chmeistry-part-2-book

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 363 concentrated sulphuric acid. The formation a yellow precipitate, insoluble in ammonium of Prussian blue colour confirms the presence of hydroxide shows the presence of iodine. nitrogen. Sodium cyanide first reacts with iron(II) sulphate and forms sodium X– + Ag+ → AgX hexacyanidoferrate(II). On heating with X represents a halogen – Cl, Br or I. concentrated sulphuric acid some iron(II) ions are oxidised to iron(III) ions which react with If nitrogen or sulphur is also present in the sodium hexacyanidoferrate(II) to produce compound, the sodium fusion extract is first iron(III) hexacyanidoferrate(II) (ferriferrocyanide) boiled with concentrated nitric acid to which is Prussian blue in colour. decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions 6CN– + Fe2+ → [Fe(CN)6]4– would otherwise interfere with silver nitrate test for halogens. 3[Fe(CN)6]4– + 4Fe3+ Fe4[Fe(CN)6]3.xH2O (B) Test for Sulphur Prussian blue (D) Test for Phosphorus (a) The sodium fusion extract is acidified with The compound is heated with an oxidising acetic acid and lead acetate is added to it. agent (sodium peroxide). The phosphorus A black precipitate of lead sulphide present in the compound is oxidised to indicates the presence of sulphur. phosphate. The solution is boiled with nitric acid and then treated with ammonium S2– + Pb2+ → PbS molybdate. A yellow colouration or precipitate Black indicates the presence of phosphorus. (b) On treating sodium fusion extract with Na3PO4 + 3HNO3 → H3PO4+3NaNO3 sodium nitroprusside, appearance of a H3PO4 + 12(NH4)2MoO4 + 21HNO3 → violet colour further indicates the presence of sulphur. Ammonium molybdate S2– + [Fe(CN)5NO]2– → [Fe(CN)5NOS]4– Violet (NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O Ammonium In case, nitrogen and sulphur both are present in an organic compound, sodium phosphomolybdate thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free 12.10 QUANTITATIVE ANALYSIS cyanide ions. Quantitative analysis of compounds is very Na + C + N + S → NaSCN important in organic chemistry. It helps Fe3+ +SCN– → [Fe(SCN)]2+ chemists in the determination of mass per cent of elements present in a compound. You have Blood red learnt in Unit-1 that mass per cent of elements If sodium fusion is carried out with excess is required for the determination of emperical of sodium, the thiocyanate decomposes to yield and molecular formula. cyanide and sulphide. These ions give their usual tests. The percentage composition of elements present in an organic compound is determined NaSCN + 2Na → NaCN+Na2S by the following methods: (C) Test for Halogens 12.10.1 Carbon and Hydrogen The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white Both carbon and hydrogen are estimated in precipitate, soluble in ammonium hydroxide one experiment. A known mass of an organic shows the presence of chlorine, a yellowish compound is burnt in the presence of excess precipitate, sparingly soluble in ammonium of oxygen and copper(II) oxide. Carbon and hydroxide shows the presence of bromine and hydrogen in the compound are oxidised to carbon dioxide and water respectively. CxHy + (x + y/4) O2 → x CO2 + (y/2) H2O 2019-20

364 CHEMISTRY Fig.12.14 Estimation of carbon and hydrogen. Water and carbon dioxide formed on oxidation of substance are absorbed in anhydrous calcium chloride and potassium hydroxide solutions respectively contained in U tubes. The mass of water produced is determined 12.10.2 Nitrogen by passing the mixture through a weighed There are two methods for estimation of nitrogen: (i) Dumas method and (ii) Kjeldahl’s U-tube containing anhydrous calcium method. (i) Dumas method: The nitrogen containing chloride. Carbon dioxide is absorbed in organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, another U-tube containing concentrated yields free nitrogen in addition to carbon dioxide and water. solution of potassium hydroxide. These tubes CxHyNz + (2x + y/2) CuO → are connected in series (Fig.12.14). The increase x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu in masses of calcium chloride and potassium Traces of nitrogen oxides formed, if any, hydroxide gives the amounts of water and are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The carbon dioxide from which the percentages of mixture of gases so produced is collected over an aqueous solution of potassium hydroxide carbon and hydrogen are calculated. which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated Let the mass of organic compound be tube (Fig.12.15). m g, mass of water and carbon dioxide Let the mass of organic compound = m g produced be m1 and m2 g respectively; Volume of nitrogen collected = V1 mL Room temperature = T1K Percentage of carbon= 12 × m2 ×100 Volume of nitrogen at STP= p1V1 × 273 44 × m 760 × T1 Percentage of hydrogen = 2 × m1 ×100 (Let it be V mL) 18 × m Where p and V are the pressure and volume Problem 12.20 11 On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon of nitrogen, p is different from the atmospheric dioxide and 0.1014g of water. Determine 1 the percentage composition of carbon and hydrogen in the compound. pressure at which nitrogen gas is collected. The value of p is obtained by the relation; Solution 1 Percentage of carbon =12 × 0.198 ×100 44 × 0.246 p1= Atmospheric pressure – Aqueous tension 22400 mL N2 at STP weighs 28 g. = 21.95% 2 × 0.1014 ×100 Percentage of hydrogen = 18 × 0.246 = 4.58%

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 365 Fig.12.15 Dumas method. The organic compound yields nitrogen gas on heating it with copper(II) oxide in the presence of CO2 gas. The mixture of gases is collected over potassium hydroxide solution in which CO2 is absorbed and volume of nitrogen gas is determined. V mL N2 at STP weighs = 28 × V g 41.9 mL of nitrogen weighs= 28 × 41.9 g 22400 22400 28 × V ×100 Percentage of nitrogen = 28 × 41.9 ×100 Percentage of nitrogen = 22400 × m 22400 × 0.3 Problem 12.21 =17.46% In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound (ii) Kjeldahl’s method: The compound gave 50mL of nitrogen collected at 300K containing nitrogen is heated with concentrated temperature and 715mm pressure. sulphuric acid. Nitrogen in the compound gets Calculate the percentage composition of converted to ammonium sulphate (Fig. 12.16). nitrogen in the compound. (Aqueous The resulting acid mixture is then heated with tension at 300K=15 mm) excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of Solution standard solution of sulphuric acid. The amount of ammonia produced is determined Volume of nitrogen collected at 300K and by estimating the amount of sulphuric acid 715mm pressure is 50 mL consumed in the reaction. It is done by Actual pressure = 715-15 =700 mm estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with Volume of nitrogen at STP = 273 × 700 × 50 standard alkali solution. The difference between 300 × 760 the initial amount of acid taken and that left = 41.9 mL 22,400 mL of N2 at STP weighs = 28 g 2019-20

366 CHEMISTRY Fig.12.16 Kjeldahl method. Nitrogen-containing compound is treated with concentrated H2SO4 to get ammonium sulphate which liberates ammonia on treating with NaOH; ammonia is absorbed in known volume of standard acid. after the reaction gives the amount of acid reacted Percentage of 14 × M× 2 ( V-V1 /2) × 100 with ammonia. N= 1000 m Organic compound + H2SO4 → (NH4)2SO4 2NaOH → Na2SO4 + 2NH3 + 2H2O =1.4 × M × 2 (V -V1 / 2) 2NH3 + H2SO4 → (NH4)2SO4 m Let the mass of organic compound taken = m g Kjeldahl method is not applicable to Volume of H2SO4 of molarity, M, compounds containing nitrogen in nitro and taken = V mL azo groups and nitrogen present in the ring Volume of NaOH of molarity, M, used for (e.g. pyridine) as nitrogen of these compounds titration of excess of H2SO4 = V1 mL does not change to ammonium sulphate V1mL of NaOH of molarity M under these conditions. = V1 /2 mL of H2SO4 of molarity M Problem 12.22 Volume of H2SO4 of molarity M unused = (V - V1/2) mL During estimation of nitrogen present in (V- V1/2) mL of H2SO4 of molarity M an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g = 2(V-V1/2) mL of NH3 solution of of the compound in Kjeldahl’s estimation molarity M. of nitrogen, neutralized 10 mL of 1 M 1000 mL of 1 M NH3 solution contains H2SO4. Find out the percentage of 17g NH3 or 14 g of N nitrogen in the compound. 2(V-V1/2) mL of NH3 solution of molarity M contains: Solution 14 × M × 2 ( V − V1 / 2) g N 1 M of 10 mL H2SO4=1M of 20 mL NH3 1000 mL of 1M ammonia contains 14 g 1000 nitrogen 20 mL of 1M ammonia contains 2019-20

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 367 14 × 20 g nitrogen Percentage of halogen 1000 = atomic mass of X × m1 ×100 molecular mass of AgX × m Percentage of nitrogen = 14 × 20×100 = 56.0% Problem 12.23 1000× 0.5 In Carius method of estimation of halogen, 0.15 g of an organic compound 12.10.3 Halogens gave 0.12 g of AgBr. Find out the percentage of bromine in the compound. Carius method: A known mass of an organic Solution compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard Molar mass of AgBr = 108 + 80 glass tube known as Carius tube, (Fig.12.17) = 188 g mol-1 188 g AgBr contains 80 g bromine 0.12 g AgBr contains 80 × 0.12 g bromine 188 Percentage of bromine= 80 × 0.12 ×100 188 × 0.15 = 34.04% 12.10.4 Sulphur Fig. 12.17 Carius method. Halogen containing A known mass of an organic compound is organic compound is heated with fuming heated in a Carius tube with sodium peroxide nitric acid in the presence of silver or fuming nitric acid. Sulphur present in the nitrate. compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding in a furnace. Carbon and hydrogen present in excess of barium chloride solution in water. the compound are oxidised to carbon dioxide The precipitate is filtered, washed, dried and and water. The halogen present forms the weighed. The percentage of sulphur can be corresponding silver halide (AgX). It is filtered, calculated from the mass of barium sulphate. washed, dried and weighed. Let the mass of organic Let the mass of organic compound taken = m g compound taken = m g Mass of AgX formed = m1 g 1 mol of AgX contains 1 mol of X and the mass of barium Mass of halogen in m1g of AgX sulphate formed = m1g 1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur = atomic mass of X × m1g molecular mass of AgX m1 g BaSO4 contains 32 × m1 g sulphur 233 Percentage of sulphur= 32 × m1 ×100 233 ×m Problem 12.24 In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of 2019-20

368 CHEMISTRY barium sulphate. What is the percentage percentage composition (100) and the sum of the of sulphur in the compound? percentages of all other elements. However, oxygen can also be estimated directly as follows: Solution A definite mass of an organic compound is Molecular mass of BaSO4 = 137+32+64 decomposed by heating in a stream of nitrogen = 233 g gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the 233 g BaSO4 contains 32 g sulphur oxygen is converted to carbon monoxide. This mixture is passed through warm iodine 0.4813 g BaSO4 contains 32 × 0.4813 g pentoxide (I2O5) when carbon monoxide is 233 oxidised to carbon dioxide producing iodine. sulphur Percentage of sulphur= 32 × 0.4813 ×100 Compound heat → O2 + other gaseous 233 × 0.157 products = 42.10% 2C + O2 1373 K→ 2CO]× 5 (A) 12.10.5 Phosphorus I2O5 + 5CO → I2 + 5CO2]× 2 (B) A known mass of an organic compound is On making the amount of CO produced in heated with fuming nitric acid whereupon equation (A) equal to the amount of CO used in phosphorus present in the compound is equation (B) by multiplying the equations (A) and oxidised to phosphoric acid. It is precipitated (B) by 5 and 2 respectively; we find that each as ammonium phosphomolybdate, (NH4)3 mole of oxygen liberated from the compound will PO4.12MoO3, by adding ammonia and produce two moles of carbondioxide. ammonium molybdate. Alternatively, phosphoric acid may be precipitated as Thus 88 g carbon dioxide is obtained if 32 g MgNH4PO4 by adding magnesia mixture which oxygen is liberated. on ignition yields Mg2P2O7. Let the mass of organic compound taken Let the mass of organic compound taken be m g = m g and mass of ammonium phospho Mass of carbon dioxide produced be m1 g molydate = m1g Molar mass of (NH4)3PO4.12MoO3 = 1877g ∴ m1 g carbon dioxide is obtained from 32 × m1 g O2 88 31× m1 ×100 ∴Percentage of oxygen = 32 × m1 × 100 % 1877 ×m 88 × m Percentage of phosphorus = % If phosphorus is estimated as Mg2P2O7, The percentage of oxygen can be derived from the amount of iodine produced also. Percentage of phosphorus = 62 × m1 ×100 % 222 ×m Presently, the estimation of elements in an organic compound is carried out by using where, 222 u is the molar mass of Mg2P2O7, microquantities of substances and automatic m, the mass of organic compound taken, m1, experimental techniques. The elements, the mass of Mg2P2O7 formed and 62, the mass carbon, hydrogen and nitrogen present in a of two phosphorus atoms present in the compound are determined by an apparatus compound Mg2P2O7. known as CHN elemental analyser. The analyser requires only a very small amount 12.10.6 Oxygen of the substance (1-3 mg) and displays the values on a screen within a short time. A The percentage of oxygen in an organic compound detailed discussion of such methods is is usually found by difference between the total beyond the scope of this book. 2019-20

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 369 SUMMARY In this unit, we have learnt some basic concepts in structure and reactivity of organic compounds, which are formed due to covalent bonding. The nature of the covalent bonding in organic compounds can be described in terms of orbitals hybridisation concept, according to which carbon can have sp3, sp2 and sp hybridised orbitals. The sp3, sp2 and sp hybridised carbons are found in compounds like methane, ethene and ethyne respectively. The tetrahedral shape of methane, planar shape of ethene and linear shape of ethyne can be understood on the basis of this concept. A sp3 hybrid orbital can overlap with 1s orbital of hydrogen to give a carbon - hydrogen (C–H) single bond (sigma, σ bond). Overlap of a sp2 orbital of one carbon with sp2 orbital of another results in the formation of a carbon–carbon σ bond. The unhybridised p orbitals on two adjacent carbons can undergo lateral (side-by-side) overlap to give a pi (π) bond. Organic compounds can be represented by various structural formulas. The three dimensional representation of organic compounds on paper can be drawn by wedge and dash formula. Organic compounds can be classified on the basis of their structure or the functional groups they contain. A functional group is an atom or group of atoms bonded together in a unique fashion and which determines the physical and chemical properties of the compounds. The naming of the organic compounds is carried out by following a set of rules laid down by the International Union of Pure and Applied Chemistry (IUPAC). In IUPAC nomenclature, the names are correlated with the structure in such a way that the reader can deduce the structure from the name. Organic reaction mechanism concepts are based on the structure of the substrate molecule, fission of a covalent bond, the attacking reagents, the electron displacement effects and the conditions of the reaction. These organic reactions involve breaking and making of covalent bonds. A covalent bond may be cleaved in heterolytic or homolytic fashion. A heterolytic cleavage yields carbocations or carbanions, while a homolytic cleavage gives free radicals as reactive intermediate. Reactions proceeding through heterolytic cleavage involve the complimentary pairs of reactive species. These are electron pair donor known as nucleophile and an electron pair acceptor known as electrophile. The inductive, resonance, electromeric and hyperconjugation effects may help in the polarisation of a bond making certain carbon atom or other atom positions as places of low or high electron densities. Organic reactions can be broadly classified into following types; substitution, addition, elimination and rearrangement reactions. Purification, qualitative and quantitative analysis of organic compounds are carried out for determining their structures. The methods of purification namely : sublimation, distillation and differential extraction are based on the difference in one or more physical properties. Chromatography is a useful technique of separation, identification and purification of compounds. It is classified into two categories : adsorption and partition chromatography. Adsorption chromatography is based on differential adsorption of various components of a mixture on an adsorbent. Partition chromatography involves continuous partitioning of the components of a mixture between stationary and mobile phases. After getting the compound in a pure form, its qualitative analysis is carried out for detection of elements present in it. Nitrogen, sulphur, halogens and phosphorus are detected by Lassaigne’s test. Carbon and hydrogen are estimated by determining the amounts of carbon dioxide and water produced. Nitrogen is estimated by Dumas or Kjeldahl’s method and halogens by Carius method. Sulphur and phosphorus are estimated by oxidising them to sulphuric and phosphoric acids respectively. The percentage of oxygen is usually determined by difference between the total percentage (100) and the sum of percentages of all other elements present. 2019-20

370 CHEMISTRY 12.1 EXERCISES 12.2 12.3 What are hybridisation states of each carbon atom in the following compounds ? 12.4 CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6 Indicate the σ and π bonds in the following molecules : C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4- one. Give the IUPAC names of the following compounds : (a) (b) (c) (d) (e) (f) Cl2CHCH2OH 12.5 Which of the following represents the correct IUPAC name for the compounds 12.6 concerned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7- 12.7 Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne. 12.8 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2 Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for : (a) 2,2,4-Trimethylpentane (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (c) Hexanedial Identify the functional groups in the following compounds (a) (b) (c) 12.9 Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and 12.10 why ? 12.11 Explain why alkyl groups act as electron donors when attached to a π system. 12.12 12.13 Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. + (a) C6H5OH (b) C6H5NO2 (c) CH 3CH=CHCHO (d) C6H5–CHO (e) C6H5 −CH2 + (f) CH3CH = CHC H2 What are electrophiles and nucleophiles ? Explain with examples. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) CH3COOH + HO– → CH3COO− + H2O 2019-20

ORGANIC CHEMISTRY – SOME BASIC PRINCIPLES AND TECHNIQUES 371 (b) – (CH3 ) C ( CN ) ( OH ) 2 CH3COCH3 + C N → + (c) C6H6 + CH3 C O → C6H5COCH3 12.14 Classify the following reactions in one of the reaction type studied in this unit. 12.15 (a) CH3CH2Br + HS− → CH3CH2SH + Br − (b) (CH3 ) C = CH2 + HCl → (CH3 ) ClC − CH3 2 2 (c) CH3CH2Br + HO− → CH2 = CH2 + H2O + Br − (d) (CH3 ) C − CH2OH + HBr → (CH3 ) CBrCH2CH2CH3 + H2O 3 2 What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or resonance contributors ? (a) (b) (c) 12.16 For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion. (a) (b) (c) (d) 12.17 Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? 12.18 12.19 (a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH 12.20 (b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography Describe the method, which can be used to separate two compounds with different solubilities in a solvent S. What is the difference between distillation, distillation under reduced pressure and steam distillation ? 2019-20

372 CHEMISTRY 12.21 Discuss the chemistry of Lassaigne’s test. 12.22 12.23 Differentiate between the principle of estimation of nitrogen in an organic 12.24 compound by (i) Dumas method and (ii) Kjeldahl’s method. 12.25 12.26 Discuss the principle of estimation of halogens, sulphur and phosphorus present 12.27 in an organic compound. 12.28 12.29 Explain the principle of paper chromatography. 12.30 12.31 Why is nitric acid added to sodium extract before adding silver nitrate for testing 12.32 halogens? 12.33 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens. 12.34 12.35 Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor. 12.36 Explain, why an organic liquid vaporises at a temperature below its boiling point 12.37 in its steam distillation ? 12.38 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give 12.39 reason for your answer. 12.40 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound? Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test? An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion. A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in 50 ml of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound. In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound. In the organic compound CH2 = CH – CH2 – CH2 – C ≡ CH, the pair of hydridised orbitals involved in the formation of: C2 – C3 bond is: (a) sp – sp2 (b) sp – sp3 (c) sp2 – sp3 (d) sp3 – sp3 In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: (a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]4 Which of the following carbocation is most stable ? + + ++ (a) (CH3)3C. C H2 (b) (CH3)3 C (c) CH3CH2 C H2 (d) CH3 C H CH2CH3 The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography The reaction: CH3CH2I + KOH(aq) → CH3CH2OH + KI is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition 2019-20

HYDROCARBONS 373 UNIT 13 HYDROCARBONS Hydrocarbons are the important sources of energy. After studying this unit, you will be able to • name hydrocarbons according to The term ‘hydrocarbon’ is self-explanatory which means compounds of carbon and hydrogen only. Hydrocarbons IUPAC system of nomenclature; play a key role in our daily life. You must be familiar with the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the • recognise and write structures abbreviated form of liquified petroleum gas whereas CNG stands for compressed natural gas. Another term ‘LNG’ of isomers of alkanes, alkenes, (liquified natural gas) is also in news these days. This is also a fuel and is obtained by liquifaction of natural gas. alkynes and aromatic Petrol, diesel and kerosene oil are obtained by the fractional distillation of petroleum found under the earth’s crust. hydrocarbons; Coal gas is obtained by the destructive distillation of coal. • learn about various methods of Natural gas is found in upper strata during drilling of oil wells. The gas after compression is known as compressed preparation of hydrocarbons; natural gas. LPG is used as a domestic fuel with the least pollution. Kerosene oil is also used as a domestic fuel but • distinguish between alkanes, it causes some pollution. Automobiles need fuels like petrol, diesel and CNG. Petrol and CNG operated automobiles alkenes, alkynes and aromatic cause less pollution. All these fuels contain mixture of hydrocarbons, which are sources of energy. Hydrocarbons hydrocarbons on the basis of are also used for the manufacture of polymers like polythene, polypropene, polystyrene etc. Higher physical and chemical properties; hydrocarbons are used as solvents for paints. They are also • draw and differentiate between used as the starting materials for manufacture of many dyes and drugs. Thus, you can well understand the various conformations of ethane; importance of hydrocarbons in your daily life. In this unit, you will learn more about hydrocarbons. • appreciate the role of 13.1 CLASSIFICATION hydrocarbons as sources of Hydrocarbons are of different types. Depending upon the energy and for other industrial types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated applications; • predict the formation of the addition products of unsymmetrical alkenes and alkynes on the basis of electronic mechanism; • comprehend the structure of benzene, explain aromaticity and understand mechanism of electrophilic substitution reactions of benzene; • predict the directive influence of substituents in monosubstituted benzene ring; • learn about carcinogenicity and toxicity. 2019-20

374 CHEMISTRY (ii) unsaturated and (iii) aromatic general formula for alkane family or hydrocarbons. Saturated hydrocarbons homologous series? If we examine the contain carbon-carbon and carbon-hydrogen formula of different alkanes we find that the single bonds. If different carbon atoms are general formula for alkanes is CnH2n+2. It joined together to form open chain of carbon represents any particular homologue when n atoms with single bonds, they are termed as is given appropriate value. Can you recall the alkanes as you have already studied in structure of methane? According to VSEPR Unit 12. On the other hand, if carbon atoms theory (Unit 4), methane has a tetrahedral form a closed chain or a ring, they are termed structure (Fig. 13.1), in which carbon atom lies as cycloalkanes. Unsaturated hydrocarbons at the centre and the four hydrogen atoms lie contain carbon-carbon multiple bonds – at the four corners of a regular tetrahedron. double bonds, triple bonds or both. Aromatic All H-C-H bond angles are of 109.5°. hydrocarbons are a special type of cyclic compounds. You can construct a large number Fig. 13.1 Structure of methane of models of such molecules of both types In alkanes, tetrahedra are joined together (open chain and close chain) keeping in mind in which C-C and C-H bond lengths are that carbon is tetravalent and hydrogen is 154 pm and 112 pm respectively (Unit 12). You monovalent. For making models of alkanes, have already read that C–C and C–H σ bonds you can use toothpicks for bonds and are formed by head-on overlapping of sp3 plasticine balls for atoms. For alkenes, alkynes hybrid orbitals of carbon and 1s orbitals of and aromatic hydrocarbons, spring models can hydrogen atoms. be constructed. 13.2.1 Nomenclature and Isomerism You have already read about nomenclature 13.2 ALKANES of different classes of organic compounds in Unit 12. Nomenclature and isomerism in As already mentioned, alkanes are saturated alkanes can further be understood with the open chain hydrocarbons containing help of a few more examples. Common names carbon - carbon single bonds. Methane (CH4) are given in parenthesis. First three alkanes is the first member of this family. Methane is a – methane, ethane and propane have only gas found in coal mines and marshy places. If one structure but higher alkanes can have you replace one hydrogen atom of methane by more than one structure. Let us write carbon and join the required number of structures for C4H10. Four carbon atoms of hydrogens to satisfy the tetravalence of the C4H10 can be joined either in a continuous other carbon atom, what do you get? You get chain or with a branched chain in the C2H6. This hydrocarbon with molecular following two ways : formula C2H6 is known as ethane. Thus you I can consider C2H6 as derived from CH4 by replacing one hydrogen atom by -CH3 group. Butane (n- butane), (b.p. 273 K) Go on constructing alkanes by doing this theoretical exercise i.e., replacing hydrogen atom by –CH3 group. The next molecules will be C3H8, C4H10 … These hydrocarbons are inert under normal conditions as they do not react with acids, bases and other reagents. Hence, they were earlier known as paraffins (latin : parum, little; affinis, affinity). Can you think of the 2019-20

HYDROCARBONS 375 II structures, they are known as structural isomers. It is also clear that structures I and 2-Methylpropane (isobutane) III have continuous chain of carbon atoms but (b.p.261 K) structures II, IV and V have a branched chain. Such structural isomers which differ in chain In how many ways, you can join five of carbon atoms are known as chain isomers. carbon atoms and twelve hydrogen atoms of Thus, you have seen that C4H10 and C5H12 C5H12? They can be arranged in three ways as have two and three chain isomers respectively. shown in structures III–V III Problem 13.1 Pentane (n-pentane) Write structures of different chain isomers (b.p. 309 K) of alkanes corresponding to the molecular formula C6H14. Also write their IUPAC names. Solution (i) CH3 – CH2 – CH2 – CH2– CH2– CH3 n-Hexane IV 2-Methylpentane 2-Methylbutane (isopentane) 3-Methylpentane (b.p. 301 K) 2,3-Dimethylbutane V 2,2 - Dimethylbutane 2,2-Dimethylpropane (neopentane) Based upon the number of carbon atoms (b.p. 282.5 K) attached to a carbon atom, the carbon atom is termed as primary (1°), secondary (2°), tertiary Structures I and II possess same (3°) or quaternary (4°). Carbon atom attached molecular formula but differ in their boiling to no other carbon atom as in methane or to points and other properties. Similarly only one carbon atom as in ethane is called structures III, IV and V possess the same primary carbon atom. Terminal carbon atoms molecular formula but have different are always primary. Carbon atom attached to properties. Structures I and II are isomers of two carbon atoms is known as secondary. butane, whereas structures III, IV and V are Tertiary carbon is attached to three carbon isomers of pentane. Since difference in atoms and neo or quaternary carbon is properties is due to difference in their attached to four carbon atoms. Can you identify 1°, 2°, 3° and 4° carbon atoms in structures I 2019-20

376 CHEMISTRY to V ? If you go on constructing structures for compounds. These groups or substituents are known as alkyl groups as they are derived from higher alkanes, you will be getting still larger alkanes by removal of one hydrogen atom. General formula for alkyl groups is CnH2n+1 number of isomers. C6H14 has got five isomers (Unit 12). and C7H16 has nine. As many as 75 isomers are possible for C10H22. Let us recall the general rules for nomenclature already discussed in Unit 12. In structures II, IV and V, you observed Nomenclature of substituted alkanes can further be understood by considering the that –CH3 group is attached to carbon atom following problem: numbered as 2. You will come across groups like –CH3, –C2H5, –C3H7 etc. attached to carbon atoms in alkanes or other classes of Problem 13.2 Write structures of different isomeric alkyl groups corresponding to the molecular formula C5H11. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain. Solution Corresponding alcohols Name of alcohol Structures of – C5H11 group (i) CH3 – CH2 – CH2 – CH2– CH2 – CH3 – CH2 – CH2 – CH2– CH2 – OH Pentan-1-ol (ii) CH3 – CH – CH2 – CH2 – CH3 CH3 – CH – CH2 – CH2– CH3 Pentan-2-ol | | OH (iii) CH3 – CH2 – CH – CH2 – CH3 CH3 – CH2 – CH – CH2– CH3 Pentan-3-ol | | OH CH3 CH3 3-Methyl- | | butan-1-ol (iv) CH3 – CH – CH2 – CH2 – CH3 – CH – CH2 – CH2– OH 2-Methyl- butan-1-ol CH3 CH3 | | 2-Methyl- butan-2-ol (v) CH3 – CH2 – CH – CH2 – CH3 – CH2 – CH – CH2– OH 2,2- Dimethyl- CH3 CH3 propan-1-ol | | 3-Methyl- (vi) CH3 –C – CH2 – CH3 CH3 – C – CH2 – CH3 butan-2-ol | | OH CH3 CH3 | | (vii) CH3 – C – CH2 – CH3 – C – CH2OH | | CH3 CH3 CH3 CH3 OH || || (viii) CH3 – CH – CH –CH3 CH3 – CH – CH –CH3 2019-20

HYDROCARBONS 377 Table 13.1 Nomenclature of a Few Organic Compounds Structure and IUPAC Name Remarks (a) 1CH3–2CH – 3CH2 – 4CH – 5CH2 – 6CH3 Lowest sum and (4 – Ethyl – 2 – methylhexane) alphabetical arrangement (b) 8CH3 – 7CH2 – 6CH2 – 5CH – 4CH – 3C – 2CH2 – 1CH3 Lowest sum and alphabetical arrangement (3,3-Diethyl-5-isopropyl-4-methyloctane) sec is not considered while arranging (c) 1CH3–2CH2–3CH2–4CH–5CH–6CH2–7CH2–8CH2–9CH2–10CH3 alphabetically; 5-sec– Butyl-4-isopropyldecane isopropyl is taken as one word (d) 1CH3–2CH2–3CH2–4CH2–5CH–6CH2–7CH2–8CH2–9CH3 Further numbering to the substituents of the side chain 5-(2,2– Dimethylpropyl)nonane Alphabetical (e) 1CH3 – 2CH2 – 3CH – 4CH2 – 5CH – 6CH2 – 7CH3 priority order 3–Ethyl–5–methylheptane Problem 13.3 important to write the correct structure from Write IUPAC names of the following the given IUPAC name. To do this, first of all, compounds : the longest chain of carbon atoms (i) (CH3)3 C CH2C(CH3)3 corresponding to the parent alkane is written. (ii) (CH3)2 C(C2H5)2 Then after numbering it, the substituents are (iii) tetra – tert-butylmethane attached to the correct carbon atoms and finally valence of each carbon atom is satisfied by Solution putting the correct number of hydrogen atoms. (i) 2, 2, 4, 4-Tetramethylpentane This can be clarified by writing the structure (ii) 3, 3-Dimethylpentane of 3-ethyl-2, 2–dimethylpentane in the (iii) 3,3-Di-tert-butyl -2, 2, 4, 4 - following steps : tetramethylpentane i) Draw the chain of five carbon atoms: C–C–C–C–C If it is important to write the correct IUPAC name for a given structure, it is equally ii) Give number to carbon atoms: C1– C2– C3– C4– C5 2019-20

378 CHEMISTRY iii) Attach ethyl group at carbon 3 and two Longest chain is of six carbon atoms and methyl groups at carbon 2 not that of five. Hence, correct name is CH3 3-Methylhexane. | C1 – 2C – 3C – 4C – 5C 7 6 54 32 1 || CH3 C2H5 (ii) CH3 – CH2 – CH – CH2 – CH – CH2 – CH3 iv) Satisfy the valence of each carbon atom by Numbering is to be started from the end putting requisite number of hydrogen which gives lower number to ethyl group. atoms : Hence, correct name is 3-ethyl-5- methylheptane. CH3 | 13.2.2 Preparation CH3 – C – CH – CH2 – CH3 Petroleum and natural gas are the main || sources of alkanes. However, alkanes can be CH3 C2H5 prepared by following methods : Thus we arrive at the correct structure. If 1. From unsaturated hydrocarbons you have understood writing of structure from Dihydrogen gas adds to alkenes and alkynes the given name, attempt the following in the presence of finely divided catalysts like problems. platinum, palladium or nickel to form alkanes. This process is called hydrogenation. These Problem 13.4 metals adsorb dihydrogen gas on their surfaces Write structural formulas of the following and activate the hydrogen – hydrogen bond. compounds : Platinum and palladium catalyse the reaction (i) 3, 4, 4, 5–Tetramethylheptane at room temperature but relatively higher (ii) 2,5-Dimethyhexane temperature and pressure are required with nickel catalysts. Solution CH2 = CH2 + H2 Pt/Pd/Ni→CH3 − CH3 (i) CH3 – CH2 – CH – C – CH– CH – CH3 Ethane (13.1) Ethene CH3 −CH =CH2 +H2 Pt/Pd/N→i CH3 −CH2 −CH3 (ii) CH3 – CH – CH2 – CH2 – CH – CH3 Propene Propane Problem 13.5 (13.2) Write structures for each of the following compounds. Why are the given names CH3 −C ≡C−H + 2H2 Pt/Pd/N→i CH3 −CH2 −CH3 incorrect? Write correct IUPAC names. Propyne Propane (i) 2-Ethylpentane (ii) 5-Ethyl – 3-methylheptane (13.3) Solution 2. From alkyl halides (i) CH3 – CH – CH2– CH2 – CH3 i) Alkyl halides (except fluorides) on reduction with zinc and dilute hydrochloric acid give alkanes. CH3 − Cl + H2 Zn, H+→ CH4 + HCl (13.4) Chloromethane Methane 2019-20

HYDROCARBONS 379 C2H5 − Cl + H2 Zn, H+→ C2H6 + HCl containing even number of carbon atoms Chloroethane Ethane (13.5) at the anode. 2CH3COO−Na+ + 2H2O CH3CH2CH2Cl + H2 Zn,H→+ CH3CH2CH3 + HCl Sodium acetate 1-Chloropropane Propane ↓ Electrolysis (13.6) CH3 −CH3 + 2CO2 +H2 + 2NaOH ii) Alkyl halides on treatment with sodium (13.9) metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction The reaction is supposed to follow the is known as Wurtz reaction and is used for the preparation of higher alkanes following path : containing even number of carbon atoms. O || i) 2CH3COO−Na+  2CH3 −C−O− +2Na+ CH3Br+2Na+BrCH3 dry ethe→r CH3 −CH3 +2NaBr ii) At anode: Bromomethane Ethane OO || || • • (13.7) 2CH3 −C−O– –2e→− 2CH3 +2CO2 ↑ −C−O: →2CH3 C2H5Br+2Na+BrC2H5 dry ethe→r C2H5 −C2H5 •• Bromoethane n-Butane Acetate ion Acetate Methyl free (13.8) free radical radical What will happen if two different alkyl halides •• are taken? iii) H3 C + CH3 → H3C−CH3 ↑ 3. From carboxylic acids iv) At cathode : i) Sodium salts of carboxylic acids on heating • H2O +e– → –OH +H with soda lime (mixture of sodium • hydroxide and calcium oxide) give alkanes 2H→H2 ↑ containing one carbon atom less than the carboxylic acid. This process of elimination Methane cannot be prepared by this of carbon dioxide from a carboxylic acid is known as decarboxylation. method. Why? CH3COO– Na+ + NaOHC∆a→O CH4 +Na2CO3 13.2.3 Properties Sodium ethanoate Physical properties Problem 13.6 Alkanes are almost non-polar molecules Sodium salt of which acid will be needed because of the covalent nature of C-C and C-H for the preparation of propane ? Write bonds and due to very little difference of chemical equation for the reaction. electronegativity between carbon and hydrogen atoms. They possess weak van der Solution Waals forces. Due to the weak forces, the first four members, C1 to C4 are gases, C5 to C17 are Butanoic acid, liquids and those containing 18 carbon atoms CH3CH2CH2COO− Na+ + NaOH CaO→ or more are solids at 298 K. They are colourless and odourless. What do you think about CH3CH2CH3 +Na2CO3 solubility of alkanes in water based upon non- polar nature of alkanes? Petrol is a mixture of ii) Kolbe’s electrolytic method An aqueous hydrocarbons and is used as a fuel for solution of sodium or potassium salt of a automobiles. Petrol and lower fractions of carboxylic acid on electrolysis gives alkane petroleum are also used for dry cleaning of clothes to remove grease stains. On the basis of this observation, what do you think about the nature of the greasy substance? You are correct if you say that grease (mixture of higher 2019-20

380 CHEMISTRY alkanes) is non-polar and, hence, hydrophobic reducing agents. However, they undergo the in nature. It is generally observed that in following reactions under certain relation to solubility of substances in solvents, conditions. polar substances are soluble in polar solvents, whereas the non-polar ones in non-polar 1. Substitution reactions solvents i.e., like dissolves like. One or more hydrogen atoms of alkanes can Boiling point (b.p.) of different alkanes are be replaced by halogens, nitro group and given in Table 13.2 from which it is clear that sulphonic acid group. Halogenation takes there is a steady increase in boiling point with place either at higher temperature increase in molecular mass. This is due to the (573-773 K) or in the presence of diffused fact that the intermolecular van der Waals sunlight or ultraviolet light. Lower alkanes do forces increase with increase of the molecular not undergo nitration and sulphonation size or the surface area of the molecule. reactions. These reactions in which hydrogen atoms of alkanes are substituted are known You can make an interesting observation as substitution reactions. As an example, by having a look on the boiling points of chlorination of methane is given below: three isomeric pentanes viz., (pentane, 2-methylbutane and 2,2-dimethylpropane). It Halogenation is observed (Table 13.2) that pentane having a continuous chain of five carbon atoms has the CH4 +Cl2 hν→ CH3Cl + HCl highest boiling point (309.1K) whereas (13.10) 2,2 – dimethylpropane boils at 282.5K. With Chloromethane increase in number of branched chains, the molecule attains the shape of a sphere. This CH3Cl + Cl2 hν → CH2Cl2 + HCl results in smaller area of contact and therefore Dichloromethane (13.11) weak intermolecular forces between spherical molecules, which are overcome at relatively CH2Cl2 + Cl2 hν → CHCl3 + HCl lower temperatures. Trichloromethane (13.12) Chemical properties As already mentioned, alkanes are generally CHCl3 + Cl2 hν → CCl4 + HCl inert towards acids, bases, oxidising and Tetrachloromethane (13.13) Table 13.2 Variation of Melting Point and Boiling Point in Alkanes Molecular Name Molecular b.p./(K) m.p./(K) mass/u formula Methane 111.0 90.5 Ethane 16 184.4 101.0 CH4 Propane 30 230.9 85.3 C2H6 Butane 44 272.4 134.6 C3H8 2-Methylpropane 58 261.0 114.7 C4H10 Pentane 58 309.1 143.3 C4H10 2-Methylbutane 72 300.9 113.1 C5H12 2,2-Dimethylpropane 72 282.5 256.4 C5H12 Hexane 72 341.9 178.5 C5H12 Heptane 86 371.4 182.4 C6H14 Octane 100 398.7 216.2 C7H16 Nonane 114 423.8 222.0 C8H18 Decane 128 447.1 243.3 C9H20 Eicosane 142 615.0 236.2 C10H22 282 C20H42 2019-20

HYDROCARBONS 381 CH3 -CH3 + Cl2 hν → CH3 −CH2Cl + HCl and may occur. Two such steps given below explain how more highly haloginated products Chloroethane (13.14) are formed. It is found that the rate of reaction of alkanes •• with halogens is F2 > Cl2 > Br2 > I2. Rate of CH3Cl + Cl → CH2Cl + HCl replacement of hydrogens of alkanes is : •• CH2Cl + Cl − Cl → CH2Cl2 + Cl 3° > 2° > 1°. Fluorination is too violent to be (iii) Termination: The reaction stops after some time due to consumption of reactants controlled. Iodination is very slow and a and / or due to the following side reactions : reversible reaction. It can be carried out in the presence of oxidizing agents like HIO3 or HNO3. The possible chain terminating steps are : CH4 +I2 CH3I+HI (13.15) •• HIO3 +5HI→ 3I2 +3H2O (13.16) (a) Cl + Cl → Cl −Cl Halogenation is supposed to proceed via •• free radical chain mechanism involving three steps namely initiation, propagation and (b) H3 C + CH3 → H3C−CH3 termination as given below: •• Mechanism (c) H3 C + Cl → H3C−Cl (i) Initiation : The reaction is initiated by Though in (c), CH3 – Cl, the one of the homolysis of chlorine molecule in the presence products is formed but free radicals are consumed and the chain is terminated. The of light or heat. The Cl–Cl bond is weaker than above mechanism helps us to understand the reason for the formation of ethane as a byproduct during chlorination of methane. the C–C and C–H bond and hence, is easiest to 2. Combustion break. Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon Cl −Cl homhoνlysi→s • + • dioxide and water with the evolution of large amount of heat. Cl Cl CH4(g)+2O2(g) → CO2(g)+2H2O(l); Chlorine free radicals ∆cH = − 890 kJ mol−1 (13.17) (ii) Propagation : Chlorine free radical attacks C4H10(g)+13/2 O2(g) → 4CO2(g)+5H2O(l); the methane molecule and takes the reaction ∆cH =−2875.84 kJ mol−1 (13.18) in the forward direction by breaking the C-H bond to generate methyl free radical with the formation of H-Cl. (a) CH4 + • hν→C• H 3 + H − Cl Cl The methyl radical thus obtained attacks The general combustion equation for any alkane is : the second molecule of chlorine to form CH3 – Cl with the liberation of another chlorine Cn H2 n+2 +  3n2+1O2 → nCO2 + (n +1) H2O free radical by homolysis of chlorine molecule.  (b) • H3 + Cl − Cl hν → CH3 − Cl • (13.19) C + Cl Due to the evolution of large amount of Chlorine heat during combustion, alkanes are used free radical as fuels. The chlorine and methyl free radicals During incomplete combustion of generated above repeat steps (a) and (b) alkanes with insufficient amount of air or respectively and thereby setup a chain of dioxygen, carbon black is formed which is reactions. The propagation steps (a) and (b) are used in the manufacture of ink, printer ink, those which directly give principal products, black pigments and as filters. but many other propagation steps are possible 2019-20

382 CHEMISTRY CH4 (g)+O2(g)cIonmcobmupstleioten→C(s)+2H2O(l) pressure in the presence of oxides of (13.20) vanadium, molybdenum or chromium supported over alumina get dehydrogenated 3. Controlled oxidation and cyclised to benzene and its homologues. Alkanes on heating with a regulated supply of This reaction is known as aromatization or dioxygen or air at high pressure and in the reforming. presence of suitable catalysts give a variety of oxidation products. (i) 2CH4 +O2 Cu/523K/100atm→2CH3OH Methanol (13.21) (13.26) (ii) CH4 + O2 Mo∆2O3→ HCHO + H2O Toluene (C7H8) is methyl derivative of Methanal benzene. Which alkane do you suggest for (13.22) preparation of toluene ? (iii) 2CH3CH3 +3O2 (CH3CO∆O)2 Mn→ 2 CH3COOH 6. Reaction with steam Ethanoic acid Methane reacts with steam at 1273 K in the + 2H2O presence of nickel catalyst to form carbon monoxide and dihydrogen. This method is (13.23) used for industrial preparation of dihydrogen gas (iv) Ordinarily alkanes resist oxidation but alkanes having tertiary H atom can be CH4 + H2O N∆i→ CO + 3H2 (13.27) oxidized to corresponding alcohols by potassium permanganate. (CH3 )3 CH OKxMidnaOtio4n→ (CH3 )3 COH 7. Pyrolysis 2 - Methylpropane 2 - Methylpropan - 2 - ol Higher alkanes on heating to higher (13.24) temperature decompose into lower alkanes, 4. Isomerisation alkenes etc. Such a decomposition reaction n-Alkanes on heating in the presence of anhydrous aluminium chloride and hydrogen into smaller fragments by the application of chloride gas isomerise to branched chain alkanes. Major products are given below. Some heat is called pyrolysis or cracking. minor products are also possible which you can think over. Minor products are generally (13.28) not reported in organic reactions. CH3(CH2 )4 CH3 Anhy. AlCl3/HC→l Pyrolysis of alkanes is believed to be a n -Hexane free radical reaction. Preparation of oil gas or CH3CH−(CH2 )2 −CH3 +CH3CH2 −CH−CH2 −CH3 petrol gas from kerosene oil or petrol involves || the principle of pyrolysis. For example, CH3 CH3 dodecane, a constituent of kerosene oil on 2-Methylpentane 3-Methylpentane heating to 973K in the presence of platinum, (13.25) palladium or nickel gives a mixture of heptane and pentene. 5. Aromatization C12 H26 Pt/Pd/N→i C7H16 + C5H10 + other n-Alkanes having six or more carbon atoms products on heating to 773K at 10-20 atmospheric 973K Dodecane Heptane Pentene (13.29) 2019-20

HYDROCARBONS 383 13.2.4 Conformations 1. Sawhorse projections In this projection, the molecule is viewed along Alkanes contain carbon-carbon sigma (σ) the molecular axis. It is then projected on paper bonds. Electron distribution of the sigma by drawing the central C–C bond as a molecular orbital is symmetrical around the somewhat longer straight line. Upper end of internuclear axis of the C–C bond which is the line is slightly tilted towards right or left not disturbed due to rotation about its axis. hand side. The front carbon is shown at the This permits free rotation about C–C single lower end of the line, whereas the rear carbon bond. This rotation results into different is shown at the upper end. Each carbon has spatial arrangements of atoms in space which three lines attached to it corresponding to three can change into one another. Such spatial hydrogen atoms. The lines are inclined at an arrangements of atoms which can be angle of 120° to each other. Sawhorse projections converted into one another by rotation around of eclipsed and staggered conformations of a C-C single bond are called conformations ethane are depicted in Fig. 13.2. or conformers or rotamers. Alkanes can thus have infinite number of conformations by Fig. 13.2 Sawhorse projections of ethane rotation around C-C single bonds. However, 2. Newman projections it may be remembered that rotation around In this projection, the molecule is viewed at the a C-C single bond is not completely free. It is C–C bond head on. The carbon atom nearer to hindered by a small energy barrier of the eye is represented by a point. Three 1-20 kJ mol–1 due to weak repulsive hydrogen atoms attached to the front carbon interaction between the adjacent bonds. Such atom are shown by three lines drawn at an a type of repulsive interaction is called angle of 120° to each other. The rear carbon torsional strain. atom (the carbon atom away from the eye) is represented by a circle and the three hydrogen Conformations of ethane : Ethane atoms are shown attached to it by the shorter molecule (C2H6) contains a carbon – carbon lines drawn at an angle of 120° to each other. single bond with each carbon atom attached The Newman’s projections are depicted in to three hydrogen atoms. Considering the Fig. 13.3. ball and stick model of ethane, keep one carbon atom stationary and rotate the other Fig. 13.3 Newman’s projections of ethane carbon atom around the C-C axis. This rotation results into infinite number of spatial arrangements of hydrogen atoms attached to one carbon atom with respect to the hydrogen atoms attached to the other carbon atom. These are called conformational isomers (conformers). Thus there are infinite number of conformations of ethane. However, there are two extreme cases. One such conformation in which hydrogen atoms attached to two carbons are as closed together as possible is called eclipsed conformation and the other in which hydrogens are as far apart as possible is known as the staggered conformation. Any other intermediate conformation is called a skew conformation.It may be remembered that in all the conformations, the bond angles and the bond lengths remain the same. Eclipsed and the staggered conformations can be represented by Sawhorse and Newman projections. 2019-20

384 CHEMISTRY Relative stability of conformations: As ethylene or ethene (C2H4) was found to form an mentioned earlier, in staggered form of ethane, oily liquid on reaction with chlorine. the electron clouds of carbon-hydrogen bonds are as far apart as possible. Thus, there are 13.3.1 Structure of Double Bond minimum repulsive forces, minimum energy Carbon-carbon double bond in alkenes and maximum stability of the molecule. On the consists of one strong sigma (σ) bond (bond other hand, when the staggered form changes enthalpy about 397 kJ mol–1) due to head-on into the eclipsed form, the electron clouds of overlapping of sp2 hybridised orbitals and one the carbon – hydrogen bonds come closer to weak pi (π) bond (bond enthalpy about 284 kJ each other resulting in increase in electron mol–1) obtained by lateral or sideways cloud repulsions. To check the increased overlapping of the two 2p orbitals of the two repulsive forces, molecule will have to possess carbon atoms. The double bond is shorter in more energy and thus has lesser stability. As bond length (134 pm) than the C–C single bond already mentioned, the repulsive interaction (154 pm). You have already read that the pi (π) between the electron clouds, which affects bond is a weaker bond due to poor sideways stability of a conformation, is called torsional overlapping between the two 2p orbitals. Thus, strain. Magnitude of torsional strain depends the presence of the pi (π) bond makes alkenes upon the angle of rotation about C–C bond. behave as sources of loosely held mobile This angle is also called dihedral angle or electrons. Therefore, alkenes are easily attacked torsional angle. Of all the conformations of by reagents or compounds which are in search ethane, the staggered form has the least of electrons. Such reagents are called torsional strain and the eclipsed form, the electrophilic reagents. The presence of maximum torsional strain. Therefore, weaker π-bond makes alkenes unstable staggered conformation is more stable than the molecules in comparison to alkanes and thus, eclipsed conformation. Hence, molecule largely alkenes can be changed into single bond remains in staggered conformation or we can compounds by combining with the say that it is preferred conformation. Thus it electrophilic reagents. Strength of the double may be inferred that rotation around C–C bond bond (bond enthalpy, 681 kJ mol–1) is greater in ethane is not completely free. The energy than that of a carbon-carbon single bond in difference between the two extreme forms is of ethane (bond enthalpy, 348 kJ mol–1). Orbital the order of 12.5 kJ mol–1, which is very small. diagrams of ethene molecule are shown in Even at ordinary temperatures, the ethane Figs. 13.4 and 13.5. molecule gains thermal or kinetic energy sufficient enough to overcome this energy Fig. 13.4 Orbital picture of ethene depicting barrier of 12.5 kJ mol–1 through intermolecular σ bonds only collisions. Thus, it can be said that rotation about carbon-carbon single bond in ethane is 13.3.2 Nomenclature almost free for all practical purposes. It has For nomenclature of alkenes in IUPAC system, not been possible to separate and isolate the longest chain of carbon atoms containing different conformational isomers of ethane. the double bond is selected. Numbering of the chain is done from the end which is nearer to 13.3 ALKENES Alkenes are unsaturated hydrocarbons containing at least one double bond. What should be the general formula of alkenes? If there is one double bond between two carbon atoms in alkenes, they must possess two hydrogen atoms less than alkanes. Hence, general formula for alkenes is CnH2n. Alkenes are also known as olefins (oil forming) since the first member, 2019-20

HYDROCARBONS 385 Fig. 13.5 Orbital picture of ethene showing formation of (a) π-bond, (b) π-cloud and (c) bond angles and bond lengths the double bond. The suffix ‘ene’ replaces ‘ane’ Solution (i) 2,8-Dimethyl-3, 6-decadiene; of alkanes. It may be remembered that first (ii) 1,3,5,7 Octatetraene; (iii) 2-n-Propylpent-1-ene; member of alkene series is: CH2 (replacing n (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene; by 1 in CnH2n) known as methene but has a very short life. As already mentioned, first Problem 13.8 Calculate number of sigma (σ) and pi (π) stable member of alkene series is C2H4 known bonds in the above structures (i-iv). as ethylene (common) or ethene (IUPAC). Solution IUPAC names of a few members of alkenes are σ bonds : 33, π bonds : 2 σ bonds : 17, π bonds : 4 given below : σ bonds : 23, π bond : 1 σ bonds : 41, π bond : 1 Structure IUPAC name CH3 – CH = CH2 Propene But – l - ene CH3 – CH2 – CH = CH2 But-2-ene Buta – 1,3 - diene CH3 – CH = CH–CH3 2-Methylprop-1-ene CH2 = CH – CH = CH2 3-Methylbut-1-ene CH2 = C – CH3 | CH3 CH2 = CH – CH – CH3 13.3.3 Isomerism | Alkenes show both structural isomerism and CH3 geometrical isomerism. Problem 13.7 Structural isomerism : As in alkanes, ethene (C2H4) and propene (C3H6) can have only one Write IUPAC names of the following structure but alkenes higher than propene compounds: have different structures. Alkenes possessing C4H8 as molecular formula can be written in (i) (CH3)2CH – CH = CH–CH2– CH the following three ways:  CH3 – CH – CH I. 1 23 4 | CH2 = CH – CH2 – CH3 C2H5 (ii) But-1-ene (C4H8) (iii) CH2 = C (CH2CH2CH3)2 (iv) CH3 CH2 CH2 CH2 CH2CH3 II. 1 2 34 || CH3 – CH = CH – CH3 CH3 – CHCH = C – CH2 – CHCH3 But-2-ene | (C4H8) CH3 2019-20

386 CHEMISTRY III. 1 23 In (a), the two identical atoms i.e., both the X or both the Y lie on the same side of the CH2 = C – CH3 double bond but in (b) the two X or two Y lie | across the double bond or on the opposite sides of the double bond. This results in CH3 different geometry of (a) and (b) i.e. disposition of atoms or groups in space in the two 2-Methyprop-1-ene arrangements is different. Therefore, they are stereoisomers. They would have the same (C4H8) geometry if atoms or groups around C=C bond can be rotated but rotation around C=C bond Structures I and III, and II and III are the is not free. It is restricted. For understanding examples of chain isomerism whereas this concept, take two pieces of strong structures I and II are position isomers. cardboards and join them with the help of two nails. Hold one cardboard in your one hand Problem 13.9 and try to rotate the other. Can you really rotate the other cardboard ? The answer is no. The Write structures and IUPAC names of rotation is restricted. This illustrates that the different structural isomers of alkenes restricted rotation of atoms or groups around corresponding to C5H10. the doubly bonded carbon atoms gives rise to different geometries of such compounds. The Solution stereoisomers of this type are called geometrical isomers. The isomer of the type (a) CH2 = CH – CH2 – CH2 – CH3 (a), in which two identical atoms or groups lie Pent-1-ene on the same side of the double bond is called cis isomer and the other isomer of the type (b) CH3 – CH=CH – CH2 – CH3 (b), in which identical atoms or groups lie on Pent-2-ene the opposite sides of the double bond is called trans isomer . Thus cis and trans isomers (c) CH3 – C = CH – CH3 have the same structure but have different | configuration (arrangement of atoms or groups CH3 in space). Due to different arrangement of atoms or groups in space, these isomers differ 2-Methylbut-2-ene in their properties like melting point, boiling point, dipole moment, solubility etc. (d) CH3 – CH – CH = CH2 Geometrical or cis-trans isomers of but-2-ene | are represented below : CH3 3-Methylbut-1-ene (e) CH2 = C – CH2 – CH3 | CH3 2-Methylbut-1-ene Geometrical isomerism: Doubly bonded carbon atoms have to satisfy the remaining two valences by joining with two atoms or groups. If the two atoms or groups attached to each carbon atom are different, they can be represented by YX C = C XY like structure. YX C = C XY can be represented in space in the following two ways : Cis form of alkene is found to be more polar than the trans form. For example, dipole moment of cis-but-2-ene is 0.33 Debye, whereas, dipole moment of the trans form is almost zero or it can be said that 2019-20

HYDROCARBONS 387 trans-but-2-ene is non-polar. This can be (ii) CH2 = CBr2 understood by drawing geometries of the two (iii) C6H5CH = CH – CH3 (iv) CH3CH = CCl CH3 forms as given below from which it is clear that in the trans-but-2-ene, the two methyl groups Solution (iii) and (iv). In structures (i) and (ii), two are in opposite directions, Threfore, dipole identical groups are attached to one of the moments of C-CH3 bonds cancel, thus making doubly bonded carbon atom. the trans form non-polar. cis-But-2-ene trans-But-2-ene 13.3.4 Preparation (µ = 0.33D) (µ = 0) 1. From alkynes: Alkynes on partial In the case of solids, it is observed that reduction with calculated amount of the trans isomer has higher melting point dihydrogen in the presence of palladised than the cis form. charcoal partially deactivated with poisons like sulphur compounds or quinoline give Geometrical or cis-trans isomerism alkenes. Partially deactivated palladised is also shown by alkenes of the types charcoal is known as Lindlar’s catalyst. XYC = CXZ and XYC = CZW Alkenes thus obtained are having cis geometry. However, alkynes on reduction Problem 13.10 with sodium in liquid ammonia form trans Draw cis and trans isomers of the alkenes. following compounds. Also write their IUPAC names : (13.30) (i) CHCl = CHCl (ii) C2H5CCH3 = CCH3C2H5 Solution iii) CH ≡CH +H2 Pd/C→ CH2 =CH2 (13.31) (13.32) Ethyne Ethene iv) CH3 −C ≡ CH+ H2 Pd/C→CH3 −CH =CH2 Propyne Propene (13.33) Problem 13.11 Will propene thus obtained show geometrical isomerism? Think for the Which of the following compounds will reason in support of your answer. show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X) (i) (CH3)2C = CH – C2H5 on heating with alcoholic potash (potassium hydroxide dissolved in alcohol, 2019-20

388 CHEMISTRY say, ethanol) eliminate one molecule of takes out one hydrogen atom from the halogen acid to form alkenes. This reaction β-carbon atom. is known as dehydrohalogenation i.e., removal of halogen acid. This is example of (13.37) β-elimination reaction, since hydrogen atom is eliminated from the β carbon atom 13.3.5 Properties (carbon atom next to the carbon to which halogen is attached). Physical properties Alkenes as a class resemble alkanes in physical (13.34) properties, except in types of isomerism and Nature of halogen atom and the alkyl difference in polar nature. The first three group determine rate of the reaction. It is members are gases, the next fourteen are observed that for halogens, the rate is: liquids and the higher ones are solids. Ethene iodine > bromine > chlorine, while for alkyl is a colourless gas with a faint sweet smell. All groups it is : tert > secondary > primary. other alkenes are colourless and odourless, 3. From vicinal dihalides: Dihalides in insoluble in water but fairly soluble in non- which two halogen atoms are attached to polar solvents like benzene, petroleum ether. two adjacent carbon atoms are known as They show a regular increase in boiling point vicinal dihalides. Vicinal dihalides on with increase in size i.e., every – CH2 group treatment with zinc metal lose a molecule added increases boiling point by 20–30 K. Like of ZnX2 to form an alkene. This reaction is alkanes, straight chain alkenes have higher known as dehalogenation. boiling point than isomeric branched chain CH2Br −CH2Br + Zn →CH2 =CH2 + ZnBr2 compounds. Chemical properties (13.35) Alkenes are the rich source of loosely held pi (π) electrons, due to which they show CH3CHBr −CH2Br + Zn → CH3CH = CH2 addition reactions in which the electrophiles + ZnBr2 add on to the carbon-carbon double bond to (13.36) form the addition products. Some reagents also add by free radical mechanism. There are 4. From alcohols by acidic dehydration: cases when under special conditions, alkenes You have read during nomenclature of also undergo free radical substitution different homologous series in Unit 12 that reactions. Oxidation and ozonolysis reactions alcohols are the hydroxy derivatives of are also quite prominent in alkenes. A brief alkanes. They are represented by R–OH description of different reactions of alkenes is where, R is CnH2n+1. Alcohols on heating given below: with concentrated sulphuric acid form 1. Addition of dihydrogen: Alkenes add up alkenes with the elimination of one water molecule. Since a water molecule is one molecule of dihydrogen gas in the eliminated from the alcohol molecule in the presence of finely divided nickel, palladium presence of an acid, this reaction is known or platinum to form alkanes (Section 13.2.2) as acidic dehydration of alcohols. This reaction is also the example of 2. Addition of halogens : Halogens like β-elimination reaction since –OH group bromine or chlorine add up to alkene to form vicinal dihalides. However, iodine does not show addition reaction under 2019-20

HYDROCARBONS 389 normal conditions. The reddish orange (13.42) colour of bromine solution in carbon tetrachloride is discharged when bromine Markovnikov, a Russian chemist made a adds up to an unsaturation site. This generalisation in 1869 after studying such reaction is used as a test for unsaturation. reactions in detail. These generalisations led Addition of halogens to alkenes is an Markovnikov to frame a rule called example of electrophilic addition reaction Markovnikov rule. The rule states that involving cyclic halonium ion formation negative part of the addendum (adding which you will study in higher classes. molecule) gets attached to that carbon atom which possesses lesser number of hydrogen (13.38) atoms. Thus according to this rule, product I i.e., 2-bromopropane is expected. In actual (ii) CH3 −CH =CH2 +Cl −Cl →CH3 −CH−CH2 practice, this is the principal product of the || reaction. This generalisation of Markovnikov rule can be better understood in terms of Cl Cl mechanism of the reaction. Propene 1,2-Dichloropropane Mechanism Hydrogen bromide provides an electrophile, H+, (13.39) which attacks the double bond to form carbocation as shown below : 3. Addition of hydrogen halides: Hydrogen halides (HCl, HBr,HI) add up to alkenes to form alkyl halides. The order of reactivity of the hydrogen halides is HI > HBr > HCl. Like addition of halogens to alkenes, addition of hydrogen halides is also an example of electrophilic addition reaction. Let us illustrate this by taking addition of HBr to symmetrical and unsymmetrical alkenes Addition reaction of HBr to symmetrical (a) less stable (b) more stable alkenes Addition reactions of HBr to symmetrical primary carbocation secondary carbocation alkenes (similar groups attached to double bond) take place by electrophilic addition (i) The secondary carbocation (b) is more mechanism. stable than the primary carbocation (a), therefore, the former predominates because CH2 = CH2 +H – Br →CH3 – CH2 – Br (13.40) it is formed at a faster rate. CH3 –CH =CH –CH3 +HBr →CH3 –CH2 – CHCH3 (ii) The carbocation (b) is attacked by Br– ion | to form the product as follows : Br (13.41) 2-Bromopropane (major product) Addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H – Br add to propene ? The two possible products are I and II. 2019-20

390 CHEMISTRY Anti Markovnikov addition or peroxide The secondary free radical obtained in the effect or Kharash effect above mechanism (step iii) is more stable than In the presence of peroxide, addition of HBr the primary. This explains the formation of to unsymmetrical alkenes like propene takes 1-bromopropane as the major product. It may place contrary to the Markovnikov rule. This be noted that the peroxide effect is not observed happens only with HBr but not with HCl and in addition of HCl and HI. This may be due Hl. This addition reaction was observed to the fact that the H–Cl bond being by M.S. Kharash and F.R. Mayo in 1933 at stronger (430.5 kJ mol–1) than H–Br bond the University of Chicago. This reaction (363.7 kJ mol–1), is not cleaved by the free is known as peroxide or Kharash effect radical, whereas the H–I bond is weaker or addition reaction anti to Markovnikov (296.8 kJ mol–1) and iodine free radicals rule. combine to form iodine molecules instead of adding to the double bond. CH3 – CH = CH2 +HBr (C6H5CO)2O2→CH3 – CH2  Problem 13.12 CH2Br Write IUPAC names of the products 1-Bromopropane obtained by addition reactions of HBr to hex-1-ene (13.43) Mechanism : Peroxide effect proceeds via free (i) in the absence of peroxide and radical chain mechanism as given below: (ii) in the presence of peroxide. (i) Solution (ii) • H5 + H – Br Homolysis→C6 H6 + • C6 Br 4. Addition of sulphuric acid : Cold concentrated sulphuric acid adds to alkenes in accordance with Markovnikov rule to form alkyl hydrogen sulphate by the electrophilic addition reaction. 2019-20

HYDROCARBONS 391 ketones and/or acids depending upon the nature of the alkene and the experimental conditions (13.49) (13.44) CH3 – CH =CH – CH3 KMnO4/H+→2CH3COOH (13.45) But -2-ene Ethanoic acid 5. Addition of water : In the presence of a (13.50) few drops of concentrated sulphuric acid alkenes react with water to form alcohols, 7. Ozonolysis : Ozonolysis of alkenes involves in accordance with the Markovnikov rule. the addition of ozone molecule to alkene to form ozonide, and then cleavage of the ozonide by Zn-H2O to smaller molecules. This reaction is highly useful in detecting the position of the double bond in alkenes or other unsaturated compounds. (13.51) (13.46) 6. Oxidation: Alkenes on reaction with cold, dilute, aqueous solution of potassium permanganate (Baeyer’s reagent) produce vicinal glycols. Decolorisation of KMnO4 solution is used as a test for unsaturation. (13.47) (13.52) (13.48) 8. Polymerisation: You are familiar with b) Acidic potassium permanganate or acidic polythene bags and polythene sheets. Polythene is obtained by the combination potassium dichromate oxidises alkenes to of large number of ethene molecules at high temperature, high pressure and in the presence of a catalyst. The large molecules thus obtained are called polymers. This reaction is known as polymerisation. The simple compounds from which polymers 2019-20

392 CHEMISTRY are made are called monomers. Other are named as derivatives of the corresponding alkenes also undergo polymerisation. alkanes replacing ‘ane’ by the suffix ‘yne’. The position of the triple bond is indicated by the n(CH2 =CH2 )HightCematpa./lypsrtessur→e —( CH2 – CH2 —)n first triply bonded carbon. Common and Polythene IUPAC names of a few members of alkyne series are given in Table 13.2. (13.53) n(CH3 – CH =CH2 )HighteCmatpa.l/ypsrtessure→ —( CH – CH2 —)n You have already learnt that ethyne and | propyne have got only one structure but there are two possible structures for butyne – CH3 (i) but-1-yne and (ii) but-2-yne. Since these two Polypropene compounds differ in their structures due to the position of the triple bond, they are known as (13.54) position isomers. In how many ways, you can Polymers are used for the manufacture of construct the structure for the next homologue plastic bags, squeeze bottles, refrigerator dishes, i.e., the next alkyne with molecular formula toys, pipes, radio and T.V. cabinets etc. C5H8? Let us try to arrange five carbon atoms Polypropene is used for the manufacture of milk with a continuous chain and with a side chain. crates, plastic buckets and other moulded Following are the possible structures : articles. Though these materials have now become common, excessive use of polythene Structure IUPAC name and polypropylene is a matter of great concern for all of us. I. 1 23 4 5 Pent–1-yne HC ≡ C– CH2 – CH2 – CH3 13.4 ALKYNES II. 12 34 5 Pent–2-yne H3 C–C ≡ C– CH2 – CH3 Like alkenes, alkynes are also unsaturated hydrocarbons. They contain at least one triple III. H3 4 – 3 – 2 ≡ 1 3-Methyl but–1-yne bond between two carbon atoms. The number of hydrogen atoms is still less in alkynes as C CH C CH compared to alkenes or alkanes. Their general formula is CnH2n–2. | The first stable member of alkyne series CH3 is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Structures I and II are position isomers and purposes in the form of oxyacetylene flame obtained by mixing acetylene with oxygen gas. structures I and III or II and III are chain Alkynes are starting materials for a large number of organic compounds. Hence, it is isomers. interesting to study this class of organic compounds. Problem 13.13 13.4.1 Nomenclature and Isomerism Write structures of different isomers corresponding to the 5th member of In common system, alkynes are named as alkyne series. Also write IUPAC names of derivatives of acetylene. In IUPAC system, they all the isomers. What type of isomerism is exhibited by different pairs of isomers? Solution 5th member of alkyne has the molecular formula C6H10. The possible isomers are: Table 13.2 Common and IUPAC Names of Alkynes (CnH2n–2) Value of n Formula Structure Common name IUPAC name 2 C2H2 H-C≡CH Acetylene Ethyne 3 C3H4 CH3-C≡CH Methylacetylene Propyne 4 C4H6 CH3CH2-C≡CH Ethylacetylene But-1-yne 4 C4H6 CH3-C≡C-CH3 Dimethylacetylene But-2-yne 2019-20

HYDROCARBONS 393 (a) HC ≡ C – CH2 – CH2 – CH2 – CH3 Hex-1-yne (b) CH3 – C ≡ C – CH2 – CH2 – CH3 Hex-2-yne (c) CH3 – CH2 – C ≡ C – CH2– CH3 Hex-3-yne 3-Methylpent-1-yne 4-Methylpent-1-yne 4-Methylpent-2-yne Fig. 13.6 Orbital picture of ethyne showing (a) sigma overlaps (b) pi overlaps. 3,3-Dimethylbut-1-yne orbitals of the other carbon atom, which Position and chain isomerism shown by different pairs. undergo lateral or sideways overlapping to form two pi (π) bonds between two carbon 13.4.2 Structure of Triple Bond Ethyne is the simplest molecule of alkyne atoms. Thus ethyne molecule consists of one series. Structure of ethyne is shown in C–C σ bond, two C–H σ bonds and two C–C π Fig. 13.6. bonds. The strength of C≡C bond (bond enthalpy 823 kJ mol-1) is more than those of Each carbon atom of ethyne has two sp C=C bond (bond enthalpy 681 kJ mol–1) and hybridised orbitals. Carbon-carbon sigma (σ) C–C bond (bond enthalpy 348 kJ mol–1). The bond is obtained by the head-on overlapping C≡C bond length is shorter (120 pm) than those of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised of C=C (133 pm) and C–C (154 pm). Electron orbital of each carbon atom undergoes overlapping along the internuclear axis with cloud between two carbon atoms is the 1s orbital of each of the two hydrogen atoms forming two C-H sigma bonds. H-C-C bond cylindrically symmetrical about the angle is of 180°. Each carbon has two unhybridised p orbitals which are internuclear axis. Thus, ethyne is a linear perpendicular to each other as well as to the plane of the C-C sigma bond. The 2p orbitals molecule. of one carbon atom are parallel to the 2p 13.4.3 Preparation 1. From calcium carbide: On industrial scale, ethyne is prepared by treating calcium carbide with water. Calcium carbide is prepared by heating quick lime with coke. Quick lime can be obtained by heating limestone as shown in the following reactions: CaCO3 ∆→ CaO + CO2 (13.55) 2019-20

394 CHEMISTRY CaO + 3C → CaC2 + CO (13.56) atoms in ethyne are attached to the sp Calcium hybridised carbon atoms whereas they are carbide attached to sp2 hybridised carbon atoms in ethene and sp3 hybridised carbons in ethane. CaC2 + 2H2O → Ca(OH)2 + C2H2 (13.57) Due to the maximum percentage of s character (50%), the sp hybridised orbitals of carbon 2. From vicinal dihalides : Vicinal atoms in ethyne molecules have highest dihalides on treatment with alcoholic electronegativity; hence, these attract the potassium hydroxide undergo shared electron pair of the C-H bond of ethyne dehydrohalogenation. One molecule of to a greater extent than that of the sp2 hydrogen halide is eliminated to form hybridised orbitals of carbon in ethene and the alkenyl halide which on treatment with sp3 hybridised orbital of carbon in ethane. sodamide gives alkyne. Thus in ethyne, hydrogen atoms can be liberated as protons more easily as compared to ethene and ethane. Hence, hydrogen atoms of ethyne attached to triply bonded carbon atom are acidic in nature. You may note that the hydrogen atoms attached to the triply bonded carbons are acidic but not all the hydrogen atoms of alkynes. 13.4.4 Properties HC ≡ CH + Na → HC ≡ C– Na+ + ½H2 Physical properties Monosodium Physical properties of alkynes follow the same ethynide trend of alkenes and alkanes. First three members are gases, the next eight are liquids (13.59) and the higher ones are solids. All alkynes are colourless. Ethyene has characteristic odour. HC ≡ C– Na+ + Na → Na+C– ≡ C– Na+ + ½H2 Other members are odourless. Alkynes are Disodium ethynide weakly polar in nature. They are lighter than water and immiscible with water but soluble (13.60) in organic solvents like ethers, carbon tetrachloride and benzene. Their melting point, CH3 – C ≡ C − H + Na+NH2– (13.61) boiling point and density increase with ↓ increase in molar mass. CH3 – C ≡ C– Na+ + NH3 Chemical properties Sodium propynide Alkynes show acidic nature, addition reactions and polymerisation reactions as follows : These reactions are not shown by alkenes and alkanes, hence used for distinction A. Acidic character of alkyne: Sodium between alkynes, alkenes and alkanes. What metal and sodamide (NaNH2) are strong bases. about the above reactions with but-1-yne and They react with ethyne to form sodium but-2-yne ? Alkanes, alkenes and alkynes acetylide with the liberation of dihydrogen gas. follow the following trend in their acidic These reactions have not been observed in case behaviour : of ethene and ethane thus indicating that ethyne is acidic in nature in comparison to i) HC ≡ CH > H2C = CH2 > CH3 –CH3 ethene and ethane. Why is it so ? Has it something to do with their structures and the ii) HC ≡ CH > CH3 – C ≡ CH >> CH3 – C ≡ C – CH3 hybridisation ? You have read that hydrogen B. Addition reactions: Alkynes contain a triple bond, so they add up, two molecules of dihydrogen, halogen, hydrogen halides etc. 2019-20

HYDROCARBONS 395 Formation of the addition product takes place according to the following steps. The addition product formed depends upon (13.66) stability of vinylic cation. Addition in unsymmetrical alkynes takes place according (iv) Addition of water to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic Like alkanes and alkenes, alkynes are also addition reactions. A few addition reactions are immiscible and do not react with water. given below: However, one molecule of water adds to alkynes (i) Addition of dihydrogen on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form carbonyl HC≡CH+H2 Pt/Pd/N→i [H2C =CH2 ]H2→CH3 –CH3 compounds. (13.62) CH3 – C ≡ CH + H2 Pt/Pd/N→i [CH3 – CH = CH2 ] Propyne Propene ↓ H2 CH3 – CH2 – CH3 Propane (13.63) (ii) Addition of halogens (13.67) (13.64) (13.68) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. (v) Polymerisation This is used as a test for unsaturation. (a) Linear polymerisation: Under suitable (iii) Addition of hydrogen halides conditions, linear polymerisation of ethyne takes place to produce polyacetylene or Two molecules of hydrogen halides (HCl, HBr, polyethyne which is a high molecular weight polyene containing repeating units of HI) add to alkynes to form gem dihalides (in (CH = CH – CH = CH ) and can be represented as —( CH = CH – CH = CH)—n Under special which two halogens are attached to the same conditions, this polymer conducts electricity. carbon atom) H – C ≡ C – H + H – Br → [CH2 = CH – Br ] → C| HBr2 Bromoethene CH3 1,1- Dibromoethane (13.65) 2019-20

396 CHEMISTRY Thin film of polyacetylene can be used as in a majority of reactions of aromatic electrodes in batteries. These films are good compounds, the unsaturation of benzene ring conductors, lighter and cheaper than the metal is retained. However, there are examples of conductors. aromatic hydrocarbons which do not contain a benzene ring but instead contain other highly (b) Cyclic polymerisation: Ethyne on passing unsaturated ring. Aromatic compounds through red hot iron tube at 873K undergoes containing benzene ring are known as cyclic polymerization. Three molecules benzenoids and those not containing a polymerise to form benzene, which is the benzene ring are known as non-benzenoids. starting molecule for the preparation of Some examples of arenes are given derivatives of benzene, dyes, drugs and large below: number of other organic compounds. This is the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (13.69) Biphenyl Problem 13.14 13.5.1 Nomenclature and Isomerism How will you convert ethanoic acid into benzene? The nomenclature and isomerism of aromatic hydrocarbons has already been discussed in Solution Unit 12. All six hydrogen atoms in benzene are equivalent; so it forms one and only one type of monosubstituted product. When two hydrogen atoms in benzene are replaced by two similar or different monovalent atoms or groups, three different position isomers are possible. The 1, 2 or 1, 6 is known as the ortho (o–), the 1, 3 or 1, 5 as meta (m–) and the 1, 4 as para (p–) disubstituted compounds. A few examples of derivatives of benzene are given below: 13.5 AROMATIC HYDROCARBON Methylbenzene 1,2-Dimethylbenzene (Toluene) (o-Xylene) These hydrocarbons are also known as ‘arenes’. Since most of them possess pleasant odour (Greek; aroma meaning pleasant smelling), the class of compounds was named as ‘aromatic compounds’. Most of such compounds were found to contain benzene ring. Benzene ring is highly unsaturated but 2019-20

HYDROCARBONS 397 Friedrich August Kekulé,a German chemist was born in 1829 at Darmsdt in Germany. He became Professor in 1856 and Fellow of Royal Society in 1875. He made major contribution to structural organic chemistry by proposing in 1858 that carbon atoms can join to one another to form chains and later in 1865,he found an answer to the challenging problem of benzene structure by suggesting that these chains can close to form rings. He gave the dynamic structural formula to benzene which forms the basis for its modern electronic structure. He described the discovery of benzene structure later as: FRIEDRICH “I was sitting writing at my textbook,but the work did not progress; my thoughts AUGUST KEKULÉ were elsewhere. I turned my chair to the fire, and dozed. Again the atoms were (7th September gambolling before my eyes. This time the smaller groups kept modestly in the 1829–13th July background. My mental eye, rendered more acute by repeated visions of this kind, could now distinguish larger structures of manifold conformations; long 1896) rows,sometimes more closely fitted together; all twisting and turning in snake like motion. But look! What was that? One of the snakes had seized hold of it’s own tail, and the form whirled mockingly before my eyes. As if by a flash of lightning I woke;.... I spent the rest of the night working out the consequences of the hypothesis. Let us learn to dream, gentlemen, and then perhaps we shall learn the truth but let us beware of making our dreams public before they have been approved by the waking mind.”( 1890). One hundred years later, on the occasion of Kekulé’s centenary celebrations a group of compounds having polybenzenoid structures have been named as Kekulenes. 1,3 Dimethylbenzene 1,4-Dimethylbenzene found to produce one and only one monosubstituted derivative which indicated (m-Xylene) ( p-Xylene) that all the six carbon and six hydrogen atoms of benzene are identical. On the basis of this 13.5.2 Structure of Benzene observation August Kekulé in 1865 proposed the following structure for benzene having Benzene was isolated by Michael Faraday in cyclic arrangement of six carbon atoms with 1825. The molecular formula of benzene, alternate single and double bonds and one C6H6, indicates a high degree of unsaturation. hydrogen atom attached to each carbon This molecular formula did not account for its atom. relationship to corresponding alkanes, alkenes and alkynes which you have studied in earlier The Kekulé structure indicates sections of this unit. What do you think about the possibility of two isomeric its possible structure? Due to its unique 1, 2-dibromobenzenes. In one of the isomers, properties and unusual stability, it took several the bromine atoms are attached to the doubly years to assign its structure. Benzene was bonded carbon atoms whereas in the other, found to be a stable molecule and found to they are attached to the singly bonded carbons. form a triozonide which indicates the presence of three double bonds. Benzene was further 2019-20

398 CHEMISTRY perpendicular to the plane of the ring as shown below: However, benzene was found to form only The unhybridised p orbital of carbon atoms one ortho disubstituted product. This problem are close enough to form a π bond by lateral was overcome by Kekulé by suggesting the overlap. There are two equal possibilities of concept of oscillating nature of double bonds forming three π bonds by overlap of p orbitals in benzene as given below. of C1 –C2, C3 – C4, C5 – C6 or C2 – C3, C4 – C5, C6 – C1 respectively as shown in the following Even with this modification, Kekulé figures. structure of benzene fails to explain unusual stability and preference to substitution Fig. 13.7 (a) reactions than addition reactions, which could later on be explained by resonance. Resonance and stability of benzene According to Valence Bond Theory, the concept of oscillating double bonds in benzene is now explained by resonance. Benzene is a hybrid of various resonating structures. The two structures, A and B given by Kekulé are the main contributing structures. The hybrid structure is represented by inserting a circle or a dotted circle in the hexagon as shown in (C). The circle represents the six electrons which are delocalised between the six carbon atoms of the benzene ring. (A) (B) (C) The orbital overlapping gives us better Fig. 13.7 (b) picture about the structure of benzene. All the six carbon atoms in benzene are sp2 hybridized. Structures shown in Fig. 13.7(a) and (b) Two sp2 hybrid orbitals of each carbon atom correspond to two Kekulé’s structure with overlap with sp2 hybrid orbitals of adjacent localised π bonds. The internuclear distance carbon atoms to form six C—C sigma bonds which are in the hexagonal plane. The remaining sp2 hybrid orbital of each carbon atom overlaps with s orbital of a hydrogen atom to form six C—H sigma bonds. Each carbon atom is now left with one unhybridised p orbital 2019-20

HYDROCARBONS 399 between all the carbon atoms in the ring has (i) Planarity been determined by the X-ray diffraction to be (ii) Complete delocalisation of the π electrons the same; there is equal probability for the p orbital of each carbon atom to overlap with the in the ring p orbitals of adjacent carbon atoms [Fig. 13.7 (iii) Presence of (4n + 2) π electrons in the ring (c)]. This can be represented in the form of two doughtnuts (rings) of electron clouds [Fig. 13.7 where n is an integer (n = 0, 1, 2, . . .). (d)], one above and one below the plane of the hexagonal ring as shown below: This is often referred to as Hückel Rule. Some examples of aromatic compounds are given below: (electron cloud) Fig. 13.7 (c) Fig. 13.7 (d) The six π electrons are thus delocalised and 13.5.4 Preparation of Benzene can move freely about the six carbon nuclei, Benzene is commercially isolated from coal tar. instead of any two as shown in Fig. 13.6 (a) or However, it may be prepared in the laboratory (b). The delocalised π electron cloud is attracted by the following methods. more strongly by the nuclei of the carbon (i) Cyclic polymerisation of ethyne: atoms than the electron cloud localised between two carbon atoms. Therefore, presence (Section 13.4.4) of delocalised π electrons in benzene makes (ii) Decarboxylation of aromatic acids: it more stable than the hypothetical cyclohexatriene. Sodium salt of benzoic acid on heating with sodalime gives benzene. X-Ray diffraction data reveals that benzene is a planar molecule. Had any one of the above (13.70) structures of benzene (A or B) been correct, two types of C—C bond lengths were expected. However, X-ray data indicates that all the six C—C bond lengths are of the same order (139 pm) which is intermediate between C— C single bond (154 pm) and C—C double bond (133 pm). Thus the absence of pure double bond in benzene accounts for the reluctance of benzene to show addition reactions under normal conditions, thus explaining the unusual behaviour of benzene. 13.5.3 Aromaticity Benzene was considered as parent ‘aromatic’ compound. Now, the name is applied to all the ring systems whether or not having benzene ring, possessing following characteristics. 2019-20

400 CHEMISTRY (iii) Reduction of phenol: Phenol is reduced (ii) Halogenation: Arenes react with halogens to benzene by passing its vapours over in the presence of a Lewis acid like anhydrous heated zinc dust FeCl3, FeBr3 or AlCl3 to yield haloarenes. (13.71) Chlorobenzene (13.73) 13.5.5 Properties (iii) Sulphonation: The replacement of a Physical properties hydrogen atom by a sulphonic acid group in Aromatic hydrocarbons are non- polar a ring is called sulphonation. It is carried out molecules and are usually colourless liquids by heating benzene with fuming sulphuric acid or solids with a characteristic aroma. You are (oleum). also familiar with naphthalene balls which are used in toilets and for preservation of clothes (13.74) because of unique smell of the compound and (iv) Friedel-Crafts alkylation reaction: the moth repellent property. Aromatic When benzene is treated with an alkyl halide hydrocarbons are immiscible with water but in the presence of anhydrous aluminium are readily miscible with organic solvents. They chloride, alkylbenene is formed. burn with sooty flame. (13.75) Chemical properties Arenes are characterised by electrophilic substitution reactions. However, under special conditions they can also undergo addition and oxidation reactions. Electrophilic substitution reactions The common electrophilic substitution reactions of arenes are nitration, halogenation, sulphonation, Friedel Craft’s alkylation and acylation reactions in which attacking reagent is an electrophile (E+) (i) Nitration: A nitro group is introduced into benzene ring when benzene is heated with a mixture of concentrated nitric acid and concentrated sulphuric acid (nitrating mixture). (13.72) (13.76) Nitrobenzene Why do we get isopropyl benzene on treating benzene with 1-chloropropane instead of n-propyl benzene? (v) Friedel-Crafts acylation reaction: The reaction of benzene with an acyl halide or acid anhydride in the presence of Lewis acids (AlCl3) yields acyl benzene. 2019-20

HYDROCARBONS 401 (13.77) (13.78) In the case of nitration, the electrophile, If excess of electrophilic reagent is used, further substitution reaction may take place + in which other hydrogen atoms of benzene ring may also be successively replaced by the nitronium ion, N O2 is produced by transfer electrophile. For example, benzene on of a proton (from sulphuric acid) to nitric acid treatment with excess of chlorine in the in the following manner: presence of anhydrous AlCl3 can be Step I chlorinated to hexachlorobenzene (C6Cl6) Step II (13.79) Mechanism of electrophilic substitution Protonated Nitronium reactions: nitric acid ion According to experimental evidences, SE (S = substitution; E = electrophilic) reactions are It is interesting to note that in the process supposed to proceed via the following three of generation of nitronium ion, sulphuric acid steps: serves as an acid and nitric acid as a base. (a) Generation of the eletrophile Thus, it is a simple acid-base equilibrium. (b) Formation of carbocation intermediate (c) Removal of proton from the carbocation (b) For mation of Carbocation (arenium ion): Attack of electrophile intermediate (a) Generation of electrophile E⊕: During results in the formation of σ-complex or chlorination, alkylation and acylation of arenium ion in which one of the carbon is sp3 bheenlpzsenine,gaennheryadtrioounsoAf ltChle3,eblcetirnogpahiLleewCils⊕,aRci⊕d, RC⊕O (acylium ion) respectively by combining hybridised. with the attacking reagent. sigma complex (arenium ion) The arenium ion gets stabilised by resonance: 2019-20

402 CHEMISTRY Sigma complex or arenium ion loses its chemical equation: aromatic character because delocalisation of CxHy + (x + y ) O2 → x CO2 + y H2O (13.83) electrons stops at sp3 hybridised carbon. 4 2 (c) Removal of proton: To restore the aromatic character, σ -complex releases proton 13.5.6 Directive influence of a functional from sp3 hybridised carbon on attack by group in monosubstituted benzene [AlCl4]– (in acylation) case of halogenation, alkylation and When monosubstituted benzene is subjected and [HSO4]– (in case of nitration). to further substitution, three possible disubstituted products are not formed in equal Addition reactions amounts. Two types of behaviour are observed. Under vigorous conditions, i.e., at high Either ortho and para products or meta temperature and/ or pressure in the presence product is predominantly formed. It has also of nickel catalyst, hydrogenation of benzene been observed that this behaviour depends on gives cyclohexane. the nature of the substituent already present in the benzene ring and not on the nature of Cyclohexane the entering group. This is known as directive (13.80) influence of substituents. Reasons for ortho/ para or meta directive nature of groups are Under ultra-violet light, three chlorine discussed below: molecules add to benzene to produce benzene hexachloride, C6H6Cl6 which is also called Ortho and para directing groups: The gammaxane. groups which direct the incoming group to ortho and para positions are called ortho and para directing groups. As an example, let us discuss the directive influence of phenolic (–OH) group. Phenol is resonance hybrid of following structures: Benzene hexachloride, It is clear from the above resonating structures that the electron density is more on (BHC) o – and p – positions. Hence, the substitution takes place mainly at these positions. However, (13.81) it may be noted that –I effect of – OH group Combustion: When heated in air, benzene also operates due to which the electron density on ortho and para positions of the benzene ring burns with sooty flame producing CO2 and is slightly reduced. But the overall electron H2O density increases at these positions of the ring due to resonance. Therefore, –OH group C6H6 + 15 O2 → 6CO2 + 3H2O (13.82) activates the benzene ring for the attack by 2 General combustion reaction for any hydrocarbon may be given by the following 2019-20

HYDROCARBONS 403 an electrophile. Other examples of activating In this case, the overall electron density on groups are –NH2, –NHR, –NHCOCH3, –OCH3, benzene ring decreases making further –CH3, –C2H5, etc. substitution difficult, therefore these groups are also called ‘deactivating groups’. The In the case of aryl halides, halogens are electron density on o – and p – position is moderately deactivating. Because of their comparatively less than that at meta position. strong – I effect, overall electron density on Hence, the electrophile attacks on benzene ring decreases. It makes further comparatively electron rich meta position substitution dif ficult. However, due to resulting in meta substitution. resonance the electron density on o– and p– positions is greater than that at the m-position. 13.6 CARCINOGENICITY AND TOXICITY Hence, they are also o– and p – directing groups. Resonance structures of chlorobenzene are Benzene and polynuclear hydrocarbons given below: containing more than two benzene rings fused together are toxic and said to possess cancer producing (carcinogenic) property. Such polynuclear hydrocarbons are formed on incomplete combustion of organic materials like tobacco, coal and petroleum. They enter into human body and undergo various biochemical reactions and finally damage DNA and cause cancer. Some of the carcinogenic hydrocarbons are given below (see box). Meta directing group: The groups which direct the incoming group to meta position are called meta directing groups. Some examples of meta directing groups are –NO2, –CN, –CHO, –COR, –COOH, –COOR, –SO3H, etc. Let us take the example of nitro group. Nitro group reduces the electron density in the benzene ring due to its strong–I effect. Nitrobenzene is a resonance hybrid of the following structures. 2019-20

404 CHEMISTRY SUMMARY Hydrocarbons are the compounds of carbon and hydrogen only. Hydrocarbons are mainly obtained from coal and petroleum, which are the major sources of energy. Petrochemicals are the prominent starting materials used for the manufacture of a large number of commercially important products. LPG (liquefied petroleum gas) and CNG (compressed natural gas), the main sources of energy for domestic fuels and the automobile industry, are obtained from petroleum. Hydrocarbons are classified as open chain saturated (alkanes) and unsaturated (alkenes and alkynes), cyclic (alicyclic) and aromatic, according to their structure. The important reactions of alkanes are free radical substitution, combustion, oxidation and aromatization. Alkenes and alkynes undergo addition reactions, which are mainly electrophilic additions. Aromatic hydrocarbons, despite having unsaturation, undergo mainly electrophilic substitution reactions. These undergo addition reactions only under special conditions. Alkanes show conformational isomerism due to free rotation along the C–C sigma bonds. Out of staggered and the eclipsed conformations of ethane, staggered conformation is more stable as hydrogen atoms are farthest apart. Alkenes exhibit geometrical (cis-trans) isomerism due to restricted rotation around the carbon–carbon double bond. Benzene and benzenoid compounds show aromatic character. Aromaticity, the property of being aromatic is possessed by compounds having specific electronic structure characterised by Hückel (4n+2)π electron rule. The nature of groups or substituents attached to benzene ring is responsible for activation or deactivation of the benzene ring towards further electrophilic substitution and also for orientation of the incoming group. Some of the polynuclear hydrocarbons having fused benzene ring system have carcinogenic property. EXERCISES 13.1 How do you account for the formation of ethane during chlorination of methane ? 13.2 Write IUPAC names of the following compounds : (a) CH3CH=C(CH3)2 (b) CH2=CH-C≡C-CH3 (c) (d) –CH2–CH2–CH=CH2 (f ) CH3(CH2 )4 CH(CH2 )3CH3 | (e) CH2 −CH (CH3 )2 13.3 (g) CH3 – CH = CH – CH2 – CH = CH – CH – CH2 – CH = CH2 13.4 | C2H5 For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated : (a) C4H8 (one double bond) (b) C5H8 (one triple bond) Write IUPAC names of the products obtained by the ozonolysis of the following compounds : (i) Pent-2-ene (ii) 3,4-Dimethylhept-3-ene (iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene 2019-20

HYDROCARBONS 405 13.5 An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3- 13.6 one. Write structure and IUPAC name of ‘A’. An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’. 13.7 Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene? 13.8 Write chemical equations for combustion reaction of the following hydrocarbons: (i) Butane (ii) Pentene (iii) Hexyne (iv) Toluene 13.9 Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why? 13.10 Why is benzene extra ordinarily stable though it contains three double bonds? 13.11 What are the necessary conditions for any system to be aromatic? 13.12 Explain why the following systems are not aromatic? (i) (ii) (iii) 13.13 How will you convert benzene into (i) p-nitrobromobenzene (ii) m- nitrochlorobenzene (iii) p - nitrotoluene (iv) acetophenone? 13.14 In the alkane H3C – CH2 – C(CH3)2 – CH2 – CH(CH3)2, identify 1°,2°,3° carbon atoms and give the number of H atoms bonded to each one of these. 13.15 What effect does branching of an alkane chain has on its boiling point? 13.16 Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism. 13.17 Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekulé structure for benzene? 13.18 Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. 13.19 Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? 13.20 How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane 13.21 Write structures of all the alkenes which on hydrogenation give 2-methylbutane. 13.22 Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+ (a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene (b) Toluene, p-H3C – C6H4 – NO2, p-O2N – C6H4 – NO2. 13.23 Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why? 13.24 Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene. 13.25 Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example. 2019-20

406 CHEMISTRY ENVIRONMENTAL CHEMISTRY UNIT 14 After studying this unit, you will be The world has achieved brilliance without wisdom, power able to without conscience. Ours is a world of nuclear giants and ethical infants. • understand the meaning of environmental chemistry; You have already studied about environment in your earlier classes. Environmental studies deal with the sum of all • define atmospheric pollution, list social, economical, biological, physical and chemical reasons for global warming. green interrelations with our surroundings. In this unit the focus house effect and acid rain; will be on environmental chemistry. Environmental chemistry deals with the study of the origin, transport, • identify causes for ozone layer reactions, effects and fates of chemical species in the depletion and its effects; environment. Let us discuss some important aspects of environmental chemistry. • give reasons for water pollution and know about international 14.1 ENVIRONMENTAL POLLUTION standards for drinking water; Environmental pollution is the effect of undesirable changes • describe causes of soil pollution; in our surroundings that have harmful effects on plants, animals and human beings. A substance, which causes • suggest and adopt strategies pollution, is known as pollutant. Pollutants can be solid, for control of environmental liquid or gaseous substances present in greater pollution; concentration than in natural abundance and are produced due to human activities or due to natural • appreciate the importance of green happenings. Do you know, an average human being chemistry in day to day life. requires nearly 12-15 times more air than the food. So, even small amounts of pollutants in the air become significant compared to similar levels present in the food. Pollutants can be degradable, like discarded vegetables which rapidly break down by natural processes. On the other hand, pollutants which are slowly degradable, remain in the environment in an unchanged form for many decades. For example, substances such as dichlorodi- phenyltrichloroethane (DDT), plastic materials, heavy metals, many chemicals, nuclear wastes etc., once released into the environment are difficult to remove. These 2019-20

ENVIRONMENTAL CHEMISTRY 407 pollutants cannot be degraded by natural sulphur dioxide, is a gas that is poisonous to processes and are harmful to living organisms. both animals and plants. It has been reported In the process of environmental pollution, that even a low concentration of sulphur pollutants originate from a source and get dioxide causes respiratory diseases e.g., transported by air or water or are dumped into asthma, bronchitis, emphysema in human the soil by human beings. beings. Sulphur dioxide causes irritation to the eyes, resulting in tears and redness. High 14.2 ATMOSPHERIC POLLUTION concentration of SO2 leads to stiffness of flower buds which eventually fall off from plants. The atmosphere that surrounds the earth is Uncatalysed oxidation of sulphur dioxide is not of the same thickness at all heights. There slow. However, the presence of particulate are concentric layers of air or regions and each matter in polluted air catalyses the oxidation layer has different density. The lowest region of sulphur dioxide to sulphur trioxide. of atmosphere in which the human beings along with other organisms live is called 2SO2 (g) +O2 (g) → 2SO3(g) troposphere. It extends up to the height of ~ 10 km from sea level. Above the troposphere, The reaction can also be promoted by between 10 and 50 km above sea level lies ozone and hydrogen peroxide. stratosphere. Troposphere is a turbulent, dusty zone containing air, much water vapour SO2 (g) +O3 (g) → SO3(g) + O2 (g) and clouds. This is the region of strong air movement and cloud formation. The SO2(g) + H2O2(l) → H2SO4(aq) stratosphere, on the other hand, contains (b) Oxides of Nitrogen: Dinitrogen and dinitrogen, dioxygen, ozone and little water dioxygen are the main constituents of air. vapour. These gases do not react with each other at a normal temperature. At high altitudes when Atmospheric pollution is generally studied lightning strikes, they combine to form oxides as tropospheric and stratospheric pollution. of nitrogen. NO2 is oxidised to nitrate ion, NO3− The presence of ozone in the stratosphere which is washed into soil, where it serves as a prevents about 99.5 per cent of the sun’s fertilizer. In an automobile engine, (at high harmful ultraviolet (UV) radiations from temperature) when fossil fuel is burnt, reaching the earth’s surface and thereby dinitrogen and dioxygen combine to yield protecting humans and other animals from its significant quantities of nitric oxide (NO) and effect. nitrogen dioxide ( NO2 ) as given below: 14.2.1 Tropospheric Pollution N2 (g) + O2 (g) 1483K→ 2NO(g) Tropospheric pollution occurs due to the NO reacts instantly with oxygen to give NO2 presence of undesirable solid or gaseous 2NO (g) + O2 (g) → 2NO2 (g) particles in the air. The following are the major gaseous and particulate pollutants present in Rate of production of NO2 is faster when the troposphere: nitric oxide reacts with ozone in the stratosphere. 1. Gaseous air pollutants: These are oxides of sulphur, nitrogen and carbon, hydrogen NO (g) + O3 (g) → NO2 (g) + O2 (g) sulphide, hydrocarbons, ozone and other The irritant red haze in the traffic and oxidants. congested places is due to oxides of nitrogen. 2. Particulate pollutants: These are dust, Higher concentrations of NO2 damage the mist, fumes, smoke, smog etc. leaves of plants and retard the rate of photosynthesis. Nitrogen dioxide is a lung 1. Gaseous air pollutants irritant that can lead to an acute respiratory disease in children. It is toxic to living tissues (a) Oxides of Sulphur: Oxides of sulphur also. Nitrogen dioxide is also harmful to are produced when sulphur containing fossil various textile fibres and metals. fuel is burnt. The most common species, 2019-20

408 CHEMISTRY (c) Hydrocarbons: Hydrocarbons are atmosphere. With the increased use of fossil composed of hydrogen and carbon only and fuels, a large amount of carbon dioxide gets are formed by incomplete combustion of fuel released into the atmosphere. Excess of CO2 used in automobiles. Hydrocarbons are in the air is removed by green plants and this carcinogenic, i.e., they cause cancer. They maintains an appropriate level of CO2 in the harm plants by causing ageing, breakdown of atmosphere. Green plants require CO2 for tissues and shedding of leaves, flowers and photosynthesis and they, in turn, emit oxygen, twigs. thus maintaining the delicate balance. As you know, deforestation and burning of fossil fuel (d) Oxides of Carbon increases the CO2 level and disturb the balance in the atmosphere. The increased amount of (i ) Carbon monoxide: Carbon monoxide (CO) CO2 in the air is mainly responsible for global is one of the most serious air pollutants. It is a warming. colourless and odourless gas, highly poisonous to living beings because of its ability Global Warming and Greenhouse Effect to block the delivery of oxygen to the organs and tissues. It is produced as a result of About 75 % of the solar energy reaching the incomplete combustion of carbon. Carbon earth is absorbed by the earth’s surface, which monoxide is mainly released into the air by increases its temperature. The rest of the heat automobile exhaust. Other sources, which radiates back to the atmosphere. Some of the produce CO, involve incomplete combustion heat is trapped by gases such as carbon of coal, firewood, petrol, etc. The number of dioxide, methane, ozone, chlorofluorocarbon vehicles has been increasing over the years all compounds (CFCs) and water vapour in the over the world. Many vehicles are poorly atmosphere. Thus, they add to the heating of maintained and several have inadequate the atmosphere. This causes global warming. pollution control equipments resulting in the release of greater amount of carbon monoxide We all know that in cold places flowers, and other polluting gases. Do you know why vegetables and fruits are grown in glass carbon monoxide is poisonous? It binds to covered areas called greenhouse. Do you haemoglobin to form carboxyhaemoglobin, know that we humans also live in a which is about 300 times more stable than the greenhouse? Of course, we are not surrounded oxygen-haemoglobin complex. In blood, when by glass but a blanket of air called the the concentration of carboxyhaemoglobin atmosphere, which has kept the temperature reaches about 3–4 per cent, the oxygen on earth constant for centuries. But it is now carrying capacity of blood is greatly undergoing change, though slowly. Just as reduced. This oxygen deficiency, results into the glass in a greenhouse holds the sun’s headache, weak eyesight, nervousness and warmth inside, atmosphere traps the sun’s cardiovascular disorder. This is the reason why heat near the earth’s surface and keeps it people are advised not to smoke. In pregnant warm. This is called natural greenhouse women who have the habit of smoking the effect because it maintains the temperature increased CO level in blood may induce and makes the earth perfect for life. In a premature birth, spontaneous abortions and greenhouse, solar radiations pass through deformed babies. the transparent glass and heat up the soil and the plants. The warm soil and plants emit (ii) Carbon dioxide: Carbon dioxide (CO2) is infrared radiations. Since glass is opaque to released into the atmosphere by respiration, infrared radiations (thermal region), it partly burning of fossil fuels for energy, and by reflects and partly absorbs these radiations. decomposition of limestone during the This mechanism keeps the energy of the manufacture of cement. It is also emitted sun trapped in the greenhouse. Similarly, during volcanic eruptions. Carbon dioxide gas carbon dioxide molecules also trap heat as is confined to troposphere only. Normally it they are transparent to sunlight but not forms about 0.03 per cent by volume of the to the heat radiation. If the amount of 2019-20

ENVIRONMENTAL CHEMISTRY 409 carbon dioxide crosses the delicate proportion Think it Over of 0.03 per cent, the natural greenhouse balance may get disturbed. Carbon dioxide is What can we do to reduce the rate of global the major contributor to global warming. warming? Besides carbon dioxide, other greenhouse If burning of fossil fuels, cutting down gases are methane, water vapour, nitrous forests and trees add to greenhouse gases oxide, CFCs and ozone. Methane is produced in the atmosphere, we must find ways to naturally when vegetation is burnt, digested use these just efficiently and judiciously. or rotted in the absence of oxygen. Large One of the simple things which we can do amounts of methane are released in paddy to reduce global warming is to minimise the fields, coal mines, from rotting garbage dumps use of automobiles. Depending upon the and by fossil fuels. Chlorofluorocarbons (CFCs) situation, one can use bicycle, public are man-made industrial chemicals used in transport system, or go for carpool. We air conditioning etc. CFCs are also damaging should plant more trees to increase the the ozone layer (Section 14.2.2). Nitrous oxide green cover. Avoid burning of dry leaves, occurs naturally in the environment. In recent wood etc. It is illegal to smoke in public years, their quantities have increased places and work places, because it is significantly due to the use of chemical harmful not only for the one who is smoking fertilizers and the burning of fossil fuels. If but also for others, and therefore, we should these trends continue, the average global avoid it. Many people do not understand temperature will increase to a level which may the greenhouse effect and the global lead to melting of polar ice caps and flooding warming. We can help them by sharing the of low lying areas all over the earth. Increase information that we have. in the global temperature increases the incidence of infectious diseases like dengue, Acid rain malaria, yellow fever, sleeping sickness etc. We are aware that normally rain water has a pH of 5.6 due to the presence of H+ ions formed by the reaction of rain water with carbon Fig. 14.1 Acid deposition 2019-20

410 CHEMISTRY dioxide present in the atmosphere. Activity 1 H2O (l) + CO2 (g)  H2CO3 (aq) You can collect samples of water from H2CO3 (aq)  H +(aq) + HCO3–(aq) nearby places and record their pH values. Discuss your results in the class. Let us When the pH of the rain water drops below discuss how we can help to reduce the 5.6, it is called acid rain. formation of acid rain. Acid rain refers to the ways in which acid This can be done by reducing the from the atmosphere is deposited on the emission of sulphur dioxide and nitrogen earth’s surface. Oxides of nitrogen and dioxide in the atmosphere. We should use sulphur which are acidic in nature can be less vehicles driven by fossil fuels; use less blown by wind along with solid particles in the sulphur content fossil fuels for power atmosphere and finally settle down either on plants and industries. We should use the ground as dry deposition or in water, fog natural gas which is a better fuel than coal and snow as wet deposition. (Fig. 14.1) or use coal with less sulphur content. Catalytic converters must be used in cars Acid rain is a byproduct of a variety of to reduce the effect of exhaust fumes on human activities that emit the oxides of the atmosphere. The main component of sulphur and nitrogen in the atmosphere. As the converter is a ceramic honeycomb mentioned earlier, burning of fossil fuels (which coated with precious metals — Pd, Pt and contain sulphur and nitrogenous matter) such Rh. The exhaust gases containing unburnt as coal and oil in power stations and furnaces fuel, CO and NOx, when pass through the or petrol and diesel in motor engines produce converter at 573 K, are converted into CO2 sulphur dioxide and nitrogen oxides. SO2 and and N2. We can also reduce the acidity of NO2 after oxidation and reaction with water the soil by adding powdered limestone to are major contributors to acid rain, because neutralise the acidity of the soil. Many polluted air usually contains particulate people do not know of acid rain and its matter that catalyse the oxidation. harmful effects. We can make them aware by passing on this information and save 2SO2 (g) + O2 (g) + 2H2O (l) → 2H2SO4 (aq) the Nature. 4NO2 (g) + O2 (g)+ 2H2O (l) → 4HNO3 (aq) Taj Mahal and Acid Rain Ammonium salts are also formed and can The air around the city of Agra, where the be seen as an atmospheric haze (aerosol of fine Taj Mahal is located, contains fairly high particles). Aerosol particles of oxides or levels of sulphur and nitrogen oxides. It is ammonium salts in rain drops result in wet- mainly due to a large number of industries deposition. SO2 is also absorbed directly on and power plants around the area. Use of both solid and liquid ground surfaces and is poor quality of coal, kerosene and firewood thus deposited as dry-deposition. as fuel for domestic purposes add up to this problem. The resulting acid rain Acid rain is harmful for agriculture, trees reacts with marble, CaCO3 of Taj Mahal and plants as it dissolves and washes away (CaCO3 +H2SO4 → CaSO4 + H2O+ CO2) nutrients needed for their growth. It causes causing damage to this wonderful respiratory ailments in human beings and monument that has attracted people from animals. When acid rain falls and flows as around the world. As a result, the ground water to reach rivers, lakes etc. it affects monument is being slowly disfigured and plants and animal life in aquatic ecosystem. It the marble is getting discoloured and corrodes water pipes resulting in the leaching lustreless. The Government of India of heavy metals such as iron, lead and copper announced an action plan in early 1995 into the drinking water. Acid rain damages to prevent the disfiguring of this historical buildings and other structures made of stone monument. Mathura refinery has already or metal. The Taj Mahal in India has been taken suitable measures to check the affected by acid rain. emission of toxic gases. 2019-20

ENVIRONMENTAL CHEMISTRY 411 This plan aims at clearing the air in herbicides and insecticides that miss their the ‘Taj Trapezium’– an area that includes targets and travel through air and form the towns of Agra, Firozabad, Mathura and mists. Bharatpur. Under this plan more than 2000 polluting industries lying inside the (d) Fumes are generally obtained by the trapezium would switch over to the use of condensation of vapours during natural gas or liquefied petroleum gas sublimation, distillation, boiling and instead of coal or oil. A new natural gas several other chemical reactions. Generally, pipeline would bring more than half a organic solvents, metals and metallic million cubic metres of natural gas a day oxides form fume particles. to this area. People living in the city will also be encouraged to use liquefied The effect of particulate pollutants are petroleum gas in place of coal, kerosene or largely dependent on the particle size. Air- firewood. Vehicles plying on highways in borne particles such as dust, fumes, mist etc., the vicinity of Taj would be encouraged to are dangerous for human health. Particulate use low sulphur content diesel. pollutants bigger than 5 microns are likely to lodge in the nasal passage, whereas particles 2. Particulate Pollutants of about 10 micron enter into lungs easily. Particulates pollutants are the minute solid Lead used to be a major air pollutant particles or liquid droplets in air. These are emitted by vehicles. Leaded petrol used to be present in vehicle emissions, smoke particles the primary source of air-borne lead emission from fires, dust particles and ash from in Indian cities. This problem has now been industries. Particulates in the atmosphere overcome by using unleaded petrol in most of may be viable or non-viable. The viable the cities in India. Lead interferes with the particulates e.g., bacteria, fungi, moulds, development and maturation of red blood cells. algae etc., are minute living organisms that are dispersed in the atmosphere. Human beings Smog are allergic to some of the fungi found in air. They can also cause plant diseases. The word smog is derived from smoke and fog. This is the most common example of air Non-viable particulates may be classified pollution that occurs in many cities according to their nature and size as follows: throughout the world. There are two types of smog: (a) Smoke particulates consist of solid or mixture of solid and liquid particles formed (a) Classical smog occurs in cool humid during combustion of organic matter. climate. It is a mixture of smoke, fog and Examples are cigarette smoke, smoke from sulphur dioxide. Chemically it is a burning of fossil fuel, garbage and dry reducing mixture and so it is also called leaves, oil smoke etc. as reducing smog. (b) Dust is composed of fine solid particles (b) Photochemical smog occurs in warm, dry (over 1µm in diameter), produced during and sunny climate. The main components crushing, grinding and attribution of solid of the photochemical smog result from the materials. Sand from sand blasting, saw action of sunlight on unsaturated dust from wood works, pulverized coal, hydrocarbons and nitrogen oxides cement and fly ash from factories, dust produced by automobiles and factories. storms etc., are some typical examples of Photochemical smog has high this type of particulate emission. concentration of oxidising agents and is, therefore, called as oxidising smog. (c) Mists are produced by particles of spray liquids and by condensation of vapours in Formation of photochemical smog air. Examples are sulphuric acid mist and When fossil fuels are burnt, a variety of pollutants are emitted into the earth’s 2019-20

412 CHEMISTRY troposphere. Two of the pollutants that are to produce chemicals such as formaldehyde, emitted are hydrocarbons (unburnt fuels) and acrolein and peroxyacetyl nitrate (PAN). nitric oxide (NO). When these pollutants build up to sufficiently high levels, a chain reaction 3CH4 + 2O3 → 3CH2 = O + 3H2O occurs from their interaction with sunlight in Formaldehyde which NO is converted into nitrogen dioxide (NO2). This NO2 in turn absorbs energy from CH2=CHCH=O CH3COONO2 sunlight and breaks up into nitric oxide and Acrolein free oxygen atom (Fig. 14.2).  O Peroxyacetyl nitrate (PAN) NO2(g) NO(g) + O(g) (i) Effects of photochemical smog Oxygen atoms are very reactive and The common components of photochemical smog are ozone, nitric oxide, acrolein, combine with the O2 in air to produce ozone. formaldehyde and peroxyacetyl nitrate (PAN). Photochemical smog causes serious health O(g) + O2 (g)  O3 (g) (ii) problems. Both ozone and PAN act as powerful eye irritants. Ozone and nitric oxide irritate the The ozone formed in the above reaction (ii) nose and throat and their high concentration causes headache, chest pain, dryness of the reacts rapidly with the NO(g) formed in the throat, cough and difficulty in breathing. Photochemical smog leads to cracking of reaction (i) to regenerate NO2. NO2 is a brown rubber and extensive damage to plant life. It gas and at sufficiently high levels can also causes corrosion of metals, stones, building materials, rubber and painted contribute to haze. surfaces. NO (g) + O3 (g) → NO2 (g) + O2 (g) (iii) Ozone is a toxic gas and both NO2 and O3 are strong oxidising agents and can react with the unburnt hydrocarbons in the polluted air Fig. 14.2 Photochemical smog occurs where sunlight acts on vehicle pollutants. 2019-20