Hydrocarbon. Alkane

Hydrocarbon ALKANE (i) Branched and unbranched aliphatic saturated hydro carbons are called member of alkane. The structural formula of alkane have only single bonds Ex. 1 Draw the all possible isomers of C6H14 and assign relations between them. or all bonds in alkane is only  bonds. (a) CH3—CH2—CH2—CH2—CH2—CH3 (ii) Alkanes does not reacts with chemical reagants such n–Hexane as dil. and conc. HCl, dil. & conc. H2SO4, dil. & (b) conc. HNO 3 , Caustic soda, acidic & basic K2Cr2O7, KMnO4 etc. That is why alkanes are Isohexane (2-methyl pentane) called paraffins. (Parum=little, affins = reactivity). (c) Property Characteristics of alkane General formula CnH2n+2 C—C Bond energy C—H Bond energy Bond angle 82.67 kcal/mole 98.67 kcal/mole 109°.28' C—C Bond length 1.54 A° C—H Bond length 1.112 A° Hybridisation on C sp3 shape Tetrahedral 3–Methyl pentane (d) 2,3–Dimethyl butane (e) Neohexane (2,2-Dimethyl butane) 149 Hydrocarbon  a, b and d - Chain isomers General Methods of preparation  b and c - Position isomers 1.   a, c, e d and e - Chain isomers Position isomers  a, c, d - Chain isomers From alkenes and alkynes (Sabatier and Sandrens reaction) or (By hydrogenation of alkenes and alkynes) : Alkenes and alkynes on catalytic hydrogenation gives alkanes. R—CH CH—R + H2 Ex. 2 Draw the structure of optically active alkane which have minimum carbon atom.   R—CH2—CH2—R Alkene C7H 16 Alkane R—C C—R + 2H2   R—CH2—CH2—R or Alkyne Catalyst : (a) Pd/Pt at ordinary temp. and pressure or (b) Ni, 200–300° C (sabatier) (c) Raney Nicker at room temp. (d) Raney nickel is obtained by boiling Ni/Al with NaOH. Al dissolved & Ni obtained in finally divided state. Note: Alkane generally shows chain and position isomerism (for more details refer isomerism in study material provided to you.) (e) Methane can not be prepared by this method (From unsaturated hydrocarbon). Composition : (a) CH4 is also known as Marsh gas (fire damp). 2. From alkyl Halides (By reduction) : (b) Calore gas : Mixture of n–butane and isobutane. R—X (c) LPG (Liquefied petroleum gas) : liquid propane, isobutane. Catalyst : (d) Kerosene oil : Mixture of alkanes having C12 to C16. (e) Gasoline : (Petrol): Mixture of alkanes having C6 to C8. (f) Natural gas : 80% methane + 10% ethane + 10% propane + small amounts of H2,N2, CO2 etc. (g) Water gas : CO + H2 (1:1) (h) Synthesis gas : CO + 3H2 (1:3)     R—H + HX (i) Zn + HCl (ii) Zn + CH3 COOH (iii) Zn—Cu couple in C2H5OH (iv) Red P + HI (v) Al + Hg + ethanol Mechanism : Zn  Zn+2 + 2e    – + – R + H Cl  R—H + Cl Product +2 – Zn + 2Cl  ZnCl2. 150 Hydrocarbon (a) Alkyl halides can also be reduced to alkane by H2/Pd or LiAlH4 or H2/Ni. (b) Reduction is due to the electron transfer from the metal to the substrate (R- X) 3.   Product   2Na + 2X  2NaX (c) If any alkyl halide is asked, the H-atom of any carbon atom of given alkane is removed by halogen atom. Ex. 2C2H5—I From alkyl halide (By Wurtz reaction): A solution of alkyl halide in ether on heating with sodium gives alkane. (b) Free radical mechanism : + 2Na  C2H5—C2H5 + 2NaI n–butane Na Na+ e     Product (a) Two moles of alkyl halide treated with Na in presence of dry ether. If ether is wet then we obtain alcohol. 2Na + H2O 2NaOH + H2     Free radicals also undergo Disproportionation i.e. one radical gains hydrogen at the expense of the other which loss hydrogen. CH3I + NaOH CH3OH + NaI Methanol (b) Methane can not be prepared by this method. The alkane produced is higher and symmetrical i.e. it contains double the number of carbon atoms present in the alkyl halide taken. Ethane Ethylene This explains the presence of ethylene and ethane in the butane obtained by Wurtz reaction. (c) Two different alkyl halides, on wurtz raction give all possible alkanes. Ex. 3 If two moles of Isopropyl chloride reacts with Na (d) The seperation of mixture in to individual in presence of dry ether. Which alkane is obtained. members is not easy because their B.P. are near to each other and thus wurtz reaction is not suitable for Sol. 2, 3-Dimethyl butane. the synthesis of alkanes containing odd Ex. 4 If isopropyl chloride and ethyl chloride both react number of carbon atom. with Na in presence of dry ether which alkanes are (e) This reaction generally fails with tertiary alkyl obtained. halide. Sol. n-Butane, 2-Methyl butane and 2, 3-Dimethyl butane. Mechanism : Two mechanism have been proposed for this reaction. (a) Ionic Mechanism: 2Na 2Na + 2e Ex. 5 Which of the following compound can not obtained from wurtz reaction. (1) ethane (2) butane (3) isobutane (4) hexane [Hint : In wurtz reaction unsymmetrical alkane can not be obtained.] Sol. (3) 151 Ex. 6 When ethyl chloride and n-propyl chloride undergoes wurtz reaction which is not obtained. (1) n–butane (2) n–pentane (3) n–hexane (4) isobutane Hydrocarbon Mechanisim : Decarboxylation proceeds via. the formation of carbanion intermediate as follows.   C H —C H + Sol. (4) C2H5—Cl + C3H7Cl  2 5 2 5 C3H7—C3H7 + C2H5—C3H7 (4) From Frankland Reagent: If Zn is used in place of Na, the reaction is named as Frankland reaction. R—X + 2Zn +RX R2Zn + ZnX2 Frankland reagent  R2Zn + R—X R—R + RZnX (5) From Carboxylic Acid (By decarboxylation) : Saturated monocarboxylic acid salt of sodium or potassium on dry distillation with soda lime give alkane. (Acidic)  RCOONa + NaOH   R—H + Na2CO3  Soda Lime (a) The process of elimination of Carbon-di-oxide from Carboxylic acid called decarboxylation. (b) Replacement of -COOH by hydrogen is known as decarboxylation. The alkane formed always contains one carbon atom less than the original acid. (c) This reaction is employed for stepping down a homologous series. (d) Soda lime is prepared by soaking quick lime CaO with NaOH solution and then drying the products. (e) Decarboxylation of sodium formate gives H2 Product  Na2CO3 (a) If in a compound two carboxylic groups are present and they are attached to same carbon atom then also decarboxylation of one of the carboxylic groups takes place simply on heating.  CH3COOH + CO2  (b) CH4 can be prepared by CH3COOH. (c) C2H6 can be prepared by CH3CH2COOH. (d) CH3 —CH2 —CH3 can be prepared by Butanoic acid and 2–Methyl propanoic acid. 152 Hydrocarbon Ex. 8 Give reactivity order for decarboxylation ? Ex. 7 How many acids can be taken to obtain isobutane from decarboxylation ? (1) 4 (2) 3 (3) 2 CH3—CH2—COOH I CH2 CH—COOH (4) 5 II Sol. (3) To obtain isobutane the acids are (i)  CH C —COOH III (1) I > II > III (2) III > II > I (3) III > I > II (4) None is correct Sol. (2) In decarboxylation intermediates are,  (ii) I  CH CH2 CH C II III The stability order of carbanion – III > II > I So reactivity order for acid is – III > II > I So two acids can be taken.  Presence of electron attracting group (–I) in the hydrocarbon part of the fatty acid increases the decarboxylation. If – I is more effective group then weak base may be taken. Ex. (1) (6) From carboxylic acid (By Kolbe's process) : Alkanes are formed on electrolysis of concentrated aqueous solution of sodium or potassium salt of saturated monocarboxylic acids. 2RCOONa + 2H2O            (2) Electrolysis of Sodium propionate solution give nbutane, ethylene, ethane and ethyl propionate as follows-   Mechanism : C2 H5—COONa   C 2H5 —COO+ (3) -Keto acids are decarboxylated readily  (Ionization) simply on heating (soda lime is not required)   At Anode :   C2H5— C2H5—COO   153  + CO2    + Hydrocarbon (6) From alkanol, alkanals, Alkanone and alkanoic acid (By reduction) : The reduction of either of the above in presence of red P and HI gives corresponding alkane.   C2H5—C2H5  R—H+H2O+ I2 R—OH +2HI   Product R—CHO + 4HI  R—CO—R+4HI    CH3—CH2—H + CH2 R—CH2—R+H2O + 2I2 CH2 (minor products) RCOOH + 6HI An ester is also formed. C2H5—  +     R—CH2—R+H2O + 3I2  C2H5 —COOC2H5 (minor products)    RCH3+H2O + 2I2   At cathode : Na + e Na Na + H2O  NaOH + In the above reaction I2 is formed which acts as reducing agent and may reduce alkane and form alkyl halide. So red P is added in the reaction to remove I2 formed in the reaction.   R—CH3 + I2   R—CH2—I + HI 2P + 3I2 2PI3  (a) Methane can not be prepared by this method. (7) From alkanones (By Clemmensen's method) : Carbonyl compound (Preferably ketones) may also (b) Electrolysis of an acid salt gives symmetrical be reduced with Zinc amalgam and concentrated alkane, However in case of a mixture of Carboxylic HCl (Zn—Hg/HCl), this reaction is called acid salts, all probable alkanes are formed. Clemmensen reduction. R'COOK + R"COOK  (R'—R" +  R—CO—R'+4H  R'—R' + R"—R") +2CO2 + H2 +2NaOH (c) Presence of alkyl groups in  - position decrease the yield of alkanes. (d) True aromatic acids do not undergo Kolbe's electrolytic reaction. R—CH2—R'+H2O CH4, CH3—CH3, isobutane and neopentane are not obtained from Ketones because these alkane do not contain CH2 group. (e) Free radical mechanism has been suggested for (8) From alkanals and alkanones (By Wolf Kishner Kolbe reaction. reaction) : (f) At anode alkane (major) and CO gas is formed 2 while at cathode NaOH and H2 gas is formed. (g) The concentration of NaOH in solution is increased with time so pH of solution is also increased. C O + NH2NH2 C N.NH2    Hydrazene  CH2 + N2 Hydrazone 154 Hydrocarbon (9) From G.R. : (a) Formation of alkanes with same number of C atoms : With same number of C-atoms as G.R. react with compound containing active hydrogen alkanes is obtained. Ex. 8 Which of the following reaction can not be used to obtained propane . (1) Wurtz reaction (2) Corey-house reaction (3) Decarboxylation of acid salt (4) All of them This reaction is used to determine the number of active H-atoms in the compound this is known as Zerewitnoff's method. (b) G.R. react with alkyl halide to give higher alkanes : RMgX + R'—X  R—R' + MgX2 [Hint: Wurtz reaction is used for symmetrical alkanes.] Sol. (1) (11) From metal carbide (By hydrolysis) : Only CH4 can be obtained by the hydrolysis of Be or Al carbides         Ex.7 Which of the following does not give alkane with R—Mg—X. Physical properties : (1) Ph—OH (2) Cl —NH2 (i) C1 to C4 gases, Neopentane also gas but npentane and isopentane are low B.P. liquids. (3) CH COOH (4) HCl  3 [Hint : Except Cl—NH2 all have active hydrogen, but Cl—NH2 when reacts with R—Mg—X the oduct is R—NH2.] Sol. (2) (10) Corey-House Synthesis : This method is suitable for the preparation of unsymmetrical alkanes i.e. those of type R—R' R—X (ii) Next members C5 to C17 are Colourless liquids and above C17 are Waxy solids. (iii) Density : The density of alkanes increases with increase in molecular weight and becomes constant at 0.8 g/mL. Thus all alkanes are lighter than water. (iv) Solubility : Alkanes being non polar and thus insoluble in water but soluble in non-polar solvents Ex. C6H6, CCl4 ,ether etc. • The solubility of alkanes decreases with increase in molecular weight    R—R'+RCu+LiX Lithium dialkyl copper Note: In Corey-house reaction symmetrical and unsymmetrical alkane both can be formed. • Liquid alkanes are themselves good non-polar solvents. (v) Boiling point -  molecular weight (for nalkanes)  i.e. boiling point Pentane < hexane < heptane Also boiling point  155 because the shape approaches to spherical which results in decrease in Vanderwaals forces (as surface area decreases) Thus boiling point n–Pentane > Isopentane > neopentane (vi) Melting Point : M.P. of alkanes do not show regular trend. Alkanes with even number of carbon atoms have higher M.P. than their alkanes of odd number of carbon atoms. The abnormal trend in M.P. is due to the fact that alkanes with odd carbon atoms have their carbon atom on the same side of the molecule and in even carbon atom alkane the end Carbon atom on opposite side. Thus alkanes with even carbon atoms are packed closely in crystal lattice to permit greater intermolecular attractions. Hydrocarbon (b) Incomplete oxidation : In limited supply of air gives carbon black and CO. 2CH4 + 3O2  2CO + 4H2O CH4 + O2  C + 2H2O C–black (used in printing) (c) Catalytic oxidation : (i) Alkanes are easily converted to alcohols and aldehydes under controlled catalytic oxidation.  2CH OH 2CH4 + O2  3 CH4 + O2   HCHO + H2O (ii) Alkanes on oxidation in presence of maganese acetate give fatty acids. < Odd number of carbon Even number of carbon (iii) Tertiary alkanes are oxidized to give tertiary alcohols by KMnO4. Ex.9 Alkanes are inert is nature, why ? Sol. Alkanes are quite inert substances with highly stable nature. Their inactiveness has been explained as:   (i) Alkanes have all the C—C and C—H bonds being stronger  bonds and are not influenced by acid, oxidants under ordinary conditions. (ii) The C—C bond is completely non polar and Ex.10 How many litre of Oxygen required for complete conbustion of 6.0 g ethane at NTP ? C—H is weak polar. Thus polar species i.e. electrophiles or nucleophiles are unable to attack Sol.    these bonds under ordinary conditions. 60 g ethane required O2 (at NTP) = 7 × 22.4 litre Chemical Properties : (1) Oxidation : 1 g ethane required O2 (at NTP) = (a) Complete oxidation or combustion : Burn readily with non-luminous flame in presence of air or oxygen to give CO2 and water with evolution of heat. Therefore, alkanes are used as fuels.          nCO2 + (n+1) H2O+Q; (H =–ve)   6 g ethane required O2 (at NTP) =     (2) Substitution Reactions : Substitution reaction in alkanes shows free radical mechanism. They give following substitution reaction. 156 Hydrocarbon (a) Halogenation : Replacement of H-atom by halogen atom R—H + X2  Step II Chain propagation step : R—X + HX Halogenation is made on exposure to (halogen + alkane) mixture to UV or at elevated temp. Methane Methyl radical The reactivity order for halogens shows the order. Step III Chain termination step : Reactivity order of hydrogen atom in alkane is (i) Fluorination : reacts explosively even in dark. Fluorination can be achieved without violence when alkane is treated with F2 diluted with an inert gas (like N2) By the action of HgF2 on bromo or iodo derivatives. C2H5I + HgF2  C2H5F + HgI2   A (      ,   ,  can also attack CH3Cl to form chloromethyl   ) free radical. This free radical participates further in the chain reaction to yield methylene chloride (dichloromethane). (ii) Chlorination : It takes place in the following conditions - and ° (i) At 300 C in darkness. (ii) At 100° C in presence of organic peroxides. (iii) At 150° C in the presence of Tetraethyllead. (iv) In the presence of diffused sunlight.or UV radiation. The monochloro derivative of alkane is obtained by taking alkane in large excess. When chlorine is in excess, a mixture of mono, di, tri, tetra and perchloro derivatives is obtained. Similarly, chloroform and CCl4 are obtained by further chain reaction. (iii) Bromination : Br2 reacts with alkanes in a similar manner but less vigorously. (iv) Iodination : Iodine reacts with alkanes reversibly. HI formed as the by product is a powerful reducing agent and is capable of reducing the CH3I to CH4. Iodination may be carried out in the presence of an oxidising agent such as HIO3, HIO4, HNO3, HgO etc. Which destroy HI,   Mechanism for CH4 + Cl2   CH3Cl + HCl Step I Chain initiation step :           5HI + HIO3 3I2 + 3H2O Iodination is very slow because energy of activation of the reaction is very large      157 Hydrocarbon (i) Halogenation is inhibited in presence of oxygen because oxygen reacts with alkyl free radicals to form less reactive peroxy alkyl radical R–O–O• which can not propagate the chain. CH3—CH3 + Ex. HNO3   CH3CH2NO2 + CH3NO2 + H2O (ii) Reactivity ratio of H atom for Chlorination (1° : 2° : 3° H) 1 : 3.8 : 5 Reactivity ratio of H atom for bromination (1 : 82 : 1600) Ex.11 What is the percentage of products obtained from monobromination of isobutane ?  Sol. Mechanism : (Free Radical substitution) Step – I        R–H+ Step – II   + H2O     (Product) Step – III      = % of product (I) =  % of product (II) =    (c) Sulphonation : Replacement of H atom of alkane by –SO3H is known sulphonation. Alkane react with fuming H2SO4 or oleum (H2S2O7). The branched lower alkanes and higher alkanes react to give alkane sulphonic acid. Ex.    (b) Nitration : (Vapour phase nitration) T h i s involves the substitution of a hydrogen atom of alkane with -NO2 group. At ordinary temperature, alkanes do not react with HNO3. But reacts with vapours of Conc. HNO3 at 450° C.          (minor)   × =  2–Methyl propane Mechanism : (Free Radical substitution)   Since the reaction is carried at high temp. the C—C bonds of alkanes break during the reaction and a mixture of nitroalkanes is formed.               158 Hydrocarbon (4) Pyrolysis or Cracking or thermal decomposition Lower members such as propane, butane, pentane When alkanes are heated to 500-7000C they are etc. react with SO3 in vapour phase to form decomposed i n to l ower hydrocarbon. This sulphonic acids. decomposition is called pyrolysis. In petroleum C3H8 + SO3  C3H7 – SO3H industry it is also termed as cracking.Cracking is used for the manufacture of petrol, petrol gas/oil gas etc. (d) Chlorosulphonation (Reed reaction) :  Reaction with a mixture of SO2 and Cl2 at ordinary Ex.    temp. in the presence of UV light is called      chlorosulphonation. Propane sulphonyl Chloride n-Butane  1-Butene + 2–Butene + Ethane + Ethene + Propene + CH4+ H2 The mechanism of pyrolysis occurs via free radicals. Further hydrolysis of alkane sulphonyl chloride gives alkane sulphonic acid. (5) propane sulphonic acid Sodium salt of sulphonic acid (used as detergent) Unbranched higher alkanes (from 6 to 10 carbon atoms) when heated in presence of oxides of Cr, Mo, V on Al2O3 support at 500 0 C aromatic hydrocarbons are formed.   + 4H2  (3) Isomerization: Unbranched chain alkanes on heating with AlCl3 + HCl / 2000C are converted in to branched chain alkanes     + 4H2   n-heptane Toluene     n-butane Isobutane n-octane o-xylene Branched chain alkanes converted to more branched It provides an excellent method of passing from aliphatic to aromatic series. alkane. (6) Insertion of CH2 group.   CH3CH2CH3+:CH2  CH3CH2CH2CH2CH3 Isomerisation of alkanes is of great importance in petroleum industry to increase the octane number of petrol (gasoline). n-Butane Isobutane Carbene (:CH2) obtained from Ketene (CH2 = C = O) or diazomethane (CH2N2) 159 Hydrocarbon Ex. 13 10 mL of a mixture of CH4 and C3H8 requires 41 mL of oxygen for complete combustion. What is the volume of CH4 and C3H8 in the mixture. (7) Chlorinolysis : Sol. Suppose the volume of CH4 in (CH4 + C3H8) mix = x c.c. = Volume of C3H8 will be = 10 – x c.c. IMPORTANT TIPS For CH4 1.  2.     Steam 3.  x c.c. of CH4 , 2x c.c. of O2 C + 2H2 For C3H8 (Carbon black used in prep. of printing ink.) 4.    5.      1 Vol. of CH4 requires 2 vol. of O2 for complete combustion  Synthesis gas    CH4  C3H8 +5O2  3CO2 + 4H2O   1 volume of C3H8 requires 5 ml of O2 for complete combustion   (10 – x) c.c. of C3H8 requires 5(10 – x) c.c. of O2 6. CO + H2    CH4 + H2O Total Volume of O2 = 2x + 5 (10 – x) it is equivalent to 41 7. CO2 + H2    CH 4 + 2H 2 O (according to question)    2x + 5 (10 – x) = 41  x = 3 c.c. Volume of CH4 is 3 c.c. and volume of C3H8 is 7 c.c. Ex.12 8 c.c. of gaseous hydrocarbon requires 40 c.c. of O2 for complete combustion. Identify hydrocarbon. Ex. 14 The density of one hydrocarbon at N.T.P. is 1.964 g/litre. Identify the hydrocarbon. Sol. Volume of hydrocarbon = 8 c.c. ; Volume of O = 2 40 c.c. Sol. Molecular weight of Hydrocarbon Formula No. 1, (For alkane) = density of 1 lit. x 22.4 = 1.964 x 22.4 = 44 So Molecular weight of hydrocarbon = 44 So the hydrocarbon is C3H8 (Propane). or The value of n comes in whole number from 1st formula it means hydrocarbon is Alkane and it is of 3C atom.  Hydrocarbon is C3H8 (Propane) 160