane 41. CH3 – NH – C2 H5 N – Methyl amino ethane

42. 3 3

3

CH N CH|CH

− − N, N – Dimethyl amino methane

43. 3 2 5

3

CH N C H|CH

− − N, N – Dimethyl amino ethane

44. H – CN Methane nitrile 45. CH3 – CN Ethane nitrile

46. 3

3

CH CH CN|CH

− − 2– methyl propane nitrile

47. 2 2CH CH

| |CN CN

− butane –1, 4 – Dinitrile

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18

48. 2 2 2CH CH CH

| | |CN CN CN

− − 3–cyano pentane – 1, 5 – dinitrile

49. CH2

COOCH3

COOCH3

Dimethyl propane dioate

50. CH2 COOCH3

CH2 COOC2H5 Ethyl methyl butane dioate

Common Name Structure 1. Acetic acid CH3 – COOH 2. Formic acid HCOOH 3. Acetaldehyde CH3CHO

4. Acetone 33 CH

O||

CCH −−

5. Vinyl cyanide CH2 = CH – CN 6. Vinyl alcohol CH2 = CH – OH 7. Acetonitrile CH3 – CN 8. Methyl carbinol CH3 – CH2 – OH

9. Trimethyl carbinol OH

3CH|C| 3CH

3CH −−

10. Ter – Butyl alcohol OH

3CH|C| 3CH

3CH −−

11. Acetyl chloride Cl

O||C3CH −−

12. Acetamide 2NH

O||C3CH −−

13. Methyl acetylene CH3 – C ≡ CH 14. Pyrene CCl4 15. Formaldehyde HCHO 16. Phenyl isocyanide C6H5 – NC

17. Dimethyl ketone 3 3||CH C CH

O

− −

18. Ethyl acetate CH3OOC2H5

19. Aldol CHO2CH

OH|CH3CH −−−

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19

20. Marsh gas CH4

21. 2, 2, 4 – Trimethyl pentane 3CH

3CH|

HCC2H

3CH

3CH

|

|CC3H −−−−

22. Hept – 2 – enal OHC – HC = CH – H2C –H2C – H2C – CH3

23. Pent – 3 – en – 1 – yne HC ≡ C – HC = CH – CH3

24. 2 – Hydroxy Butanoic acid 3CHC2H

OH|

HCHOOC −−−

25. 2 – methoxy propane 3CH

3OCH|CH3CH −−

26. 4 – Keto pentanal 3CH

O||CC2HC2HOHC −−−−

27. Ethyl methanoate HCOOC2H5 28. Ethoxy ethane C2H5 – O – C2H5

29. Ethanoyl chloride Cl

O||C3CH −−

30. Ethanamide 2NH

O||C3CH −−

31. Propan – 2 – amine 3CH

2NH|CH3CH −−

32. N – ethyl amino ethane C2H5 – NH – C2H5

33. N, N – diethyl amino propane 3CH2CH2CH

5H2C|

N5H2C −−−−

34. Ethane nitrile CH3 – CN 35. Methyl carbyl amine CH3 – NC

WRITING THE STRUCTURE OF THE COMPOUND WHOSE NAME IS GIVEN

(a) Observe the word root and write the continuous carbon chain. (b) Number the carbon atoms in a suitable way and attach the functional groups, substituents and

multiple bonds at their respective carbon atoms. (c) Carbon has tetravalency. Attach the required number of hydrogen atoms at each carbon atom to

satisfy its tetra valency. Now the structure is complex. Example : 3 Bromo 2 Methyl pentan 2 ol− − − − −

a) As per the word root pent we have 1 2 3 4 5C C C C C− − − −

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20

b) As per the name

1 2 3 4 5

3

|

| |

OH

C C C C C

CH Br

− − − −

c) To satisfy the tetravalency of each carbon atom hydrogen atoms ate added to each carbon atom as per requirement.

3

H OH H H H| | | ||

H C C C C C H| | | | |

H CH Br H H

− − − − − −

Hence, the structure is complete Let us see the nomenclature of cyclic compounds. Cyclohexene 3-Chlorocyclohexene Cyclohexyne Cyclohexan 1-ol Nomenclature of substituted benzene compounds

The common names of the above compounds are given in brackets wherever necessary.

Substituents are mentioned in alphabetical order giving lowest number to the first substituent. 1) Methylbenzene IUPAC 2) Methoxybenzene 3) Bromobenzene (Toluene) (Anisole) 4) 1, 2-dibromobenzene 5) 1, 3-dichlorobenzene 6) 1-chloro-4-nitrobenzene (orthodibromobenzene) (metadichlorobenzene) 7) Benzene carbaldehyde 8) Benzene carboxylic acid 9) Styrene (benzaldehyde) (benzoic acid) 10) Phenol 11) Acetophenone

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21

Dimethyl derivatives of benzene are called Xylenes. 1, 2-dimethyl benzene (O - xylene) Consider It is named as 2, 4, 6 - trinitro toluene taking toluene as the base name. Similarly is named as 4-ethyl - 2 - fluroanisole. When no simple base name other than benzene is possible, the positions are numbered so as to

give the lowest locate at the first point of difference. Example : 1-Chloro-2, 4-dinitrobenzene 4-ethyl-1-fluoro-2-nitrobenzene (not 4-chloro -1, 3-dinitrobenzene) If benzene ring is named as substituent it is named as phenyl . Similarly an arene is named as

aryl. 2-phenylethanol Benzyl Chloride