C1-Allens Made Chemistry Theory {PART-1}

ALKANEIntroduction :(i)Branched and unbranched aliphatic saturated hydro carbons are called member of alkane. Thestructural formula of alkane have only single bonds or all bonds in alkane is only bonds.(ii)Alkanes does not reacts with chemical reagants such as dil. and conc. HCl, dil. & conc. H SO ,24dil. & conc. HNO , Caustic soda, acidic & basic K Cr O , KMnO etc. That is why alkanes are32274called paraffins. (Parum=little, affins = reactivity).Prop er t yCharacteristicsProp er t yCharacteristicsof alkaneof alkaneGeneral formulaC Hn2n+2C—C Bond length1.54 A°C—C Bond energy82.67 kcal/moleC—H Bond length1.112 A°C—H Bond energy98.67 kcal/moleHybridisation on Csp3Bond angle109°.28'shapeTetrahedralGeneral Methods of Preparations :1 .From alkenes and alkynes (Sabatier and Sandrens reaction) or (By hydrogenation of alkenesand alkynes) : Alkenes and alkynes on catalytic hydrogenation gives alkanes.R—CHCH—R + H2CatalystR —C H —C H — R22Alkene AlkaneR—CC—R + 2H2CatalystR —C H —C H — R22AlkyneCatalyst :(a)Pd/Pt at ordinary temp. and pressure(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 dividedstate.(e)Methane can not be prepared by this method (From unsaturated hydrocarbon).2 .From alkyl Halides (By reduction) :R—X 2H(Nascent Hydrogen) R—H + HXCatalyst :(i) Zn + HCl(ii) Zn + CH COOH3(iii) Zn—Cu couple in C H OH25(iv) Red P + HI(v) Al + Hg + ethanolMechanism : Zn  Zn + 2e+2 R — X  2eR + X R + H Cl –+– R—H + ClZn + 2Cl +2– ZnCl .2ProductHY DROC ARB ON

(a)Alkyl halides can also be reduced to alkane by H /Pd or LiAlH or H /Ni.242(b)Reduction is due to the electron transfer from the metal to the substrate (R- X)(c)If any alkyl halide is asked, the H-atom of any carbon atom of given alkane is removed by halogenatom.3 .From alkyl halide (By Wurtz reaction):A solution of alkyl halide in ether on heating with sodium gives alkane.DryetherR — X2NaX — RR — R2NaX(a)Two moles of alkyl halide treated with Na in presence of dry ether. If ether is wet then we obtainalcohol.2Na +H O22NaOH + H2CH I +3NaOHCH OH + NaI3Methanol(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.(c)Two different alkyl halides, on wurtz raction give all possible alkanes.(d)The seperation of mixture in to individual members is not easy because their B.P. are near toeach other and thus wurtz reaction is not suitable for the synthesis of alkanes containing oddnumber of carbon atom.(e)This reaction generally fails with tertiary alkyl halide.Mechanism : Two mechanism have been proposed for this reaction.( a )Ionic Mechanism:2Na 2Na + 2e R — X  2eR + X  R — X + R R — R + X–2Na + 2X 2NaX ProductExample : 2C H —I25+2NaC H —C H + 2NaI2525n–butane( b )Free radical mechanism :Na Na + e R — X e R + X.RRR — RNaXNaXProductFree radicals also undergo Disproportionation i.e. one radical gains hydrogen at the expense of the otherwhich loss hydrogen.CH — CH + CH2225C H + CH2624...H. Ethane EthyleneThis explains the presence of ethylene and ethane in the butane obtained by Wurtz reaction.

E x .If two moles of Isopropyl chloride reacts with Na in presence of dry ether. Which alkane is obtained.S o l . 2, 3-Dimethyl butane.E x .If isopropyl chloride and ethyl chloride both react with Na in presence of dry ether which alkanes areobtained.S o l . n-Butane, 2-Methyl butane and 2, 3-Dimethyl butane.E x .Which of the following compound can not obtained from wurtz reaction.(A) ethane(B) butane(C) isobutane(D) hexaneSo l .(C) [Hint : In wurtz reaction unsymmetrical alkane can not be obtained.E x .When ethyl chloride and n-propyl chloride undergoes wurtz reaction which is not obtained.(A) n–butane(B) n–pentane(C) n–hexane(D) isobutaneS o l . (D) C H —Cl + C H Cl 2537Nadry ether C H —C H + C H —C H + C H —C H25 253737 25374 .Corey-House Synthesis :This method is suitable for the preparation of unsymmetrical alkanes i.e. those of type R—R'(i)RX + Li RLi + LiX(ii)2RLi + CuX R CuLi + LiX2(iii)R CuLi + R 'X2(1 or2 ) R—R' + RCu + LiXNote : In Corey-house reaction symmetrical and unsymmetrical alkane both can be formed.5 .From Frankland Reagent: If Zn is used in place of Na, the reaction is named as Frankland reaction.R—X + 2Zn +RX R Zn + ZnX22 Frankland reagentR Zn + R—X 2R—R + RZnX6 .From Carboxylic Acid (By decarboxylation) : Saturated monocarboxylic acid salt of sodium or potassiumon dry distillation with soda lime give alkane.RCOONa + NaOH CaoR—H + Na CO23Soda 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 theproducts.(e)Decarboxylation of sodium formate gives H22233423HCOONa + NaOH (CaO)H + Na COCH COONa + NaOH + CaOCH + Na CO

Mechanisim : Decarboxylation proceeds via. the formation of carbanion intermediate as follows.CH C O + OH(NaOH)3ONaONaOHCH 3CO ONaOHCH 3CO CH + HO C O (NaHCO)3 ONa3(Acidic)CH + H O3 COONaCH 3CO H + OONaProductCO OO Na + NaNa CO23(a)If in a compound two carboxylic groups are present and they are attached to same carbon atomthen also decarboxylation of one of the carboxylic groups takes place simply on heating.CH 2COOHCOOH CH COOH + CO32(b)CH can be prepared by CH COOH.43(c)C H can be prepared by CH CH COOH.2632(d)CH —CH —CH can be prepared by Butanoic acid and 2–Methyl propanoic acid.3 2 3E x .How many acids can be taken to obtain isobutane from decarboxylation ?(A) 4(B) 3(C) 2(D) 5S o l . (C)To obtain isobutane the acids are(i)CH 3CH CH2COOHCH 3CH 3CH CH2HCH 3(ii)COOHCH 3CCH 3CH 3HCH 3CCH 3CH 3So two acids can be taken.Reactivity of acid stability of carbanion 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.Example :(i)RCH CHCOOH2OH3 NaHCORCH CH2OHH(ii)RCH CHCOOH2NO 23 NaHCORCH CH2NO 2H

(iii)-Keto acids are decarboxylated readily simply on heating (soda lime is not required)C CHCOOH O R2C CHO R3E x .Give reactivity order for decarboxylation ?CH —CH —COOH32CH2CH—COOHCHC —COOH I IIIII(A) I > II > III (B) III > II > I(C) III > I > II(D) None is correctS o l . (B)In decarboxylation intermediates are,32 CH — CHCH2CHCHC III IIIThe stability order of carbanion – III > II > ISo reactivity order for acid is – III > II > I7 .From carboxylic acid (By Kolbe's process) :Alkanes are formed on electrolysis of concentrated aqueous solution of sodium or potassium salt of saturatedmonocarboxylic acids.2RCOONa + 2H O 2Electrolysis22At AnodeAt CathodeR — R2CO2NaOH H Electrolysis of Sodium propionate solution give n-butane, ethylene, ethane and ethyl propionate as follows-25252522electro.— COONa— C H2C HC H2CO2NaOHHMechanism :C H —COONa252electro.H OC H25—COO + Na  (Ionization)At Anode :C H —COO25 –eC H —25COOCH25OO C.Fragmentation52C H  + CO252C H + 52C H  C H —C H2525ProductCH — CH + CH2225...H.CH —CH —H + CH322CH2(minor products)An ester is also formed.C H —25CO O + 52C H  C H —COOC H2525(minor products)At cathode :Na + e NaNa + H O 2 NaOH + 122 H

(a)Methane can not be prepared by this method.(b)Electrolysis of an acid salt gives symmetrical alkane, However in case of a mixture of Carboxylicacid salts, all probable alkanes are formed.R'COOK + R\"COOKElectrolysis(R'—R\" + R'—R' + R\"—R\") +2CO + H +2NaOH22(c)Presence of alkyl groups in - position decrease the yield of alkanes.(d)True aromatic acids do not undergo Kolbe's electrolytic reaction.(e)Free radical mechanism has been suggested for Kolbe reaction.(f)At anode alkane (major) and CO gas is formed while at cathode NaOH and H gas is formed.22(g)The concentration of NaOH in solution is increased with time so pH of solution is also increased.8 .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.R—OH +2HIRe d P150 CR—H+H O+ I22R—CHO + 4HIRe d P150 CRCH +H O + 2I322R—CO—R+4HIRe d P150 CR—CH —R+H O + 2I222RCOOH + 6HIRe d P150 CR—CH —R+H O + 3I222In the above reaction I is formed which acts as reducing agent and may reduce alkane and form alkyl2halide. So red P is added in the reaction to remove I formed in the reaction.2R—CH + I32R—CH —I + HI22P + 3I22PI39 .From alkanones (By Clemmensen's method) :Carbonyl compound (Preferably ketones) may also be reduced with Zinc amalgam and concentrated HCl(Zn—Hg/HCl), this reaction is called Clemmensen reduction.R—CO—R'+4HZn / Hgcon. HClR—CH —R'+H O22CH , CH —CH , isobutane and neopentane are not obtained from Ketones because these alkane do433not contain CH group.21 0 .From alkanals and alkanones (By Wolf Kishner reaction) :CO + NH NH22CN.NH 2Glycol / KOHCH + N22HydrazineHydrazoneFrom G.R. :(a)Formation of alkanes with same number of C atoms : With same number of C-atoms asG.R. react with compound containing active hydrogen alkanes is obtained.RMgX + HOHR H + Mg (OH) XOHR H + Mg (OR) X + RNHHR H + Mg (NHR) X + RThis reaction is used to determine the number of active H-atoms in the compound this is knownas Zerewitnoff's method.(b)G.R. react with alkyl halide to give higher alkanes :RMgX + R'—XR—R' + MgX2

E x .Which of the following does not give alkane with R—Mg—X.(A) Ph—OH(B) Cl —NH2(C) CH COOH3(D) HClS o l . (B)[Hint : Except Cl—NH all have active hydrogen, but Cl—NH when reacts with R—Mg—X the product22is R—NH .]2From metal carbide (By hydrolysis) :Only CH can be obtained by the hydrolysis of Be or Al carbides4432Al C12H O344Al(OH)3CH22Be C4H O242Be(OH)CH1 1 .Physical Properties :(i)C to C gases, Neopentane also gas but n-pentane and isopentane are low B.P. liquids.14(ii)Next members C to C are Colourless liquids and above C51717are Waxy solids.(iii)Density : The density of alkanes increases with increase in molecular weight and becomes constantat 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 solventsExample : C H , CCl ,ether etc.664The solubility of alkanes decreases with increase in molecular weightLiquid alkanes are themselves good non-polar solvents.(v)Boiling point -  molecular weight (for n-alkanes)Vanderwaals force of attractionmolecular weightsurface area of molecule.i.e.boiling pointPentane < hexane < heptaneAlso boiling point1number of side chain because the shape approaches to spherical which results in decrease in Vanderwaals forces (as surfacearea decreases)Thus boiling point n–Pentane > Isopentane > neopentane(vi)Melting Point : M.P. of alkanes do not show regular trend. Alkanes with even number ofcarbon 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 carbonatom on the same side of the molecule and in even carbon atom alkane the end Carbon atomon opposite side. Thus alkanes with even carbon atoms are packed closely in crystal lattice to permitgreater intermolecular attractions.CCCCCCC<CC CCCCOdd number of carbonEven number of carbonE x .Alkanes are inert is nature, why ?S o l . 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 byacid, oxidants under ordinary conditions.(ii)The C—C bond is completely non polar and C—H is weak polar. Thus polar species i.e. electrophilesor nucleophiles are unable to attack these bonds under ordinary conditions.

1 2 .Chemical Properties :Oxidation :Complete oxidation or combustion : Burn readily with non-luminous flame in presence of airor oxygen to give CO and water with evolution of heat. Therefore, alkanes are used as fuels.2n2n 22C3n 1HO2  nCO + (n+1) H O + Q; ( H =–ve)22Incomplete oxidation : In limited supply of air gives carbon black and CO.2CH + 3O422CO + 4H O2CH + O42C + 2H O2C–black (used in printing)Catalytic oxidation :(i)Alkanes are easily converted to alcohols and aldehydes under controlled catalytic oxidation.2CH + O 42Red hot Cu or Fe tubeHigh P and T2CH OH3CH + O42230Mo O230 C,100atmHCHO + H O2(ii)Alkanes on oxidation in presence of manganese acetate give fatty acids.32 n332 n(CH COO) Mn32high TempCH (CH ) CHCH (CH ) COOH (iii)Tertiary alkanes are oxidized to give tertiary alcohols by KMnO .4C HCH 3CH 3CH 3[O]KMnO4C OHCH 3CH 3CH 3E x .How many litre of Oxygen required for complete conbustion of 6.0 g ethane at NTP ?S o l .262222C H7O4CO6H O60 g ethane required O (at NTP) = 7 × 22.4 litre21 g ethane required O (at NTP) = 27 22.4litre2 30 6 g ethane required O (at NTP) = 27 22.46 15.68 litre2 30  Substitution Reactions : Substitution reaction in alkanes shows free radical mechanism.They give following substitution reaction.( a )Halogenation : Replacement of H-atom by halogen atomR—H + X2R—X + HXHalogenation is made on exposure to (halogen + alkane) mixture to UV or at elevated temp.The reactivity order for halogens shows the order.2222 F > Cl > Br > IReactivity order of hydrogen atom in alkane isTertiary C – H > Sec. C – H > primary C – H(i)Fluorination : Reacts explosively even in dark. Fluorination can be achieved without violencewhen alkane is treated with F diluted with an inert gas (like N )22By the action of HgF on bromo or iodo derivatives.2C H I + HgF 252 C H F + HgI252

(ii)Chlorination :2222ClClClCl432234CHCH ClCH ClCHClCCl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.42Explosively CH + Cl C + HClMechanism for CH + Cl 42U VCH Cl + HCl3Step IChain initiation step :U VorCl:ClClCl Step IIChain propagation step :Cl + H : CH3H: Cl + CH3..MethaneMethyl radicalCH + Cl : Cl3CHCl + Cl3..Step IIIChain termination step :2ClClCl, 33CHClCH Cl,3333CHCHCH CHA Cl can also attack CH Cl to form chloromethyl ( CH Cl ) free radical. This free radical 32participates further in the chain reaction to yield methylene chloride (dichloromethane).C ClHCl + H:H.andC + Cl : Cl HH .ClSimilarly, chloroform and CCl are obtained by further chain reaction.4(iii)Bromination : Br reacts with alkanes in a similar manner but less vigorously.2(iv)Iodination : Iodine reacts with alkanes reversibly. HI formed as the by product is a powerfulreducing agent and is capable of reducing the CH I to CH .34Iodination may be carried out in the presence of an oxidising agent such as HIO , HIO , HNO ,343HgO etc. Which destroy HI,423CHICH I HI5HI + HIO 33I + 3H O22Iodination is very slow because energy of activation of the reaction is very large43CHIHI+ CH Note :Halogenation is inhibited in presence of oxygen because oxygen reacts with alkyl free radicalsto form less reactive peroxy alkyl radical R–O–O which can not propagate the chain.•Reactivity selectivity Principle :(i)Probability factor : The factor is based on the number of each kind of H atom in themolecule.For example is CH —CH —CH —CH there are six equivalent 1° H's and four equivalent 2°3223H's.The probability of abstracting 1° H's to 2° H's is 6 to 4. i.e., 3 to 2.

(ii)Reactivity of halogen free radical : the more reactive chlorine free radical is less selective andmore influenced by the probability factor. On the other hand, the less reactive Br radical ismore selective and less influenced by the probability factor (Reactivity selectivity principle).(iii)Reactivity of alkanes (ease of abstration of 'H' atoms) : Since the rate determining step inhalogenations is abstraction of hydrogen by a halogen atom be the formation of alkyl radical,halogenation of alkanes follows order of stability of free radical is 3° > 2° > 1° > CH .3Reactivity ratio of H atom for Chlorination(1° :2° :3° H) 1 : 3.8 : 5Reactivity ratio of H atom for bromination(1 : 82 : 1600)The above order of stability of radicals is due to the ease of their formation from thecorresponding alkane which in turn is due to difference in the value of H. CH —H3••3 CHH H 1CH —CH —H32••32 CH — CHH H 2CH —CH—CH33HCH —CH—CH +H33•• H 3CH —C—CH33HCH 3CH —C—CH +H33CH 3•• H 4Reactivity of any H-atom  number of H atoms of that kind × reactivity of that H.Thus the amount of energy required to form the various classes of radicals decreases in theorder CH > 1° > 2° > 3° ( H > H > H > H ). Therefore, it easiest to form 3° radical3 1 2 3 4and it is most difficult to form CH . We can also interpret this in an alternative way the case3of abstraction of H atoms from hydrocarbon fallows the sequence 3° > 2° > 1° CH which4should also be the case of formation of free radicals.The above order of stability is in accordance with the stability of free radicals on the basisof delocalization of odd electron. Order of stability of free radical is :Allyl, benzyl > 3° > 2° > 1° > methyl, vinyl.E x .What is the percentage of products obtained from monobromination of isobutane ?S o l .CH CH + Br CH 3CH 33 2C CH + CHBrCH 333CH 3CH CH2BrCH 3(I)(II)Pr oduct (I)Pr oduct (II)= No.of primary HNo.of tertiary H× reactivity of primary Hreactivity of tertiary H =9191 1600 1600 % of product (I) = 9100 0.56%1600 9% of product (II) = 1600100 99.44%1600 9( b )Nitration : (Vapour phase nitration) This involves the substitution of a hydrogen atom ofalkane with -NO group.2

At ordinary temperature, alkanes do not react with HNO . But reacts with vapours of Conc.3HNO at 450 C.3°0400 500 C222R — HHO — NOR — NOH OSince the reaction is carried at high temp. the C—C bonds of alkanes break during thereaction and a mixture of nitroalkanes is formed.Example : CH —CH + HNO3330450 CCH CH NO + CH NO + H O322322CHCHCH + HNO32331–Nitro propane 25%2–Nitro propane 40%Nitro ethane 10%Nitromethane 25%450°CMechanism : (Free Radical substitution)Step – I22HONOO HN OR – H + OH R  + H O2Step – II22(Pr oduct)R + HO – NOR – NO + OHStep – III22R + NOR – NOR + OR – OH (minor)2OH + NHONO ( c )Sulphonation : Replacement of H atom of alkane by –SO H is known sulphonation.3Alkane react with fuming H SO or oleum (H S O ).242 27The branched lower alkanes and higher alkanes react to give alkane sulphonic acid.Example :C H + HO SOHCH 3CH 3CH 33C SOH + HOCH 3CH 3CH 3322–Methyl propane2–methyl propane–2–sulphonic acidThe reactivity order for sulphonation istert. H > Sec. H > prim. HMechanism : (Free Radical substitution)33400 CHOSO HHOSO H 6613132C H HOHC HH O61336133C HS O HC H SO H

Lower members such as propane, butane, pentane etc. react with SO in vapourphase to3form sulphonic acids.C H + SO 383 C H – SO H373( d )Chlorosulphonation (Reed reaction) : Reaction with a mixture of SO and Cl at ordinary22temp. in the presence of UV light is called chlorosulphonation.3822372UVC HSOClC H SO ClHClPropane sulphonyl ChlorideFurther hydrolysis of alkane sulphonyl chloride gives alkane sulphonic acid.2372373H OC H SO ClC H SO HHClpropane sulphonic acid373373C H SO HNaClC H SO NaHClSodium salt of sulphonic acid (used as detergent)Isomerization :Unbranched chain alkanes on heating with AlCl + HCl / 200 C are converted in to branched chain30alkanes33223AlClHClCH — CH — CH — C HCH CHCH 3CH 33n-butane IsobutaneBranched chain alkanes converted to more branched alkane.CH CHCH 3CH 32CH 2CH 33 AlClHClCH CH CHCH CH3CH 333Isomerisation of alkanes is of great importance in petroleum industry to increase the octane numberof petrol (gasoline).Pyrolysis or Cracking or thermal decomposition :When alkanes are heated to 500-700 C they are decomposed in to lower hydrocarbon. This0decomposition is called pyrolysis. In petroleum industry it is also termed as cracking.Cracking is usedfor the manufacture of petrol, petrol gas/oil gas etc.Example : CH1000 C42CH 33222500 Cabsence of airCHCHCHCHHCHCHCH323CH 2CH + CH2 4CH CH CH + H322n-Butane Cracking1-Butene + 2–Butene + Ethane + Ethene + Propene + CH + H42The mechanism of pyrolysis occurs via free radicals.

Hydroforming or dehydrogenation or cyclisation or catalytic reforming or aromatization :Unbranched higher alkanes (from 6 to 10 carbon atoms) when heated in presence of oxides of Cr, Mo,V on Al O support at 500 C aromatic hydrocarbons are formed.230500 CCr O / Al O2 32 3nhexane+4H2232332 53500 CCr O / Al OCH (CH )CHCH 3+ 4H2n-heptane Toluene2323 O / Al O32 63500 C CrCH(CH ) – CHCH 3CH 3n-octaneo-xyleneIt provides an excellent method of passing from aliphatic to aromatic series.Chlorinolysis :32322640300 400 CPressureCHCHCHClC ClCClHCl(g)( )(s) 

SOLVED EXAMPLESEx .1Which of the following reactions can be employed for getting unsymmetrical alkanes in good yield ?(A) Wurtz reaction(B) Corey–House reaction(C) Both(D) None of theseAns.(B)Sol.Wurtz reaction is suitable for symmetrical alkanesEx .2Sodium propionate on decarboxylation with sodalime gives(A) Propane(B) Ethane(C) Butane(D) PentaneAns.(B)Sol.Decarboxylation with soda lime results in the formation of alkane with one carbon less than the startingcompoundsEx .3Which of the following alkanes cannot be produced by Kolbe electrolysis of sodium or potassium salts ofcarboxylic acids ?(A) Methane(B) Ethane(C) Butane(D) HexaneAns.(A)Sol.In Kolbe electrolysis, the alkane is formed by union of two alkyl groups. The alkane formed has, thus, twoor more carbon atoms.Ex .4The homolytic fission of hydrocarbon results in the formation of -(A) Free radicals(B) Carbocations(C) Carbanions(D) Carbenes.Ans.(A)Sol.Homolytic fission results in the formation of free radicals.Ex .5n-Heptane when heated to a temperature of about 800 K under high pressure in the presence ofCr O /Al O catalyst gives -2323(A) 1-heptene(B) 2-Methylhexane(C) Toluene(D) Xylene.Ans.(C)Sol.CH —(CH ) —CH 32 532 4HCH 3TolueneEx .6The reaction conditions leading to the best yield of C H Cl are -25(A) C H (excess) + Cl 262UV light(B) C H + Cl 262Darkroom temperature(C) C H + Cl (excess) 262UV light(D) C H + Cl 262UV lightAns.(D)Sol.C H should be used in excess, otherwise polychlorination will take place26Ex .7Number of isomer which can be theoretically obtained on monochlorination of 2-methylbutane is -(A) 1(B) 2(C) 3(D) 4Ans.(D)Sol.CH 3CH—CH—CH—CH3231 21 3 4Ex .8Complete oxidation of ethane yields -(A) Ethanol(B) Ethanoic acid(C) Ethanal(D) CO and H O2 2Ans.(D)Sol.2C H + 7O 262 4CO + 6H O22

Ex. 9In iso-pentane, the H atom that can be most easily substituted is on -CH 3CH—CH—CH—CH323 3 1 2 4 (A) C—1(B) C—2(C) C—3(D) C—4Ans.(B)Sol.Ease of substitution of various types of H atoms is 3 > 2 > 1 .000E x . 1 08 c.c. of gaseous hydrocarbon requires 40 c.c. of O for complete combustion. Identify hydrocarbon.2S o l .Volume of hydrocarbon = 8 c.c. ; Volume of O = 40 c.c.2Formula No. 1,82 = 403n+ 1(For alkane)12 = or 3n + 1 = 1053n+ 1 or3n = 10 - 1 = 9 , n = 3The value of n comes in whole number from 1st formula it means hydrocarbon is Alkane andit is of 3C atom. Hydrocarbon is C H (Propane)38E x . 1 110 mL of a mixture of CH and C H requires 41 mL of oxygen for complete combustion.438What is the volume of CH and C H in the mixture.438S o l .Suppose the volume of CH in (CH + C H ) mix = x c.c.4438= Volume of C H will be = 10 – x c.c.38For CH CH442222OCO2H O1 Vol. of CH requires 2 vol. of O for complete combustion42x c.c. of CH , 2x c.c. of O4 2For C H C H +5O 38382 3CO + 4H O221 volume of C H requires 5 ml of O for complete combustion382(10 – x) c.c. of C H requires 5(10 – x) c.c. of O382Total Volume of O = 2x + 5 (10 – x) it is equivalent to 412(according to question)2x + 5 (10 – x) = 41x = 3 c.c.Volume of CH is 3 c.c. and volume of C H is 7 c.c.438E x . 1 2If 5 g C H I reacts with Na (Metallic) in presence of ether, and the yield is 60% then how many25grams of n-butane will you get.S o l .254102C H I 2NaC H2NaIMolecular weight of 2C H I = 24 + 5 + 127 = 15625Molecular weight of C H410 = 48 + 10 = 58Two molecule of C H I are taking part in above reaction.25

We get 58 g of C H from 2 x 156 g of C H I41025We get 582x156 g C H from 1 g of C H I41025We get 58x52x156 g C H from 5 g of C H I41025yield is 60%So the quantity of C H410 will be 58x560x2x156 100 g = 0.55 gE x . 1 3The density of one hydrocarbon at N.T.P. is 1.964 g/litre. Identify the hydrocarbon.S o l .Molecular weight of Hydrocarbon= density of 1 lit. x 22.4 = 1.964 x 22.4 = 44So Molecular weight of hydrocarbon = 44So the hydrocarbon is C H (Propane).38

ALKENEIntroduction :Alkenes are hydrocarbons with carbon-carbon double bonds, Alkenes are sometimes called olefins, a termderived from olefiant gas, meaning 'oil forming gas'. Aleknes are among the most important industrial compoundsand many alkenes are also found in plants and many alkenes are also found in plants and animals. Ethyleneis the largest-volume industrial organic compound, used to make polyethylene and a variety of other industrialand consumer chemicals. Alkenes polymerises to give many important polymers.Structure and bonding in Alkenes :(a)Alkenes are unsaturated hydrocarbons having at least one double bond.(b)They are represented by general foumula (G.F.) C H (one double bond)n2n(c)In Ethene C = C bond length is 1.34 Å(d)Its bond energy is 146 kcal. mol–1(e)The hybridization of (C = C) alkenic carbon is sp2(f)The e cloud is present above and below the plane of s-bonded skeleton.–(g)They are also known as olefins since ethene, the first member of the homologous series forms oilyliquid substance when treated with halogens.(h)Compounds may exist as conjugated polyenes or as cumulated polyenes or as isolated polyenes117.2°(b)121.2°(a)1.34Å1.10ÅC — C HHH HC — C HHH HNote :That angle a < b since repulsion due to electrons (double bond - single bond repulsion > singlebond - single bond repulsion according to VSEPR theory.IUPAC Nomenclature of alkenes and alkadienes :Table - IS.No. Compound Name Type1.(CH ) C = CH3 222–MethylpropeneAlkene2.CH –CH=CH–CH –CH=CH322Hexa–1, 4–dieneIsolated diene3.CH =CH–CH = CH22Buta–1, 3–dieneConjugated diene4.CH –CH=C=CH–CH33Penta–2, 3–dieneCumulated diene5.CH–CH–CH=CH–CH–CH–CH3223Cl12345676–Chlorohept–3–eneAlkene6.CH=CH–CH=C–CH2312345OCH34–methoxypenta–1, 3–dieneConjugated diene7.CH=CH–CH–C CH2123CH 45CH 23–Ethynylpenta–1, 4–dieneIsolated diene

S.No. Compound Name Type8.CH–CH–C = C–CH–CH3223654CH CH3253213–Ethyl–4–methylhex–3–eneAlkene9.CH 3CH 31234562, 3–Dimethylcyclohex–1–eneCycloalkene10.CH = C = CH22PropadieneCumulated diene11.CH = C = O2EthenoneAlkene12.MethylenecyclopentaneAlkene13.CyclopentyletheneAlkene14.3, 7, 11–trimethyldodeca–Isolated triene1, 6, 10–trieneEx.Write IUPAC names of :(a) CH 3CH 3(b) So l. (a) 2, 3–Dimethylcyclohexene; (b) 1–(2–butenyl) cyclohex–1–eneEx.Give the structure for each of the following(a) 4–Methyl–1, 3–hexadiene ;(b) 1–IsopropenylcyclopenteneSo l. (a) ; (b) Isomerism :Alkenes show chain, Ring chain or functional, Position, Geometrical isomerism and optical isomerism.For more details refer to isomerism provided to you in study material.E x .What is relation between CH CH3CH , 2 CH 2CH 2CH 2 ?S o l . Ring chain isomerismE x .(a) CH CH CH32CH2(b) CH —CH3CH—CH3 (c) C CHCH 3CH 32(d) CH 2CH 2CH 2CH 2Define relations between a,b,c,d ?

S o l . a and b – Position isomerism ; a and c – Chain isomerisma and d – Ring chain isomerism ; b also show Geometrical isomerismCH 3C CCH 3HHCH 3C CCH 3HHcis–2–butene trans–2–buteneE x .How many minimum carbon atom persent in optically active alkene?S o l . 6C, C CH CH HCH 2CH 3CH 32*3–Methyl–1–PentenePreparation of Alkenes :( 1 )From Alcohols : Alkenes can be prepared from monohydric alcohols or alkanols by the loss ofH O and the reaction is known as dehydration reaction.2C C H OH   C C + HO2Alkene AlcoholThe dehydration can be carried with Al O or with mineral acid upon heating.23( a )Dehydration with Al O : Ethene is prepared by heating ethanol with Al O at 620 K.2323CH —CH —OH 3223 Al O620 KCH2CH +H O22 Ethanol Ethene( b )Dehydration with mineral acid : Alcohols upon heating with conc. H SO form alkenes24and the reaction is called acidic dehydration.CH —CH —OH 3224 95 % H SO440 KCH2CH +H O22 Ethanol EtheneOHCH CH CH 3324 60% H SO373K CH —CH3CH + H O22 Propan–2–ol PropeneC CHCH 3CH 33OH24 30% H SO363KC CH + HO2 2CH 3CH 32–Methylpropan–2–ol 2–Methylpropene

From the above reactions, it is clear that the order of acidic dehydration in different alcohols isTertiary > Secondary > PrimaryCycloalkenes can be prepared in the same way by the dehydration of cycloalkanols.Example : OH34H PO / heat+ HO2Cyclohexanol CyclohexeneCH 3OH34H PO / heatCH 3CH 3CH 2++(major)Regioselectivity of elimination is governed by Zaitsev's Rule.Machanism of Reaction: The acidic dehydration of alcohol proceeds through the formation of acarbocation intermediate and is explained as follows :Step I : Alcohol being a Lewis base accepts a proton (H ) from the acid in a reversible step as follows:+CH 2CH 3O H + H ....CH O H2CH 3H .. Ethanol (From acid)Protonated ethanoStep II : Due to presence of positive charge on electronegative oxygen, its electron accepting tendencyincreases. As a result C – O bond becomes weak and cleaves as follows :CH O H2CH 3H .. SlowCH —32 CH  + H O2 Ethyl carbocationThis is a slow and is regarded as rate determining step.Step III :Carbocation is unstable in nature and loses a H and changes into ethene in a fast step as follows:+H—CH —22 CH FastCH2CH + H2+EtheneNote: Dehydration of secondary and tert alcohol is best carried out by using dil. H SO . Since alkenes24produced from those alcohols have a tendency to form polymers under the influence of concentrated acid.Saytzeff Rule : When two possible alkenes are obtained by the elimination reaction than that alkene will be in good yield, containing maximum number of alkyl group on double bonded C–atomsCH CH CH2OHCH 3324 H SOCH —CH3CH—CH +CH —CH —CH332CH22–butanolmain product1–butene2–butene 80%2 0 %

CH —CH —CH —CH —OH 322224 H SOCH —CH3CH—CH + CH CH CH332CH21–butanol2–butene80% 1–butene 20% Main productMechanism : Acid catalyzed dehydration of alkanols proceeds via the formation of more stablecarbonium ion.CH CH CH —CH — O —H+3222H  CH CH CH CH OH322223222232222CH CH CH CH OHCH CH CH — CHH OPrimary Carbonium ionC C C HH HCH 3H H HRe arrangement by1, 2 hydride ion shiftC C C HHCH 3H H HH 1 Carbonium 02 Carbonium more stable 0C C C HHCH 3H H HH E lim ination ofa protonCH CH CH CH 332– butene (major Product)CH CH CH 2CH 321–butene (minor product)(Saytzeff rule)( 2 )From Alkyl halide (By dehydrohalogenation): Removal of HX from a substrate by alcoholicKOH or NaNH222KOH (Alc.)HXRCH CH XRCHCH2Example :CH CH CH2CH 3X3KOH (Alc.)HXCH CH3CH—CH + CH CH CH332CH2(Saytzeff rule)The ease of dehydrohalogenation show the orderFor alkyl grouptertiary>secondary>primaryFor halogen in halideIodide > Bromide > Chlorideflouride It is single step and synchronous process. Removal of proton, the formation of multiple bondbetween C and C and the release of the leaving group X take place simultaneously.(E mechanism)2Example :O +CH C 3H HC H25H BrC H CH 3HC C H +C HOH + BrH2532225Rate of reaction[CH CH CH Br] [C H O] 

Example :3222KOH (Alc)CH CH CH CH Br3221 Bute n eCH CH CHCHExample :CHCHCHCH323BrKOH (Alc)332 Bute n e (major)CH CHCHCHPrimary and secondary alkyl halides undergo elimination reaction by E mechanism. E elimination21reactions are shown by tertiary alkyl halides which are capable of producing stable (tert) Carboniumion on show ionization.CH 3CH 3C ClCH 3            CH 3CH 3C + ClCH 3HO +H CHCH 3CCH 32CH 2CH 3C + HO2CH 3E mechanism : Those alkyl halides which do not give Stable Carbonium ion on ionization show2E elimination.2HOClCH CH HCH 2BaseCH 32CH CH + HO + Cl CH 2CH 322( A )Dehalogenation of vicinal dihalides :There are two types of dihalides namely gem (or geminal) dihalides in which the two halogen atomsare attached to the same carbon atom and vicinal dihalides in which the two halogen atoms areattached to the adjacent carbon atoms.Dehalogenation of vicinal dihalides can be effected either by NaI in acetone or zinc in presence ofacetic acid or ethanol.General Reaction(i)Br–C–C–Br3NaIor Zn,CH COOHC=CMech.Br–C–C–Br Zn–C––C–Br••–C=C–+ZnBr2(ii)CH –CHBr–CH Br 32325Zn dustCH COOH orC H OH as solvent CH –CH=CH32Mech.With NaI in acetone :I X••– C – C –XC=C+ IXIt involves an of halogen atomsantielimination

Remarks :(i)Both are E elimination.2(ii)Both are stereospecific antielimination.Example : CH –CHBr–CHBr–CH 33NaIAcetone CH –CH=CH–CH33E x .Identify the product in the following reactions :(a)CH 3HHBrBrCH 3NaI / acetone(b)CH 3HH BrBrCH 3NaI / acetoneS o l . (a) HCH 3H HC3 ; (b) HCH 3HCH 3( B )From gem dihalide : Higher alkene obtainedCH CH 322 X + 2Zn + XCHCH 3 CH —CH3CH—CH + ZnX32( 3 )By Pyrolysis of ester :CH 3OC O CH2CH RH 400 500 CCH COOH + CH32CHRHoffmann's Rule : Less substituted or less stable alkene is major product.Example : CH 3OC O CH CH2CH 3CH 3CH 3CH 2COOH + CH CH2CH + CH CH CH33CH 3(Major)(Minor)In the reaction to form an alkene a -hydrogen from alkyl ester is attracted by oxygen atom ofketo group & Hoffmann's alkene will be the major product.( 4 )By Pyrolysis of tetra alkyl ammonium ion :CH 3CH 3NCH 3CH 2CH 2HOH CH 3CH 3N+ CHCH 32CH + HO22Example :CH 3CH CH3N CH 3CH CH23CH 3—OH —H O2CH 3CH CH3N CH 3CH CH22CH 3CH CH2CH 3CH 21° more stablemajor productCH 3CH CH3N CH 3CH CH CH33CH CH CH3CH minor product32° less stable

Example :CH 3CH CH2N CH CH CH 3223CH 3—OH —H O2CH 2CH CH (minor)3CH 2CH (major)2(a)In this reaction –hydrogen of tetra-alkyl ammonium ion is attracted by a base and alkeneis formed.(b)In this reaction intermediate is carbanion.So yield of product depends on stability ofcarbanion.(c)In this reaction Hoffmann's Rule is followed.( 6 )By Kolbe's method:Electrolysis of potassium or sodium salt of saturated dicarboxylic acid gives alkene.CHCOONa2CHCOONa2CH + 2CO+ NaOH + H2 2 2CH 2electrolysisAt AnodeAt CathodeCH 3CHCOONa2CH COONaElectrolysisCH CH+2CO + NaOH + H3 CH 22 2At AnodeAt CathodeCH 3CHCOONaCH 3CHCOONaCH CH+2CO + NaOH + H3 CH2 2CH 3 electrolysisAt AnodeAt Cathode( 7 )Pyrolysis of Tri alkyl amine Oxide : (Cope Reaction)CH 3CH 2CH 3N OCH 2R150R—CHCH2 CH 3CH 3+ N OH( 8 )Hydrogenation of alkyne :By partial reduction of Alkynes -(a)By catalytic Hydrogenation of Alkynes in presence of poisoned catalyst (A SynAddition of Hydrogen : Synthesis of cis-Alkenes : This is performed by)(i)Lindlar's catalyst : Metallic palladium deposited on calcium carbonate conditioned with leadacetate and quinoline.(ii)P- catalyst (Ni B nickel boride)22General reaction R–C C–R (Lindlar's catalyst)quinoline 23 H Pd/CaCORHC=CRHMechanism of hydrogenation :H – H + – C C – + H – HH H –C = C– H HHmetal surfaceadsorptiondesorption(1)(2)HC = CHH

Steps : The reactant alkyne molecules and hydrogen molecules get adsorbed at the surface of metalcatalyst. It is chemical adsorption (chemisorption).In this state, the reactants lie very close to each other and so the hydrogen atoms start formingbond with carbon. Two hydrogen atoms are added to two triply bonded carbon atoms from thesame side of bond and a cis or syn addition product is formed. The product alkene nowescapes away from the surface of the catalyst. Quinoline occupies the metal surface inhibiting furtherreduction to alkanes Quinoline therefore is called catalyst poison and such palladium is calleddectivated catalyst or poisoned catalyst.Example :CHCHC CCHCH3223H/Ni B(P-2)or H/Pd/CaCO 2223(syn addition)3-HexyneC = C CHCH32HHCHCH23(Z)-3-Hexene(cis-3-Hexene)(97%)(b)Birch Reduction : (Anti addition of hydrogen : synthesis of trans-alkenes)General reactionNa/LiLiq. NH3C = C RHRHR–C C–R Mechanism : Reagents Na (or Li, K) + liq NH Na + e– (sollvated electron)3+R – C R  NaRC = CRH – NH2C = C RHR•Na •NaNH +2C = C RHRH(~100%)(trans alkene)H N–H2Na C = C RHRExample : : CH –CH –C C–CH –CH 32233Na / NH ( )C = C CHCH32HCHCH23Htranshex-3-eneNote :This process of reduction is not eligible when terminal alkynes are taken (R–C CH) becauseterminal alkynes form sodium salt with Na metal.CH –C CH + Na/NH 33 CH –C C–Na + [H]3++E x .Identify the reagent for following synthesisOCH–C C–CHCH223(A)?Ocis-JasmoneS o l .H /Lindlar's catalyst2A 2 HLindlar 's catalystcis - Jasmone

E x .Identify the product in the following reaction :CH–C CCH233 Na / NHS o l .C = C CH 2HCH 3H( 9 )Wittig Reaction :The aldehydes and ketones are converted into alkenes by using a special class of compounds calledphosphorus ylides, also called Wittig reagents.The Triphenyl group of phosphorane has a strong tendency to pull oxygen atom of the aldehydeor ketone via a cyclic transition state forming an alkene.R\"R\"R\"R\"R\"R\"C=OC=CC – OR–C–PPh3R'ylideR–C – PPh3R'R'R'+ Ph P = O3(R, R', R\" and R''' may be hydrogen or any alkyl group)e.g. Ph P: + CH – Br [Ph P –CH] Br ]33 33 Ph P –CHPh P=O +323Bu–Li..MeMeC = OMeMeC = CH2Product alkeneYlideMethyltriphenylphosphorium saltPhysical Properties of Alkenes / Hydrocarbons :Physical propertiesHomologus seriesI s o m e r s 1.Physical stateC – C gases13C – C liquids420> C 0 : solids2 2.Dipolemoment ( )cis > trans 3.Polarity_cis > trans (for C =C type of alkenes)abab 4.Melting pointincreases with M.W.trans > cis(due to more packing capacity 5.Boiling pointincreases with M.W.cis > trans# branching decreases B.P.CC–C=C < C – C = C – CPolarity increases, boiling point increases 6.SolubilityPractically insoluble incis > transwater but fairly solublePolarity increases, solubility in polarin nonpolar solventssolvents increases.like benzene petroleumether, etc. 7.Stabilitytrans > cis (cis isomers has moreVander Waals repulsion

Chemical Properties :Alkenes are more reactive than alkane this is because -(a)Theelectrons of double bond are located much far from the carbon nuclei and are thus lessfirmly bound to them.(b)bond is weaker thanbond and more easily broken.The reactivity order for alkenes -CH2CH > R—CH2CH > R C2 2CH2  RCHCHR > R C2CHR > R C2CR2 (Trans < Cis)The reactivity order of alkenes has been delt in terms of heat of hydrogenation of alkene, more is theheat of hydrogenation ( H = –ve), more is the reactivity, the reactivity of alkene is however alsorelated to(i) Steric hinderence(ii) Hyperconjugation(iii) Heat of Combustion.All four butenes may be compared, since all give the same products on combustion viz. 4CO +4H O22Alkenes give the following type of reactions :(a) Addition reaction(b) Oxidation reaction.(c) Substitution reaction.(d) Polymerization Reaction.(e) IsomerisationAlk eneHea t of com bus tion (k J/m ol)Hea t of hydrogena tion (k ca l/m ol)1-Butene271930.3Isobutene270327.2Cis-2-butene271228.6trans-2-butene270727.6( A )Addition Reaction :[ A ] free radical addition :1Addition of H :2R—CHCH +H 22Ni,Pt or PdR—CH —CH +23Heat of Hydrogenation.Not e:(a) Reaction is exothermic, It is called heat of hydrogenation.(b) 211Stability of alkeneheat of hydrogenationreactivity of alkene with H(c) The process is used to obtain vegetable (saturated fats) ghee from hydrogenation of oil.( B )[ A ] Electrophilic addition reactions :2Because of the presence of >CC< bond in molecules, alkenes generally take part in the additionreact ions.C C + ABC C ABAlkeneAttacking moleculeAddition product

From mechanism point of view, the addition in alkenes is generally electrophilic in nature whichmeans that attacking reagent which carries the initial attack is an electrophile (E+). This is quiteexpected also as there is high electron density in the double bond. The mechanism proceeds in twosteps.Step I : The –electron cloud of the double bond causes the polarisation of the attacking molecule(E–Nu) which cleaves to release the electrophile (E ) for the attack. The double bond simultaneouslyundergoes electromeric effect and the attack by the electrophile is accomplished in slow step (alsocalled rate determining step) to form a carbocation intermediate.C C (Slow )Rate determining step (RDS)Addition productStep II : The nucleophile (: Nu ) released in the slow step combines with the carbocation to give–the desired addition product in the fast step.1 .Addition of Halogen : It is a electrophilic addition reaction.R CH CH + XCH CH XR2 2 2X(Vicinal halides)Not e:(a)Reactivity order of halogen is : Cl > Br > I222(b)Addition of F is exothermic reaction so it is difficult to control.2(c)The addition of Br on alkenes provides a useful test for unsaturation in molecule.The brown2colour of the bromine being rapidly discharged. Thus decolarization of 5% Br in CCl by a24compound suggest unsaturation in it. Colourless dibromo compound is formed.(d)I reacts slowly with alkenes to form Vicinal di-iodides which are unstable and eliminated I22molecule very readily to give original alkene due to large size of Iodine they overlap.CH —CH3CH + I 22CH CH 2IICH 3UnstableMechanism : It is interesting to note that product which is mainly formed as a result of additionis trans in nature whereas the cis isomer is obtained in relatively smaller proportions. Sincecarbocation intermediate is planar (sp hybridised), both cis and trans addition products must be2formed almost in equal proportions. The trans product can be justified in case a cyclic ion is formedby the initial electrophile attack.CH = CH + Br—Br2 2 + –(Slow)CH 2BrCH 2 (Halonium ion)The attack of Br ion on the cyclic ion takes place from the side opposite to side where bromine–atom is present in order to minimise steric hindrance.

Br  + HC2HC2Br(Fast)HC2BrCH Br21,2,–Dibromoethane,For 2, 3-dimethylbut-2-ene :Step I :H C3H C3CH 3CH 3BrBr•••••••••••• + –H C3CH 3CH 3Br••••CH 3+Bromonium ionBr ••••••+••Bromide ionA bromine molecule becomes polarized as it approaches the alkene. The polarized bromine moleculetransfers a positive bromine atom (with six electrons in its valance shell) to the alkene resulting inthe formaiton of bromonium ion.Step II :H C3CH 3H C3CH 3vic-dibromideBr ••••••+••Bromide ionC—CH C3Br•••• +Bromonium ionCH 3C—CCH 3H C3Br ••••••Br ••••••Ex. 22EthyleneCHCH + Br 2aq.NaClProducts, what are the products?S o l .CH 2BrBrCH 2CH + Br2 2BrCH 2CH 2BrClClCH 2CH 2CH 2BrBrOH 2OHH O2CH 2CH 2CH 2BrCH 2–HSimilarly CH2CH + Br2 2 CH 2BrNal CH + Br2BrCH 2CH 2IBrCH 2CHBr + BrCHCH222NO 3NaNO32 .Addition of halogen acid :R—CHCH—R + HX CH 2CH R XRR—CHCH + HX 2CH CH3XR

Not e:(i)The order of reactivity of hydrogen halide is : HI > HBr > HCl > HF(ii)Their addition is an example of electrophilic addition.(iii)Addition on alkene proceeds via the formation of more stable carbonium ion.(iv)Addition of HX on unsymmetrical alkenes (R—CHCH ) takes place according to2Markownikoff's rule.MARKOWNIKOFF'S RULE :( a )First Rule : When molecule of a HX add up on unsymmetrical unsaturated hydrocarbon, thehalogen atom goes to the unsaturated carbon atom bearing lesser number of hydrogen atoms.CH 3CH CH + HX2 CH 3CH CH X2HMechanism : It is electrophilic addition and is illustrated by the action of HCl to propene.CH 3CH CH + H Cl2Slow–33Secondary carbocationCH — C H — CHClCl + CH–3CH CH3FastCH 3CH CHCl32–ChloropropanePrimary carbocation (CH —CH —322 CH ) is formed but only in very small proportion since it is lessstable than the secondary carbocation. Markownikff's rule can also be stated as :The electrophilic addition to unsymmetrical alkenes always occurs through the formationof a more stable carbocation intermediate.( b )Second Rule : In the addition of HX to vinyl halide and analogous compounds, the halogenattaches itself to the carbon atom, on which the halogen atom is already present.CH2CH—Cl+HCl  CH 3CH ClClEthylidene chlorideMechanism :CH 2CH Cl :CH 2CH Cl...... ....: CH 3CH ClCl H ClCH 2CH Cl.. ....In vinyl chloride two effects operate simultaneously in opposite direction-(i) Inductive effect – electron attracting (-I) effect of chlorine.(ii) Resonance effect – electron pair releasing (+R) effect of chlorine.The resonance effect is much more than the -I effect of Chlorine at the time of attack.This createscentres of +ve and –ve charges.

All polar reagents of the general structure Y Z (such as H X, H OH H SOH,XOH3) add onunsymmetrical unsaturated compound in accordance with Markownikoff's rules. Such additions arecalled normal Markownikoff's rule, where as additions in the opposite manner are reffered to asabnormal or antimarkownikoff's additions.ANTI MARKONIFF'S RULE OR PEROXIDE EFFECT OR KHARASCH RULE :(i)In the presence of oxygen of peroxides the addition of HBr on unsaturated unsymmetricalcompound takes place contrary to Markownikoff's rule. This is called peroxide effect and isdue to the difference in the mechanism of the addition.(ii)In the normal Markownikoff's addition the mechanism is ionic.(iii)In the presence of peroxide the addition of HBr takes place via free radicals.CH 3CH CH2CH 3CH CHBr3Isopropyl bromideMarkownikoff's addition.CH 3n–Propyl bromideAnti Markownikoff's additionHBrR O O RHBrCH—CH—Br22Mechanism :( i )Chain initiation -(a)R—O—O—R  2RO•(b)HBr + RO • ROH + Br•( ii )Chain propagationCH 3CH CH + Br2CH 3CH CHBrBr22° free radical more stable1° free radical less stable..HBrCHCHCHBr + Br322.(major)CH 3CH CH2.HBrBrCHCHCH + Br33 .(iii) Chain termination :RRR—RRBrR—BrBrBrBr—BrE x .Why HCl and HI do not give antimarkownikoff products in the presence of peroxides.?S o l . (a)The H—Cl bond is stronger than H—Br.(b)The H-I bond is weaker than H—Br bond. It is broken by the alkoxy free radicals obtainedfrom peroxides, but the addition of iodine atom on alkene is endothermic as compared toBr atom therefore iodine atoms so formed combine with each other to yield iodine.

3 .Addition of Hypohalous acid (or X /H O, or HOX) : It is a electrophilic addition and follows22Markownikoff's rule.CH 2CH 2ClCH + Cl2Cl Cl + HC2Slow–CarbocationCH 2CH + H O H2Cl....(Fast)CH 2CH 2Cl HO H.. –HCH 2ClCH 2OH–Ethylene chlorohydrinIn the fast step, there is competition between –Clion and H O molecule to act as nucleophile but2H O is a better nucleophile.2Re activity order isHOCl > HOBr > HOI4 .Addition of H SO : Alkene react with conc. H SO to produce alkyl hydrogen sulphate. Which2424gives alcohols on hydrolyses.This reaction used to seprate alkene from a mixture of alkane andalkene.CH+ HOSOH23CH 3OSOH3CH CHCH CH33 HO2CH 3OHCH CH + HSO324Isopropyl alcoholCH 2CH + HSO2 24CHCHHSO324HO2C H + HSO24 24C H OH + HSO25 24Ethyl hydrogen sulphate give ethylene when heated 430-440K while ethanol is obtained on boilingit with water.5 .Addition of water (Hydration of alkenes) : Propene and higher alkenes react with water inthe presence of acid to form alcohol. This reaction is known as the hydration reaction .Intermediate in this reaction is carbo cation, so rearrangement will take place.(i)CH —CH=CH + H O32 2CH —CH=CH33 OH PropeneH +Propan-2-ol(ii)CH 3C CH + HO22CH 32–MethylpropeneH CH 3C CH3CH 3OH2-Methylpropan-2-olMechanism :CH CH + H2+CH 3(Slow)CH CH 3CH 3 Carbocation (2°)

CH 3CH CH + H O H3 ....(Fast)H O H CH 3CH CH3..–HO H CH 3CH CH3Propan-2-ol6 .Addition of NOCl (Tilden reagant) : CH CH + NOCl2CH 3ClCH CH CH3   NOPropylene nitrosochloride7 .Hydroboration : It obeys markoni'koff's rule.Diborane readily reacts with alkenes giving trialkylboranes. The reaction is called hydroboration.CH CH + BH2262R2(R CH2CH) BH22CH CH2RCH 2CH) B2 3(RCH CH R2CH 2CH)2 22(RBH+ –+ –TrialkylboraneBH does not exist or stable as monomer so a solvent THF (tetra hydro furane) is used.3Example :CH CH + BH23CH3HHTHF(CH —CH —CH ) B322 3BHR also can be taken.2Example :CH CH + BHR22CH 3  CH 2CH 3CH 2BR 23CH CH3HO/H22(CH3CH 2CH) B2 3CH + HBO333PropaneCH CH3H O /OH2 22CH OH2Propanol(1° alcohal)CH CH3Cl2CH NH + NaCl + HBO2(1° amine)233PropanamineNH 2NaOHTripropyl Borane+8 .Oxymercuration – demercuration : Mercuric acetate in tetrahydro furan (THF) is treated withan alkene.The addition product on reduction with sodium Boro hydride in aqueous NaOH solutiongives alcohol. It follows the markonikoff's rule.CH —CH3CH2OH CH 3CH CH3(i)(AcO) Hg/H O (Mercuric acetate) or (CH COO) Hg/H O2232 2(ii)NaBH /NaOH4Mechanism :CH 3COOCH 3COOHgH o2CH —COO3— + CH —COOHg (Electrophile)3+

CH CH + HgOOCCH CH 3HgOOCCH32 3CH CH2CH 3++:HgOOCCH3CH CH2CH 3(cyclic cation)HO2OH  HCH CH CH 32HgOOCCH3OHCH CH CH 32HgOOCCH3(Oximercuration)N/4OHCH CH + CHCOOHg CH 323H(Product) +OH H::–HO3+Note : Intermediate is cyclic cation so their is no rearrangement.CH 3CH 3OHCH 3H /HO2C CHCH 3CH CH3(i)BH/THF3CH CH CHHClCHCH CHCH 322CH 3with rearrangementmarkownikoff's rule(ii)HO/ OH22(i) (AcO) Hg/HO22(ii) NaBH/NaOH4HBrPeroxide2OH3without rearrangementanti-markownikoff's rulewith rearrangememtmarkonikoff's rulewithout rearrangementmarkownikoff's rulewithout rearrangementanti - markownikoff's ruleCH 3CH 3ClC CH CH23CH 3CH 3CH CH CH2OHHCH 3HCH CH CH2BrCH 3+–9 .Hydroformylation or Oxo reaction : Alkenes react with Carbon monoxide and hydrogen at100 – 150°C temperature and high pressure (200 atm) in the presence of Cobalt catalyst to producean aldehyde.It does not follows markonikoff's rule.The net reaction is the addition of a H–atom to one of the Olefinic bond and a formyl (–CHO)group to the other.R CH CH + CO + H22Co/150°Chigh pressureR CH2CHO CH 2R CH CH3CHO

1 0 .Alkenylation (Addition of alkene) In presence of H SO or H PO at 80°C dimerisation of2434isobutylene take place gives two isomer of octene.2CH3C CH2CH 3CH 2C CH2CH 3CH 3C CH + CH33CH 3C CH C CH H SO , 80°C24CH 3CH 33CH 3Mechanism :CH 3C CH + H2CH 3C CH 3CH 3 CH 3CH 3C CH + C CH 2 CH 3CH 33CH 3C CH2CH 3CH 3C CH 3CH 3 CH 3CH 2C CH2CH 3CH 3C CH 3CH 3CH 3C CH C CH CH 3CH 33CH 3–H 1 1 .Addition of Carbene : The addition of carbene to alkene is always carried by diazomethaneCH N22Carbene group obtained from diazomethane is added to alkene and give cycloalkanes.CH2CH + CH N222 CH CH + N222 CH 2CH 3CH 3CH 2CHCH CH + CH..2 2: CH 2Since : CH is an electrophile (neutral) and there is more electron density on double bond so first2attack of :CH will be at double bond.21 2 .Addition of HCN: CH 3CH CH + HCN2CH 3CH CH3CN   ( B )OXIDATION REACTION :Alkenes are easily oxidised by oxidising agents. Oxidising agents attack on double bond and productformed during oxidation depends on oxidising agents.( 1 )alkene on combustion gives CO and H O22C H + n2n3n2O 2nCO + nH O22One mole of alkene requires 3n2 moles of O for complete combustion.2

Ex : 90 mL of oxygen is required for complete combustion of unsatuarated 20 mL gaseous hydrocarbon,hydrocarbon is ?S o l . Following two formulae can be used for solution of the above asked question.2Volume of Hydrocarbon2Volume of O3n (for Alkene)2Volume of Hydrocarbon2Volume of O3n 1 (for Alkyne)By putting the values in above formulae we can find the hydrocarbon for which n is natural number.202903n  n = 3 So hydrocarbon is Propene [C H ].36E x .How many mole of oxygen is required for complete combustion of 1 mole of Alkene.S o l . 2C Hn2n2223nO2nCO2nH Okeeping in mind, the above equation.for 2 mole of Alkene, 3n mole of O is required for combustion.2for 1 mole of Alkene, 3n2 mole of O is required for combustion.2= 1.5n mole of O2E x .30 mL mixture of ethylene and Butylene is burnt in presence of oxygen then 150 mL of oxygenis required, what is the volume of Ethylene & Butylene in mixture.S o l . Let the volume of C H = x mL24So volume of Butylene = (30–x) mLFor C H242422C H3O2C2H O  from equationfor 1 volume C H , 3 volume of O is required.242for x mL vol. of C H , 3x ml volume of O is required.242For C H48C H + 6O 482 4CO + 4H O22for 1 volume C H , 6 volume of O is required.482for (30–x) mL\"\" , 6 (30–x) mL of O is required.2Total volume of O = 3x + 6 (30–x) mL = 150 mL(Given)2x = 10Volume of C H in mixture is 10 mL24Volume of C H in mixture is 20 mL24( 2 )Ozonolysis : (A test for unsaturation in molecule)(i)The addition of ozone on the double bonds and subsequent a reductive hydrolysis of theozonide formed is termed as ozonolysis.(ii)When ozone is passed through an alkene in an inert solvent, it adds across the double bondto form an ozonide. Ozonides are explosive compound they are not isolated.(iii)On warming with Zn and H O, ozonides cleave at the site of the double bond, the products2are carbonyl compound (aldehyde or ketone) depending on the nature of the alkene.

Example :CH 3C CH CH3CH 3OzonolysisCH 3C O + CHCHO3CH 3(iv)Ozonolysis of alkenes helps in locating the position of double bond in an alkene. It can beachieved by joining together the carbon atoms of the two carbonyl compounds formed as theproducts of ozonolysis with double bond.Example :CH 3C O + O C CH HH3CH 3CH CH CH31234EthanalBut-2-eneExample :H C O + O C CHHH2CH 2CH CH2CH 31234CH 3MethanalPropanalBut-1-eneExample :CH 3C O + O C CH 3CH 3C C CH3CH 3CH CH33CH 3 Propanone 2,3–Dimethyl but–2–eneIt may be noted that reaction with bromine water or Baeyer's reagent detects the presence of doublebond (or unsaturation) in an alkene while ozonolysis helps in locating the position of the doublebond. In an reduction of ozonide by LiAlH or NaBH gives corresponding alcohols.44R' CH CH R\"OOOLiAlH4R'CHOH + R\"CHOH22(Alcohols)( 3 )Hydroxylation : Oxidation of carbon-carbon double bond to OH OHC C is known as hydroxylation.( a )Oxidation by Baeyer's reagent (A test for unsaturation) : Alkenes on passing through dilutealkaline 1% cold KMnO (i.e., Baeyer's reagent) decolourise the pink colour of KMnO and gives44brown ppt MnO and glycol.2CC + HO + [O]24OHKMnO–OH OHC CGlycol(cis–addition)( b )By OsO4 :R CHR CH+ OsO4R CH OR CH OOsOOHO2R CH OHR CH OH+ HOsO24cis–addition

( c )By peracid :OH C O O H> CC < +HCOOH>C C<O2H OHO H >CC<OHtrans glycol( 4 )Epoxidation :(a)Alkenes reacts with oxygen in the presence of Ag catalyst at 250°–400° C to form epoxide.CH2CH221O2 AgOCH 2CH 22H OOHOHCH 2CH 2(anti addition)(b)Prileschiaev reaction: When an alkene is treated with perbezoic acid an epoxide is formed.Such an epoxidation is known as Prileschiave reactions.RCHCH + C H COOOH 265RCH 2CHO + CHCOOH65 EpoxideEmmons have found that perbenzoic oxy trifluoroacetic acid (CF COO H) is a very good reagent32for epoxidation and hydroxylation.Anti additionSyn additioncis alkeneMeso compoundRacemic mixtureAnti additionSyn additiontrans alkeneMeso compoundRacemic mixture22422222 Syn addition on alkeneH ,O ,Baeyer 's reagent,OSO / H OAnti addition on alkeneX ,HOX,RCOOH / H O, Ag O / H O Example :CCH CH 3CH H3Br2Anti additionD2Syn additionHBrBrHCH 3CH 3HCH 3CH 3DHDcis–alkene+Racemic mixtureHBrBrH CH 3CH 3Meso compound(i)CCH CH 3CH 3HBaeyer'sReagentRCOOOH/H O2Anti additionOHHHOHCH 3CH 3HCH 3CH 3OHHOHtrans-alkene+Racemic mixtureOHHHOH CH 3CH 3Meso compoundSyn addition(ii)

( 5 )Oxidation by strong oxidising agent (Oxidative cleavage) : The alkenes themselves are readilyoxidised to acid or ketone by means of acid permagnate or acid dichromate.If HCOOH is formed,it further oxidized to CO and H O. Keep it in mind that no further oxidation of ketones will takes22place.CH2CH +4[O] 22HCOOH2[O]2CO + H O22CH CH3CH25[O]CH COOH+CO +H O322CH CH3CHCH 34[O]2CH COOH3CH 3CH 3C CH24[O]CH 3CH 3C O + CO + HO22( 6 )Oxidation with retention of Carbon-Carbon bond - (Waker process) :CH2CH +H O2222l2PdClCuCCH CHO3( C )SUBSTITUTION REACTION (Allylic substitution) :When alkenes are treated with Cl or Br at high temp., one of their allylic hydrogen is replaced22by halogen atom. Allylic position is the carbon adjacent to one of the unsaturated carbon atoms.Itis free radical substitution.CH —CH3CH + Cl 22500 C0ClCH —CH2CH + HCl2Allyl chloride(3-Chloro-1-propene)N-Bromosuccinimide (NBS) is an important reagent used for allylic bromination and benzlic substitution.OCH + CH22CCH 3CHCONBrCH 2(NBS)OCH 2CCONH + BrCH 2CH 2CH CH2Substitution reaction is not given by ethene.Example : (i)CH 3NBSCH 2Br(ii)CH 3HNBSCH 3.Br CH 3Br3° more stable(D) POLYMERIZATION :(i)Two or more than two molecules of same compound unit with each other to form a longchain molecule with same empirical formula. This long chain molecule having repeatingstructural units called polymer, and the starting simple molecule as monomer and process iscalled addition polymerization.

(ii)Molecular weight of polymer is simple multiple of monomer.(iii)Polymerization can be carried out by free radical or ionic mechanism.(iv)The presence of oxygen initiates free radical mechanism.(v)Addition polymerization can also be carried out by ionic mechanism by using Ziegler - NattaCatalysts (R Al+TiZCl )34Name ofStructure ofStructure of Properties Uses Propertiesp o ly m e rm o n o m e rP o l y m e r1.PolyvinylCH =CH–Cl2( HC–HC ) –2nCl–PiliableUsed in handbag,chloride(easily moulded)raincoats, vinyl(PVC)flooring, good elec-trical insulatorfor wires2.Polytetrafluor-F C=CF22( F C–CF ) –22 n–Flexible and inertFor making non-stickethylene orto solvents, boilingutensils coatingTeflon (PTFE)acids,even aquaregiastable upto 598K.3.Natural rubberCH=CH–C=CH22CH 3isopreneWaxy and non-elasticUsed as raw materialfor making vulcanisedrubber which is strongand elastic vulcanisedrubber is used inmaking tyres hose,pipes etc.4.Orlonacrylonitrile( HC–HC) –2n–CNFibrousUsed in makingFabrics5.Poly methylCH 3CH=C2COOCH3Methyl methacrylateCH 3–CH C2–C=OOCH3nmethacrylate(PMMA)( E )ISOMERISATION :Alkene on heating to 500° to 700 °C or on heating in presence of catalyst [AlCl or Al (SO ) ]324 3undergo isomerisation.CH CH —CH32CH2CatalystCH —CH3CH—CH 3+CH 3C CH2CH 31–Butene 2–ButeneIsobutylene

Uses :(a)In plastic formation.(b)In oxy ethylene welding(c)As food preservatives and ripening fruits.(d)As general anaesthetic (C H with 10% O )242(e)In preparation of mustard gasS CH+SCH 2CH 2 + S Cl + 22CH 2CHCl CHCl22CH 2CH 22 2,2' or ( , ') dichloro diethyl-sulphide (mustard gas)Laboratory test of alkene :Fu nctional Group Reag entObser vat ionReactionRem ark s(1) Bayer's Reagent alk. dil. cold KMnO4Pink colour disappearsDihydroxylation(2) Br /H O22Red colour decolourisesDibrominaiton(3) O (ozone)3OzonolysisCH =CH +H O+O222CH –CH22alk. KMnO4OH OHBr +CH =CH222CH –CH22BrBrWhite ppt.CH =CH +O2232HCHO Zn/H O2C = CC=OCompounds: DIENES :Dienes are the unsaturated hydrocarbons with carbon-carbon double bonds in their molecules. Theseare represented by the general formula n2n 2 C H which means that they are isomeric with alkyness(functional isomers). However, their properties are quite different from those of alkynes. Dependingupon the relative positions of the two double bond, dienes are classified in three types :Isolated dienes or non conjugated dienes : In an isolated diene, the two double bonds areseparated by more than one single bond. For example,CH CH 2CH 2CH 2CH12345CH 3CH CH CH 2CH 2CH12345Penta–1,4–diene3–Methylpenta–1, 4–dieneCunjugated dienes : In a conjugated diene, the two double bonds are present in the conjugatedor alternate position and are separated by a single bond.CH CH CH 2CH12324CH 3CH CH CH 3CH 2C12345Buta-1,3-diene 2-Methylpenta-1,3-diene

Commulate dienes : In this case, the two double bonds in the molecules are present at adjacentpositions. For example,CH 2CH 2C 123CH CH3CH 2C1234Propa-1,2–diene Buta–1, 2–dieneComparison of relative stabilities of isolated and conjugated dienes :Resonance Theory : The relative stabilities of the two types dienes can also be justifid on thebasis of the theory of resonance. Penta-1,3-diene (conjugated diene) is a hybrid of the followingcontributing structures.CH CH CH 3CH 2CHCH CH CHCH 2CH3CH CH CHCH 2CH3The delocalisation of -electron charge because of resonanace decreases the energy of the moleculeor increases its stability.Penta-1, 4–diene (isolated diene) has only two contributing structures.CH CH 2CH 2CH 2CH12345CH CH 2CH 2CH 2CH12345(I)(II)Since the carbon atom C is not involved in any resonance, the contributing structures are less in3number are compared to the conjugated diene. The isolated diene is, therefore, less stable than aconjugated diene.Properties of Conjugated Dienes :The properties of the isolated dienes are similar to those of simple alkene but those of conjugateddienes are somewhtat modified because of delocalisation of the -electron charge. However, they alsoparticipate in the addition reactions. The important chemical characteristics of the conjugated dienesare briefly discussed.1 .Addition Reaction : Conjugated or 1,3–dienes take part in the addition reactions which canproceed by electrophilic as well as free radical mechanism depending upon the nature of theattacking reagent and the reaction conditions.( A )Electrophilic Addition Reactions : The electrophilic addition is illustrated by the attack of halogenand halogen acid on buta–1,3–diene, a conjugated diene.( a )Addition of halogen : If one mole of halogen attacks per mole of the diene, two typesof addition products are formed. There are 1, 2 and 1, 4 addition products. For example,CH 2CH 2CH CH CHCHCH CH221234Br212Br3Br43,4–Dibromobut–1–ene(1,2–Addition)CH 2CH CH CH212Br3Br41,4–Dibromobut–2–ene(1,4–Addition)Buta–1, 3 diene1, 2–addition is a normal addition in which one mole of halogen has been added to one of thedouble bond. But 1, 4-addition is somewhat unexpected.Mechanism : The addition is electrophilic in nature and the halogen molecule (bromine) providesthe electrophilie for the attack.

CH CH + BrBrCH 2CH2CH CHBr + BrCH 2CH2Buta-1, 3-dieneCarbocation (2°)The 2° carbocation get stablised by resonance as follows –CH CHCH 2CH2BrCH CHCH 2CH2BrIIIThe attack of Br  ion on carbocation ( I and II )CH 2CH CH CH Br2BrBr(1,2–Addition)CH 2CH CH CH2BrBrCH 2CH CH CH Br2CH 2CH CH CH Br21,4–addition (Main product)( b )Addition of H – X :CH2CH—CHCH2HBr?Mechanism : The addition is electrophilic in nature as H  ion is the electrophile.CH CH + HBrCH 2CH2CH CH+ BrCH 2CH3–Carbocation (2°)The carbocation gets resonance stabilised as follows :CH CHCH 2CH3CH CH CH2CH 3(I)(II)The attack of Br ion on the carbocation (I) gives 1,2-addition product whereas the attack on the–carbocation (II) yields 1,4-addition product.Br + CH2CH CH CH3–BrCH 2CH CH CH31,2–Addition productBr + CH2CH CH CH3 –BrCH 2CH CH CH31,4–Addition product( B )Free Radical Addition Reaction : The addition to conjugated dienes can also proceed by freeradical mechanism provided it is carried in the presence of a suitable reagent which can help informing a free radical. However, the addition also yields 1,2 and 1,4 addition products. The freeradical addition is illustratated by the attack of bromotrichloromethane (BrCCl ) on buta–1,3–diene3in the presence of an organic peroxide such as benzoyl peroxide.CH 2CH CH CH + BrCCl231234BenzoylperoxideBrCl CCHCH CH CH + Cl CCH32232CH CHCH 243214321Buta–1,3–dieneBromotri-chloromethane

Mechanism : The mechanism of addition is free radical in nature which is initiated by benzoyl freeradical. It is explained in the following steps :Step IGeneration of free radicalC H CO—O—O—COC H 6565HomolysisC H —CO— O O —COC H6565C H —CO—65O 65C H +CO 2653C HBrCClC H —Br + CCl653 Trichloromethyl free radicalStep IIAttack of free radical on buta-1, 3-diene3C Cl +CH 2CH—CHCH 2Cl C—CH —32C H —CHCH2 Free radical (secondary)The free radical gets resonance stabilisedCl C—CH —32C H —CHCH 2Cl C—CH —CH32CH— C H2Step IIIChange of free radical into addition productThe free radicals take up Br from the attacking reagent to give the desired addition products.i.e. 1,2 and 1,4 addition products.BrCCl + Cl C CH332CH CH CH .2BrCl C CH32CH CHCH 2 (1,2-Addition product)BrCl C CH32CH CHCH 2BrCCl + Cl C33CH 2CH CH CH.2(1,4–Addition product)2 .Cyclo-Addition Reaction (Diel Alder Reactions) :Cyclo-addition reactions are one of the most important reactions of conjugated dienes. Cyclo-addition involvesthe combination between a conjugated diene (4 -electron system) and a compound containing a double bond(2 -electron system) called dienophile which means a diene loving or attracting molecule. As a result, a sixmembered ring gets formed and the reaction is therefore known as cyclo-addition reaction. It is quite oftentermed as (4 +2) cyclo-addition reaction because four 4 -electron system adds to a two 2 electron system.The reactions of this type are known as : Diel Alder Reaction.The addition product is called Diel Alder Adduct. For example.HCHCCH 2CH 2+CH 2CH 2HCHCCH 2CH 2CH 2CH 2Buta–1,3–dieneEtheneCyclohexene (Diel Alder adduct)

HCHCCH 2CH 2+CH CHOCH 2HCHCCH 2CH 2CH CHOCH 2Buta–1,3–dienePropenal3-cyclohexene carbaldehyde(Acrolein)E x .The density of a hydrocarbon at N.T.P. 2.5 gram/litre. What is hydrocarbon?S o l . Density of 1 litre hydrocarbon = 2.5 gram/litre molecular weight of hydrocarbon = 2.5 x 22.4 = 56After molecular weight , we calculate the molecular formula.n2n 2 C H= molecular weight (Alkane) or 14n+2 = molecular weightn2n C H= molecular weight (alkene) or 14n = molecular weightn2n 2 C Hn2n 2 C H= molecular weight (Alkyne) or 14n–2 = molecular weightwith the help of above three formulae, we can identify the given hydrocarbon14n = 56 (Alkene) Hydrocarbon is C H (Butene).48 n = 4Volume of hydrocarbon will be given and volume of O for complete combustion will also be2given.What is hydrocarbon is to be asked ?The above question may be solved with the help of following three formulae.Formula No. 12Volumeof H.C.2Volume of O3n+1 (for Alkane)Formula No. 22Volume of H.C.2Volume of O3n (for Alkene)Formula No. 32Volume of H.C.2Volume of O3n–1 (for Alkyne)E x .How much propanol is required for dehydration to get 2.24 litre of Propene at N.T.P. if yield is100%.S o l .382436224C H OH SOC HH OH SOMolecular weight of propanol = 60from the equation given above we can see that from dehydration of 1 mole or 60 gram ofpropanol we get 1 mole (22.4lit.) of propene as product. 22.4 litre of C H can be get from dehydration of 60 g of propanol.36 1 litre of propene can be get from dehydration of 6022.4 g of propanol 2.24 litre of propene can be get from dehydration of 60x2.2422.4 g of propanolAns. = 6 g



Introduction :A triple bond gives an alkyne four fewer hydrogen atoms than the corresponding alkane. Therefore, the triplebond contribute two degree of unsaturation (DU).Alkynes are not as common in nature as alkenes, but some plants do use alkynes to protect themselves againstdisease or predators. Acetylene is by far the most important commercial alkyne. Acetylene is an importantindustrial feedstock, but its largest use is as the fuel for the oxyacetylene welding torch.Structure and bonding in Alkynes(a)Alkynes are hydrocarbons that contain carbon-carbon triple bond.(b)Alkynes are also called acetylenes because they are derivatives of acetylene(c)The general formula is : C Hn2n-2. (one triple bond)(d)In alkyne C C bond length is 1.20 Å(e)Its bond energy is 192 kcal. mol–1(f)The hybridization of carbon atoms having triple bond (C C) in alkynes is sp.(g)Overlapping of these sp hybrid orbitals with each other and with the hydrogen orbitals gives the sigmabond framework which is linear (180°) structure.(h)Two bonds result form overlap of the two remaining unhybridized p orbitals on each carbon atom.theseorbitals overlap at right angles (90°) to each other, forming one bond with electron density aboveand below the C-C sigma bond, and the other with electron density in front and in back of the sigmabond. This result in a cylindrical electron cloud around bonded structure.180°1.20Å 1.06ÅHHH – C – C – HNote : Any type of stereoisomerism does not arise in acetylenic bond due to linearity of C C bond.IUPAC Nomenclature of Alkynes :SN .Comp oundN a m e1.CH CHEthyne2.CH – C CH3Propyne3.HC C – CH – CH23But-1-yne4.CH –C C – CH33But-2-yne5.CH–CH–C C–CH–CH–CH323CH 3Br6-Bromo-2-methylhept-3-yneIsomerism in AlkynesTyp e Categary Examples(i) Chain isomerismCH–CH–CH–C CH & CH–CH–C CH3223CH 3Structural(ii) Positional isomerismCH –CH –CH –C CH & CH –CH –C C–CH322323Isomerism(iii) Functional group isomerism CH –CH –C C–CH & CH –CH=C=CH–CH & 32333ALKYNE

Ex.Cis-trans isomerism is not possible in alkynes because of :-Sol. 180° bond-angle at the carbon-carbon triple bond.Ex.Draw the geometrical isomers of hept-2-en-5-yne ?Sol.C = C MeC CCH2HMeH(trans) , C = C MeC C–CH2H MeH(Cis)General Methods of Preparation :1 .From Gem dihalides ( by dehydrohalogenation) : Dehydrohalogenation agents are : NaNH (Sodamide)2or Alc. KOH or ROH + RONa.XRCXCHHHalc.KOHHXRCXCHH2 NaNH R—CCH + NaX + NH3(Stable by resonance)(Vinyl halide)(a)Due to stability of vinyl halide by resonance there is partial double bond in which elimination does nottake place by alc. KOH so stronger base NaNH is used.2(b)Basic strength : 2 N H  is stronger base then RO(c)Trans elimination takes place in forming of alkynes.2 .From Vicinal dihalides (by dehydrohalogenation) :RX CC HHXHalc.KOHHXRCHC HX2 NaNHHX R—CCH(a)Elimination of Vic. dihalides gives also alkadiene (1, 2 and 1, 3 alkadienes) but the major product is alkyne.Example :X CHCCHHHXHH2 NaNH CH2CCH2+ CH—C3C—H(Major)(b)Non terminal gem dihalide gives 2-Alkyne in presence of alc. KOH while gives 1-alkyne in presence ofNaNH .2Example :H CH 3CC HXXCHHHAlc. KOHNaNH2CH C C3CH (major)3CH CH32CCH (major)3 .Dehalogenation of tetrahalo alkane : By heating 1, 1, 2, 2 - tetra halo alkane with Zn dust.RX C CXHXX2Zn R—CCH + 2ZnX2

4 .From Kolbe's electrolysis : By the electrolysis of aqueous solution of sodium or potassium fumarate ormaleate, acetylene is formed at anode.CH COOKCH COOKElectrolysisCHCH+CO2Mechanism :CH COOKCH COOKlonizationCH COOCH COO——+2K+at anode (Alkyl and CO gas is formed)2CH COOCH COO——–2eCH COOCH COO..(Oxygen free radical)CH C OOCH C O......O2 2COCHCHat cathode (KOH and H gas is formed)22K + 2e+—2K2K + 2H O 22KOH + H2Ex.Is PH of solution changed in Kolbe's electrolysis.Sol. The concentration of NaOH solution increased so pH of solution is increased with time.5 .Preparation of higher alkynes by Grignard reagent : By this method lower alkyne is converted in tohigher alkyneCH3C — H CH — Mg —Br CCHMgBr+ CH 4 RI MgBrI+ R —CCH R—CCH + CH Mg—Br 3CMgBrCR + CH 4 R I'R' —CC—R + MgBrI6 .Preparation of Ethyne or Acetylene:( a )From Metal carbide [ Laboratory method] : Acetylene is prepared in the laboratory by the actionof water on calcium carbide.CaC + 2H O 22CHCH + Ca(OH)2Ca + +2C C  + 2H + 2OH +–CHCH + Ca(OH)2( b )Manufacture : Acetylene is manufactured by heating methane or natural gas at 1500 C in an electric arc02CH4Electric arcCHCH + 3H2( c )Berthelot's process : Acetylene is synthesized by striking an electric arc between carbon electrodes inpresence of hydrogen.2C + H 21200 CCHCH

( d )From haloform [CHI , CHCl ] : Pure acetylene is obtained when iodoform or chloroform is heated33with Silver powderCHI + 6Ag + I CH 33CHCH + 6AgI( e )Partial oxidation of methane : A recent method for manufacturing of acetylene is the controlledpartial oxidation of methane at high temperature.4CH + 3O 421500 C2CHCH + 6H O2Physical Properties :( a )Alkynes are relatively nonopolar (w.r.t. alkyl halides and alcohols) and nearly insoluble in water (but theyare more polar than alkenes and alkanes). They are quite soluble in most organic solvents, (acetone,ether, emthylene chloride, chloroform and alcohols).( b )Acetylene, propyne, and the butynes are gases at room temperature, just like the corresponding alkanesand alkanes. In fact, the boiling points of alkynes are nearly the same as those of alkanes and alkeneswith same number of carbon atoms.Chemical Properties :The chemical properties of alkynes are due to two factors( a )Presence of electrons : Due to presence of loosely bonded electrons, alkynes like alkenes, undergoeasily electrophilic addition reaction.Carbon-carbon triple bond is less reactive than the carbon-carbon double bondtowards electrophilic addition reactions.In addition to electrophilic additons, alkynes also undergo nucleophilic addition with nucleophiles( b )Presence of acidic hydrogen atom : The hydrogen atom attached to the triple bonded carbon canbe easily removed by a strong base and hence acetylene and 1-alkynes are considered as weak acids.Explanation : The amounts of s-character in various types of C—H bonds is as-C—HC—H— C — H5 0 %3 3 %2 5 %Since s electrons are closer to the nucleus than the p electrons, the electrons present in a bond having more s-character will be more closer to nucleus. Due to high s-character of the C—H bond in alkyne (s50%) theelectrons constituting this bond are more strongly held by the carbon nucleus, with the result the H present onCH can be easily removed as proton.The acidic nature of the three types of –C–H bonds as C—H > C—H > —C—H spsp 2sp 3Relative acidic orderHO > ROH > HCCH > HNH > CHCH > CH—CH222 233Addition Reaction :( 1 )Electrophilic addition : Addition reactions where the addition is initiated by electrophile (positive group).Thecharacteristic reaction of alkynes is electrophilic addition but the reactivity of alkynes towards electrophilicaddition is less than alkenes because in CC, the  electrons are tightly held by carbon nuclei and so they areless easily available for reaction with electrophiles.Reactivity order of hydrocarbons for electrophilic additionAlkenes > Alkynes > Alkanes.