Hydrocarbons PDF - Unit 13

HYDROCARBONS 373 UNIT 13 HYDROCARBONS Hydrocarbons are the important sources of...

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 IUPAC system of nomenclature; compounds of carbon and hydrogen only. Hydrocarbons recognise and write structures play a key role in our daily life. You must be familiar with of isomers of alkanes, alkenes, the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the alkynes and aromatic abbreviated form of liquified petroleum gas whereas CNG hydrocarbons; stands for compressed natural gas. Another term ‘LNG’ learn about various methods of preparation of hydrocarbons; (liquified natural gas) is also in news these days. This is distinguish between alkanes, also a fuel and is obtained by liquifaction of natural gas. alkenes, alkynes and aromatic Petrol, diesel and kerosene oil are obtained by the fractional hydrocarbons on the basis of distillation of petroleum found under the earth’s crust. physical and chemical properties; Coal gas is obtained by the destructive distillation of coal. draw and differentiate between Natural gas is found in upper strata during drilling of oil various conformations of ethane; wells. The gas after compression is known as compressed appreciate the role of natural gas. LPG is used as a domestic fuel with the least hydrocarbons as sources of pollution. Kerosene oil is also used as a domestic fuel but energy and for other industrial it causes some pollution. Automobiles need fuels like petrol, applications; predict the formation of the diesel and CNG. Petrol and CNG operated automobiles addition products of cause less pollution. All these fuels contain mixture of unsymmetrical alkenes and hydrocarbons, which are sources of energy. Hydrocarbons alkynes on the basis of electronic are also used for the manufacture of polymers like mechanism; polythene, polypropene, polystyrene etc. Higher comprehend the structure of hydrocarbons are used as solvents for paints. They are also benzene, explain aromaticity used as the starting materials for manufacture of many and understand mechanism dyes and drugs. Thus, you can well understand the of electrophilic substitution importance of hydrocarbons in your daily life. In this unit, reactions of benzene; you will learn more about hydrocarbons. predict the directive influence of substituents in monosubstituted 13.1 CLASSIFICATION benzene ring; learn about carcinogenicity and Hydrocarbons are of different types. Depending upon the toxicity. types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated 2022-23 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 C nH 2n+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 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. Fig. 13.1 Structure of methane In alkanes, tetrahedra are joined together 13.2 ALKANES in which C-C and C-H bond lengths are As already mentioned, alkanes are saturated 154 pm and 112 pm respectively (Unit 12). You open chain hydrocarbons containing have already read that C–C and C–H σ bonds 3 carbon - carbon single bonds. Methane (CH4) are formed by head-on overlapping of sp is the first member of this family. Methane is a hybrid orbitals of carbon and 1s orbitals of gas found in coal mines and marshy places. If hydrogen atoms. you replace one hydrogen atom of methane by 13.2.1 Nomenclature and Isomerism carbon and join the required number of hydrogens to satisfy the tetravalence of the You have already read about nomenclature other carbon atom, what do you get? You get of different classes of organic compounds in C 2 H 6. This hydrocarbon with molecular Unit 12. Nomenclature and isomerism in formula C2H6 is known as ethane. Thus you alkanes can further be understood with the can consider C2H6 as derived from CH4 by help of a few more examples. Common names replacing one hydrogen atom by -CH3 group. are given in parenthesis. First three alkanes Go on constructing alkanes by doing this – methane, ethane and propane have only theoretical exercise i.e., replacing hydrogen one structure but higher alkanes can have atom by –CH3 group. The next molecules will more than one structure. Let us write be C3H8, C4H10 … structures for C4H10. Four carbon atoms of C4H10 can be joined either in a continuous chain or with a branched chain in the following two ways : I 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 Butane (n- butane), (b.p. 273 K) 2022-23 HYDROCARBONS 375 II structures, they are known as structural isomers. 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. 2-Methylpropane (isobutane) Thus, you have seen that C4H10 and C5H12 (b.p.261 K) have two and three chain isomers respectively. In how many ways, you can join five Problem 13.1 carbon atoms and twelve hydrogen atoms of Write structures of different chain isomers C5H12? They can be arranged in three ways as of alkanes corresponding to the molecular shown in structures III–V formula C6H14. Also write their IUPAC III names. Solution (i) CH3 – CH2 – CH2 – CH2– CH2– CH3 n-Hexane Pentane (n-pentane) (b.p. 309 K) IV 2-Methylpentane 3-Methylpentane 2-Methylbutane (isopentane) (b.p. 301 K) 2,3-Dimethylbutane V 2,2 - Dimethylbutane Based upon the number of carbon atoms attached to a carbon atom, the carbon atom is 2,2-Dimethylpropane (neopentane) termed as primary (1°), secondary (2°), tertiary (b.p. 282.5 K) (3°) or quaternary (4°). Carbon atom attached Structures I and II possess same to no other carbon atom as in methane or to molecular formula but differ in their boiling only one carbon atom as in ethane is called points and other properties. Similarly primary carbon atom. Terminal carbon atoms structures III, IV and V possess the same are always primary. Carbon atom attached to molecular formula but have different two carbon atoms is known as secondary. properties. Structures I and II are isomers of Tertiary carbon is attached to three carbon butane, whereas structures III, IV and V are atoms and neo or quaternary carbon is isomers of pentane. Since difference in attached to four carbon atoms. Can you identify properties is due to difference in their 1°, 2°, 3° and 4° carbon atoms in structures I 2022-23 376 CHEMISTRY to V ? If you go on constructing structures for compounds. These groups or substituents are higher alkanes, you will be getting still larger known as alkyl groups as they are derived from number of isomers. C6H14 has got five isomers alkanes by removal of one hydrogen atom. and C7H16 has nine. As many as 75 isomers General formula for alkyl groups is CnH2n+1 are possible for C10H22. (Unit 12). In structures II, IV and V, you observed Let us recall the general rules for that –CH3 group is attached to carbon atom nomenclature already discussed in Unit 12. numbered as 2. You will come across groups Nomenclature of substituted alkanes can like –CH3, –C2H5, –C3H7 etc. attached to carbon further be understood by considering the atoms in alkanes or other classes of following problem: 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 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 2022-23 HYDROCARBONS 377 Table 13.1 Nomenclature of a Few Organic Compounds Structure and IUPAC Name Remarks Lowest sum and 1 2 3 4 5 6 (a) CH3– CH – CH2 – CH – CH2 – CH3 alphabetical (4 – Ethyl – 2 – methylhexane) arrangement Lowest sum and 8 (b) CH3 – 7CH2 – 6CH2 – 5CH – 4CH – 3 C – 2CH2 – 1CH3 alphabetical arrangement (3,3-Diethyl-5-isopropyl-4-methyloctane) sec is not considered 1 (c) CH3–2CH2–3CH2–4CH–5CH–6CH2–7CH2–8CH2–9CH2–10CH3 while arranging alphabetically; isopropyl is taken 5-sec– Butyl-4-isopropyldecane as one word 1 (d) CH3–2CH2–3CH2–4CH2–5CH–6CH2–7CH2–8CH2–9CH3 Further numbering to the substituents of the side chain 5-(2,2– Dimethylpropyl)nonane 1 (e) CH3 – 2CH2 – 3CH – 4CH2 – 5CH – 6CH2 – 7CH3 Alphabetical priority order 3–Ethyl–5–methylheptane Problem 13.3 important to write the correct structure from the given IUPAC name. To do this, first of all, Write IUPAC names of the following the longest chain of carbon atoms compounds : corresponding to the parent alkane is written. (i) (CH3)3 C CH2C(CH3)3 Then after numbering it, the substituents are (ii) (CH3)2 C(C2H5)2 attached to the correct carbon atoms and finally (iii) tetra – tert-butylmethane valence of each carbon atom is satisfied by putting the correct number of hydrogen atoms. Solution This can be clarified by writing the structure (i) 2, 2, 4, 4-Tetramethylpentane of 3-ethyl-2, 2–dimethylpentane in the (ii) 3, 3-Dimethylpentane following steps : (iii) 3,3-Di-tert-butyl -2, 2, 4, 4 - i) Draw the chain of five carbon atoms: tetramethylpentane C–C–C–C–C If it is important to write the correct IUPAC ii) Give number to carbon atoms: 1 2 3 4 5 name for a given structure, it is equally C –C –C –C –C 2022-23 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. | 7 6 5 4 3 2 1 1 2 3 4 5 C – C– C– C– C (ii) CH3 – CH2 – CH – CH2 – CH – CH2 – CH3 | | CH 3 C2 H5 iv) Satisfy the valence of each carbon atom by putting requisite number of hydrogen Numbering is to be started from the end atoms : which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5- CH3 | methylheptane. CH3 – C – CH – CH2 – CH3 | | 13.2.2 Preparation CH3 C2H5 Petroleum and natural gas are the main Thus we arrive at the correct structure. If sources of alkanes. However, alkanes can be you have understood writing of structure from prepared by following methods : the given name, attempt the following problems. 1. From unsaturated hydrocarbons Dihydrogen gas adds to alkenes and alkynes Problem 13.4 in the presence of finely divided catalysts like Write structural formulas of the following platinum, palladium or nickel to form alkanes. compounds : This process is called hydrogenation. These metals adsorb dihydrogen gas on their surfaces (i) 3, 4, 4, 5–Tetramethylheptane and activate the hydrogen – hydrogen bond. (ii) 2,5-Dimethyhexane Platinum and palladium catalyse the reaction at room temperature but relatively higher Solution temperature and pressure are required with nickel catalysts. Pt /Pd/Ni (i) CH3 – CH2 – CH – C – CH– CH – CH3 CH2 = CH2 + H2 → CH3 − CH3 (13.1) Ethene Ethane Pt/Pd/Ni CH3 − CH = CH2 + H2  →CH3 − CH2 − CH3 Propene Propane (ii) CH3 – CH – CH2 – CH2 – CH – CH3 (13.2) Problem 13.5 Pt/Pd/Ni CH3 − C ≡ C − H + 2H2  → CH3 − CH2 − CH3 Write structures for each of the following compounds. Why are the given names Propyne Propane incorrect? Write correct IUPAC (13.3) names. 2. From alkyl halides (i) 2-Ethylpentane i) Alkyl halides (except fluorides) on (ii) 5-Ethyl – 3-methylheptane reduction with zinc and dilute hydrochloric Solution acid give alkanes. (i) CH3 – CH – CH2– CH2 – CH3 + Zn, H CH 3 − Cl + H 2   → CH 4 + HCl (13.4) Chloromethane Methane 2022-23 HYDROCARBONS 379 + Zn, H C 2 H 5 − Cl + H 2   → C 2 H 6 + HCl containing even number of carbon atoms Chloroethane Ethane (13.5) at the anode. Zn,H+ 2CH3 COO− Na+ + 2H2 O CH3 CH2CH2 Cl + H2 → CH3 CH2 CH3 + HCl Sodium acetate 1-Chloropropane Propane (13.6) ↓ Electrolysis CH3 − CH3 + 2CO2 + H2 + 2NaOH (13.9) ii) Alkyl halides on treatment with sodium metal in dry ethereal (free from moisture) The reaction is supposed to follow the solution give higher alkanes. This reaction following path : is known as Wurtz reaction and is used O for the preparation of higher alkanes || − − containing even number of carbon i) 2CH3 COO Na +
2CH3 − C − O + 2Na + atoms. dry ether ii) At anode: CH3Br +2Na +BrCH3  →CH3 −CH3 +2NaBr O O Bromomethane Ethane || || − (13.7) –2e 2CH3 −C −O  – →2CH3 − C−O:  →2CH3 + 2CO2 ↑ dry ether C2 H5 Br + 2Na + BrC2H5  →C2 H5 −C2 H5 Acetate ion Acetate Methyl free Bromoethane n-Butane free radical radical (13.8) iii) H3 C + CH3  → H3C −CH3 ↑ What will happen if two different alkyl halides are taken? iv) At cathode : 3. From carboxylic acids H2O + e– → – OH + H i) Sodium salts of carboxylic acids on heating 2H  → H2 ↑ with soda lime (mixture of sodium Methane cannot be prepared by this hydroxide and calcium oxide) give alkanes method. Why? containing one carbon atom less than the carboxylic acid. This process of elimination 13.2.3 Properties of carbon dioxide from a carboxylic acid is Physical properties known as decarboxylation. Alkanes are almost non-polar molecules – because of the covalent nature of C-C and C-H CH3 COO Na+ + NaOH →CH4 + Na2 CO3 CaO ∆ bonds and due to very little difference of Sodium ethanoate electronegativity between carbon and hydrogen atoms. They possess weak van der Problem 13.6 Waals forces. Due to the weak forces, the first Sodium salt of which acid will be needed four members, C1 to C4 are gases, C5 to C17 are for the preparation of propane ? Write liquids and those containing 18 carbon atoms chemical equation for the reaction. or more are solids at 298 K. They are colourless and odourless. What do you think about Solution solubility of alkanes in water based upon non- Butanoic acid, polar nature of alkanes? Petrol is a mixture of − CaO CH3 CH2 CH2 COO Na + + NaOH → hydrocarbons and is used as a fuel for automobiles. Petrol and lower fractions of CH3 CH2 CH3 + Na2 CO3 petroleum are also used for dry cleaning of clothes to remove grease stains. On the basis ii) Kolbe’s electrolytic method An aqueous of this observation, what do you think about solution of sodium or potassium salt of a the nature of the greasy substance? You are carboxylic acid on electrolysis gives alkane correct if you say that grease (mixture of higher 2022-23 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, 1. Substitution reactions whereas the non-polar ones in non-polar solvents i.e., like dissolves like. One or more hydrogen atoms of alkanes can be replaced by halogens, nitro group and Boiling point (b.p.) of different alkanes are sulphonic acid group. Halogenation takes given in Table 13.2 from which it is clear that 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 hν CH4 + Cl2  → CH3 Cl + HCl continuous chain of five carbon atoms has the highest boiling point (309.1K) whereas Chloromethane (13.10) 2,2 – dimethylpropane boils at 282.5K. With hν increase in number of branched chains, the CH 3 Cl + Cl2  → CH 2 Cl2 + HCl molecule attains the shape of a sphere. This Dichloromethane (13.11) results in smaller area of contact and therefore weak intermolecular forces between spherical hν CH2 Cl2 + Cl2  → CHCl3 + HCl molecules, which are overcome at relatively Trichloromethane (13.12) lower temperatures. Chemical properties hν CHCl3 + Cl2  → CCl4 + HCl As already mentioned, alkanes are generally Tetrachloromethane (13.13) inert towards acids, bases, oxidising and Table 13.2 Variation of Melting Point and Boiling Point in Alkanes Molecular Name 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 2022-23 HYDROCARBONS 381 hν and may occur. Two such steps given below CH3 -CH3 + Cl2  → CH3 − CH2 Cl + HCl explain how more highly haloginated products Chloroethane (13.14) are formed. It is found that the rate of reaction of alkanes CH3 Cl + Cl → CH2Cl + HCl with halogens is F2 > Cl2 > Br2 > I2. Rate of replacement of hydrogens of alkanes is : CH2 Cl + Cl − Cl → CH 2Cl 2 + Cl 3° > 2° > 1°. Fluorination is too violent to be (iii) Termination: The reaction stops after controlled. Iodination is very slow and a some time due to consumption of reactants reversible reaction. It can be carried out in the and / or due to the following side reactions : presence of oxidizing agents like HIO3 or HNO3. The possible chain terminating steps are : CH4 + I2
CH3 I + HI (13.15) (a) Cl + Cl → Cl − Cl HIO3 + 5HI → 3I2 + 3H2 O (13.16) (b) H3 C + CH3 → H3 C − CH3 Halogenation is supposed to proceed via free radical chain mechanism involving three (c) H3 C + Cl → H3 C − Cl steps namely initiation, propagation and termination as given below: Though in (c), CH3 – Cl, the one of the products is formed but free radicals are Mechanism consumed and the chain is terminated. The (i) Initiation : The reaction is initiated by above mechanism helps us to understand the homolysis of chlorine molecule in the presence reason for the formation of ethane as a of light or heat. The Cl–Cl bond is weaker than 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 hν Cl − Cl  → homolysis Cl + Cl dioxygen are completely oxidized to carbon Chlorine free radicals dioxide and water with the evolution of large amount of heat. (ii) Propagation : Chlorine free radical attacks the methane molecule and takes the reaction CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(l); in the forward direction by breaking the C-H ∆c H V = − 890 kJ mol−1 bond to generate methyl free radical with the (13.17) formation of H-Cl. C4 H10 (g)+13/2 O2 (g) → 4CO2 (g) +5H2O(l); hν ∆c HV =−2875.84 kJ mol−1 ( a) CH4 + Cl  →CH + H − Cl 3 (13.18) The methyl radical thus obtained attacks The general combustion equation for any the second molecule of chlorine to form alkane is : CH3 – Cl with the liberation of another chlorine  3n +1 free radical by homolysis of chlorine molecule. Cn H2n+2 +  O2 → nCO2 + (n +1) H2 O  2  hν (b) C H 3 + Cl − Cl  → CH3 − Cl + C l (13.19) Chlorine Due to the evolution of large amount of free radical heat during combustion, alkanes are used as fuels. The chlorine and methyl free radicals generated above repeat steps (a) and (b) During incomplete combustion of respectively and thereby setup a chain of alkanes with insufficient amount of air or reactions. The propagation steps (a) and (b) are dioxygen, carbon black is formed which is those which directly give principal products, used in the manufacture of ink, printer ink, but many other propagation steps are possible black pigments and as filters. 2022-23 382 CHEMISTRY Incomplete pressure in the presence of oxides of CH4 (g) + O2 (g)  combustion→C(s) + 2H2 O(l) vanadium, molybdenum or chromium (13.20) supported over alumina get dehydrogenated and cyclised to benzene and its homologues. 3. Controlled oxidation This reaction is known as aromatization or Alkanes on heating with a regulated supply of reforming. dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products. Cu/523K/100atm (i) 2CH4 + O2 → 2CH3 OH Methanol (13.21) Mo2O3 (ii) CH4 + O2 → HCHO + H2 O (13.26) ∆ Methanal Toluene (C7 H8) is methyl derivative of (13.22) benzene. Which alkane do you suggest for (CH3COO)2 Mn preparation of toluene ? (iii) 2CH3 CH3 + 3O2  → 2 CH3 COOH ∆ Ethanoic acid 6. Reaction with steam + 2H2 O Methane reacts with steam at 1273 K in the (13.23) presence of nickel catalyst to form carbon monoxide and dihydrogen. This method is (iv) Ordinarily alkanes resist oxidation but used for industrial preparation of dihydrogen alkanes having tertiary H atom can be gas oxidized to corresponding alcohols by Ni potassium permanganate. CH4 + H2O  → CO + 3H2 (13.27) ∆ KMnO (CH3 )3 CH → 4 Oxidation (CH3 )3 COH 7. Pyrolysis 2-Methylpropane 2-Methylpropan-2-ol Higher alkanes on heating to higher (13.24) temperature decompose into lower alkanes, alkenes etc. Such a decomposition reaction 4. Isomerisation into smaller fragments by the application of n-Alkanes on heating in the presence of heat is called pyrolysis or cracking. anhydrous 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. (13.28) Anhy. AlCl3 /HCl Pyrolysis of alkanes is believed to be a CH3 (CH2 )4 CH3 → free radical reaction. Preparation of oil gas or n -Hexane petrol gas from kerosene oil or petrol involves CH3CH −(CH2 )2 − CH3 + CH3 CH2 −CH −CH2 −CH3 the principle of pyrolysis. For example, | | dodecane, a constituent of kerosene oil on CH3 CH3 heating to 973K in the presence of platinum, 2-Methylpentane 3-Methylpentane palladium or nickel gives a mixture of heptane (13.25) and pentene. 5. Aromatization C12 H26  973K → C7 H16 + C5H10 + other Pt/Pd/Ni products n-Alkanes having six or more carbon atoms Dodecane Heptane Pentene on heating to 773K at 10-20 atmospheric (13.29) 2022-23 HYDROCARBONS 383 13.2.4 Conformations 1. Sawhorse projections Alkanes contain carbon-carbon sigma (σ) In this projection, the molecule is viewed along bonds. Electron distribution of the sigma the molecular axis. It is then projected on paper molecular orbital is symmetrical around the by drawing the central C–C bond as a internuclear axis of the C–C bond which is somewhat longer straight line. Upper end of not disturbed due to rotation about its axis. the line is slightly tilted towards right or left This permits free rotation about C–C single hand side. The front carbon is shown at the bond. This rotation results into different lower end of the line, whereas the rear carbon spatial arrangements of atoms in space which is shown at the upper end. Each carbon has can change into one another. Such spatial three lines attached to it corresponding to three arrangements of atoms which can be hydrogen atoms. The lines are inclined at an converted into one another by rotation around angle of 120° to each other. Sawhorse projections a C-C single bond are called conformations of eclipsed and staggered conformations of or conformers or rotamers. Alkanes can thus ethane are depicted in Fig. 13.2. 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 1-20 kJ mol due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Conformations of ethane : Ethane Fig. 13.2 Sawhorse projections of ethane molecule (C2H6) contains a carbon – carbon 2. Newman projections single bond with each carbon atom attached to three hydrogen atoms. Considering the In this projection, the molecule is viewed at the ball and stick model of ethane, keep one C–C bond head on. The carbon atom nearer to carbon atom stationary and rotate the other the eye is represented by a point. Three carbon atom around the C-C axis. This hydrogen atoms attached to the front carbon rotation results into infinite number of spatial atom are shown by three lines drawn at an arrangements of hydrogen atoms attached to angle of 120° to each other. The rear carbon one carbon atom with respect to the hydrogen atom (the carbon atom away from the eye) is atoms attached to the other carbon atom. represented by a circle and the three hydrogen These are called conformational isomers atoms are shown attached to it by the shorter (conformers). Thus there are infinite number lines drawn at an angle of 120° to each other. of conformations of ethane. However, there are The Newman’s projections are depicted in two extreme cases. One such conformation in Fig. 13.3. 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. Fig. 13.3 Newman’s projections of ethane 2022-23 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 13.3.1 Structure of Double Bond are as far apart as possible. Thus, there are 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 –1 other hand, when the staggered form changes enthalpy about 397 kJ mol ) due to head-on 2 into the eclipsed form, the electron clouds of overlapping of sp hybridised orbitals and one the carbon – hydrogen bonds come closer to weak pi (π) bond (bond enthalpy about 284 kJ –1 each other resulting in increase in electron mol ) 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 weaker π-bond makes alkenes unstable maximum torsional strain. Therefore, molecules in comparison to alkanes and thus, staggered conformation is more stable than the alkenes can be changed into single bond eclipsed conformation. Hence, molecule largely compounds by combining with the remains in staggered conformation or we can electrophilic reagents. Strength of the double say that it is preferred conformation. Thus it –1 bond (bond enthalpy, 681 kJ mol ) is greater may be inferred that rotation around C–C bond than that of a carbon-carbon single bond in in ethane is not completely free. The energy –1 ethane (bond enthalpy, 348 kJ mol ). Orbital difference between the two extreme forms is of –1 diagrams of ethene molecule are shown in the order of 12.5 kJ mol , which is very small. Figs. 13.4 and 13.5. Even at ordinary temperatures, the ethane molecule gains thermal or kinetic energy sufficient enough to overcome this energy –1 barrier of 12.5 kJ mol 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. 13.3 ALKENES Alkenes are unsaturated hydrocarbons Fig. 13.4 Orbital picture of ethene depicting containing at least one double bond. What σ bonds only should be the general formula of alkenes? If there is one double bond between two carbon atoms 13.3.2 Nomenclature in alkenes, they must possess two hydrogen For nomenclature of alkenes in IUPAC system, atoms less than alkanes. Hence, general formula the longest chain of carbon atoms containing for alkenes is CnH2n. Alkenes are also known as the double bond is selected. Numbering of the olefins (oil forming) since the first member, chain is done from the end which is nearer to 2022-23 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’ of alkanes. It may be remembered that first Solution member of alkene series is: CH2 (replacing n (i) 2,8-Dimethyl-3, 6-decadiene; by 1 in CnH2n) known as methene but has a (ii) 1,3,5,7 Octatetraene; very short life. As already mentioned, first (iii) 2-n-Propylpent-1-ene; stable member of alkene series is C2H4 known as ethylene (common) or ethene (IUPAC). (iv) 4-Ethyl-2,6-dimethyl-dec-4-ene; IUPAC names of a few members of alkenes are Problem 13.8 given below : Calculate number of sigma (σ) and pi (π) Structure IUPAC name bonds in the above structures (i-iv). CH3 – CH = CH2 Propene CH3 – CH2 – CH = CH2 But – l - ene Solution CH3 – CH = CH–CH3 But-2-ene σ bonds : 33, π bonds : 2 CH2 = CH – CH = CH2 Buta – 1,3 - diene σ bonds : 17, π bonds : 4 CH2 = C – CH3 2-Methylprop-1-ene σ bonds : 23, π bond : 1 | σ bonds : 41, π bond : 1 CH3 CH2 = CH – CH – CH3 3-Methylbut-1-ene 13.3.3 Isomerism | Alkenes show both structural isomerism and CH3 geometrical isomerism. Structural isomerism : As in alkanes, ethene Problem 13.7 (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 (i) (CH3)2CH – CH = CH – CH2 – CH C4H8 as molecular formula can be written in y the following three ways: CH3 – CH – CH | I. 1 2 3 4 C2H5 CH2 = CH – CH2 – CH3 (ii) But-1-ene (C4H8) (iii) CH2 = C (CH2CH2CH3)2 (iv) CH3 CH2 CH2 CH2 CH2CH3 II. 1 2 3 4 | | CH3 – CH = CH – CH3 CH3 – CHCH = C – CH2 – CHCH3 | But-2-ene CH3 (C4H8) 2022-23 386 CHEMISTRY III. 1 2 3 In (a), the two identical atoms i.e., both the CH2 = C – CH3 X or both the Y lie on the same side of the | double bond but in (b) the two X or two Y lie CH3 across the double bond or on the opposite 2-Methyprop-1-ene sides of the double bond. This results in different geometry of (a) and (b) i.e. disposition (C4H8) of atoms or groups in space in the two Structures I and III, and II and III are the arrangements is different. Therefore, they are examples of chain isomerism whereas stereoisomers. They would have the same structures I and II are position isomers. geometry if atoms or groups around C=C bond can be rotated but rotation around C=C bond Problem 13.9 is not free. It is restricted. For understanding Write structures and IUPAC names of this concept, take two pieces of strong different structural isomers of alkenes cardboards and join them with the help of two corresponding to C5H10. nails. Hold one cardboard in your one hand Solution and try to rotate the other. Can you really rotate the other cardboard ? The answer is no. The (a) CH2 = CH – CH2 – CH2 – CH3 rotation is restricted. This illustrates that the Pent-1-ene restricted rotation of atoms or groups around (b) CH3 – CH=CH – CH2 – CH3 the doubly bonded carbon atoms gives rise to Pent-2-ene different geometries of such compounds. The stereoisomers of this type are called (c) CH3 – C = CH – CH3 geometrical isomers. The isomer of the type | (a), in which two identical atoms or groups lie CH3 on the same side of the double bond is called 2-Methylbut-2-ene cis isomer and the other isomer of the type (d) CH3 – CH – CH = CH2 (b), in which identical atoms or groups lie on | the opposite sides of the double bond is called CH3 trans isomer. Thus cis and trans isomers 3-Methylbut-1-ene have the same structure but have different configuration (arrangement of atoms or groups (e) CH2 = C – CH2 – CH3 in space). Due to different arrangement of | atoms or groups in space, these isomers differ CH3 in their properties like melting point, boiling 2-Methylbut-1-ene point, dipole moment, solubility etc. Geometrical or cis-trans isomers of but-2-ene Geometrical isomerism: Doubly bonded are represented below : 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 2022-23 HYDROCARBONS 387 trans-but-2-ene is non-polar. This can be (ii) CH2 = CBr2 understood by drawing geometries of the two forms as given below from which it is clear that (iii) C6H5CH = CH – CH3 in the trans-but-2-ene, the two methyl groups (iv) CH3CH = CCl CH3 are in opposite directions, Threfore, dipole moments of C-CH3 bonds cancel, thus making Solution the trans form non-polar. (iii) and (iv). In structures (i) and (ii), two identical groups are attached to one of the doubly bonded carbon atom. 13.3.4 Preparation 1. From alkynes: Alkynes on partial reduction with calculated amount of cis-But-2-ene trans-But-2-ene dihydrogen in the presence of palladised (µ = 0.33D) (µ = 0) charcoal partially deactivated with poisons In the case of solids, it is observed that like sulphur compounds or quinoline give the trans isomer has higher melting point alkenes. Partially deactivated palladised than the cis form. charcoal is known as Lindlar’s catalyst. Alkenes thus obtained are having cis Geometrical or cis-trans isomerism geometry. However, alkynes on reduction is also shown by alkenes of the types XYC = CXZ and XYC = CZW with sodium in liquid ammonia form trans alkenes. Problem 13.10 Draw cis and trans isomers of the following compounds. Also write their IUPAC names : (i) CHCl = CHCl (13.30) (ii) C2H5CCH3 = CCH3C2H5 Solution (13.31) Pd/C iii) CH ≡ CH + H2  → CH2 = CH2 (13.32) Ethyne Ethene Pd/C iv) CH3 − C ≡ CH + H2  →CH3 − CH = CH2 Propyne Propene (13.33) Will propene thus obtained show geometrical isomerism? Think for the Problem 13.11 reason in support of your answer. Which of the following compounds will show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X) on heating with alcoholic potash (i) (CH3)2C = CH – C2H5 (potassium hydroxide dissolved in alcohol, 2022-23 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 β-elimination reaction, since hydrogen atom is eliminated from the β carbon atom (carbon atom next to the carbon to which halogen is attached). (13.37) 13.3.5 Properties Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and (13.34) difference in polar nature. The first three members are gases, the next fourteen are Nature of halogen atom and the alkyl liquids and the higher ones are solids. Ethene group determine rate of the reaction. It is is a colourless gas with a faint sweet smell. All observed that for halogens, the rate is: other alkenes are colourless and odourless, iodine > bromine > chlorine, while for alkyl insoluble in water but fairly soluble in non- groups it is : tert > secondary > primary. polar solvents like benzene, petroleum ether. 3. From vicinal dihalides: Dihalides in They show a regular increase in boiling point which two halogen atoms are attached to with increase in size i.e., every – CH2 group two adjacent carbon atoms are known as added increases boiling point by 20–30 K. Like vicinal dihalides. Vicinal dihalides on alkanes, straight chain alkenes have higher treatment with zinc metal lose a molecule boiling point than isomeric branched chain of ZnX2 to form an alkene. This reaction is compounds. known as dehalogenation. Chemical properties CH2 Br − CH2 Br + Zn  →CH2 = CH2 + ZnBr2 Alkenes are the rich source of loosely held (13.35) pi (π) electrons, due to which they show addition reactions in which the electrophiles CH3 CHBr − CH2 Br + Zn  → CH3CH = CH2 add on to the carbon-carbon double bond to + ZnBr2 form the addition products. Some reagents also add by free radical mechanism. There are (13.36) cases when under special conditions, alkenes 4. From alcohols by acidic dehydration: also undergo free radical substitution You have read during nomenclature of reactions. Oxidation and ozonolysis reactions different homologous series in Unit 12 that are also quite prominent in alkenes. A brief alcohols are the hydroxy derivatives of description of different reactions of alkenes is alkanes. They are represented by R–OH given below: where, R is CnH2n+1. Alcohols on heating 1. Addition of dihydrogen: Alkenes add up with concentrated sulphuric acid form one molecule of dihydrogen gas in the alkenes with the elimination of one water presence of finely divided nickel, palladium molecule. Since a water molecule is or platinum to form alkanes (Section 13.2.2) eliminated from the alcohol molecule in the presence of an acid, this reaction is known 2. Addition of halogens : Halogens like as acidic dehydration of alcohols. This bromine or chlorine add up to alkene to reaction is also the example of form vicinal dihalides. However, iodine β-elimination reaction since –OH group does not show addition reaction under 2022-23 HYDROCARBONS 389 normal conditions. The reddish orange colour of bromine solution in carbon tetrachloride is discharged when bromine adds up to an unsaturation site. This reaction is used as a test for unsaturation. Addition of halogens to alkenes is an example of electrophilic addition reaction involving cyclic halonium ion formation (13.42) which you will study in higher classes. Markovnikov, a Russian chemist made a generalisation in 1869 after studying such reactions in detail. These generalisations led Markovnikov to frame a rule called Markovnikov rule. The rule states that (13.38) negative part of the addendum (adding molecule) gets attached to that carbon atom (ii) CH3 − CH = CH2 + Cl − Cl →CH3 − CH − CH2 which possesses lesser number of hydrogen | | atoms. Thus according to this rule, product I Cl Cl i.e., 2-bromopropane is expected. In actual Propene 1,2-Dichloropropane practice, this is the principal product of the (13.39) reaction. This generalisation of Markovnikov 3. Addition of hydrogen halides: rule can be better understood in terms of Hydrogen halides (HCl, HBr,HI) add up to mechanism of the reaction. alkenes to form alkyl halides. The order of Mechanism reactivity of the hydrogen halides is + Hydrogen bromide provides an electrophile, H , HI > HBr > HCl. Like addition of halogens which attacks the double bond to form to alkenes, addition of hydrogen halides is carbocation as shown below : 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 alkenes Addition reactions of HBr to symmetrical (a) less stable (b) more stable alkenes (similar groups attached to double primary carbocation secondary carbocation bond) take place by electrophilic addition mechanism. (i) The secondary carbocation (b) is more stable than the primary carbocation (a), CH2 = CH2 + H – Br  →CH3 – CH2 – Br (13.40) therefore, the former predominates because CH3 –CH = CH –CH3 + HBr  →CH3 –CH2 – CHCH3 it is formed at a faster rate. – | (ii) The carbocation (b) is attacked by Br ion Br to form the product as follows : (13.41) Addition reaction of HBr to unsymmetrical alkenes (Markovnikov Rule) How will H – Br add to propene ? The two 2-Bromopropane possible products are I and II. (major product) 2022-23 390 CHEMISTRY Anti Markovnikov addition or peroxide effect or Kharash effect In the presence of peroxide, addition of HBr to unsymmetrical alkenes like propene takes place contrary to the Markovnikov rule. This happens only with HBr but not with HCl and The secondary free radical obtained in the Hl. This addition reaction was observed above mechanism (step iii) is more stable than by M.S. Kharash and F.R. Mayo in 1933 at the primary. This explains the formation of the University of Chicago. This reaction 1-bromopropane as the major product. It may is known as peroxide or Kharash effect be noted that the peroxide effect is not observed or addition reaction anti to Markovnikov in addition of HCl and HI. This may be due rule. to the fact that the H–Cl bond being –1 (C H CO) O stronger (430.5 kJ mol ) than H–Br bond CH3 –CH = CH2 + HBr  6 5 2 2 →CH3 –CH2 –1 (363.7 kJ mol ), is not cleaved by the free x radical, whereas the H–I bond is weaker –1 CH2 Br (296.8 kJ mol ) and iodine free radicals 1-Bromopropane combine to form iodine molecules instead of (13.43) adding to the double bond. Mechanism : Peroxide effect proceeds via free radical chain mechanism as given below: Problem 13.12 Write IUPAC names of the products (i) obtained by addition reactions of HBr to hex-1-ene (i) in the absence of peroxide and (ii) in the presence of peroxide. Solution Homolysis (ii) C6 H5 + H – Br  →C6 H6 + 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. 2022-23 HYDROCARBONS 391 ketones and/or acids depending upon the nature of the alkene and the experimental conditions (13.49) (13.44) KMnO4 /H+ CH3 – CH = CH – CH3  →2CH3COOH But -2-ene Ethanoic acid (13.50) 7. Ozonolysis : Ozonolysis of alkenes involves 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 (13.45) the position of the double bond in alkenes or other unsaturated compounds. 5. Addition of water : In the presence of a few drops of concentrated sulphuric acid alkenes react with water to form alcohols, in accordance with the Markovnikov rule. (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. (13.52) (13.47) 8. Polymerisation: You are familiar with polythene bags and polythene sheets. Polythene is obtained by the combination of large number of ethene molecules at high temperature, high pressure and in the presence of a catalyst. The large molecules (13.48) thus obtained are called polymers. This b) Acidic potassium permanganate or acidic reaction is known as polymerisation. The potassium dichromate oxidises alkenes to simple compounds from which polymers 2022-23 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 High temp./pressure position of the triple bond is indicated by the n(CH2 = CH2 ) Catalyst → —( CH2 –CH2 — )n first triply bonded carbon. Common and Polythene IUPAC names of a few members of alkyne series (13.53) are given in Table 13.2. High temp./pressure You have already learnt that ethyne and n(CH3 – CH = CH2 )  Catalyst → — ( CH– CH2 — )n propyne have got only one structure but there | are two possible structures for butyne – CH3 Polypropene (i) but-1-yne and (ii) but-2-yne. Since these two compounds differ in their structures due to the (13.54) position of the triple bond, they are known as Polymers are used for the manufacture of position isomers. In how many ways, you can plastic bags, squeeze bottles, refrigerator dishes, construct the structure for the next homologue toys, pipes, radio and T.V. cabinets etc. i.e., the next alkyne with molecular formula Polypropene is used for the manufacture of milk C5H8? Let us try to arrange five carbon atoms crates, plastic buckets and other moulded with a continuous chain and with a side chain. articles. Though these materials have now Following are the possible structures : become common, excessive use of polythene Structure IUPAC name and polypropylene is a matter of great concern for all of us. 1 2 3 4 5 I. HC ≡ C– CH – CH – CH Pent–1-yne 2 2 3 13.4 ALKYNES 1 2 3 4 5 II. H C–C ≡ C– CH – CH Pent–2-yne 3 2 3 Like alkenes, alkynes are also unsaturated hydrocarbons. They contain at least one triple 4 3 2 1 III. H C– CH – C ≡ CH 3-Methyl but–1-yne 3 bond between two carbon atoms. The number | of hydrogen atoms is still less in alkynes as CH3 compared to alkenes or alkanes. Their general formula is CnH2n–2. Structures I and II are position isomers and structures I and III or II and III are chain The first stable member of alkyne series isomers. is ethyne which is popularly known as acetylene. Acetylene is used for arc welding Problem 13.13 purposes in the form of oxyacetylene flame Write structures of different isomers obtained by mixing acetylene with oxygen gas. th corresponding to the 5 member of Alkynes are starting materials for a large alkyne series. Also write IUPAC names of number of organic compounds. Hence, it is all the isomers. What type of isomerism is interesting to study this class of organic exhibited by different pairs of isomers? compounds. 13.4.1 Nomenclature and Isomerism Solution th In common system, alkynes are named as 5 member of alkyne has the molecular derivatives of acetylene. In IUPAC system, they 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 2022-23 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. orbitals of the other carbon atom, which undergo lateral or sideways overlapping to 3,3-Dimethylbut-1-yne form two pi (π) bonds between two carbon atoms. Thus ethyne molecule consists of one Position and chain isomerism shown by C–C σ bond, two C–H σ bonds and two C–C π different pairs. bonds. The strength of C≡C bond (bond -1 enthalpy 823 kJ mol ) is more than those of –1 13.4.2 Structure of Triple Bond C=C bond (bond enthalpy 681 kJ mol ) and –1 Ethyne is the simplest molecule of alkyne C–C bond (bond enthalpy 348 kJ mol ). The series. Structure of ethyne is shown in C≡C bond length is shorter (120 pm) than those Fig. 13.6. of C=C (133 pm) and C–C (154 pm). Electron cloud between two carbon atoms is Each carbon atom of ethyne has two sp cylindrically symmetrical about the hybridised orbitals. Carbon-carbon sigma (σ) internuclear axis. Thus, ethyne is a linear bond is obtained by the head-on overlapping molecule. of the two sp hybridised orbitals of the two carbon atoms. The remaining sp hybridised 13.4.3 Preparation orbital of each carbon atom undergoes 1. From calcium carbide: On industrial overlapping along the internuclear axis with scale, ethyne is prepared by treating calcium the 1s orbital of each of the two hydrogen atoms carbide with water. Calcium carbide is forming two C-H sigma bonds. H-C-C bond prepared by heating quick lime with coke. angle is of 180°. Each carbon has two Quick lime can be obtained by heating unhybridised p orbitals which are limestone as shown in the following perpendicular to each other as well as to the reactions: plane of the C-C sigma bond. The 2p orbitals of one carbon atom are parallel to the 2p CaCO3 ∆→ CaO + CO2 (13.55) 2022-23 394 CHEMISTRY atoms in ethyne are attached to the sp CaO + 3C → CaC2 + CO (13.56) hybridised carbon atoms whereas they are Calcium 2 attached to sp hybridised carbon atoms in carbide 3 ethene and sp hybridised carbons in ethane. CaC2 + 2H2O  → Ca(OH)2 + C2 H2 (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 2 to a greater extent than that of the sp hydrogen halide is eliminated to form hybridised orbitals of carbon in ethene and the alkenyl halide which on treatment with 3 sp 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. – HC ≡ CH + Na → HC ≡ C Na + + ½H2 13.4.4 Properties Monosodium Physical properties ethynide Physical properties of alkynes follow the same trend of alkenes and alkanes. First three (13.59) members are gases, the next eight are liquids – – and the higher ones are solids. All alkynes are HC ≡ C – Na + + Na → Na + C ≡ C Na + + ½H 2 colourless. Ethyene has characteristic odour. Disodium ethynide Other members are odourless. Alkynes are (13.60) weakly polar in nature. They are lighter than water and immiscible with water but soluble CH3 – C ≡ C − H + Na + NH2– in organic solvents like ethers, carbon ↓ tetrachloride and benzene. Their melting point, boiling point and density increase with CH3 – C ≡ C – Na + + NH3 (13.61) increase in molar mass. Sodium propynide Chemical properties These reactions are not shown by alkenes Alkynes show acidic nature, addition reactions and alkanes, hence used for distinction and polymerisation reactions as follows : between alkynes, alkenes and alkanes. What A. Acidic character of alkyne: Sodium about the above reactions with but-1-yne and metal and sodamide (NaNH2) are strong bases. but-2-yne ? Alkanes, alkenes and alkynes They react with ethyne to form sodium follow the following trend in their acidic acetylide with the liberation of dihydrogen gas. behaviour : These reactions have not been observed in case i) HC ≡ CH > H2 C = CH2 > CH3 –CH3 of ethene and ethane thus indicating that ethyne is acidic in nature in comparison to ii) HC ≡ CH > CH3 – C ≡ CH >> CH3 –C ≡ C – CH3 ethene and ethane. Why is it so ? Has it B. Addition reactions: Alkynes contain a something to do with their structures and the triple bond, so they add up, two molecules of hybridisation ? You have read that hydrogen dihydrogen, halogen, hydrogen halides etc. 2022-23 HYDROCARBONS 395 Formation of the addition product takes place according to the following steps. The addition product formed depends upon stability of vinylic cation. Addition in unsymmetrical alkynes takes place according to Markovnikov rule. Majority of the reactions of alkynes are the examples of electrophilic (13.66) addition reactions. A few addition reactions are (iv) Addition of water given below: Like alkanes and alkenes, alkynes are also (i) Addition of dihydrogen immiscible and do not react with water. Pt/Pd/Ni H2 However, one molecule of water adds to alkynes HC ≡ CH+ H2  →[H2C = CH2 ] →CH3 –CH3 on warming with mercuric sulphate and dilute (13.62) sulphuric acid at 333 K to form carbonyl Pt/Pd/Ni compounds. CH3 – C ≡ CH + H2 →[CH3 – CH = CH2 ] Propyne Propene ↓ H2 CH3 – CH2 – CH3 Propane (13.63) (ii) Addition of halogens (13.67) (13.64) Reddish orange colour of the solution of bromine in carbon tetrachloride is decolourised. This is used as a test for unsaturation. (13.68) (iii) Addition of hydrogen halides (v) Polymerisation Two molecules of hydrogen halides (HCl, HBr, HI) add to alkynes to form gem dihalides (in (a) Linear polymerisation: Under suitable which two halogens are attached to the same conditions, linear polymerisation of ethyne carbon atom) takes place to produce polyacetylene or H – C ≡ C – H + H – Br → [CH2 = CH – Br] → CHBr2 polyethyne which is a high molecular weight Bromoethene | polyene containing repeating units of CH3 (CH = CH – CH = CH ) and can be represented 1,1-Dibromoethane as —( CH = CH – CH = CH)— n Under special (13.65) conditions, this polymer conducts electricity. 2022-23 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 (b) Cyclic polymerisation: Ethyne on passing a benzene ring but instead contain other highly through red hot iron tube at 873K undergoes unsaturated ring. Aromatic compounds cyclic polymerization. Three molecules containing benzene ring are known as polymerise to form benzene, which is the benzenoids and those not containing a starting molecule for the preparation of benzene ring are known as non-benzenoids. derivatives of benzene, dyes, drugs and large Some examples of arenes are given number of other organic compounds. This is below: the best route for entering from aliphatic to aromatic compounds as discussed below: Benzene Toluene Naphthalene (13.69) Problem 13.14 How will you convert ethanoic acid into Biphenyl benzene? 13.5.1 Nomenclature and Isomerism Solution The nomenclature and isomerism of aromatic hydrocarbons has already been discussed in 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