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Basic concepts of chemistry, alkanes, alkenes, alkynes, benzene, their preparation methods, properties and uses are explained. Isomerism in alkanes and alkynes also discussed.

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  1. 1. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net HYDROCARBONS INTRODUCTION Compounds containing only two elements carbons & hydrogen are known as hydrocarbons. On the basis of structure & properties, hydrocarbons are divided into two main classes viz. aliphatic & aromatic. Aliphatic hydrocarbons are further divided into different families namely alkanes, alkenes, alkynes and their cyclic analogs (cycloalkanes, cycloakenes and cycloalkynes etc.) ALKANES: Alkanes are open chain (acyclic) hydrocarbons comprising the homologous series with the general formula n 2n 2C H where ‘n’ is an integer. They have only single bonds & therefore are said to be saturated. Because of their low chemical activity, alkanes are also known as Paraffins. NOMENCLATURE: According to IUPAC system of nomenclature open chain aliphatic compounds are named as al- kanes. CH3 CH3 CH3 CH3 CH3 CH3 CH3CH3 ethane propane 2,2-Dimethyl propane H5C2 CH3 CH3 CH3 CH3 CH3 2,2,4,4-tetramethylhexane CH3 CH3 CH3 C2H5 3-ethyl-2-methylpentane CH3 CH3 CH3 CH3 CH3 3,4,7-trimethylnonane CH3CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 2, 3-dimethylpentane 2,5, 6-trimethyloctane Names of several common alkyl groups
  2. 2. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms CH3 CH3 CH3 CH3 CH3 CH3 isopropyl isobutyl sec-butyl CH3 CH3 CH3 CH3 CH3 CH3 tert butyl neopentyl CONFORMATIONS OF ALKANES The different arrangement of atoms in space that results from the free rotation of the groups about C – C bond axis are called conformers or conformational isomers and the phenomenon is known as conformational isomerism. Conformations of Ethane Of the infinite number of possible arrangements of ethane two conformations represent the extremes. These are called the eclipsed conformation (I) and the staggered conformation (II).Anyother arrangement which will be between these two extreme positions are known as Gauche or Skew form. Sawhorse Projection H H H H H H H HH H H H Eclipsed form (I) Staggered form (II) Newman Projection H H H H HH H H HH H H Eclipsed form (I) Staggered form (II) dihedral angle = 0 dihedral angle= 180° Relative Stabilities of the conformations of Ethane The potential energy of staggered form is minimum and that of eclipsed form is maximum for ethane. The difference in energy content between the eclipsed and staggered form is 12 kJ mol–1 . This small barrier to rotation is called Torsional Barrier. This energy is not large enough to prevent the rotation. Hence the conformers keep on changing from one form to another. The eclipsed conformation is least stable where as the staggered conformation is most stable.
  3. 3. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net H H H H HH H H HH H H Eclipsed 12 kJ mol-1 Staggered H H H H HH Staggered Rotation PE Conformations of n-Butane n-Butane molecule can be considered as a dimethyl derivative of ethane in which one H-atom of each carbon atom is replaced by a methyl group as shown below: CH3 C H H C H H CH3 1 2 3 4 If one of these central carbon atom (C2 or C3 ) is fixed and the other is rotated round the C2 – C3 bond, the following important conformations are obtained. CH3 H H H HCH3 CH3 H HH CH3 H Staggered (Anti) (I) CH3 H H CH3 HH Eclipsed (II) Gauche (III) CH3 H HH H CH3 Fully eclipsed (IV) CH3 H H H CH3H Gauche (V) Eclipsed (VI) CH3 H H CH3 H H
  4. 4. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms Relative stability of the conformation of n-Butane Anti  Gauche  Eclipsed  Fully eclipsed CH3 H H H HCH3 16 kJ mol-1 CH3 H H CH3 HH Rotation PE CH3 H H H CH3H CH3 H H CH3 HH CH3 H HH CH3 H CH3 H HH H CH3 CH3 H H CH3 H H 3.8 kJ mol-1 19 kJ mol-1 3.8 kJ mol-1 16 kJ mol-1 0 60° 120° 180° 240° 300° 360° PREPARATION OF ALKANES: Reduction of Alkenes and Alkynes Reduction of alkenes: alkenes are reduced into alkanes by hydrogen in the presence of catalyst [Pt, Pd, Ni/S. 2PtO ]    catalyst 2 2 2R — CH CH — R H R — CH — CH — R When catalyst is Ni the reaction is known as Sabatier - Sanderen’s reaction Reduction of alkynes: Alkynes also reduced by hydrogen in presence of catalyst.    catalyst 2 2 2R — C C — R 2H R — CH — CH — R First alkene forms thereafter alkane. Reduction of alkyl halides: Alkyl halides undergo reduction with nascent hydrogen to form alkanes R X 2H R H HX     Alkyl halide Alkane The yields are generally good and the hydrocarbons obtained are pure. The nascent hydrogen for reduc- tion may be obtained by using any one of the following : (a) Zinc and dilute hydrochloric acid (b) Zinc and acetic acid; Zn and NaOH (c) Zinc-copper couple in ethanol (d) Red phosphorus and hydrogen iodide
  5. 5. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net (e)Al-Hg in ethanol 2 5 Zn HCL 3 4orZn Cu / C H OH CH I 2H CH HI     2 5 Zn HCL 2 5 2 6orZn Cu / C H OH C H I 2H C H HI     0 Re dP 3 2 3 3 2150 C CH CH Br 2HI CH CH HBr I    The purpose of use of red phosphorus is to remove iodine. alkyl halides may also be reduced catalyti- cally by hydrogen using catalyst like nickel, palladium, platinum etc. Pd 2R X H R H H X     Primary and secondary alkyl halides may be conveniently reduced with lithium aluminium hadride in a dry organic solvent Organic 4 4 3Sovent 4RX LiAlH 4RH LiAlX (LiX AlX )    4LiAlH is not useful for 0 3 alkyl halide which is converted into alkenes. In such case 4NaBH or TPH is used. 3CH3CH 3CH C Cl 4LiAlH  2CH3CH 3CH C Cl 4NaBH   0 3 2-methylpropane Alkyl halides  0 0 0 1 ,2 ,3 can also be reduced to alkanes with TPH (Triphenyl tin hydride, 3Ph SnH) . Reduction of alcohols: Alcohols can be reduced to corresponding alkane as  2P/I / R — OH R — H 2P /I / 3 2 3 2CH CH OH CH CH I     Wurtz Reaction: When alkyl halide is treated with sodium in presence of dry ether, higher alkane is formed. Na dry ether R — X X — R R — R 2NaX         R — X R — X R — R R — R R — R 4CH Cannot be prepared by this method.
  6. 6. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms 3° R—X do not give this reaction When a mixture of two different alkyl halides is used a mixture of three alkanes is obtained    Ether 3 2 5 3 2 5CH Br 2Na BrC H CH C H + 2NaBr Ether 3 3 3 3CH Br 2Na BrCH CH CH 2NaBr     Ether 2 5 2 5 2 5 2 5C H Br 2Na BrC H C H C H 2NaBr     The separation of the mixture into individual members is not always easy. Thus Wurtz reation is not suitable for the synthesis of alkanes containing odd nubmer of carbon atoms but the method is useful for the preparation of symmetrical alkanes Mechanism : . Mechanism: R — X Na R NaX   R R R — R  Frankland’s reaction This method is similar to Wurtz reaction. When alkyl halides is heated with zinc in inert solvent higher alkanes are formed.    2 5Zn/C H OH 2R — X R — X R — R ZnX    2 5Zn/C H OH 3 3 3 3 2CH — Br CH — Br CH — CH ZnBr Decarboxylation of Salts of Carboxylic Acids Sodium salt of carboxylic acids undergo decarboxylation when heated with sodalime. R– COONa + NaOH    CaO 2 3R H Na CO CH3 —COONa + NaOH   CaO 4 2 3CH Na CO Mechanism OH R C O O R O O OH R  H O C O O 2 3CO R H    Corey-House synthesis A superior method for coupling is the corey-house synthesis which could be employed for obtaining alkanes containing odd number of carbon atoms.An alkyl halide (R–X) which may be primary, secondary or tertiary is first converted into alkyl lithium by treating with lithium. The alkyl lithium is then reacted with cuprous halide to get lithium dialkyl cuprate. The complex is then treated with another alkyl halide (R—X) which must be preferably primary. The reaction follows SN 2 mechanism. With secondary alkyl halides the reaction leads to partly substitution forming alkane and partly elimination forming alkene, with tert alkyl halide only elimination takes place.   dry ether R — X 2Li R — Li LiX   22R — Li CuI R CuLi LiI
  7. 7. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net     2R CuLi R — X R — R R — Cu LiX Li CuI 3 3 Methyl bromide Methyllithium CH Br CH Li    CH3 CuLi CH3 Lithium dimethylcopper 2 5C H Br 3 2 3 Propane CH CH CH   CH3 Cl CH3 sec-Butyl chloride Li CuI   (CH3CH2CH-)2CuLi CH3 CH3 Br n-Pentyl bromide CH3 CH3 CH3 3-methyloctane Illustration - 1 : Prepare 2-methylbutane from chloroethane and 2-chloropropane using Corey-House synthesis. Solution: 3 2CH CH Cl1. Li 3 3 2 3 2 32. cuI CH CHCl (CH CH) CuLi CH CHCH CH  3CH 3CH 3CH 2-Chloropropane Lithiumdiisopropyl cuprate From organo metallic compounds: Grignard reagents and alkyl lithium reacts with water and other compounds having acidic hydrogen to give hydrocarbon corresponding to alkyl group of the organo metallic compounds.  R — Mg — X H — OH R — H  R — Li H — OH R — H Kolbe Electrolysis Alkanes can be prepared by the electrolysis of a concentrated solution of sodium or potassium salt of carboxylic acid. 2H O electrolysis 2 2R — C — O — K R — C — O — K R — R 2CO H 2KOH     O O 4CH can not be prepared by this method.
  8. 8. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms – –e Anode (Oxidation) R C O O R C O O 2R CO  R R R — R  PHYSICAL PROPERTIES Physical State i) All are colourless and possess no characteristic odour. ii) Lower alkanes (C1 to C4 ) are gases, middle one (C5 to C17 ) are liquids and highers are solids. Boiling Points i) The boiling point of alkanes increases with increase in molecular weight due to increase in van der Waals forces with increase in molecular weight i.e.Boiling point: pentane  hexane  heptane ii)Also the branching in alkanes gives a decrease in surface area (as the shape approaches to spherical) which results in decrease in van der Waals forces. The boiling point of isomeric alkanes show the order: pentane  isopentane  neopentane. Melting Points The melting points of alkanes do not show a regular trend. Alkanes with even number of carbon atoms have higher melting point than their adjacent alkanes to odd number of carbon atoms. Melting order:        ( 172.0 C) ( 182.5 C)( 187.7 C) propane ethane methane The abnormal trend in melting point is probably due to the fact that alkanes with odd carbon atoms have their end carbon atom on the same side of the molecule and with 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. Density: The density of alkanes increases with increase in molecular weight and becomes constant at 0.76g/ml. Thus all alkanes are lighter than water. Solubility: i) Alkanes being non polar and thus insoluble in water but soluble in non polar solvents e.g. C6 H6 , CCl4 , ether etc. ii) The solubility of alkanes decreases with increase in molecular weight. iii) Liquid alkanes are themselves good, non polar solvents. CHEMICAL PROPERTIES OF ALKANES: Halogenations: One of the important chemical reactions of alkanes is halogenation which is defined as the replacement of hydrogen atom of the molecule by halogen atom. A mixture of alkanes and chlorine remains unaffected in the dark but a vigorous reaction occurs when it is exposed to light or at a higher temperature.    hv or 2R — H Cl R — Cl HCl As pointed above the reactions takes place by free radical chain mechanism which proceeds in the following three distinct steps. a) Chain initiations hv or Cl — Cl 2Cl [ H 58 kcal/ mole]     b) Chain propagation 3 3Cl H — CH H — Cl CH [ H –1kcal/ mole]    
  9. 9. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net 3 3CH Cl — Cl CH — Cl Cl [ H –23 kcal/ mole]     c) Chain termination Cl — Cl Cl — Cl 3 3CH Cl CH — Cl  3 3 3 3CH CH CH — CH  The ease of substitution at various carbon atoms is of the order tertiary > secondary > primary which is the same as the stability of various alkyl radicals. Bromination of alkanes has close similarities to chlorination except the rate of reaction is slow. Fluorination of alkane is so violent that it results in the cleavage of C—C as well as C—H bonds.  Reactivity of   2 2 2 2 2X :F Cl Br I The potential energy curve for the halogenation (chlorination) of alkane is shown as Potentialenergy Progress of reaction R H Cl   2R Cl  R Cl Cl   act E REACTIVITY SELECTIVITY PRINCIPLE: In more complex alkanes, the abstraction of each different kind of H atom gives a different isomeric product. Three factors determine the relatives yields of isomeric products are 1. Probability factor: This factor is based on the number of each kind of H atom in the molecule. For example, in CH3 CH2 CH2 CH3 there are six equivalent 1° H’s and four equivalent 2° H’s. The ratio of abstracting a 1°H are thus 6 to 4, or 3 to 2. 2. Reactivity of H* : The order of reactivity of H is 3°  2°  1° 3. Reactivity of X* : The more reactive Cl* is less selective and more influenced by the probability factor. The less reactive Br* is more selective and less influenced by the probability factor, as summarized by the Reactivity-Selectivity Principle. If the attacking species is more reactive, it will be less selective, and the yields will be closer to those expected from the probability factor.
  10. 10. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms 2 o Cl /hv 25 C  CH3 CH3 CH3 Cl CH3 CH3 Cln-butane 28% 72% CH3 CH3 CH3 CH3 CH3 Cl CH3 CH3 CH3 Cl 36%64% 2 o Cl /hv 25 C  CH3 CH3 CH3 Br CH3 CH3 Br   2 o Br , light 127 C  2% 98% CH3 Br CH3 CH3 CH3 CH3 BrCH3 CH3 CH3 2 o Br , light 127 C  trace over 99% The rate of abstraction of hydrogen atoms is always found to follow the sequence 3°  2°  1°. At room temperature, for example, the relative rate per hydrogen atom are 5.0 : 3.8: 1.0. Using these values we can predict quite well the ratio of isomeric chlorination products from a given alkane. For example: CH3 Cl CH3 CH3 Cl 2 o Cl /hv 25 C CH3 CH3     o o o o n butylchloride no. of 1 H reactivity of 1 H sec butylchloride no. of 2 H reactivity of 2 H    6 1.0 6 28% equivalent to 4 3.8 15.2 72% Inspite of these differences in reactivity, chorination rarely yields a great excess of any single isomer. The same sequence of reactivity, 3°  2°  1°, is found in bromination, but with enormously larger reactivity ratios. At 127°C, for example, the relative rates per hydrogen atom are 1600: 82:1. Here, differences in reactivity are so marked as vastly to outweight probability factors. Hence bromination gives selective product. In bromination of isobutene at 127°C,
  11. 11. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net    o o o o Isobutylbromide no. of 1 H reactivity of 1 H tert butylbromide no. of 3 H reactivity of 3 H   9 1 1 1600  9 1600 Hence, tert-butyl bromide happens to be the exclusive product (over 99%). How many monochlorinated products are obtained by the chlorination of isopentane and what is the percentage of each assuming the reactivity ratio of 3°H : 2°H : 1°H = 5 : 3.8 : 1. Solution: Isopentane on chlorination in presence of sun light gives four different monochlorinated products. hv 3 2 3 2 2 2 3CH – CH – CH – CH Cl CH Cl – CH – CH – CH   3CH 3CH (A) 3 2 3CH — CCl — CH — CH  3CH 3CH(B) * 3 3CH — CH — CHCl — CH (C) 3 CH 3 2 2 CH—CH—CH—CH Cl (D) A and C are optical isomers. Product Reactivity Factor  Probability factor = Number of parts Percentage (A) 1  6 = 6 27.8% (B) 5  1 = 5 23.2% (C) 3.8  2 = 7.6 35.2% (D) 1  3 = 3 13.8% Total number of parts = 21.6 Nitration The displacement of an atom of hydrogen by a nitro group ( 2–NO ) is called nitration. Alkanes can be nitrated with nitric acid in the gas phase generally at 400–500°C.    400–500 C 2 2 2R — H HO — NO R — NO H O
  12. 12. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms 2HNO 3 2 3 400–500 C CH — CH — CH   3 2 2 2 3 3CH — CH — CH — NO CH — CH — CH 2NO 3 2 2 3 2CH — CH — NO CH — NO Sulphonation Sulphonation of a hydrogen atom of the alkane bysulphuric acid group 3(–SO H) is called sulphonation. The ease of substitution is tertiary  secondary > primary.. 2H OCH3 CH3 CH3 H CH3 CH3 CH3 SO3H2 4H SO  Isomerisation Heating of straight chain alkanes with anhydrous aluminum chloride (lewis acid) affords branched chain hydrocarbons. This process is termed isomerisation. 3AlCl 3 2 2 3CH — CH — CH — CH   CH3 H CH3 CH3 The process of isomerisation has been of immense utility in petroleum industry for raising the octane number of a particular petroleum fraction. Example 3AlCl (RCl) 25 C  Rearrangement of alkanes are thermodynamically controlled Aromatisation Aromatisation of n-alkanes containing six or more carbon atoms into benzene and its homologues takes place at high temperature (600° C) in presence Cr2 O3 —Al2 O3 as a catalyst.
  13. 13. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net 2 3 2 3 o Cr O Al O 600 C   23H  2 3 2 3 o Cr O Al O 600 C   23H  n-hexane cyclohexane benzene n-octane ethyl cyclohexane ethyl benzene Oxidation All alkanes readily burn in excess of air or oxygen to form carbon dioxide and water. Cn H2n+2 +    2 2 2 (3n 1) (2n 2) O (g) nCO (g) H O(l) 2 2 On the other hand, controlled oxidation under various conditions, leads to different products. Extensive oxidation gives a mixture of acids consisting of the complete range of C1 to Cn carbon atoms. Less extensive oxidation gives a mixture of products in which no chain fission has occurred. Under moderate conditions mixed ketones are the major products and oxidation in the presence of boric acid produces a mixture of secondary alcohols. The oxidation of alkanes in the vapour state occurs via free radicals, e.g., alkyl (R* ), alkylperoxy (ROO* ) and alkoxy(RO* ). Oxidising agents such as potassium permanganate readily oxidizes a tertiary hydrogen atom to a hydroxyl group. For example, isobutene is oxidized to tert butyl alcohol. (CH3 )3 CH + [O] 4KMnO 3 3(CH ) COH ALKENES: Alkenes are unsaturated hydrocarbons having carbon-carbon double bonds. The general molecular formula for alkene is n 2nC H . Alkenes and cycloalkanes are isomeric compounds because both have same general molecular formula. NOMENCLATURE: The IUPAC names of alkenes are given by replacing the –ane ending of the corresponding alkane with –ene.  2 2 IUPAC Ethene Commen Name: ethylene H C CH 3 2 IUPAC : propene Commen Name: Pr opylene CH — CH CH    1 2 3 4 2 2 3 1 Butene (not 1 2 butene) H C CH — CH — CH   6 5 4 3 2 1 3 2 2 3 2 Hexene (not 4 hexene) CH — CH — CH — CH CH — CH
  14. 14. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms 4 3 2 1 3 2CH — CH — CH CH 3CH 3-methyl-1-butene 6 5 4 3 2 2 2 2 2 3CH — CH — CH — CH — CH — CH — CH 2 1 2CH CH 6-Bromo-3-propyl-1-hexene Br STRUCTURE AND BONDING IN ALKENES: 117.2° 121.4° 134 pm 110 pm H H H H Ethylene is planar, each carbon is sp2 hybridized and the double bond is considered to have a  component and  component. The component arises from overlap of 2 sp hybrid orbital along a line connecting the two carbon atoms, the  component via a side by side overlap of two p orbital. Regions of high electron density attributed to  electrons, appear above and below the plane of the molecule and are clearly evident in the electrostatic potential map. RELATIVE STABILITIES OF ALKENE: The stabilities of alkene are governed by — a) Degree of substitution: Double bonds can be classified as monosubstituted, disubstituted, trisubstituted and tetrasubstituted. According to the number of carbon atoms directly attached to the C=C structural unit. Alkenes with more highly substituted double bonds are more stable than isomers with less substituted double bond. b) Vander Waal’s strain: Alkenes are more stable when large substituents are trans to each other than when they are cis. When two or more atoms are close enough in space that a repulsion occurs between them is one type of steric effect and therefore trans isomers are more stable than cis isomer. c) Heat of hydrogenation: The addition of hydrogen is called hydrogenation. Hydrogenation reactions are exothermic (negative H0 ). CH3 CH3 CH3 2-methylbut-2-ene Pt /C 2H  CH3 CH3 CH3 0 H 26.9 kcal/mol   CH2 CH3 CH3 Pt /C 2H  CH3 CH3 CH3 0 H 28.5 kcal/mol   2-methylbut-1-ene
  15. 15. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net CH3 CH2 CH3 Pt /C 2H  CH3 CH3 CH3 0 H 30.3 kcal/mol   3-methylbut-1-ene The preceding three catalytic hydrogenation reaction all from the same alkane. So the energy of the product is the same for the each of three reactions. However, three reactants must have different energies. The reaction associated with the least negative  H° has the most stable reactant and the reaction with the most negative has the least stable reactant. CH3 CH2 CH3 0 H 30.3 kcal/mol   CH2 CH3 CH3 CH3 CH3 CH3 PotentialEnergy 0 H 28.5 kcal/mol   0 H 26.9 kcal/mol   The greater the number of alkyl groups attached to the doubly bonded carbon atoms the more stable the alkene. Stability of alkenes R2 C = CR2  R2 C = CHR  R2 C = CH2  RCH = CHR  RCH = CH2  CH2 = CH2 PREPARATION OF ALKENES: Dehydration of Alcohols: Alcohols when heated in presence of H2 SO4 , H3 PO4 , P2 O5 orAl2 O3 undergo loss of water molecule with the formation of alkene. H 2H — C — C — OH C C H O     Alcohol Alkene    2 4H SO 3 2 2 2 2160 C EthyleneEthyl alcohol CH — CH — OH H C CH H O 2 4H SO 140 C  OH 2H O
  16. 16. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms 2 4H SO  CH3 C CH3 OH CH3 CH2 CH3 CH3 2H O 2 4H SO and 3 4H PO are the acids most frequently used in alcohol dehydration. Regioselectivity in Dehydration of Alcohol 2 4H SO 80 C CH3 CH3 OH CH2 CH3 CH2 CH3 CH3 CH3 CH3 CH3  2-Methyl-1-butene (10%) 2-Methyl-2-butene (90%) Dehydration of this alcohol is selective in respect to its direction 3 4H PO   CH3 OH CH3 CH3  84% 16% Regioselectivity of  elimination is governed by Zaitsev’s Rule. Stereoselectivity in Alcohol Dehydration: In addition to being regioselective, alcohol dehydration are stereoselective. A stereoselective reaction is one which a single starting material can yield two or more sterioisomeric products, but give one of them in greater amount than other. 2 4H SO   OH CH3CH3 CH3 H H CH3  CH3 H H CH3 cis-2-pentene (25%) trans-2-pentene (75%) Illustration - 2 : Predict the major product of dehydration of each of: (A) (CH3 )2 C(OH)CH2 CH3 (B) (CH3 )2 CHCH(OH)CH3 (C) (CH3 )C(OH)CH(CH3 )2 Solution: CH3 C CH3 CHCH3 C C CH3 CH3 CH3 CH3 (A) and (B) (C) Dehydrohalogenations of Alkyl Halide: Dehydrohalogenation is the loss of hydrogen and a halogen to form an alkene.
  17. 17. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net H — C — C — X C C HX   The reaction is carried out in presence of strong base such as sodium ethoxide 2 5(NaOC H ) in ethyl alcohol as solvent. 2 5 2 5H — C — C — X NaOC H C C C H OH NaX     2 3NaOCH CH  H H Cl H H H Similarly 3 3 3NaOCH ,KOC(CH ) are the preferred bases used in this process.   3 3KOC(CH ) 3 2 15 2 2 3 2 15 2DMSO 25 C CH – (CH ) – CH – CH – Cl CH – (CH ) – CH CH The regeoselectivityof dehydrohalogenation of alkyl halide follow zaitsev rule of elimination predominates the direction. CH3 CH3 Br CH3 CH2 CH3 CH3  CH3 CH3 CH3 2-methyl-1-butene (29%) 2-methyl-2-butene (71%) Illustration - 3 : Predict the various product formed by dehydrohalogenation of CH3 CH3 Br Solution: CH3 CH3 CH3 CH3 CH2 CH3 (major) (minor) (minor) Dehalogenation of Vicinal Dihalides There are two types of dihalides namely gem (or geminal) dihalides in which the two halogen atoms are
  18. 18. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms attached to the same carbon atom and vic. (or vicinal) dihalides in which the two halogen atoms are attached to the adjacent carbon atoms. Dehalogenation of vic dihalides can be effected by either NaI in acetone or zinc in presence of acetic acid or ethanol.      3 2 5 Zn dust 3 2 3 2CH COOH or C H OH CH CHBr CH Br CH CH CH       NaI 3 3 3 3Acetone CH CHBr CHBr CH CH CH CH CH The reaction mechanism involves loss of the two halogen atoms in two steps. The two halogen atoms align themselves at 180° and in the same plane before they are lost. i) With NaI in acetone: X   C X C X C CC C X I IX I ii) With Zn-dust and acetic acid Zn  Zn2+ + 2e– X  C X C X C C XC C X2e Partial Reduction of Alkynes Reduction of alkyne to alkene is brought about by any one of the following reducing agents. i) Alkali metal dissolved in liquid ammonia. ii) Hydrogen in presence of palladium poisoned with BaSO4 or CaCO3 along with quinoline (Lindlar’s catalyst). iii) Hydrogen in presence of Ni2 B (nickel boride). R—CH2 —C  CH (i), (ii) or (iii) R—CH2 CH = CH2 Alkali metal dissolved in liquid ammonia produces nearly100% trans alkene bythe following mechanism: Na + liq. NH3  Na+ + es (solvated electron)
  19. 19. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net R C C R    C C R R 2H NH  C C R R H se se C C R R H 2H N H C R R H H 2NH  ( - 100%) (trans-alkene) The dissolution of alkali metal in liquid NH3 produces solvated electrons. The reaction is initiated by the attack of sp hybridized carbon atom of alkyne molecule with a solvated electron (es ) when the - electrons move to the other sp hybridized carbon atom. In order to acquire greater stability the single electron on one carbon atom and the pair of electrons on the adjacent carbon atom orient themselves or far away as possible forcing the two alkyl groups to acquire the farthest position. The carbanion then picks up a proton from NH3 to produce a vinylic radical.Attack by another solvated electon gives vinylic anion which produces trans-alkenes by picking up a proton from NH3 . Hydrogenation of alkynes by Lindlar’s catalyst or nickel boride produces nearly 100% cis-alkene. The catalyst provides a heterogeneous surface on which alkyne molecules get absorbed. Hydrogen molecules collide with adsorbed alkyne to produce cis-alkene in which both the hydrogen atom come from the same side. 2 Lindlar 's cat. 2 or Ni B R C R H     H R R H Hofmann Degradation Method Alkenes can be prepared by heating quaternary ammonium hydroxide under reduced pressure at a temperature between 100°C and 200°C 3 3 2 2 2OH (CH ) N CH CH H O    CH3 N CH3 CH3 CH2 CH2 H   This is the final step of the overall three step reaction called Hofmann degradation. In the first step, primary, secondary or tertiaryamine is treated with enough CH3 I to convert it to the quaternary ammonium iodide. In the second step, the iodide in converted into hydroxide by treatment with Ag2 O. 32CH I  2Ag O    N H CH3 N CH3 CH3 CH3 N CH2 CH3 CH3 H OH N CH3 CH3 I I
  20. 20. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms The OH ion invariably removes proton from  -carbon atom. If two or more alkyl groups have -carbon atoms, OH removes proton from that  -carbon atom which gives more stable carbonion. That means if tetra alkyl ammonium halide contains ethyl group as one of the alkyl groups, then ethene will be the major product in this reaction. For example OH   2 2CH CH CH3 CH CH3 CH2N CH3 H2C CH2CH3 CH2CH3 CH3 CH CH3 CH2N CH3 H2C CH2CH3 PHYSICAL PROPERTIES OF ALKENE: a) All are colouress and have no characteristic odour. Ethene has pleasant smell. b) Lower members (C2 to C4 ) are gases, middle one (C5 to C17 ) are liquids, highers are solids. c) The boiling points, melting points, and specific gravities show a regular increase with increase in molecular weight, however less volatile than corresponding alkanes. d) A cis isomer has high boiling and melting point than trans because of more polar nature. e) Like alkanes, these too are soluble in non polar solvents. f) Alkenes are weak polar. The polaritiy of cis isomer is more than trans which are either non polar or less polar (e.g. trans butene-2- is non polar; trans pentene-2-is weakly polar). CHEMICAL PROPERTIES OF ALKENES: Addition of Hydrogen: Alkenes reacts with hydrogen in the presence of platinum, palladium, rhodium or nickel catalyst to form the corresponding alkane. Pt,Pd,Rh or Ni 2 2 2 2 2R C CR H R CH CHR    2H Pt  Addition of Halogens: Bromine and chlorine adds to alkenes to form vicinal dihalide.Acyclic halonium ion is an intermediate stereospecific anti addition is observed. 2 2 2R C CR X   R R X X R R  CH2 CH3 2Br Br Br CH3
  21. 21. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net Mechanism Step 1: CH3 CH3 CH3 CH3 Br + CH3 CH3 CH3 CH3  – Br Br Br Bromonium ion Bromide ion A bromine molecule becomes polarized as it approaches the alkene. The polarized bromine molecule transfers a positive bromine atom (with six electrons in its valance shell) to the alkene resulting in the formation of bromonium ion. Step 2: Br Bromonium ion vic-dibromide C C Br + CH3 CH3 CH3 CH3 C C CH3 CH3 CH3 CH3 Br Br Bromide ion A bromide ion attacks at the back side of one carbon (or the other) of the brominium ion in an SN 2 reaction causing the ring to open and resulting in the formation of vic-dibromide. Addition of Hydrogen Halide Hydrogen halides (HCl, HBr and HI) add to the alkenes to form alkyl halides. The addition of HX to alkenes is an electrophilic addition reaction. The order of reactivity of hydrogen halides is HI  HBr  HCl. HX  R X R H Mechanism Step – 1 slow RDS H X  C C R R  X Step – 2
  22. 22. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms Fast X   C R C HX C R C The addition takes place according to Markownikov’s rule. In the ionic addition of an unsymmetrical reagent to a double bond, the positive part of the reagent attaches itself to the carbon atom of the double bond so as to yield more stable carbocation as intermediate. Because, this is the step that occurs first, it is the step that determines the overall orientation of the reaction. The reaction in which rearrangement of carbocation occurs, the overall addition of HX to alkenes does not follow Markownikov’s rule. For example the addition of HBr to 3-methylbut-1-ene gives a mixture of 2-Bromo-2-methylbutane and 2-Bromo-3-methylbutane. Br HBr     CH2 CH3 CH3 CH3 CH CH3 CH3 H shift  Br  CH3 CH3 CH3 Br CH3 C CH3 CH3 Br  CH3 CH3Br CH3 2-Bromo-3-methyl butane 2-bromo-2-methylbutane Addition of HBr in presence of peroxide (Peroxide effect) Addition of HBr to unsymmetrical alkenes in presence of organic peroxide is an anti Markownikov’s addition ROOR HBr  CH3 H H H CH3 CH2 CH2 Br Mechanism In presence of peroxide addition of HBr to alkenes follows free radical mechanism       R O O R RO OR       R O HBr R OH Br
  23. 23. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net 3 2CH CH CH Br      CH3 CH CH2 Br CH3 CH CH2 Br (more stable) (less stable) H Br  CH3 CH CH2 Br CH3 CH CH2 Br Br  (2°) (1°) Note:– Peroxide effect is not observed by HCl and HI Addition of sulphuric Acid: Alkene reacts with sulphuric acid to form alkyl hydrogen sulphates. A proton and a hydrogen sulphate ion adds to double bond in accordance with Markovnikov’s rule. 2 2 2 2 2R C CR HOSO OH R — CH — CR   2OSO OH Halohydrin formation: When treated with bromine or chlorine in aqueous solution alkenes are converted to vicinal halohydrin. A halonium ion is an intermediate. The halogen adds to the carbon that has the greater number of hydrogens. Addition is anti. 2 2 2R — CH CR X H O    OH R R R X HX 2 2 Br H O  CH2 OH Br Mechanism     2 2H O X HO X HX CH3 CH3 CH3 CH3 X CH3 CH3 CH3 CH3    – X X X Halide ion C C X CH3 CH3 CH3 CH3 C C CH3 CH3 CH3 CH3 O X H H O H H  O H H
  24. 24. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms Here however a water molecule is a nucleophile and attacks a carbon of the ring causing the formation of a protonated halohydrin. C C CH3 CH3 CH3 CH3 O X H O + H H H The protonated halohydrin loses a proton (it is transferred to molecule water). This step produces the halohydrin and hydronium ion. Epoxidation: Peroxy acids transfer oxygen to the double bond of alkenes to yield epoxides the reaction is a sterospecific syn addition. 2 2 2 2R C CR R — C — OOH R C — CR R — COH    O O O 3CH COOH CH3 CH3 O3CH — COOOH  Mechanism  C C CH3CH3 CH3 CH3  O O H O R C C O CH3 CH3 CH3 CH3  C O H RO Alkene Peroxy acid Epoxide Carboxylic Acid The peroxy acid transfers an oxygen atom to the alkene in a cyclic single step mechanism. The result is the syn addition of the oxygen to the alkene, with formation of an epoxide and a carboxylic acid. Acid-Catalyzed Hydration: Addition of water to the double bond of an alkene takes place in aquous acid.Addition occurs according to Markovnikov’s rule.Acarbocation is an intermediate and is captured by a molecule of water acting as a nucleophile. H 2 2 2 2RCH CR H O R — CH — CR     OH    H 2 3 2 2 3 3H C C(CH ) H O (CH ) COH Mechanism
  25. 25. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net Step 1: CH2 CH3 CH3 H O + H H CH3 C CH3 H  O H H slow The alkene accepts a proton to form more stable 3° carbocation. Step 2: C CH3 CH3 CH3  O H H CH3 C CH3 CH3 O + H Hfast The carbocation reacts with a molecule of water to form a protonated alcohol. Step 3: fast  O H H CH3 C CH3 CH3 O H CH3 C CH3 CH3 O + H H H O + H H A transfer of proton to a molecule of water leads to a product. Hydroboration-Oxidation: The two step sequence achieves hydration of alkene in a stereospecific syn manner with a regeoselectivity opposite to Markovnikov’s rule. An organoborane is formed by electrophilic addition of diborane to an alkene. Oxidation of organoborane intermediate with hydrogen peroxide completes the process. Rearrangement do not occur. 2 6 – 2 2 B H 2 2H O , OH R — CH CR R — CH — CH — CR  OH  3 – 2 2 H B. THF 3 2 2 3 2 2 2H O . OH (CH ) — CH — CH CH (CH ) — CH — CH — CH — OH
  26. 26. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms Mechanism CH3 CH3 H H pi-complex four centered transition state CH3 CH3 H H B HH H  B HH H CH3 CH3 H H H - B H H  –  CH3 CH3 H H H B H H 22 – 22 BH HO,OH  CH3 CH3 H H H OH     Addition takes place through the initial formation of a  complex which changes into a cyclic four center transition state with the boron atom adding to less hindered carbon atom. The dashed bonds in the transition state represent bonds that are partially formed or partially broken. The transition state posses over to become an alkylborane. The other B—H bonds of alkylborane can undergo similar additions, leading finally to a tri-alkylborane. - 3 2 2 2(BH ) H O , OH   CH3 methylcyclopentane CH3 H H OH trans-2-Methyl-1-cyclopentanol syn-Addition Oxymercuration-demercuration Alkenes react with mercuric acetate in the presence of water to give hydroxymercurial compounds which on reduction yield alcohols. 2 2Hg(OAc) H O   OH HgOAc 4NaBH  OH H Markovnikov addition Mercuric acetate The first stage, oxymercuration, involves addition of –OH and HgOAc to the C–C double bond. Then, in demercuration, HgOAc is replaced by H. The reaction sequence amounts to hydration of the alkene, but is much more widely applicable than direct hydration. Oxymercuration-demercuration gives alcohols corresponding to Markovnikov addition of water to the
  27. 27. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net carbon-carbon double bond. (anti addition). For example: 2 2 4Hg(OAc) , H O NaBH   Norbornene OH exo-Norborneol Syn-Hydroxylation Hydroxylation with permanganate is carried out by reaction at room temperature of the alkene and aqueous permanganta solution; either neutral or slightly alkaline. Hydroxylation is a very good method for the synthesis of 1, 2-diols. 2 2 4 23CH CH 2KMnO 4H O 3    22MnO2KOH CH2 CH2 OH OH (ethylene glycol) 4 32 (1)OsO (2)NaHSO/HO  CH2 CH2 OH OH CH3 (propylene glycol) CH – CH = CH3 2 The mechanism for the formation of glycols by permanganate ion and osmium tetroxide involves the formation of cyclic intermediates. Then in several steps cleavage at the oxygen-metal bond takes place ultimately producing the glycol and MnO2 or Os metal. The course of these reactions is syn- hydroxylation. Hydroxylation can also be carried out with peroxy acids such as peroxy formic acid (HCO2 OH) mixture of hydrogen peroxide and formic acid allowed to stand for few hours. Sequently the product is heated with water give glycol.
  28. 28. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms 2 cold 4 H O/OH MnO     O Mn O O O 2 OH H O Several steps   2MnO OH OH 2 cold 4 H O/OH MnO     OH H OH H Oxidative cleavage of Alkenes Alkenes are oxidatively cleaved by hot alkaline permanganate solution or higher conc.of KMnO4 or acid medium (bayer’s reagent) The terminal CH2 group of 1-alkene is completely oxidized to CO2 and water. Adisubstituted atom of a double bond becomes C = O group of a ketone. A monosubstituted atom of a double bond becomes aldehyde group which is further oxidized to salts of carboxylic acids. = CH2 part get oxidized to CO2 + H2 O = CHR part get oxidized to RCOOH = CR R part get oxidized to RCOR Example 4KMnO , OH 3 3 heat (cis or trans) CH CH CHCH 2  H 3 22CH CO H  CH3 O O 4KMnO , OH H 2 2CO H O CH3 CH2 CH3 CH3 O CH3 Ozonolysis of Alkenes A more widely used method for locating the double bond of an alkene involves the use of ozone (O3 ). Ozone reacts vigorouslywith alkenes to form unstable compounds called initial ozonides, which rearranges spontaneouslyto form compounds known as ozonides. Ozonides, themselves are unstable and reduced directly with Zn and water. The reduction produces carbonyl compounds that can be isolated and identified. 3O   2Zn/H O 2O O O O O O O Ozonolysis can be of either of reductive type or of oxidative type. The difference lies in the fact that products of reductive ozonolysis are aldehydes and/ore ketones while in oxidative ozonolysis, the
  29. 29. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net products are carboxylic acids and/ore ketones (this is because H2 O2 formed would oxidise aldehydes to carboxylic acids but ketones are not oxidized). In reductive ozonolysis, we add zinc which reduces H2 O2 to H2 O and thus H2 O2 is not present to oxidise any aldehyde formed. Zn + H2 O2  ZnO + H2 O For example, (CH3 )2 CH – CH = CH2 3 2 2 2 (i) O , CH Cl (ii) Zn/H O (CH3 )2 CHCHO + HCHO (CH3 )2 C = CH –CH3 3 2 2 2 (i) O , CH Cl (ii) Zn/H O (CH3 )2 C = O + CH3 CHO Illustration - 4 : Identify the products (x, y) of following reaction:- 3 2O /H O,Zn (X) (Y) CH3 CH3 CH3 CH3 CH3 Solution: CH3 C CH3 O H C O C CH3 CH3 CH3(X) : (Y) : Substitution Reactions at Allylic Position Cl2 + H2 C = CHCH3 High temperature H2 C = CHCH2 Cl + HCl Br2 + H2 C = CH–CH3  2 Low concentration of Br CH2 = CH – CH2 Br + HBr These halogenations are like free radical substitution of alkanes. The order of reactivity of H-abstraction is allyl  3°  2°  1°  vinyl. Allylic substitution by chlorine is carried out using Cl2 at high temperature and alkene (with  -carbon) in gaseous phase.Allylic bromination can be carried out using N-Bromosuccinimide. Propene undergoes allylic bromination when it is treated with N-bromosuccinimide (NBS) in CCl4 in the presence of peroxides or light. 4 LightorROOR 2 2CCl N Br CH CH CH Br    CH = CH – CH2 3  O O O O NH N-Bromosuccinimide (NBS) 3-Bromopropene (allyl bromide) Succinimide Dimerization
  30. 30. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms acid CH3 CH2 CH3  CH3 CH2 CH3 CH3 CH3 CH3 CH3 CH3 2,4,4-trimethylpent-2-ene  CH3 CH3 CH3 CH2 CH3 2,4,4-trimethylpent-1-ene ALKYNES: Introduction: Unsaturated hydrocarbons having a carbon-carbon triple bond are called alkynes. They correspond to general formula n 2n–2C H i.e. they have smaller proportion of hydrogen than alkene.Alkynes are isomeric with cycloalkenes and alkadienes. NOMENCLATURE:. HC C– CH – CH = CH2 2 pent1-ene4-yne CH3 5 4 3 2 CH 1 CH3 4-methylpent-1-yne CH3 1 2 3 4 5 CH3 6 Cl 2-chlorohex-3-yne CH3 CH CH3 CH Cyclopentylethyne 5-methylhex-1-yne PREPATATIONS OF ALKYNES: Dehydrohalogenation of vic dihalides: Vicinal dihalides having hydrogens on carbon gives alkynes with strong base. 2/NaNH R —CH—CH—R R —C C—R   X X Kolbe hydrocarbon synthesis: Potassium salt of maleic acid and its alkyl derivatives gives alkynes on electrolysis. R —C—COOK R —C—COOK electrolysis 2 2R —C C—R 2CO 2KOH H    
  31. 31. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net Hydrolysis of Carbides Some of the lower members of alkyne series can be synthesized by the hydrolysis of carbides. For example, calcium carbide on hydrolysis gives acetylene and magnesium carbide on hydrolysis gives propyne. CaC2 + 2H2 O  HC  CH + Ca(OH)2 Mg2 C3 + 4H2 O  CH3 –C CH + 2Mg(OH)2 The difference in the behaviour of calcium carbide and magnesium carbide is due to the differences in their structures. Both the carbides are ionic in nature. In calcium carbide, the anion exists as C C while in magnesium carbide, the anion exists as    3 C C C. It is pertient to note that aluminium carbide (Al4 C3 ) and beryllium carbide (Be2 C) do not form any alkyne on hydrolysis, instead they form methane on hydrolysis. This is due to the fact that their anions exist as C4– . PHYSICAL PROPERTIES: i) Lower members (C2 to C4 ) are gases; middle one (C5 to C12 ) are liquids; highers are solids. ii) The boiling point, melting point and specific gravity of alkynes show a regular increase with increase in molecular weight; however less volatile than alkene. The order of boiling point in hydrocarbons has been explained in terms of polarity. Alkynes possess more polarity and thus have higher boiling point. alkyne  Alkene  Alkane iii) All are colourless and possess no characteristic odour; however C2 H2 has garlic odour due to the impurities of PH3 , H2 S etc. Pure C2 H2 has ethereal odour. iv) Soluble in organic solvents like acetone, alcohol and sparingly soluble in water. v) The boiling point of acetylene is –84°C. Liquid acetylene is dangerously explosive and therefore storage and transportation of liquid acetylene is prohibited by law. That is why acetylene is stored and transported by dissolving it in acetone soaked on porous material like asbestos packed in steel cylinders, under high pressure. CHEMICAL PROPERTIES OF ALKYNES: Acidic Character of alkynes The hydrogen bonded to the carbon of terminal alkyne is considerably more acidic than those bonded to carbons of alkenes and alkanes. The pKa values for ethyne, ethene and ethane illustrate this point H H H H H H H H H H H C C H   pKa 25 pKa 44 pKa 50 The order of basicity of their anions is opposite that of their relative acidity Relative Basicity CH3 CH2 –  CH2 = CH–  HC C– From the pKa values it can be concluded that there is a vast difference in the stability of the carbanions. This difference can be readily explained interms of the character of the orbital occupied by the lone pair electrons in three anions. In ethyl anion the lone pair electrons are in sp3 hybrid orbitals, in vinyl anion the lone pair is in the sp2 hybrid orbital and in acetylide ion the lone pair is in sp hybrid orbital. The percentage s-character of the orbital increases in the order sp3  sp2  sp. Greater is s-character of the hybrid orbital containing a lone pair of electrons, the less basic is that anion and the more acidic the corresponding conjugate acid.
  32. 32. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms Relative Acidity                2 2 2 3 16 17 2515.7 38 44 50 pKa :H OH H OR H C C H H NH H CH CH H CH CH Relative Basicity         2 2 2 3OH OR C C H NH CH CH CH CH Reactions due to Acidic Hydrogen i) Reaction with Sodium : Terminal alkynes react with sodium in liquid ammonia or sodamide to form sodium alkynide R – C C – H + Na 3NH (l) R – C C Na + 2 1 H 2 R – C C – H + NaNH2 3NH (l) R – C CNa + NH3 This reaction is utilized for the preparation of higher alkynes. R – C C – H 2 3 NaNH NH (l) R – C C – Na R – C CNa  R X R – C C - R + NaX ii) Reaction with ammonical cuprous chloride solution :- When acetylene is passed through ammonical cuprous chloride solution, a red precipitate of copper acetylide is formed. H – C C – H + Cu2 Cl2 + 2NH4 OH     copper acetylide (Red ppt) Cu C C Cu + 2NH4 Cl + 2H2 O R – C C – H + Cu2 Cl2 + 2NH4 OH     Copper alkynide R C C Cu + 2NH4 Cl + 2H2 O iii) Reaction with ammonical silver nitrate solution :- When acetylene is passed through ammonical silver nitrate solution (Tollen’s reagent) a white precipitate of silver acetylide is formed. H – C C – H + 2AgNO3 + 2NH4 OH     Silver acetylide (white ppt) Ag C C Ag + 2NH4 NO3 + 2H2 O The alkynes can be regenerated from the insoluble salts and the overall process serves as a method for purifying terminal alkynes. iv) Reaction with Grignard reagent: These two reagents reacts with terminal alkyne to form hydrocarbon and new organometallic compound respectively. R — C C — H 3 4CH MgBr R — C C — MgBr CH   3CH Li 4R — C C — Li CH   Addition of Hydrogen Alkynes can be reduced directly to alkanes by the addition of 2H in the presence of Ni,Pt or Pd as a catalyst. The addition reaction takes place in two steps. It is not possible to isolate the intermediate alkene under the above reaction conditions. By using Lindlar’s catalyst , nickel boride or palladised charcoal, alkynes can be partially hydrogenated to alkenes Ni Pt or Pd 3 2 3 2 3CH C CH 2H CH CH CH   
  33. 33. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net Lindlar 's catalyst 3 2 3 2CH C CH H CH CH CH      2-butyne when reduced with Lindlar’s catalyst gives nearly 100% cis-isomer while Na in liquid ammonia gives nearly 100% trans-isomer. 3CH 3CH 3CH 3CH 3CH 3CH C C C C C C H H H H Na in 3liq .NH + 2H (Cis) (trans) lindlar’s catalyst Addition of Halogen Acids Addition of halogen acids to alkynes occur in accordance with Markovnikoff’s rule. Addition of one molecule of halogen acid gives an unsaturated halide, which then adds another molecule of hydrogen halide to form gem dihalides. For example, additon of HI to propyne first gives 2-iodopropane and then 2,2-iodopropane 3CH C CH HI    3 2CH C CH- = I HI  3 3CH C CH- - I I 2-iodopropene 2,2-diiodopropane The order of reactivity of halogen acids towards addition reaction is H I H Br H Cl     Addition of Halogens One or two molecules of halogens can be added to alkynes giving dihalides and tetra halides respec- tively. Chloride and bromide add readily to the triple bond while iodine reacts rather slowly 2Cl 2 2 2CH CH Cl CHCl CHCl CHCl CHCl      2Br 3 2 3 3 2 2CH C CH Br CH CBr CHBr CH CBr CHBr        Addition of water (Hydration) Alkynes cannot be hydrated more easily than alkenes because of their low reactivity towards electro- philic addition reactions. Further when acetylene is passed through 40% 2 4H SO containing 1% 4HgSO at 0 80 C , a water molecule adds upto give acetaldehyde.
  34. 34. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms CH CH + HOH 2 4 4 40%H SO 1%HgSO 2CH CHOH Vinyl alcohol Tautomerises 3CH CHO Acetaldehyde   R C CH HOH   2 4 H SO4 dil HgSO CR 2CH OH (Enolic) Tautomerises CR O 3CH ketone Addition of Boron hydrides Diborane, the simplest hydride of boron reacts with alkyne to form trialkenylborane. Diborane splits into two 3BH units and the addition of 3BH takes place following Markovnikoff’s rule. The addition contin- ues as long as hydrogen is attached to boron atom. 2 6 22R C CH B H 2R CH CH BH        2 2 R C CH R CH CH BH R CH CH BH            2 3 R C CH R CH CH BH R CH CH B        Trialkenylborane on hydrolysis gives alkene    3CH COOH 2Hydrolysis3 3 R CH CH B 3R CH CH B OH      Internal alkynes give rise to alkene where geometrical isomerism is possible. Hydroboration followed by hydrolysis of alkynes gives cis alkene as the major product.  2 6 3 2B H CH CO H 3 R C C R RCH CR B      C C H H(cis) R R Oxidation of trialkenylborane with alkaline 2 2H O results in the formation of carbonyl compounds. Terminal alkynes give rise to aldehydes whereas internal alkynes gives rise to ketones.
  35. 35. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net   2 2H O /NaOH 3 R CH CH B   CH CH OH R 2R CH CHO     2 2H O /NaOH 3 R CH CHR B CH CH OH R O CR R2CH Oxidation Alkynes are oxidised by allkaline 4KMnO . which causes cleavage of C C   resulting in the formation of salts of carboxylic acids. The salts on acidification are converted into acids. internal alkynes give mixture of carboxylic acids while terminal alkynes give a carboxylic acid and the terminal C atom is oxidised to 2CO and 2H O . 4(i)KMnO / OH 3 3 2 2(ii)H CH C CH CH COOH CO H O       4(i)KMnO / OH 3 2 3 3 3 2(ii)H CH C C CH CH CH COOH CH CH COOH       Ozonolysis Reaction of alkynes with 3O gives rise to the formation of ozonide. Hydrolysis of ozonide with 2H O gives a mixture of two carboxylic acids. This is called oxidative ozonolysis. 3 3CH C CH O    O O O 3CH C CH 2H O  3CH COOH HCOOH However, if ozonide is hydrolysed with Zn and 2H O , a diketone is formed. This is called reductive ozonolysis 3 2 3 3CH C C CH CH O     O O O 3CH C C 2 3CH CH 2Zn /H O  O O 3CH C C 2 3CH CH Ethane behaves differently. Ozonolysis followed by oxidative hydrolysis of ethyne gives a mixture of glyoxal and formic acid.
  36. 36. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms 3 2 1.O 2.H O HC CH CHO CHO HCOOH    polymerisation : Acetylene undergoes polymerisation yielding different types of polymeric compounds under different conditions: (i) Cyclic polymerisation : When acetylene is passed through a red-hot metallic tube at 0 600 C , cyclic polymerisation takes place with the formation of benzene 3HC CH 0 Heat 600 C  Benzene Propyne on heating trimerises under similar conditions and forms mesitylene (1,3,5-trimethyl benzene). Cu Tube 3 Heat 3CH C CH   3CH 3CH3CH Mesitylene (ii) Acetylene dissolved in tetrahydrofuran polymerises into cyclo-octa-1,3.5,7-tetrene in presence of  2 Ni CN and under high pressure.  2 Anhydrous Ni CN Tetrahydrofuran (Solvent) 4HC CH  CH CH CH CH CH CH CH CH Cyclo-octatetrene (iii) Linear polymerisation : When acetylene is passed into cuprous chloride solution containing 4NH Cl linear polymerisation occurs forming monovinyl acetylene ad divinyl acetylene 2 2 4 Cu Cl 2NH Cl HC CH HC CH CH CH C CH       Monovinyl acetylene HC CH 3rd molecule 2 2CH CH C C CH CH     Divinyl acetylene Vinyl acetylene on reduction with 2H /Pt in presence of 4BaSO forms buta-1,3-diene.
  37. 37. HYDROCARBONS CHEMISTRY gdvms www.chemadda.net   2 4 H / Pt 2 2 2BaSO H C CH C CH H C CH CH CH       Vinyl acetylene 1,3-Butadiene KEY CONCEPTS Alkanes  Alkanes are saturated hydrocarbons. They are called paraffins. General formula of alkanes is Cn H2n+2  Hydrogenation of alkenes or alkynes in presence of Nickel catalyst at 250°C gives alkanes.  Alkyl halides on reduction with Zn/HCl, Zn/CH3 COOH or red P/HI or of subjected to Wurtz reaction or Frankland reaction give alkanes.  Alkanes are formed by the reduction of alcohols, aldehydes, ketones & carboxylic acids with HI & Red P.  Decarboxylation of sodium or potassium salt of carboxylic acid gives alkanes.  Grignard’s reagent react with compound containing active hydrogen like H2 O; C2 H5 OH, CH  CH, NH3 etc give alkanes.  Alkanes containing odd number of carbon atoms can be prepared by corey-house synthesis.  Melting points, boiling point and density of alkanes increase with the increase in molecular weights.  Straight chain alkanes have higher boiling point than branched chain alkanes.  Due to strong and non polar C—C and C—H bond alkanes are not very reactive.  Halogenation, nitration and sulphonation of alkanes follow free radical mechanism.  In the halogenation of alkanes the reactivity with respect to halogens follows the order F2  Cl2  Br2  I2 . Reactivity of with respect to C—H follows the order teretiary hydrogen  secondary hydrogen  pri mary hydrogen. Alkenes  Alkenes are unsaturated hydrocarbons having one carbon-carbon double bond.  Alkenes are called olefins. General formula of alkenes is Cn H2n .  Dehydration of alcohols leads to the formation of alkenes. Dehydrating agents are H2 SO4 , P2 O5 ,Al2 O3 , H3 PO4 and ZnCl2 . The ease of dehydration follows the order 3°  2°  1°.  Dehydrohalogenation of alkyl halides gives alkenes. The ease of dehydro-halogenation is 3°  2° 1°.  Alkenes can be prepared by the controlled partial hydrogenation of alkynes with Lindlar’s catalyst.  Dehalogenation of vic-dihalides with Zinc dust gives alkenes.  The formation of more substituted alkene as major product is called Saytzeff elimination.  The formation of less substituted alkene as major product is called Hofmann elimation.  The stability of alkenes follow the order R2 C = CR2  R2 C = CHR  R2 C = CH2  RCH = CHR  R—CH = CH2  CH2 = CH2 .  Trans-but-2-ene is more stable than cis-but-2-ene.  Olefines show electrophilic addition reaction.  Alkenes give addition product with halogens to form vic-dihalides.  Alkenes decolourised by i) Baeyer’s reagent ii) Br2 water.
  38. 38. HYDROCARBONS CHEMISTRY www.chemadda.net gdvms These reactions are the test of unsaturation.  Alkenes react with HX to form alkyl halides. The order of reactivity of HX is HI  HBr  HCl  HF.  Addition of HX to unsymmetrical alkenes takes place according to Markownikov’s rule.  Addition of HBr to unsymmetrical alkenes in presence of peroxide follows anti-Markownikov’s addition.  HCl and HI do not give anti Markownikov’s addition.  Alkenes on reductive ozonolysis give carbonyl compounds.  Hydroxylation of alkenes by cold aqueous alkaline KMnO4 solution (Baeyer’s reagent is a syn-addition.)  Hydroboration-oxidation of alkene is an anti markownikov’s addition of water to a double bond.  Oxymercuration-demercuration is a markownikov’s addition of water to a double bond.  In Oxymercuration-demercuration reaction and hydroboration-oxidation reaction no rearrangement takes place.  Oxidation of alkenes by acidic KMnO4 or acidic K2 Cr2 O7 give ketone or carboxylic acid or both.  Periodic acid (HIO4 ) or leadtetra acetate [(CH3 COO)4 Pb] oxidizes alkenes into glycols and finally gives aldehydes or ketones depending upon the nature of alkene.  Oxidation of alkenes by SeO2 affect the allylic position.  NBS (N-Bromosceccinimide) is used for the bromination of alkenes at the allylic position. Alkynes  Alkynes are unsaturated hydrocarbons with one carbon-carbon triple bond. General formula Cn H2n-2 .  Dehydrohalogenation of vic-dihalides give alkynes.  Dehalogenation of 1, 1, 2, 2-tetrahalides with Zn dust and alcohol give alkynes.  Acetylene can be prepared by the action of water on calcium carbide.  Magnesium carbide on hydrolysis give propyne.  The acidic character of Ethyne, Ethene and Ethane follows the order H – C  C – H  CH2 = CH2  CH3 – CH3 .  The basic character of their conjugate base follow the order CH  C–  CH2 = CH–  CH3 —CH2 –  Alkynes are less reactive towards electrophilic addition as compared to alkenes.  Acetylene and terminal alkynes react with ammonical silver nitrate solution (Tollen’s reagent) to give corresponding alkynides (white ppt.).  Acetylene and terminal alkynes react with ammonical cuprous chloride solution to give a red precipitate of corresponding alkylnides.  Hydrogenation of alkynes in presence of Lindlar’s catalyst give cis-alkene where as hydrogenation by Na/Liq. NH3 give trans alkenes. O Ozonolysis [O] 3 2 2 3 3 2 (F) (D) CH CH — C — CCH CH CH CH COOH  O (E)