Hydrocarbons PDF

Hydrocarbons 295 Unit 9 Hydrocarbons...

Hydrocarbons 295 Unit 9 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 of isomers of alkanes, with the terms ‘LPG’ and ‘CNG’ used as fuels. LPG is the alkenes, alkynes and aromatic abbreviated form of liquified petroleum gas whereas CNG hydrocarbons; learn about various methods of stands for compressed natural gas. Another term ‘LNG’ 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 draw and differentiate between coal. Natural gas is found in upper strata during drilling various conformations of ethane; of oil wells. The gas after compression is known as appreciate the role of compressed natural gas. LPG is used as a domestic fuel hydrocarbons as sources of with the least pollution. Kerosene oil is also used as a energy and for other industrial applications; domestic fuel but it causes some pollution. Automobiles predict the formation of need fuels like petrol, diesel and CNG. Petrol and CNG the addition products of operated automobiles cause less pollution. All these fuels unsymmetrical alkenes and contain mixture of hydrocarbons, which are sources of alkynes on the basis of electronic energy. Hydrocarbons are also used for the manufacture mechanism; of polymers like polythene, polypropene, polystyrene etc. comprehend the structure of Higher hydrocarbons are used as solvents for paints. They benzene, explain aromaticity are also used as the starting materials for manufacture and understand mechanism of many dyes and drugs. Thus, you can well understand of electrophilic substitution the importance of hydrocarbons in your daily life. In this reactions of benzene; predict the directive influence of unit, you will learn more about hydrocarbons. substituents in monosubstituted 9.1 CLASSIFICATION benzene ring; learn about carcinogenicity and Hydrocarbons are of different types. Depending upon toxicity. the types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated 2024-25 Unit 9.indd 295 10/10/2022 10:37:52 AM  296 chemistry (ii) unsaturated and (iii) aromatic of the general formula for alkane family hydrocarbons. Saturated hydrocarbons or homologous series? If we examine the contain carbon-carbon and carbon-hydrogen formula of different alkanes we find that single bonds. If different carbon atoms are the general formula for alkanes is CnH2n+2. It joined together to form open chain of carbon represents any particular homologue when n atoms with single bonds, they are termed is given appropriate value. Can you recall the as alkanes as you have already studied in structure of methane? According to VSEPR Unit 8. 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. 9.1), 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. 9.1 Structure of methane In alkanes, tetrahedra are joined together 9.2 ALKANES in which C-C and C-H bond lengths are As already mentioned, alkanes are saturated 154 pm and 112 pm respectively (Unit 8). open chain hydr ocarbons containing You have already read that C–C and C–H σ carbon - carbon single bonds. Methane (CH4) bonds are formed by head-on overlapping of is the first member of this family. Methane is 3 sp hybrid orbitals of carbon and 1s orbitals a gas found in coal mines and marshy places. of hydrogen atoms. If you replace one hydrogen atom of methane by carbon and join the required number of 9.2.1 Nomenclature and Isomerism hydrogens to satisfy the tetravalence of the You have already read about nomenclature other carbon atom, what do you get? You of different classes of organic compounds get C2H6. This hydrocarbon with molecular in Unit 8. Nomenclature and isomerism formula C2H6 is known as ethane. Thus you in alkanes can further be understood with can consider C2H6 as derived from CH4 by the help of a few more examples. Common replacing one hydrogen atom by -CH3 group. names are given in parenthesis. First three Go on constructing alkanes by doing this alkanes – methane, ethane and propane have theoretical exercise i.e., replacing hydrogen only one structure but higher alkanes can atom by –CH3 group. The next molecules will have more than one structure. Let us write be C3H8, C4H10 … structures for C4H10. Four carbon atoms of H H H C4H10 can be joined either in a continuous replace any H by - CH3 chain or with a branched chain in the H—C—H H—C—C—H or C2H6 following two ways : H H H 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 Butane (n- butane), (b.p. 273 K) 2024-25 Unit 9.indd 296 11/10/2022 15:24:01  Hydrocarbons 297 II 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. Thus, you have seen that C4H10 2-Methylpropane (isobutane) and C5H12 have two and three chain isomers (b.p.261 K) respectively. In how many ways, you can join five Problem 9.1 carbon atoms and twelve hydrogen atoms of Write structures of different chain C5H12? They can be arranged in three ways isomers of alkanes corresponding to the as shown in structures III–V molecular formula C6H14. Also write their III IUPAC 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 molecular to no other carbon atom as in methane or to formula but differ in their boiling points and only one carbon atom as in ethane is called other properties. Similarly structures III, IV primary carbon atom. Terminal carbon and V possess the same molecular formula atoms are always primary. Carbon atom but have different properties. Structures I and attached to two carbon atoms is known as II are isomers of butane, whereas structures secondary. Tertiary carbon is attached to III, IV and V are isomers of pentane. Since three carbon atoms and neo or quaternary difference in properties is due to difference in carbon is attached to four carbon atoms. Can their structures, they are known as structural you identify 1°, 2°, 3° and 4° carbon atoms in 2024-25 Unit 9.indd 297 10/10/2022 10:37:53 AM  298 chemistry structures I to V ? If you go on constructing compounds. These groups or substituents structures for higher alkanes, you will be are known as alkyl groups as they are derived getting still larger number of isomers. C6H14 from alkanes by removal of one hydrogen has got five isomers and C7H16 has nine. As atom. General formula for alkyl groups is many as 75 isomers are possible for C10H22. CnH2n+1 (Unit 8). 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 8. numbered as 2. You will come across groups Nomenclature of substituted alkanes can like –CH 3, –C 2H 5, –C 3H 7 etc. attached to further be understood by considering the carbon atoms in alkanes or other classes of following problem: Problem 9.2 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 2024-25 Unit 9.indd 298 11/10/2022 15:24:36  Hydrocarbons 299 Table 9.1 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 7 6 5 4 3 2 1 (b) CH3 – CH2 – CH2 – CH – CH – C – CH2 – CH3 alphabetical arrangement (3,3-Diethyl-5-isopropyl-4-methyloctane) 1 2 3 4 5 6 7 8 9 10 sec is not considered (c) CH3– CH2– CH2– CH– CH– CH2– CH2– CH2– CH2– CH3 while arranging alphabetically; isopropyl is taken 5-sec– Butyl-4-isopropyldecane as one word 1 2 3 4 5 6 7 8 9 (d) CH3– CH2– CH2– CH2– CH– CH2– CH2– CH2– CH3 Further numbering to the substituents of the side chain 5-(2,2– Dimethylpropyl)nonane 1 2 3 4 5 6 7 (e) CH3 – CH2 – CH – CH2 – CH – CH2 – CH3 Alphabetical priority order 3–Ethyl–5–methylheptane Problem 9.3 important to write the correct structure from the given IUPAC name. To do this, first Write IUPAC names of the following of all, 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 (iii) tetra – tert-butylmethane finally valence of each carbon atom is satisfied by putting the correct number of hydrogen Solution atoms. This can be clarified by writing the (i) 2, 2, 4, 4-Tetramethylpentane structure of 3-ethyl-2, 2–dimethylpentane in (ii) 3, 3-Dimethylpentane the 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 ii) Give number to carbon atoms: 1 2 3 4 5 IUPAC name for a given structure, it is equally C –C –C –C –C 2024-25 Unit 9.indd 299 10/10/2022 10:37:53 AM  300 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 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 iv) Satisfy the valence of each carbon atom by putting requisite number of hydrogen Numbering is to be started from the atoms : end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5- methylheptane. CH3 – C – CH – CH2 – CH3 9.2.2 Preparation Petroleum and natural gas are the main Thus we arrive at the correct structure. sources of alkanes. However, alkanes can be If you have understood writing of structure prepared by following methods : from the given name, attempt the following problems. 1. From unsaturated hydrocarbons Dihydrogen gas adds to alkenes and alkynes Problem 9.4 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 (i) 3, 4, 4, 5–Tetramethylheptane metals adsorb dihydrogen gas on their surfaces and activate the hydrogen – hydrogen (ii) 2,5-Dimethyhexane bond. Platinum and palladium catalyse the Solution reaction at room temperature but relatively higher temperature and pressure are required with nickel catalysts. (i) CH3 – CH2 – CH – C – CH– CH – Pt/Pd/Ni CH 2 =CH 2 + H 2 CH 3 −CH 3 CH3 EthenePropane (9.1) Pt/Pd/Ni CH 2–CH=CH 2 + H 2 CH 3−CH 2CH 3 PropanePropane     (ii) CH3 – CH – CH2 – CH2 – CH – CH3 (9.2) Pt/Pd/Ni Problem 9.5 CH 3–C≡ C–H + 2H CH 3−CH 2CH 3 Writ e stru ctu r es fo r each o f t h e PropynePropane following compounds. Why are the given (9.3) names incorrect? Write correct IUPAC names. 2. From alkyl halides (i) 2-Ethylpentane i) Alkyl halides (except fluorides) on reduction (ii) 5-Ethyl – 3-methylheptane with zinc and dilute hydrochloric acid give Solution alkanes. (i) CH3 – CH – CH2– CH2 – CH3 + Zn,H CH–C1+H2 CH4+HC1 (9.4) Chloromethane Methane 2024-25 Unit 9.indd 300 11/10/2022 15:25:29  Hydrocarbons 301 C2H5–C1+H2 Zn,H + C2H6+HC1 alkane containing even number of Chloroethane Ethane (9.5) carbon atoms at the anode. − + 2CH3COO Na + 2H2O CH3CH2CH3+CH1 + CH3CH2CH2C1 + H2 Zn,H 1-Chloropropane Propane Sodium acetate (9.6) ↓ Electrolysts ii) Alkyl halides on treatment with sodium CH3 − CH3 + 2CO2 + H2 + 2NaOH (9.9) metal in dry ethereal (free from moisture) solution give higher alkanes. This reaction The reaction is supposed to follow the is known as Wurtz reaction and is used following path : O for the preparation of higher alkanes – + – + containing even number of carbon i) 2CH3COO Na 2CH3 – C – O +2Na atoms. ii) At anode: CH3Br+2Na+BrCH3 dry ether CH3+2Na O O Bromomenthane Ethane – –02e–. (9.7) 2CH3 –C–O 2CH3 – C – 2CH3+2CO2↑ C2H5Br+2Na+BrC2H5 dry ether C2H5–C2H Acetate ion Acetate Methyl free free radical radical Bromoethane n–Butane iii) H3C + CH3 H3C–CH3↑ (9.8) iv) At cathode : What will happen if two different alkyl halides – are taken? H2O+e →–OH+ 2 →H2↑ 3. From carboxylic acids i) Sodium salts of carboxylic acids on Methane cannot be prepared by this heating with soda lime (mixture of sodium method. Why? hydroxide and calcium oxide) give alkanes 9.2.3 Properties containing one carbon atom less than the carboxylic acid. This process of elimination Physical properties of carbon dioxide from a carboxylic acid is Alkanes are almost non-polar molecules known as decarboxylation. because of the covalent nature of C-C and – + CaO C-H bonds and due to very little difference CH3COO Na +NaOH ∆ CH4+Na2CO3 of electronegativity between carbon and Sodium ethanoate hydrogen atoms. They possess weak van der Waals forces. Due to the weak forces, the first Problem 9.6 four members, C1 to C4 are gases, C5 to C17 Sodium salt of which acid will be needed are liquids and those containing 18 carbon for the preparation of propane ? Write atoms or more are solids at 298 K. They are chemical equation for the reaction. colourless and odourless. What do you think about solubility of alkanes in water based Solution upon non-polar nature of alkanes? Petrol Butanoic acid, is a mixture of hydrocarbons and is used – + CH3CH2CH2COO Na + NaOH CaO as a fuel for automobiles. Petrol and lower fractions of petroleum are also used for dry CH3CH2CH3+Na2CO3 cleaning of clothes to remove grease stains. On the basis of this observation, what do you think about the nature of the greasy ii) Kolbe’s electrolytic method: An aqueous substance? You are correct if you say that solution of sodium or potassium salt of grease (mixture of higher alkanes) is non- a carboxylic acid on electrolysis gives 2024-25 Unit 9.indd 301 03-05-2024 17:16:25  302 chemistry polar and, hence, hydrophobic in nature. It is reducing agents. However, they undergo generally observed that in relation to solubility the following reactions under certain of substances in solvents, polar substances conditions. are soluble in polar solvents, whereas the 1. Substitution reactions 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 Boiling point (b.p.) of different alkanes are and sulphonic acid group. Halogenation given in Table 9.2 from which it is clear that there is a steady increase in boiling point with takes 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 forces increase with increase of the molecular do 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 three isomeric pentanes viz., (pentane, chlorination of methane is given below: 2-methylbutane and 2,2-dimethylpropane). It Halogenation is observed (Table 9.2) that pentane having hv a continuous chain of five carbon atoms has CH2 + C1 CH3C1 + HC1 the highest boiling point (309.1K) whereas Chloromethane (9.10) 2,2 – dimethylpropane boils at 282.5K. With increase in number of branched chains, CH3C1 + hv CH2 C12 + HC1 the molecule attains the shape of a sphere. Dichloromethane (9.11) This results in smaller area of contact and therefore weak intermolecular forces between CH2C12 hv CHC13 + HC1 spherical molecules, which are overcome at Trichloromethane (9.12) relatively lower temperatures. hv Chemical properties CHC13 + C12 CC14 + HC1 As already mentioned, alkanes are generally Tetrachloromethane (9.13) inert towards acids, bases, oxidising and Table 9.2 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 2024-25 Unit 9.indd 302 10/10/2022 10:37:54 AM  Hydrocarbons 303 hv steps are possible and may occur. Two such CH3–CH3 + C12 CH3–CH2C1 + HC1 Chloroethane (9.14) steps given below explain how more highly haloginated products are formed. It is found that the rate of reaction of.. alkanes with halogens is F2 > Cl2 > Br2 > I2. CH3C1 + C  1 → C  H2C1 + HC1.. Rate of replacement of hydrogens of alkanes is : C  H2C1 + C1– C1 → CH2C12 + C  1 3° > 2° > 1°. Fluorination is too violent to be controlled. Iodination is very slow and a (iii) Termination: The reaction stops after reversible reaction. It can be carried out in the some time due to consumption of reactants presence of oxidizing agents like HIO3 or HNO3. and / or due to the following side reactions : The possible chain terminating steps are: CH4+I2 CH3I+HI(9.15).. (a) C  1 + C  1 → C1–C1 HIO3+5HI→312+3H2O(9.16).. (b) H3 C   + C   H3 → H3C– CH3.. Halogenation is supposed to proceed via (c) H3 C  1 + C  1 → H3C–C1 free radical chain mechanism involving three Though in (c), CH3 – Cl, the one of the steps namely initiation, propagation and products is formed but free radicals are termination as given below: consumed and the chain is terminated. The Mechanism above mechanism helps us to understand (i) Initiation : The reaction is initiated the reason for the formation of ethane as a by homolysis of chlorine molecule in the byproduct during chlorination of methane. presence of light or heat. The Cl–Cl bond 2. Combustion is weaker than the C–C and C–H bond and Alkanes on heating in the presence of air or hence, is easiest to break. dioxygen are completely oxidized to carbon hv. dioxide and water with the evolution of large C1–C1 homolysis C H3 + C1 amount of heat. Chlorine free radicals CH4 (g) + 202 (g) → CO2 (g) + 2H2O(1); (ii) Propagation : Chlorine free radical Äc Hè − 890kJ mol-1 attacks the methane molecule and takes the (9.17) reaction in the forward direction by breaking C4H10 (g) + 13/2O2 (g) → 4CO2 (g) + 5H2O(1) the C-H bond to generate methyl free radical with the formation of H-Cl. Äc Hè = −2875.84kJ mol-1 + (a) CH4 + C1 hv + C H3 + H–C1 (9.18) The general combustion equation for any The methyl radical thus obtained attacks alkane is : the second molecule of chlorine to form  3n + 1 CnH2n+2 +  O → nCO2 + (n + 1)H2O CH3 – Cl with the liberation of another chlorine  2  2 free radical by homolysis of chlorine molecule. (9.19) (b) CH3 + C1–C1 hv CH3 – C1 + C1 Due to the evolution of large amount of heat during combustion, alkanes are used as The chlorine and methyl free radicals fuels. generated above repeat steps (a) and (b) During incomplete combustion of alkanes respectively and thereby setup a chain of with insufficient amount of air or dioxygen, reactions. The propagation steps (a) and carbon black is formed which is used in (b) are those which directly give principal the manufacture of ink, printer ink, black products, but many other propagation pigments and as filters. 2024-25 Unit 9.indd 303 10/10/2022 10:37:54 AM  304 chemistry CH4(g) + O2(g) incomplete C(s)+2H2 O(1) pressure in the presence of oxides of vanadium, combustion molybdenum or chromium supported over (9.20) alumina get dehydrogenated and cyclised to 3. Controlled oxidation benzene and its homologues. This reaction is Alkanes on heating with a regulated supply known as aromatization or reforming. of dioxygen or air at high pressure and in the presence of suitable catalysts give a variety of oxidation products. (i) 2CH4 + O2 Cu/523K/100atm 2CH3 OH Methanol (9.21) Mo2O3 (9.26) (ii) CH4 + O2 HCHO + H2O ∆ Toluene (C 7H8) is methyl derivative of Methanal benzene. Which alkane do you suggest for preparation of toluene ? (9.22) (CH3COO)Mn 6. Reaction with steam (iii) 2CH3CH3 + 3O2 2CH3COOH ∆ Methane reacts with steam at 1273 K in the  Ethanoic acid presence of nickel catalyst to form carbon  + 2H2O monoxide and dihydrogen. This method is (9.23) used for industrial preparation of dihydrogen (iv) Ordinarily alkanes resist oxidation but gas alkanes having tertiary H atom can be Ni oxidized to corresponding alcohols by CH4 + H2IO CO + 3H2(9.27) ∆ potassium permanganate. KMnO4 7. Pyrolysis (iCH3)3 CH Oxidation (CH3)3 COH Higher alkanes on heating to higher 2-Methylpropane2-Methylpropane-2-01 temperature decompose into lower alkanes, alkenes etc. Such a decomposition reaction (9.24) into smaller fragments by the application of 4. Isomerisation heat is called pyrolysis or cracking. n-Alkanes on heating in the presence of 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 (9.28) generally not reported in organic reactions. Pyrolysis of alkanes is believed to be a free radical reaction. Preparation of oil gas CH3(CH)2)4CH3 Anhy, AICI / HCI3 or petrol gas from kerosene oil or petrol n-Hexane involves the principle of pyrolysis. For CH3CH–(CH2)2–CH3+CH3CH2–CH–CH2–CH3 example, dodecane, a constituent of kerosene oil on heating to 973K in the presence of CH3 CH3 platinum, palladium or nickel gives a mixture 2-Methylpentane 3-Methylpenatone of heptane and pentene. (9.25) 5. Aromatization C12H26 Pt/Pd/Ni 973K C7H16 + C5H10 + Other Products n-Alkanes having six or more carbon atoms Dodecane Heptane Pentene on heating to 773K at 10-20 atmospheric (9.29) 2024-25 Unit 9.indd 304 03-05-2024 16:28:29  Hydrocarbons 305 9.2.4 Conformations 1. Sawhorse projections Alkanes contain carbon-carbon sigma (σ) In this projection, the molecule is viewed bonds. Electron distribution of the sigma along the molecular axis. It is then projected molecular orbital is symmetrical around the on paper by drawing the central C–C bond internuclear axis of the C–C bond which is as a somewhat longer straight line. Upper not disturbed due to rotation about its axis. end of the line is slightly tilted towards This permits free rotation about C–C single right or left hand side. The front carbon is bond. This rotation results into different shown at the lower end of the line, whereas spatial arrangements of atoms in space the rear carbon is shown at the upper end. which can change into one another. Such Each carbon has three lines attached to it spatial arrangements of atoms which can corresponding to three hydrogen atoms. be converted into one another by rotation The lines are inclined at an angle of 120° to around a C-C single bond are called each other. Sawhorse projections of eclipsed conformations or conformers or rotamers. and staggered conformations of ethane are Alkanes can thus have infinite number of depicted in Fig. 9.2. 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 –1 energy barrier of 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. 9.2 Sawhorse projections of ethane molecule (C2H6) contains a carbon – carbon single bond with each carbon atom attached 2. Newman projections 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 carbon atom stationary and rotate the other to 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 Fig. 9.3. 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. Fig. 9.3 Newman’s projections of ethane 2024-25 Unit 9.indd 305 10/10/2022 10:37:54 AM  306 chemistry Relative stability of conformations: As the first member, ethylene or ethene (C2H4) mentioned earlier, in staggered form of ethane, was found to form an oily liquid on reaction the electron clouds of carbon-hydrogen bonds with chlorine. are as far apart as possible. Thus, there are 9.3.1 Structure of Double Bond minimum repulsive forces, minimum energy and maximum stability of the molecule. On the Carbon-carbon double bond in alkenes other hand, when the staggered form changes consists of one strong sigma (σ) bond (bond –1 into the eclipsed form, the electron clouds of the enthalpy about 397 kJ mol ) due to head-on 2 carbon – hydrogen bonds come closer to each overlapping of sp hybridised orbitals and other resulting in increase in electron cloud one weak pi (π) bond (bond enthalpy about –1 repulsions. To check the increased repulsive 284 kJ mol ) obtained by lateral or sideways forces, molecule will have to possess more overlapping of the two 2p orbitals of the two energy and thus has lesser stability. As already carbon atoms. The double bond is shorter in mentioned, the repulsive interaction between bond length (134 pm) than the C–C single the electron clouds, which affects stability of bond (154 pm). You have already read that a conformation, is called torsional strain. the pi (π) bond is a weaker bond due to poor Magnitude of torsional strain depends upon sideways overlapping between the two 2p the angle of rotation about C–C bond. This orbitals. Thus, the presence of the pi (π) bond angle is also called dihedral angle or torsional makes alkenes behave as sources of loosely angle. Of all the conformations of ethane, the held mobile electrons. Therefore, alkenes are easily attacked by reagents or compounds staggered form has the least torsional strain which are in search of electrons. Such and the eclipsed form, the maximum torsional reagents are called electrophilic reagents. strain. Therefore, staggered conformation is The presence of weaker π-bond makes alkenes more stable than the eclipsed conformation. unstable molecules in comparison to alkanes Hence, molecule largely remains in staggered and thus, alkenes can be changed into single conformation or we can say that it is preferred bond compounds by combining with the conformation. Thus it may be inferred that electrophilic reagents. Strength of the double rotation around C–C bond in ethane is not –1 bond (bond enthalpy, 681 kJ mol ) is greater completely free. The energy difference between than that of a carbon-carbon single bond in the two extreme forms is of the order of 12.5 –1 –1 ethane (bond enthalpy, 348 kJ mol ). Orbital kJ mol , which is very small. Even at ordinary diagrams of ethene molecule are shown in temperatures, the ethane molecule gains Figs. 9.4 and 9.5. thermal or kinetic energy sufficient enough to –1 overcome this energy 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. 9.3 Alkenes Alkenes are unsaturated hydrocarbons Fig. 9.4 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 9.3.2 Nomenclature atoms in alkenes, they must possess two For nomenclature of alkenes in IUPAC system, hydrogen atoms less than alkanes. Hence, the longest chain of carbon atoms containing general formula for alkenes is CnH2n. Alkenes the double bond is selected. Numbering of the are also known as olefins (oil forming) since chain is done from the end which is nearer to 2024-25 Unit 9.indd 306 11/10/2022 15:26:35  Hydrocarbons 307 Fig. 9.5 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 (i) 2,8-Dimethyl-3, 6-decadiene; n by 1 in CnH2n) known as methene but has (ii) 1,3,5,7 Octatetraene; a 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 Problem 9.8 are 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 9.3.3 Isomerism | CH3 Alkenes show both structural isomerism and geometrical isomerism. Structural isomerism : As in alkanes, ethene Problem 9.7 (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  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) 2024-25 Unit 9.indd 307 10/10/2022 10:37:55 AM  308 chemistry III. 1 2 3 In (a), the two identical atoms i.e., both CH2 = C – CH3 the X or both the Y lie on the same side | of the double bond but in (b) the two X or CH3 two Y lie across the double bond or on the 2-Methylprop-1-ene opposite sides of the double bond. This (C4H8) results in different geometry of (a) and (b) i.e. disposition of atoms or groups in space in Structures I and III, and II and III are the two arrangements is different. Therefore, the examples of chain isomerism whereas they are stereoisomers. They would have the structures I and II are position isomers. same geometry if atoms or groups around Problem 9.9 C=C bond can be rotated but rotation around C=C bond is not free. It is restricted. For Write structures and IUPAC names of understanding this concept, take two pieces different structural isomers of alkenes of strong cardboards and join them with the corresponding to C5H10. help of two nails. Hold one cardboard in your Solution one hand and try to rotate the other. Can (a) CH2 = CH – CH2 – CH2 – CH3 you really rotate the other cardboard ? The answer is no. The rotation is restricted. This Pent-1-ene illustrates that the restricted rotation of atoms (b) CH3 – CH=CH – CH2 – CH3 or groups around the doubly bonded carbon Pent-2-ene atoms gives rise to different geometries of (c) CH3 – C = CH – CH3 such compounds. The stereoisomers of this | type are called geometrical isomers. The CH3 isomer of the type (a), in which two identical 2-Methylbut-2-ene atoms or groups lie on the same side of the double bond is called cis isomer and the (d) CH3 – CH – CH = CH2 other isomer of the type (b), in which identical | atoms or groups lie on the opposite sides of CH3 the double bond is called trans isomer. Thus 3-Methylbut-1-ene cis and trans isomers have the same structure (e) CH2 = C – CH2 – CH3 but have different configuration (arrangement | of atoms or groups in space). Due to different CH3 arrangement of atoms or groups in space, these isomers differ in their properties like 2-Methylbut-1-ene melting point, boiling point, dipole moment, Geometrical isomerism: Doubly bonded solubility etc. Geometrical or cis-trans isomers carbon atoms have to satisfy the remaining of but-2-ene are represented below : 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 2024-25 Unit 9.indd 308 03-05-2024 16:07:55  Hydrocarbons 309 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 (iii) C6H5CH = CH – CH3 that in the trans-but-2-ene, the two methyl (iv) CH3CH = CCl CH3 groups are in opposite directions, Threfore, dipole moments of C-CH3 bonds cancel, thus Solution making 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. 9.3.4 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 the like sulphur compounds or quinoline give trans isomer has higher melting point than alkenes. Partially deactivated palladised 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 9.10 Draw cis and trans isomers of the following compounds. Also write their IUPAC names : (i) CHCl = CHCl (9.30) (ii) C2H5CCH3 = CCH3C2H5 Solution (9.31) Pd/C iii) CH≡ CH+H2 CH2 =CH2 (9.32) Ethyne Ethene Pd/C CH3–C≡ CH+H2 CH3–CH =CH2 iv) Propyne Propene (9.33) Wi l l p r o p e n e t h u s o b t a i n e d s h o w Problem 9.11 geometrical isomerism? Think for the Which of the following compounds will reason in support of your answer. show cis-trans isomerism? 2. From alkyl halides: Alkyl halides (R-X) on (i) (CH3)2C = CH – C2H5 heating with alcoholic potash (potassium hydroxide dissolved in alcohol, say, 2024-25 Unit 9.indd 309 10/10/2022 10:37:55 AM  310 chemistry ethanol) eliminate one molecule of halogen takes out one hydrogen atom from the acid to form alkenes. This reaction is β-carbon atom. 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). (9.37) 9.3.5 Properties Physical properties Alkenes as a class resemble alkanes in physical properties, except in types of isomerism and difference in polar nature. (9.34) The first three members are gases, the next Nature of halogen atom and the alkyl group fourteen are liquids and the higher ones are determine rate of the reaction. It is observed solids. Ethene is a colourless gas with a faint that for halogens, the rate is: iodine > sweet smell. All other alkenes are colourless bromine > chlorine, while for alkyl groups and odourless, insoluble in water but fairly it is : tertiary > secondary > primary. soluble in non-polar solvents like benzene, petroleum ether. They show a regular increase 3. From vicinal dihalides: Dihalides in in boiling point with increase in size i.e., every which two halogen atoms are attached – CH2 group added increases boiling point by to two adjacent carbon atoms are known 20–30 K. Like alkanes, straight chain alkenes as vicinal dihalides. Vicinal dihalides on have higher boiling point than isomeric treatment with zinc metal lose a molecule branched chain compounds. of ZnX2 to form an alkene. This reaction is known as dehalogenation. Chemical properties Alkenes are the rich source of loosely held CH2Br–CH2Br + Zn CH2=CH2+ ZnBr2 pi (π) electrons, due to which they show (9.35) addition reactions in which the electrophiles CH3CHBr–CH2Br + Zn CH3CH=CH2 add on to the carbon-carbon double bond to form the addition products. Some reagents +ZnBr2 also add by free radical mechanism. There (9.36) are cases when under special conditions, 4. From alcohols by acidic dehydration: alkenes also undergo free radical substitution You have read during nomenclature of reactions. Oxidation and ozonolysis reactions different homologous series in Unit 12 are also quite prominent in alkenes. A brief that alcohols are the hydroxy derivatives description of different reactions of alkenes of alkanes. They are represented by R–OH is given below: where, R is CnH2n+1. Alcohols on heating 1. Addition of dihydrogen: Alkenes add with concentrated sulphuric acid form up one molecule of dihydrogen gas in alkenes with the elimination of one water the presence of finely divided nickel, molecule. Since a water molecule is palladium or platinum to form alkanes eliminated from the alcohol molecule in (Section 9.2.2) the presence of an acid, this reaction is 2. Addition of halogens : Halogens like known as acidic dehydration of alcohols. bromine or chlorine add up to alkene to This reaction is also the example of form vicinal dihalides. However, iodine β-elimination reaction since –OH group does not show addition reaction under 2024-25 Unit 9.indd 310 03-05-2024 16:08:27  Hydrocarbons 311 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 (9.42) 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 negative part of the addendum (adding (9.38) molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. Thus according to this rule, product I i.e., 2-bromopropane is expected. In actual practice, this is the principal product of the reaction. This generalisation of Markovnikov (9.39) rule can be better understood in terms of mechanism of the reaction. 3. Addition of hydrogen halides: Hydrogen halides (HCl, HBr,HI) add up to alkenes Mechanism to for m alkyl halides. The order of Hydrogen bromide provides an electrophile, + reactivity of the hydrogen halides is H , which attacks the double bond to form HI > HBr > HCl. Like addition of halogens carbocation as shown below : to alkenes, addition of hydrogen halides is also an example of electrophilic addition reaction. Let us illustrate this by taking addition of HBr to symmetrical and unsymmetrical alkenes Addition reaction of HBr to symmetrical alkenes (a) less stable (b) more stable Addition reactions of HBr to symmetrical primary carbocation secondary carbocation alkenes (similar groups attached to double bond) take place by electrophilic addition (i) The secondary carbocation (b) is more mechanism. stable than the primary carbocation (a), therefore, the former predominates CH2=CH2+H–Br CH3–CH2–Br (9.40) because it is formed at a faster rate. – CH3–CH=CH–CH3+HBr CH3–CH–CHCH3 (ii) The carbocation (b) is attacked by Br ion to form the product as follows : Br (9.41) 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) 2024-25 Unit 9.indd 311 10/27/2022 2:19:43 PM  312 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 The secondary free radical obtained in the and Hl. This addition reaction was observed above mechanism (step iii) is more stable than by M.S. Kharash and F.R. Mayo in 1933 the primary. This explains the formation of at 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 (C6H5CO)2O2 stronger (430.5 kJ mol ) than H–Br bond CH3 – CH=CH2+HBr CH3–CH2 –1 (363.7 kJ mol ), is not cleaved by the free radical, whereas the H–I bond is weaker CH2Br –1 (296.8 kJ mol ) and iodine free radicals 1–Bromopropane combine to form iodine molecules instead of adding to the double bond. (9.43) Mechanism : Peroxide effect proceeds via Problem 9.12 free radical chain mechanism as given below: Write IUPAC names of the products obtained by addition reactions of HBr to (i) hex-1-ene (i) in the absence of peroxide and (ii) in the presence of peroxide. Solution. Homolysis. (ii) C 6H5+H–Br C6H3+ B r 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. 2024-25 Unit 9.indd 312 10/10/2022 10:37:56 AM  Hydrocarbons 313 ketones and/or acids depending upon the nature of the alkene and the experimental conditions (9.49) + KMnO4/H CH3 – CH=CH–CH3 2CH3COOH (9.44) But-2-ene Ethanoic acid (9.50) 7. Ozonolysis : Ozonolysis of alkenes involves the addition of ozone molecule to alkene to form ozonide, and then cleavage of the ozonide by Zn-H 2O to smaller molecules. This reaction is highly useful in detecting the position of the double (9.45) bond in alkenes or other unsaturated 5. Addition of water : In the presence of a compounds. few drops of concentrated sulphuric acid alkenes react with water to form alcohols, in accordance with the Markovnikov rule. (9.51) (9.46) 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. (9.52) (9.47) 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 thus obtained are called (9.48) polymers. This reaction is known as b) Acidic potassium permanganate or acidic polymerisation. The simple compounds potassium dichromate oxidises alkenes to from which polymers are made are called 2024-25 Unit 9.indd 313 03-05-2024 16:08:51  314 chemistry monomers. Other alkenes also undergo are named as derivatives of the corresponding polymerisation. alkanes replacing ‘ane’ by the suffix ‘yne’. High temp./pressure n(CH2 =CH2) Catalyst —( CH2–CH2 — ) The position of the triple bond is indicated Polythene by the first triply bonded carbon. Common and IUPAC names of a few members of alkyne (9.53) series are giv