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6CN – + Fe2+ → [Fe(CN)6]4– 3[Fe(CN)6]4– + 4Fe3+ Fe4[Fe(CN)6]3.xH2O Prussian blue (B) Test for Sulphur (a) The sodium fusion extract is acidified with acetic acid and lead acetate is added to it. A black precipitate of lead sulphide indicates the pr esence of sulphur. S2– + Pb2+ → PbS Black (b) On treating sodium fusion extract with sodium nitroprusside, appearance of a violet colour further indicates the presence of sulphur. S2– + [Fe(CN)5NO]2– → [Fe(CN)5NOS]4– Violet In case, nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed. It gives blood red colour and no Prussian blue since there are no free cyanide ions. Na + C + N + S → NaSCN Fe3+ +SCN– → [Fe(SCN)]2+ Blood red If sodium fusion is carried out with excess of sodium, the thiocyanate decomposes to yield cyanide and sulphide. These ions give their usual tests. NaSCN + 2Na → NaCN+Na2S (C) Test for Halogens The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate. A white precipitate, soluble in ammonium hydroxide shows the presence of chlorine, a yellowish precipitate, sparingly soluble in ammonium hydroxide shows the presence of bromine and a yellow precipitate, insoluble in ammonium hydroxide shows the presence of iodine. X– + Ag+ → AgX X represents a halogen – Cl, Br or I.
If nitr ogen or sulphur is also present in the compound, the sodium fusion extract is first boiled with concentrated nitric acid to decompose cyanide or sulphide of sodium formed during Lassaigne’s test. These ions would otherwise interfere with silver nitrate test for halogens. (D) Test for Phosphorus The compound is heated with an oxidising agent (sodium peroxide). The phosphorus present in the compound is oxidised to phosphate. The solution is boiled with nitric acid and then treated with ammonium molybdate. A yellow colouration or precipitate indicates the presence of phosphorus. Na 3PO4 + 3HNO3 → H3PO4+3NaNO3 H3PO4 + 12(NH4)2MoO4 + 21HNO3 → Ammonium molybdate (NH4)3PO4.12MoO3 + 21NH4NO3 + 12H2O Ammonium phosphomolybdate 8.10 quantitati Ve analysis Quantitative analysis of compounds is very important in organic chemistry. It helps chemists in the determination of mass per cent of elements present in a compound. You have learnt in Unit-1 that mass per cent of elements is required for the determination of emperical and molecular formula. The percentage composition of elements present in an organic compound is determined by the following methods: 8.10.1 carbon and Hydrogen Both carbon and hydr ogen are estimated in one experiment.
A known mass of an organic compound is burnt in the presence of excess of oxygen and copper(II) oxide. Carbon and hydrogen in the compound are oxidised to carbon dioxide and water respectively. C xHy + (x + y/4) O2 → x CO2 + (y/2) H2O Unit 8.indd 285 10/10/2022 10:37:37 AM 2024-25 286 chemistry The mass of water produced is determined by passing the mixture through a weighed U-tube containing anhydrous calcium chloride. Carbon dioxide is absorbed in another U-tube containing concentrated solution of potassium hydroxide. These tubes are connected in series (Fig. 8.14). The increase in masses of calcium chloride and potassium hydroxide gives the amounts of water and carbon dioxide from which the percentages of carbon and hydrogen are calculated. Let the mass of organic compound be m g, mass of water and carbon dioxide produced be m 1 and m2 g respectively; Percentage of carbon= 2 12 100 44m m×× × Percentage of hydrogen = 1 2 100 18m m×× × Problem 8.20 On complete combustion, 0.246 g of an organic compound gave 0.198g of carbon dioxide and 0.1014g of water . Determine the percentage composition of carbon and hydrogen in the compound. solution 12 0.198 100Percentageof carbon44 0.246 21.95%××=× = Percentage of hydrogen 2 0.1014 100 18 0.246 4.58%××=× =8.10.2 nitrogen There are two methods for estimation of nitr ogen: (i) Dumas method and (ii) Kjeldahl’s method.
(i) Dumas method: The nitrogen containing organic compound, when heated with copper oxide in an atmosphere of carbon dioxide, yields free nitrogen in addition to carbon dioxide and water. C xHyNz + (2x + y/2) CuO → x CO2 + y/2 H2O + z/2 N2 + (2x + y/2) Cu Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over a heated copper gauze. The mixture of gases so produced is collected over an aqueous solution of potassium hydroxide which absorbs carbon dioxide. Nitrogen is collected in the upper part of the graduated tube (Fig.8.15). Let the mass of organic compound = m gVolume of nitrogen collected = V 1 mL Room temperature = T1K 11 1273Volumeof nitrogen at STP760 (Let it be mL)PV T V×=× Where p1 and V1 are the pressure and volume of nitrogen, p1 is different from the atmospheric pressure at which nitrogen gas is collected. The value of p1 is obtained by the relation; p1= Atmospheric pressure – Aqueous tension 22400 mL N2 at STP weighs 28 g.Fig.8.14 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. Unit 8.indd 286 10/10/2022 10:37:38 AM 2024-25 287 organic chemistry – some basic principles and techniques 228mL N at STP weighs g22400VV×= Percentage of nitrogen = 28 100 22400V m×× × Problem 8.21 In Dumas’ method for estimation of nitrogen, 0.3g of an organic compound gave 50mL of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. (Aqueous tension at 300K=15 mm) solution Volume of nitrogen collected at 300K and 715mm pressure is 50 mLActual pressure = 715-15 =700 mm 273 700 50Volumeof nitrogen at STP300 760 41.9mL×× × 22,400 mL of N2 at STP weighs = 28 g28 41.941.9mLof nitrogen weighs22400g×= 28 41.9 100Percentageof nitrogen22400 0.3 17.46%××=× =Fig.
8.15 Dumas method. The organic compound yields nitrogen gas on heating it with copper(II) oxide in the presence of CO 2 gas. The mixture of gases is collected over potassium hydroxide solution in which CO2 is absorbed and volume of nitrogen gas is determined. (ii) Kjeldahl’s method: The compound containing nitrogen is heated with concentrated sulphuric acid. Nitrogen in the compound gets converted to ammonium sulphate (Fig. 8.16). The resulting acid mixture is then heated with excess of sodium hydroxide. The liberated ammonia gas is absorbed in an excess of standard solution of sulphuric acid. The amount of ammonia produced is determined by estimating the amount of sulphuric acid consumed in the reaction. It is done by estimating unreacted sulphuric acid left after the absorption of ammonia by titrating it with standard alkali solution.
The difference between the initial amount of acid Unit 8.indd 287 10/10/2022 10:37:38 AM 2024-25 288 chemistry taken and that left after the reaction gives the amount of acid reacted with ammonia. Organic compound + H2SO4 → (NH4)2SO4 Na2SO4 + 2NH3 + 2H2O 2NH3 + H2SO4 → (NH4)2SO4 Let the mass of organic compound taken = m g Volume of H2SO4 of molarity, M, taken = V mL Volume of NaOH of molarity, M, used for titration of excess of H2SO4 = V1 mL V1mL of NaOH of molarity M = V1 /2 mL of H2SO4 of molarity M Volume of H2SO4 of molarity M unused = (V - V1/2) mL (V- V1/2) mL of H2SO4 of molarity M = 2(V -V1/2) mL of NH3 solution of molarity M. 1000 mL of 1 M NH3 solution contains 17g NH3 or 14 g of N 2(V-V1/2) mL of NH3 solution of molarity M contains: 1 14 M 2(V V / 2)gN1000×× − 1 14 M 2(V V / 2) 100Percentageof N1000 1.4 M 2(V V / 2)m m×× −=× ×× −= Kjeldahl method is not applicable to compounds containing nitrogen in nitro and azo groups and nitrogen present in the ring (e.g. pyridine) as nitrogen of these compounds does not change to ammonium sulphate under these conditions. Problem 8.22 During estimation of nitrogen present in an organic compound by Kjeldahl’s method, the ammonia evolved from 0.5 g of the compound in Kjeldahl’s estimation of nitrogen, neutralized 10 mL of 1 M H 2SO4. Find out the percentage of nitrogen in the compound. solution 1 M of 10 mL H2SO4=1M of 20 mL NH3 1000 mL of 1M ammonia contains 14 g nitrogen 20 mL of 1M ammonia contains Fig.8.16 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. Unit 8.indd 288 10/10/2022 10:37:39 AM 2024-25 289 organic chemistry – some basic principles and techniques 14 20 1000× g nitrogen 14×20×100Percentageof nitrogen =56.0%1000×0.5= 8.10.3 halogens Car ius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Fig.8.17) Percentage of halogen 1 atomic mass of X m g molecular mass of AgX×= Problem 8.23 In Carius method of estimation of halogen, 0.15 g of an organic compound gave 0.12 g of AgBr. Find out the percentage of bromine in the compound. solution Molar mass of AgBr = 108 + 80 = 188 g mol-1 188 g AgBr contains 80 g bromine 0.12 g AgBr contains ×80 0.12g188 g bromine ××= =80 0.12 100Percentage of bromine 188×0.15 34.04% 8.10.4 sulphur A known mass of an organic compound is heated in a Carius tube with sodium peroxide or fuming nitric acid.
Sulphur present in the compound is oxidised to sulphuric acid. It is precipitated as barium sulphate by adding excess of barium chloride solution in water. The precipitate is filtered, washed, dried and weighed. The percentage of sulphur can be calculated from the mass of barium sulphate. Let the mass of organic compound taken = m g and the mass of barium sulphate formed = m1g 1 mol of BaSO4 = 233 g BaSO4 = 32 g sulphur m1 g BaSO4 contains 1 32 100 233m m×× × g sulphur 1 32 100Percentageof sulphur233m m××=×Fig. 8.17 Carius method. Halogen containing organic compound is heated with fuming nitric acid in the presence of silver nitrate. in a furnace. Carbon and hydrogen present in the compound are oxidised to carbon dioxide and water. The halogen present forms the corresponding silver halide (AgX).
It is filtered, washed, dried and weighed.Let the mass of organic compound taken = m g Mass of AgX formed = m 1 g 1 mol of AgX contains 1 mol of X Mass of halogen in m1g of AgX 1 atomic mass of X m g molecular mass of AgX×=Problem 8.24In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of Unit 8.indd 289 10/10/2022 10:37:39 AM 2024-25 290 chemistry percentage composition (100) and the sum of the percentages of all other elements. However, oxygen can also be estimated directly as follows: A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas. The mixture of gaseous products containing oxygen is passed over red-hot coke when all the oxygen is converted to carbon monoxide. This mixture is passed through warm iodine pentoxide (I 2O5) when carbon monoxide is oxidised to carbon dioxide producing iodine. Compound heat→ O2 + other gaseous products 2C + O2 1373 K→ 2CO]× 5 (A) I2O5 + 5CO → I2 + 5CO2]× 2 (B) On making the amount of CO produced in equation (A) equal to the amount of CO used in equation (B) by multiplying the equations (A) and (B) by 5 and 2 respectively; we find that each mole of oxygen liberated from the compound will produce two moles of carbondioxide. Thus 88 g carbon dioxide is obtained if 32 g oxygen is liberated. Let the mass of organic compound taken be m g Mass of carbon dioxide produced be m 1 g ∴ m1 g carbon dioxide is obtained from 1 232gO88m× ∴Percentage of oxygen = 1 32 100%88m m×× × The percentage of oxygen can be derived from the amount of iodine produced also. Presently, the estimation of elements in an organic compound is carried out by using microquantities of substances and automatic experimental techniques. The elements, carbon, hydrogen and nitrogen present in a compound are determined by an apparatus known as CHN elemental analyser. The analyser requires only a very small amount of the substance (1-3 mg) and displays the values on a screen within a short time.
A detailed discussion of such methods is beyond the scope of this book.barium sulphate. What is the percentage of sulphur in the compound? solution Molecular mass of BaSO4 = 137+32+64 = 233 g 233 g BaSO4 contains 32 g sulphur 0.4813 g BaSO4 contains 32 0.4813g233× g sulphur 32 0.4813 100Percentageof sulphur233 0.157 42.10%××=× = 8.10.5 Phosphorus A known mass of an organic compound is heated with fuming nitric acid whereupon phosphorus present in the compound is oxidised to phosphoric acid. It is precipitated as ammonium phosphomolybdate, (NH 4)3 PO4.12MoO3, by adding ammonia and ammonium molybdate. Alternatively, phosphoric acid may be precipitated as MgNH 4PO4 by adding magnesia mixture which on ignition yields Mg2P2O7. Let the mass of organic compound taken = m g and mass of ammonium phospho molydate = m 1g Molar mass of (NH4)3PO4.12MoO3 = 1877g Percentage of phosphorus 1 31 100%1877m m××=× If phosphorus is estimated as Mg2P2O7, Percentage of phosphorus 1 62 100%222m××=× where, 222 u is the molar mass of Mg2P2O7, m, the mass of organic compound taken, m1, the mass of Mg2P2O7 formed and 62, the mass of two phosphorus atoms present in the compound Mg 2P2O7. 8.10.6 Oxygen The percentage of oxygen in an organic compound is usually found by difference between the total Unit 8.indd 290 10/10/2022 10:37:39 AM 2024-25 291 organic chemistry – some basic principles and techniques 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 sp 3, 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 sp 3 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 ar e 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 hydr ogen 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. Unit 8.indd 291 10/10/2022 10:37:39 AM 2024-25 292 chemistry ExERCISES 8.1 What are hybridisation states of each carbon atom in the following compounds ?
CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6 8.2 Indicate the σ and π bonds in the following molecules : C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3 8.3 Write bond line formulas for : Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one. 8.4 Give the IUPAC names of the following compounds : (a) (b) (c) (d) (e) (f) Cl2CHCH2OH 8.5 Which of the following represents the correct IUP AC name for the compounds concerned ? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-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. 8.6 Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH 3COCH3 (c) H–CH=CH2 8.7 Give condensed and bond line structural formulas and identify the functional group(s) pr esent, if any, for : (a) 2,2,4-Trimethylpentane (b) 2-Hydroxy-1,2,3-propanetricarboxylic acid (c) Hexanedial 8.8 Identify the functional groups in the following compounds (a) (b) (c) 8.9 Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why? 8.10 Explain why alkyl groups act as electron donors when attached to a π system. 8.11 Draw the resonance structures for the following compounds. Show the electron shift using curved-arr ow notation. (a) C6H5OH (b) C6H5NO2 (c) CH3CH=CHCHO (d) C6H5–CHO (e) C6H5–C+ H2 (f) CH3CH=CH C+ H2 8.12 What are electrophiles and nucleophiles ? Explain with examples.
8.13 Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: (a) CH3COOH + HO– → CH3COO–+H2O Unit 8.indd 292 10/10/2022 10:37:40 AM 2024-25 293 organic chemistry – some basic principles and techniques (b) CH3COCH3+ C– N → (CH3)2C(CN)(OH) (c) C6H6 + CH3C+ O → C6H5COCH3 8.14 Classify the following reactions in one of the reaction type studied in this unit. (a) CH3CH2Br + HS– → CH3CH2SH + Br– (b) (CH3)2C = CH2 + HCI → (CH3)2CIC – CH3 (c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br– (d) (CH3)3C– CH2OH + HBr → (CH3)2CBrCH2CH2CH3 + H2O 8.15 What is the relationship between the members of following pairs of structures ? Are they structural or geometrical isomers or r esonance contributors ? (a) (b) (c) 8.16 For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify r eactive intermediate produced as free radical, carbocation and carbanion. (a) (b) (c) (d) 8.17 Explain the terms Inductive and Electromeric effects. Which electr on displacement effect explains the following correct orders of acidity of the carboxylic acids? (a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH (b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3C.COOH 8.18 Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography 8.19 Describe the method, which can be used to separate two compounds with different solubilities in a solvent S. 8. 20 What is the difference between distillation, distillation under reduced pr essure and steam distillation ?
Unit 8.indd 293 10/10/2022 10:37:40 AM 2024-25 294 chemistry 8.21 Discuss the chemistry of Lassaigne’s test. 8.22 Differentiate between the principle of estimation of nitrogen in an or ganic compound by (i) Dumas method and (ii) Kjeldahl’s method. 8.23 Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound. 8.24 Explain the principle of paper chromatography. 8.25 Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens? 8.26 Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens. 8.27 Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor. 8.28 Explain, why an organic liquid vaporises at a temperatur e below its boiling point in its steam distillation ? 8.29 Will CCl4 give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.
8.30 Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound? 8.31 Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test? 8.32 An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water pr oduced when 0.20 g of this substance is subjected to complete combustion. 8.33 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 H 2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound. 8.34 0.3780 g of an organic chlor o compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.
8.35 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. 8.36 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 8.37 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 8.38 Which of the following carbocation is most stable ? (a) (CH3)3C. C+ H2 (b) (CH3)3C+ (c) CH3CH2C+ H2 (d) CH3C+ H CH2CH3 8.39 The best and latest technique for isolation, purification and separation of organic compounds is: (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography 8.40 The reaction: CH3CH2I + KOH(aq) → CH3CH2OH + KI is classified as : (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition Unit 8.indd 294 10/10/2022 10:37:40 AM 2024-25 Unit 7 7.25 15 g Unit 8 8.32 Mass of carbon dioxide formed = 0.505 g Mass of water formed = 0.0864 g 8.33 % fo nitrogen = 56 8.34 % of chlorine = 37.57 8.35 % of sulphur = 19.66 Unit 9 9.1 Due to the side reaction in ter mination step by the combination of two C. H3 free radicals. 9.2 (a) 2-Methyl-but-2-ene (b) Pent-1-ene-3-yne (c) Buta-1, 3-diene (d) 4-Phenylbut-1-ene (e) 2-Methylphenol (f) 5-(2-Methylpropyl)-decane (g) 4-Ethyldeca –1,5,8- triene 9.3 (a) (i) CH2 = CH – CH2 – CH3 But-1-ene (ii) CH3 – CH2 = CH – CH3 But-2-ene (iii) CH2 = C – CH3 2-Methylpropene | CH3 (b) (i) HC ≡ C – CH2 – CH2 – CH3 Pent-1-yne (ii) CH3 – C ≡ C – CH2 – CH3 Pent-2-yne (iii) CH3 – CH – C ≡ CH 3-Methylbut-1-yne | CH3 9.4 (i) Ethanal and propanal (ii) Butan-2-one and pentan-2-one (iii) Methanal and pentan-3-one (iv) Propanal and benzaldehyde 9.5 3-Ethylpent-2-ene 9.6 But-2-ene 9.7 4-Ethylhex-3-ene CH3 – CH2– C = CH – CH2–CH3 | CH2–CH3Answer to Some Selected Problems Answers.indd 328 10/27/2022 2:32:20 PM 2024-25 329 answers 9.8 (a) C4H10(g)+13/202(g) ∆ 4CO2(g) + 5H2O(g) (b) C5H10(g)+15/202(g) ∆ 5CO2(g) + 5H2O(g) (c) C5H10(g) + 17/2 O2(g) ∆ 6CO2(g) + 5H2O(g) (d) C7H8(g) + 902(g) ∆ 7CO2(g) + 4H2O(g) cis-Hex-2-ene trans-Hex-2-ene The cis form will have higher boiling point due to more polar nature leading to stronger intermolecular dipole–dipole interaction, thus requiring more heat energy to separate them. 9.10 Due to resonance 9.11 Planar , conjugated ring system with delocalisation of (4n+2) π electrons, where, n is an integer 9.12 Lack of delocalisation of (4n +2) π electr ons in the cyclic system. 9.13 (i) (ii) Answers.indd 329 10/10/2022 11:12:18 AM 2024-25 330 chemistry (iii) (iv) 9.14 15 H attached to 1° carbons 4 H attached to 2° carbons 1 H attached to 3° carbons 9.15 More the branching in alkane, lower will be the boiling point. 9.16 Refer to addition reaction of HBr to unsymmetrical alkenes in the text. All the three products cannot be obtained by any one of the Kekulé’s structures.
This shows that benzene is a resonance hybrid of the two resonating structures. 9.18 H – C ≡ C – H > C6H6 > C6H14. Due to maximum s orbital character in enthyne (50 per cent) as compared to 33 per cent in benzene and 25 per cent in n-hexane. 9.19 Due to the presence of 6 π electrons, benzene behaves as a rich source of electrons thus being easily attacked by reagents deficient in electrons. Answers.indd 330 10/10/2022 11:12:18 AM 2024-25 331 answers 9.20 (i) (ii) C2H4 Br2 CH2–CH2 alc, KOH CH2=CHNr NaNH2 Br BR K (iii) CH3 | 9.21 CH2 = C – CH2 – CH3 2-Methylbut-1-ene CH3 | CH3 – C = CH – CH3 2-Methylbut-2-ene CH3 | CH3 – CH –CH = CH2 3-Methylbut-1-ene 9.22 (a) Chlo robenzene>p -nitrochlorobenzene> 2,4 – dinitrochlorobenzene (b) Toluene > p-CH3-C6H4-NO2 > p-O2N–C6H4–NO2 9.23 Toleune undergoes nitration most easily due to electron releasing nature of the methyl group. 9.24 FeCl3 9.25 Due to the for mation of side products. For example, by starting with 1-bromopropane and 1-bromobutane, hexane and octane are the side products besides heptane. Answers.indd 331 03-05-2024 16:40:47 2024-25 Notes Answers.indd 332 12/1/2022 11:01:10 2024-25 295 Hydrocarbons Unit 9 After studying this unit, you will be able to • name hydrocarbons according to IUP AC system of nomenclature; • recognise and write structures of isomers of alkanes, alkenes, alkynes and aromatic hydrocarbons; • learn about various methods of preparation of hydrocarbons; • distinguish between alkanes, alkenes, alkynes and aromatic hydr ocarbons on the basis of physical and chemical properties; • draw and differentiate between various confor mations of ethane; • appreciate the r ole of hydrocarbons as sources of energy and for other industrial applications; • predict the format ion of the addition products of unsymmetrical alkenes and alkynes on the basis of electronic mechanism; • comp rehend the structure of benzene, explain aromaticity and understand mechanism of electrophilic substitution reactions of benzene; • predict the dir ective influence of substituents in monosubstituted benzene ring; • learn about carcinogenicity and toxicity.Hydrocarbons The term ‘hydrocarbon’ is self-explanatory which means compounds of carbon and hydrogen only. Hydrocarbons play a key role in our daily life. You must be familiar with the terms ‘LPG’ and ‘CNG’ used as fuels.
LPG is the abbreviated form of liquified petroleum gas whereas CNG stands for compressed natural gas. Another term ‘LNG’ (liquified natural gas) is also in news these days. This is also a fuel and is obtained by liquifaction of natural gas. Petrol, diesel and kerosene oil are obtained by the fractional distillation of petroleum found under the earth’s crust. Coal gas is obtained by the destructive distillation of coal. Natural gas is found in upper strata during drilling of oil wells. The gas after compression is known as compressed natural gas. LPG is used as a domestic fuel with the least pollution. Kerosene oil is also used as a domestic fuel but it causes some pollution. Automobiles need fuels like petrol, diesel and CNG.
Petrol and CNG operated automobiles cause less pollution. All these fuels contain mixture of hydrocarbons, which are sources of energy. Hydrocarbons are also used for the manufacture of polymers like polythene, polypropene, polystyrene etc. Higher hydrocarbons are used as solvents for paints. They are also used as the starting materials for manufacture of many dyes and drugs. Thus, you can well understand the importance of hydrocarbons in your daily life. In this unit, you will learn more about hydrocarbons. 9.1 cLassiFication Hydr ocarbo ns are of different types. Depending upon the types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated Hydrocarbons are the important sources of energy. Unit 9.indd 295 10/10/2022 10:37:52 AM 2024-25 296 chemistry (ii) unsaturated and (iii) aromatic hydrocarbons.
Saturated hydrocarbons contain carbon-carbon and carbon-hydrogen single bonds. If different carbon atoms are joined together to form open chain of carbon atoms with single bonds, they are termed as alkanes as you have already studied in Unit 8. On the other hand, if carbon atoms form a closed chain or a ring, they are termed as cycloalkanes. Unsaturated hydrocarbons contain carbon-carbon multiple bonds – double bonds, triple bonds or both. Aromatic hydrocarbons are a special type of cyclic compounds. You can construct a large number of models of such molecules of both types (open chain and close chain) keeping in mind that carbon is tetravalent and hydrogen is monovalent. For making models of alkanes, you can use toothpicks for bonds and plasticine balls for atoms. For alkenes, alkynes and aromatic hydrocarbons, spring models can be constructed. 9.2 ALKANES As already mentioned, alkanes are saturated open chain hydrocarbons containing carbon - carbon single bonds. Methane (CH4) is the first member of this family.
Methane is a gas found in coal mines and marshy places. If you replace one hydrogen atom of methane by carbon and join the required number of hydrogens to satisfy the tetravalence of the other carbon atom, what do you get? You get C 2H6. This hydrocarbon with molecular formula C2H6 is known as ethane. Thus you can consider C2H6 as derived from CH4 by replacing one hydrogen atom by -CH3 group. Go on constructing alkanes by doing this theoretical exercise i.e., replacing hydrogen atom by –CH 3 group. The next molecules will be C3H8, C4H10 … of the general formula for alkane family or homologous series? If we examine the formula of different alkanes we find that the general formula for alkanes is C nH2n+2. It represents any particular homologue when n is given appropriate value. Can you recall the structure of methane?
According to VSEPR theory (Unit 4), methane has a tetrahedral structure (Fig. 9.1), in which carbon atom lies at the centre and the four hydrogen atoms lie at the four corners of a regular tetrahedron. All H-C-H bond angles are of 109.5°. In alkanes, tetrahedra are joined together in which C-C and C-H bond lengths are 154 pm and 112 pm respectively (Unit 8). You have already read that C–C and C–H σ bonds are formed by head-on overlapping of sp 3 hybrid orbitals of carbon and 1s orbitals of hydrogen atoms. 9.2.1 Nomenclature and Isomerism You have already read about nomenclature of different classes of organic compounds in Unit 8. Nomenclature and isomerism in alkanes can further be understood with the help of a few more examples. Common names are given in parenthesis. First three alkanes – methane, ethane and propane have only one structure but higher alkanes can have more than one structure. Let us write structures for C 4H10.
Four carbon atoms of C4H10 can be joined either in a continuous chain or with a branched chain in the following two ways :Fig. 9.1 Structure of methane Butane (n- butane), (b.p. 273 K)I H H H H—C—H replace any H by - CH3 H—C—C—H or C2H6 H H H These hydrocarbons are inert under normal conditions as they do not r eact with acids, bases and other reagents. Hence, they were earlier known as paraffins (latin : parum, little; affinis, affinity). Can you think Unit 9.indd 296 11/10/2022 15:24:01 2024-25 297 Hydrocarbons In how many ways, you can join five carbon atoms and twelve hydrogen atoms of C5H12? They can be arranged in three ways as shown in structures III–Visomers. It is also clear that structures I and III have continuous chain of carbon atoms but structures II, IV and V have a branched chain. Such structural isomers which differ in chain of carbon atoms are known as chain isomers. Thus, you have seen that C 4H10 and C5H12 have two and three chain isomers respectively. Problem 9.1 Write structures of different chain isomers of alkanes corresponding to the molecular formula C 6H14.
Also write their IUPAC names. solution (i) CH3 – CH2 – CH2 – CH2– CH2– CH3 n-Hexane 2-Methylpentane 3-Methylpentane 2,3-Dimethylbutane 2,2 - Dimethylbutane Based upon the number of carbon atoms attached to a carbon atom, the carbon atom is t ermed as primary (1°), secondary (2°), tertiary (3°) or quaternary (4°). Carbon atom attached to no other carbon atom as in methane or to only one carbon atom as in ethane is called primary carbon atom. Terminal carbon atoms are always primary. Carbon atom attached to two carbon atoms is known as secondary. Tertiary carbon is attached to three carbon atoms and neo or quaternary carbon is attached to four carbon atoms. Can you identify 1°, 2°, 3° and 4° carbon atoms in II 2-Methylpropane (isobutane) (b.p.261 K) Structures I and II possess same molecular formula but dif fer in their boiling points and other properties. Similarly structures III, IV and V possess the same molecular formula but have different properties. Structures I and II are isomers of butane, whereas structures III, IV and V are isomers of pentane. Since difference in properties is due to difference in their structures, they are known as structural III Pentane (n-pentane) (b.p.
309 K) 2-Methylbutane (isopentane) (b.p. 301 K)IV 2,2-Dimethylpropane (neopentane) (b.p. 282.5 K)V Unit 9.indd 297 10/10/2022 10:37:53 AM 2024-25 298 chemistry structures I to V ? If you go on constructing structures for higher alkanes, you will be getting still larger number of isomers. C 6H14 has got five isomers and C7H16 has nine. As many as 75 isomers are possible for C10H22. In structures II, IV and V, you observed that –CH3 group is attached to carbon atom numbered as 2. You will come across groups like –CH 3, –C2H5, –C3H7 etc. attached to carbon atoms in alkanes or other classes of compounds. These groups or substituents are known as alkyl groups as they are derived from alkanes by removal of one hydrogen atom.
General formula for alkyl groups is C nH2n+1 (Unit 8). Let us recall the general rules for nomenclatur e already discussed in Unit 8. Nomenclature of substituted alkanes can further be understood by considering the following problem: Problem 9.2 Write structures of different isomeric alkyl groups corresponding to the molecular formula C 5H11. Write IUPAC names of alcohols obtained by attachment of –OH groups at different carbons of the chain. Solution Structures of – C5H11 group Corresponding alcohols Name of alcohol (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 CH3 CH3 2-Methyl- | | butan-1-ol (v) CH3 – CH2 – CH – CH2 – CH3 – CH2 – CH – CH2– OH CH3 CH3 2-Methyl- | | butan-2-ol (vi) CH3 – C – CH2 – CH3 CH3 – C – CH2 – CH3 | | OH CH3 CH3 2,2- Dimethyl- | | propan-1-ol (vii) CH3 – C – CH2 – CH3 – C – CH2OH | | CH3 CH3 CH3 CH3 OH 3-Methyl- | | | | butan-2-ol (viii) CH3 – CH – CH –CH3 CH3 – CH – CH –CH3 Unit 9.indd 298 11/10/2022 15:24:36 2024-25 299 Hydrocarbons remarks Lowest sum and alphabetical arrangement Lowest sum and alphabeticalarrangement sec is not considered while arrangingalphabetically; isopropyl is taken as one wor d Further numbering to the substituentsof the side chain Alphabetical priority ordertable 9.1 nomenclature of a Few organic compounds important to write the correct structure from the given IUPAC name. To do this, first of all, the longest chain of carbon atoms corresponding to the parent alkane is written. Then after numbering it, the substituents are attached to the correct carbon atoms and finally valence of each carbon atom is satisfied by putting the correct number of hydrogen atoms. This can be clarified by writing the structure of 3-ethyl-2, 2–dimethylpentane in the following steps : i) Draw the chain of five carbon atoms: C – C – C – C – C ii) Give number to carbon atoms: C1– C2– C3– C4– C5structure and iUPac name (a) 1CH3–2CH – 3CH2 – 4CH – 5CH2 – 6CH3 (4 – Ethyl – 2 – methylhexane) (b) 8CH3 – 7CH2 – 6CH2 – 5CH – 4CH – 3C – 2CH2 – 1CH3 (3,3-Diethyl-5-isopropyl-4-methyloctane) (c) 1CH3–2CH2–3CH2–4CH–5CH–6CH2–7CH2–8CH2–9CH2–10CH3 5-sec– Butyl-4-isopr opyldecane (d) 1CH3–2CH2–3CH2–4CH2–5CH–6CH2–7CH2–8CH2–9CH3 5-(2,2– Dimethylpropyl)nonane (e) 1CH3 – 2CH2 – 3CH – 4CH2 – 5CH – 6CH2 – 7CH3 3–Ethyl–5–methylheptane Problem 9.3 Write IUPAC names of the following compounds : (i) (CH3)3 C CH2C(CH3)3 (ii) (CH3)2 C(C2H5)2 (iii) tetra – tert-butylmethane solution (i) 2, 2, 4, 4-Tetramethylpentane (ii) 3, 3-Dimethylpentane (iii) 3,3-Di-tert-butyl -2, 2, 4, 4 - tetramethylpentane If it is important to write the correct IUPAC name for a given structur e, it is equally Unit 9.indd 299 10/10/2022 10:37:53 AM 2024-25 300 chemistry iii) Attach ethyl group at carbon 3 and two methyl groups at carbon 2 C1 – 2C – 3C – 4C – 5C iv) Satisfy the valence of each carbon atom by putting requisite number of hydrogen atoms : CH3 – C – CH – CH2 – CH3 Thus we arrive at the correct structure. If you have understood writing of structure from the given name, attempt the following pr oblems. Problem 9.4 Write structural formulas of the following compounds : (i) 3, 4, 4, 5–Tetramethylheptane (ii) 2,5-Dimethyhexane Solution (i) CH3 – CH2 – CH – C – CH– CH – CH3 (ii) CH3 – CH – CH2 – CH2 – CH – CH3 Problem 9.5Write structures for each of the following compounds.
Why are the given names incorrect? Write correct IUPAC names. (i) 2-Ethylpentane (ii) 5-Ethyl – 3-methylheptane Solution(i) CH3 – CH – CH2– CH2 – CH3 Longe st chain is of six carbon atoms and not that of five. Hence, correct name is 3-Methylhexane. 7 6 5 4 3 2 1 (ii) CH3 – CH2 – CH – CH2 – CH – CH2 – CH3 Numbering is to be started from the end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5-methylheptane. 9.2.2 Preparation Petroleum and natural gas are the main sources of alkanes. However, alkanes can be prepared by following methods : 1. From unsaturated hydrocarbons Dihydrogen gas adds to alkenes and alkynes in the presence of finely divided catalysts like platinum, palladium or nickel to form alkanes. This process is called hydrogenation.
These metals adsorb dihydrogen gas on their surfaces and activate the hydrogen – hydrogen bond. Platinum and palladium catalyse the reaction at room temperature but relatively higher temperature and pressure are required with nickel catalysts. CH 2 =CH2 + H2 Pt/Pd/Ni CH3−CH3 Ethene Propane (9.1) CH2–CH=CH2 + H2 Pt/Pd/Ni CH3−CH2CH3 Propane Propane (9.2) CH3–C≡ C–H + 2H Pt/Pd/Ni CH3−CH2CH3 Propyne Propane (9.3) 2. From alkyl halides i) Alkyl halides (except fluorides) on reduction with zinc and dilute hydrochloric acid give alkanes. CH–C1+H 2 Zn,H+ CH4+HC1 (9.4) Chloromethane Methane Unit 9.indd 300 11/10/2022 15:25:29 2024-25 301 Hydrocarbons C2H5–C1+H2 Zn,H+ C2H6+HC1 Chloroethane Ethane (9.5) CH3CH2CH2C1 + H2 Zn,H+ CH3CH2CH3+CH1 1-Chloropropane Propane (9.6) ii) Alkyl halides on treatment with sodium metal in dry ethereal (free from moistur e) solution give higher alkanes. This reaction is known as Wurtz reaction and is used for the preparation of higher alkanes containing even number of carbon atoms. CH 3Br+2Na+BrCH3 dry ether CH3+2Na Bromomenthane Ethane (9.7) C2H5Br+2Na+BrC2H5 dry ether C2H5–C2H Bromoethane n–Butane (9.8) What will happen if two different alkyl halides are taken? 3. Fr om carboxylic acids i) Sodium salts of carboxylic acids on heating with soda lime (mixture of sodium hydroxide and calcium oxide) give alkanes containing one carbon atom less than the carboxylic acid. This process of elimination of carbon dioxide from a carboxylic acid is known as decarboxylation.
CH3COO– N a++NaOH ∆CaO CH4+Na2CO3 Sodium ethanoate Problem 9.6 Sodium salt of which acid will be needed for the preparation of propane ? Write chemical equation for the r eaction. Solution Butanoic acid, CH3CH2CH2COO–Na++ NaOHCaO CH3CH2CH3+Na2CO3 ii) Kolbe’s electrolytic method: An aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis gives alkane containing even number of carbon atoms at the anode. + 32 3 3 22Electrolysts2CH COO Na 2H O Sodium acetate CH CH 2CO H 2NaOH−+ ↓ − + ++ (9.9) The reaction is supposed to follow the following path : O i) 2CH3COO–Na+ 2CH3 – C – O–+2Na+ ii) At anode: O O 2CH3 –C–O– –02e– 2CH3 – C – 2C. H3+2CO2↑ Acetate ion Acetate Methyl free free radical radical iii) H3C + CH3 H3C–CH3↑ iv) At cathode : H2O+e–→–OH+ 2→H2↑ Methane cannot be prepared by this method. Why? 9.2.3 Properties Physical properties Alkanes are almost non-polar molecules because of the covalent nature of C-C and C-H bonds and due to very little difference of electronegativity between carbon and hydrogen atoms. They possess weak van der Waals forces. Due to the weak forces, the first four members, C 1 to C4 are gases, C5 to C17 are liquids and those containing 18 carbon atoms or more are solids at 298 K. They are colourless and odourless. What do you think about solubility of alkanes in water based upon non-polar nature of alkanes?
Petrol is a mixture of hydrocarbons and is used as a fuel for automobiles. Petrol and lower fractions of 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 alkanes) is non- Unit 9.indd 301 03-05-2024 17:16:25 2024-25 302 chemistry polar and, hence, hydrophobic in nature. It is generally observed that in relation to solubility of substances in solvents, polar substances are soluble in polar solvents, whereas the non-polar ones in non-polar solvents i.e., like dissolves like. Boiling point (b.p.) of different alkanes are given in Table 9.2 from which it is clear that there is a steady increase in boiling point with increase in molecular mass. This is due to the fact that the intermolecular van der Waals forces increase with increase of the molecular size or the surface area of the molecule. You can make an interesting observation by having a look on the boiling points of three isomeric pentanes viz., (pentane, 2-methylbutane and 2,2-dimethylpropane). It is observed (Table 9.2) that pentane having a continuous chain of five carbon atoms has the highest boiling point (309.1K) whereas 2,2 – dimethylpropane boils at 282.5K.
With increase in number of branched chains, the molecule attains the shape of a sphere. This results in smaller area of contact and therefore weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperatures. Chemical properties As already mentioned, alkanes are generally inert towards acids, bases, oxidising and reducing agents. However, they undergo the following reactions under certain conditions. 1. Substitution reactions One or more hydrogen atoms of alkanes can be replaced by halogens, nitro group and sulphonic acid group. Halogenation takes place either at higher temperature (573-773 K) or in the presence of diffused sunlight or ultraviolet light. Lower alkanes do not undergo nitration and sulphonation reactions. These reactions in which hydrogen atoms of alkanes are substituted are known as substitution reactions. As an example, chlorination of methane is given below: Halogenation CH 2 + C1 hv CH3C1 + HC1 Chloromethane (9.
10) CH3C1 + hv CH2 C12 + HC1 Dichloromethane (9. 11) CH2C12 hv CHC13 + HC1 Trichloromethane (9. 12) CHC13 + C12 hv CC14 + HC1 Tetrachloromethane (9. 13) table 9.2 Variation of Melting Point and boiling Point in alkanes Molecular n ame Molecular b.p./(K) m.p./(K) formula mass/u CH4 Methane 16 111.0 90.5 C2H6 Ethane 30 184.4 101.0 C3H8 Propane 44 230.9 85.3 C4H10 Butane 58 272.4 134.6 C4H10 2-Methylpropane 58 261.0 114.7 C5H12 Pentane 72 309.1 143.3 C5H12 2-Methylbutane 72 300.9 113.1 C5H12 2,2-Dimethylpropane 72 282.5 256.4 C6H14 Hexane 86 341.9 178.5 C7H16 Heptane 100 371.4 182.4 C8H18 Octane 114 398.7 216.2 C9H20 Nonane 128 423.8 222.0 C10H22 Decane 142 447.1 243.3 C20H42 Eicosane 282 615.0 236.2 Unit 9.indd 302 10/10/2022 10:37:54 AM 2024-25 303 Hydrocarbons CH3–CH3 + C12 hv CH3–CH2C1 + HC1 Chloroethane (9. 14) It is found that the rate of reaction of alkanes with halogens is F2 > Cl2 > Br2 > I2. Rate of replacement of hydrogens of alkanes is : 3° > 2° > 1°. Fluorination is too violent to be controlled. Iodination is very slow and a reversible reaction. It can be carried out in the presence of oxidizing agents like HIO 3 or HNO3. CH4+I2 CH3I+HI (9.15) HIO3+5HI→312+3H2O (9.16) Halogenation is supposed to proceed via free radical chain mechanism involving thr ee steps namely initiation, propagation and termination as given below: Mechanism (i) Initiation : The r eaction is initiated by homolysis of chlorine molecule in the presence of light or heat.
The Cl–Cl bond is weaker than the C–C and C–H bond and hence, is easiest to break. C1–C1 hv homolysis C. H3 + C1 Chlorine free radicals (ii) Propagation : Chlorine free radical attacks the methane molecule and takes the reaction in the forward direction by breaking the C-H bond to generate methyl free radical with the formation of H-Cl. (a) CH 4 + C+ 1 hv C+ H3 + H–C1 The methyl radical thus obtained attacks the second molecule of chlorine to for m CH3 – Cl with the liberation of another chlorine free radical by homolysis of chlorine molecule. (b) CH 3 + C1–C1 hv CH3 – C1 + C1 The chlorine and methyl free radicals generated above r epeat steps (a) and (b) respectively and thereby setup a chain of reactions. The propagation steps (a) and (b) are those which directly give principal products, but many other propagation steps are possible and may occur. Two such steps given below explain how more highly haloginated products are formed. CH 3C1 + C. 1 → C. H2C1 + HC1 C. H2C1 + C1– C1 → CH2C12 + C. 1 (iii) Termination: The r eaction stops after some time due to consumption of reactants and / or due to the following side reactions : The possible chain terminating steps are: (a) C. 1 + C. 1 → C1–C1 (b) H3 C. + C. H3 → H3C– CH3 (c) H3 C. 1 + C. 1 → H3C–C1 Though in (c), CH3 – Cl, the one of the products is formed but free radicals are consumed and the chain is terminated. The above mechanism helps us to understand the reason for the formation of ethane as a byproduct during chlorination of methane. 2. Combustion Alkanes on heating in the presence of air or dioxygen are completely oxidized to carbon dioxide and water with the evolution of large amount of heat.
42 2 2 è -1 c 4 10 2 2 2 è -1 cCH (g) 20 (g) CO (g) 2H O(1); Ä H 890kJmol (9.17) C H (g) 13/2O (g) 4CO (g) 5H O(1) Ä H 2875.84kJmol (9.18)+→ + − + →+ =− The general combustion equation for any alkane is : n 2n+2 2 2 23n 1C H O nCO (n 1)H O2+ + → ++ (9.19) Due to the evolution of large amount of heat during combustion, alkanes ar e used as fuels. During incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. Unit 9.indd 303 10/10/2022 10:37:54 AM 2024-25 304 chemistry CH4(g) + O2(g) incomplete combustion C(s)+2H2 O(1) (9.20) 3. Controlled oxidation Alkanes on heating with a regulated supply of dioxygen or air at high pressur e and in the presence of suitable catalysts give a variety of oxidation products. (i) 2CH4 + O2 Cu/523K/100atm 2CH3 OH Methanol (9.21) (ii) CH4 + O2 Mo2O3 ∆ HCHO + H2O Methanal (9.22) (iii) 2CH3CH3 + 3O2 (CH3COO)Mn ∆ 2CH3COOH Ethanoic acid + 2H2O (9.23) (iv) Ordinarily alkanes resist oxidation but alkanes having tertiary H atom can be oxidized to corr esponding alcohols by potassium permanganate. (iCH3)3 CH KMnO4 Oxidation (CH3)3 COH 2-Methylpropane 2-Methylpropane-2-01 (9.24) 4. Isomerisation n-Alkanes on heating in the presence of anhydr ous aluminium chloride and hydrogen chloride gas isomerise to branched chain alkanes. Major products are given below. Some minor products are also possible which you can think over. Minor products are generally not reported in organic reactions.
CH3(CH)2)4CH3Anhy, AICI3/ HCI n-Hexane CH 3CH–(CH2)2–CH3+CH3CH2–CH–CH2–CH3 CH3 CH3 2-Methylpentane 3-Methylpenatone (9.25) 5. Aromatization n-Alkanes having six or more carbon atoms on heating to 773K at 10-20 atmospheric pressure in the presence of oxides of vanadium, molybdenum or chromium supported over alumina get dehydrogenated and cyclised to benzene and its homologues. This reaction is known as aromatization or reforming. (9.26) Toluene (C7H8) is methyl derivative of benzene. Which alkane do you suggest for preparation of toluene ? 6. Reaction with steam Methane reacts with steam at 1273 K in the presence of nickel catalyst to form carbon monoxide and dihydrogen. This method is used for industrial preparation of dihydrogen gas CH 4 + H2IO Ni ∆ CO + 3H2 (9.27) 7. Pyrolysis Higher alkanes on heating to higher temperature decompose into lower alkanes, alkenes etc. Such a decomposition reaction into smaller fragments by the application of heat is called pyrolysis or cracking.
(9.28) Pyrolysis of alkanes is believed to be a free radical reaction. Preparation of oil gas or petr ol gas from kerosene oil or petrol involves the principle of pyrolysis. For example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of platinum, palladium or nickel gives a mixture of heptane and pentene. C 12H26 Pt/Pd/Ni 973K C7H16 + C5H10 + Other Products Dodecane Heptane Pentene (9.29) Unit 9.indd 304 03-05-2024 16:28:29 2024-25 305 Hydrocarbons 9.2.4 confor mations Alkanes contain carbon-carbon sig ma (σ ) bonds. Electron distribution of the sigma molecular orbital is symmetrical around the internuclear axis of the C–C bond which is not disturbed due to rotation about its axis. This permits free rotation about C–C single bond. This rotation results into different spatial arrangements of atoms in space which can change into one another. Such spatial arrangements of atoms which can be converted into one another by rotation around a C-C single bond are called conformations or conformers or rotamers. Alkanes can thus have infinite number of conformations by rotation around C-C single bonds. However, it may be remembered that rotation around a C-C single bond is not completely free.
It is hindered by a small energy barrier of 1-20 kJ mol –1 due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Conformations of ethane : Et hane molecule (C2H6) contains a carbon – carbon single bond with each carbon atom attached to three hydrogen atoms. Considering the ball and stick model of ethane, keep one carbon atom stationary and rotate the other 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 .1. Sawhorse projections In this projection, the molecule is viewed along the molecular axis. It is then projected on paper by drawing the central C–C bond as a somewhat longer straight line. Upper end of the line is slightly tilted towards right or left hand side. The front carbon is shown at the lower end of the line, whereas the rear carbon is shown at the upper end. Each carbon has three lines attached to it corresponding to three hydrogen atoms. The lines are inclined at an angle of 120° to each other. Sawhorse projections of eclipsed and staggered conformations of ethane are depicted in Fig. 9.2. 2.
Newman projections In this projection, the molecule is viewed at the C–C bond head on. The carbon atom nearer to the eye is represented by a point. Three hydrogen atoms attached to the front carbon atom are shown by three lines drawn at an angle of 120° to each other. The rear carbon atom (the carbon atom away from the eye) is represented by a circle and the three hydrogen atoms are shown attached to it by the shorter lines drawn at an angle of 120° to each other. The Newman’s projections are depicted in Fig. 9.3. Fig. 9.2 Sawhorse projections of ethane Fig. 9.3 Newman’s projections of ethane Unit 9.indd 305 10/10/2022 10:37:54 AM 2024-25 306 chemistry Fig. 9.4 Orbital picture of ethene depicting σ bonds onlyRelative stability of conformations: As mentioned earlier, in staggered form of ethane, the electron clouds of carbon-hydrogen bonds are as far apart as possible.
Thus, there are minimum repulsive forces, minimum energy and maximum stability of the molecule. On the other hand, when the staggered form changes into the eclipsed form, the electron clouds of the carbon – hydrogen bonds come closer to each other resulting in increase in electron cloud repulsions. To check the increased repulsive forces, molecule will have to possess more energy and thus has lesser stability. As already mentioned, the repulsive interaction between the electron clouds, which affects stability of a conformation, is called torsional strain. Magnitude of torsional strain depends upon the angle of rotation about C–C bond. This angle is also called dihedral angle or torsional angle. Of all the conformations of ethane, the staggered form has the least torsional strain and the eclipsed form, the maximum torsional strain. Therefore, staggered conformation is more stable than the eclipsed conformation. Hence, molecule largely remains in staggered conformation or we can say that it is preferred conformation. Thus it may be inferred that rotation around C–C bond in ethane is not completely free.
The energy difference between the two extreme forms is of the order of 12.5 kJ mol –1, which is very small. Even at ordinary temperatures, the ethane molecule gains thermal or kinetic energy sufficient enough to overcome this energy barrier of 12.5 kJ mol –1 through intermolecular collisions. Thus, it can be said that rotation about carbon-carbon single bond in ethane is almost free for all practical purposes. It has not been possible to separate and isolate different conformational isomers of ethane. 9.3 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 C nH2n. Alkenes are also known as olefins (oil forming) since the first member, ethylene or ethene (C2H4) was found to form an oily liquid on reaction with chlorine. 9.3.1 structure of Double Bond Carbon-carbon double bond in alkenes consists of one strong sigma ( σ) bond (bond enthalpy about 397 kJ mol–1) due to head-on overlapping of sp2 hybridised orbitals and one weak pi (π) bond (bond enthalpy about 284 kJ mol –1) obtained by lateral or sideways overlapping of the two 2p orbitals of the two carbon atoms.
The double bond is shorter in bond length (134 pm) than the C–C single bond (154 pm). You have already read that the pi (π) bond is a weaker bond due to poor sideways overlapping between the two 2p orbitals. Thus, the presence of the pi (π) bond makes alkenes behave as sources of loosely held mobile electrons. Therefore, alkenes are easily attacked by reagents or compounds which are in search of electrons. Such reagents are called electrophilic reagents. The presence of weaker π -bond makes alkenes unstable molecules in comparison to alkanes and thus, alkenes can be changed into single bond compounds by combining with the electrophilic reagents. Strength of the double bond (bond enthalpy, 681 kJ mol –1) is greater than that of a carbon-carbon single bond in ethane (bond enthalpy, 348 kJ mol –1). Orbital diagrams of ethene molecule are shown in Figs. 9.4 and 9.5. 9.3.2 nomenclature For nomenclature of alkenes in IUPAC system, the longest chain of carbon atoms containing the double bond is selected.
Numbering of the chain is done from the end which is nearer to Unit 9.indd 306 11/10/2022 15:26:35 2024-25 307 Hydrocarbons the double bond. The suffix ‘ene’ replaces ‘ane ’ of alkanes. It may be remembered that first member of alkene series is: CH2 (replacing n by 1 in CnH2n) known as methene but has a very short life. As already mentioned, first stable member of alkene series is C 2H4 known as ethylene (common) or ethene (IUPAC). IUPAC names of a few members of alkenes are given below : structure iUPac name CH3 – CH = CH2 Propene CH3 – CH2 – CH = CH2 But – l - ene CH3 – CH = CH–CH3 But-2-ene CH2 = CH – CH = CH2 Buta – 1,3 - diene CH2 = C – CH3 2-Methylprop-1-ene | CH3 CH2 = CH – CH – CH3 3-Methylbut-1-ene | CH3 Problem 9.7 Write IUPAC names of the following compounds: (i) (CH3)2CH – CH = CH – CH2 – CH  CH3 – CH – CH | C2H5 (ii) (iii) CH2 = C (CH2CH2CH3)2 (iv) CH3 CH2 CH2 CH2 CH2CH3 | | CH3 – CHCH = C – CH2 – CHCH3 | CH3solution (i) 2,8-Dimethyl-3, 6-decadiene; (ii) 1,3,5,7 Octatetraene; (iii) 2-n-Pr opylpent-1-ene; (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene; Problem 9.8 Calculate number of sigma (σ ) and pi (π) bonds in the above structures (i-iv). solution σ bonds : 33, π bonds : 2 σ bonds : 17, π bonds : 4 σ bonds : 23, π bond : 1 σ bonds : 41, π bond : 1 9.3.3 isomerism Alkenes show both structural isomerism and geometrical isomerism. Structural isomerism : As in alkanes, ethene (C2H4) and propene (C3H6) can have only one structure but alkenes higher than propene have different structures. Alkenes possessing C 4H8 as molecular formula can be written in the following three ways: I. 1 2 3 4 CH2 = CH – CH2 – CH3 But-1-ene (C4H8) II. 1 2 3 4 CH3 – CH = CH – CH3 But-2-ene (C4H8)Fig.
9.5 Orbital picture of ethene showing formation of (a) π -bond, (b) π-cloud and (c) bond angles and bond lengths Unit 9.indd 307 10/10/2022 10:37:55 AM 2024-25 308 chemistry III. 1 2 3 CH2 = C – CH3 | CH3 2-Methylprop-1-ene (C4H8) Structures I and III, and II and III are the examples of chain isomerism whereas structures I and II are position isomers. Problem 9.9 Write structures and IUPAC names of different structural isomers of alkenes corresponding to C 5H10. Solution (a) CH2 = CH – CH2 – CH2 – CH3 Pent-1-ene (b) CH3 – CH=CH – CH2 – CH3 Pent-2-ene (c) CH3 – C = CH – CH3 | CH3 2-Methylbut-2-ene (d) CH3 – CH – CH = CH2 | 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 gr oups. 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 : In (a), the two identical atoms i.e., both the X or both the Y lie on the same side of the 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 different geometry of (a) and (b) i.e. disposition of atoms or groups in space in the two arrangements is different. Therefore, they are stereoisomers. They would have the same geometry if atoms or groups around C=C bond can be rotated but rotation around C=C bond is not free.