Alkene Notes IIT-JEE Main + NEET / ISC / CBSE / NCERT / Class XI, Study notes of Chemistry

Download Alkene Notes IIT-JEE Main + NEET / ISC / CBSE / NCERT / Class XI and more Study notes Chemistry in PDF only on Docsity! A LKEN E  Introduct ion : Alkenes are hydrocarbons with carbon-carbon double bonds, Alkenes are sometimes called olefins, a term derived from olefiant gas, meaning 'oil forming gas'. Aleknes are among the most important industrial compounds and many alkenes are also found in plants and many alkenes are also found in plants and animals. Ethylene is the largest-volume industrial organic compound, used to make polyethylene and a variety of other industrial and consumer chemicals. Alkenes polymerises to give many important polymers.  Structure and bonding i n Alkenes : (a) Alkenes are unsaturated hydrocarbons having at least one double bond. (b) They are represented by general foumula (G.F.) C n H 2n (one double bond) (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 oily liquid substance when treated with halogens. (h) Compounds may exist as conjugated polyenes or as cumulated polyenes or as isolated polyenes 117.2° (b) 121.2° (a) 1.34Å 1.10Å C — C H H H H C — C H H H H Note : 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 a lkene s and alkadienes : Table - I S .No . Compound Name Type 1. (CH 3 ) 2 C = CH 2 2–Methylpropene Alkene 2. CH 3 –CH=CH–CH 2 –CH=CH 2 Hexa–1, 4–diene Isolated diene 3. CH 2 =CH–CH = CH 2 Buta–1, 3–diene Conjugated diene 4. CH 3 –CH=C=CH–CH 3 Penta–2, 3–diene Cumulated diene 5. CH–CH–CH=CH–CH–CH–CH3 2 2 3 Cl 1 2 3 4 5 6 7 6–Chlorohept–3–ene Alkene 6. CH=CH–CH=C–CH2 3 1 2 3 4 5 OCH3 4–methoxypenta–1, 3–diene Conjugated diene 7. CH=CH–CH–C CH2  1 2 3 CH 4 5 CH2 3–Ethynylpenta–1, 4–diene Isolated diene S .No . Compound Name Type 8. CH–CH–C = C–CH–CH3 2 2 3 6 5 4 CH3CH2 5 3 2 1 3–Ethyl–4–methylhex–3–ene Alkene 9. CH3 CH3 1 2 3 4 5 6 2, 3–Dimethylcyclohex–1–ene Cycloalkene 10. CH 2 = C = CH 2 Propadiene Cumulated diene 11. CH 2 = C = O Ethenone Alkene 12. Methylenecyclopentane Alkene 13. Cyclopentylethene Alkene 14. 3, 7, 11–trimethyldodeca– Isolated triene 1, 6, 10–triene E x . Write IUPAC names of : (a) CH3 CH3 (b) S o l . (a) 2, 3–Dimethylcyclohexene; (b) 1–(2–butenyl) cyclohex–1–ene E x . Give the structure for each of the following (a) 4–Methyl–1, 3–hexadiene ;(b) 1–Isopropenylcyclopentene S o 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 CH3CH CH2 , CH2 CH2 CH2 ? S o l . Ring chain isomerism E x . (a) CH3CH2CH CH2 (b) CH3—CH CH—CH3 (c) C CH3 CH3 CH2 (d) CH2 CH2 CH2 CH2 Define relations between a,b,c,d ? CH3—CH2—CH2—CH2—OH 2 4H SO  CH3—CH CH—CH3 + CH3CH2CH CH2 1–butanol 2–butene80% 1–butene 20% Main product Mechanism : Acid catalyzed dehydrat ion of alkanols proceeds via the format ion of more stable carbonium ion. CH3CH2CH2—CH2— O   —H+ H   CH3CH2CH2CH2 2OH  3 2 2 2 2 3 2 2 2 2CH CH CH CH OH CH CH CH — CH H O     Primary Carbonium ion C H CH3 H C H H C H  H Re arrangement by 1, 2 hydride ion shift  C H CH3 H C H H C H  H 10 Carbonium 20 Carbonium more stable C H CH3 H C H H C H  H E lim ination of a proton  CH CH3 CH CH 3 2– butene (major Product) CH 2CH3 CH CH 2 1–butene (minor product) (Saytzeff rule) ( 2 ) From Alkyl hal ide (By dehydrohalogenat ion): Removal of HX from a substrate by alcoholic KOH or NaNH2 2 2 KOH (Alc.) HX RCH CH X    RCH CH2 Example : CH 2CH3 CH X CH3 KOH (Alc.) HX    CH3CH CH—CH3 + CH3CH2CH CH2 (Saytzeff rule) The ease of dehydrohalogenation show the order For alkyl group tertiary > secondary > primary For halogen in halide Iodide > Bromide > Chloride flouride It is single step and synchronous process. Removal of proton, the format ion of mult iple bond between C and C and the release of the leaving group X take place simultaneously. (E2 mechanism) Example : O +CH 3 H CH2 5 H C H Br H CH3  C H C C H H +CHOH + Br2 5    3 2 2 2 5Rate of reaction [CH CH CH Br] [C H O]  Example : 3 2 2 2 KOH (Alc)CH CH CH CH Br  3 2 2 1 Bute n e CH CH CH CH   Example : CHCHCHCH3 2 3 Br  KOH (Alc) 3 3 2 Bute n e (major ) CH CH CHCH   Primary and secondary alkyl halides undergo elimination react ion by E2 mechanism. E1 elimination reactions are shown by tertiary alkyl halides which are capable of producing stable (tert) Carbonium ion on show ionization. CH3 CH3 C Cl CH3               CH3 CH3 C + Cl CH3  HO +H CH3 C CH3  CH2    CH2 CH3 C + HO2 CH3 E2 mechanism : Those alkyl halides which do not give Stable Carbonium ion on ionization show E2 elimination. HO Cl CH H  CH2   Base CH3 CH2 CHCH2CH3 CH + HO + Cl2 2     ( A ) Dehalogenation of vic inal dihal ides : There are two types of dihalides namely gem (or geminal) dihalides in which the two halogen atoms are attached to the same carbon atom and vicinal dihalides in which the two halogen atoms are attached to the adjacent carbon atoms. Dehalogenation of vicinal dihalides can be effected either by NaI in acetone or zinc in presence of acetic acid or ethanol.  General Reaction (i) Br –C–C– Br 3 NaI or Zn,CH COOH  C=C Mech. Br –C–C– Br Zn –C––C– Br ••  –C=C–+ZnBr2 (ii) CH3–CHBr–CH2Br 3 2 5 Zn dust CH COOH or C H OH as solvent       CH3–CH=CH2 Mech. With NaI in acetone : I X• • – C – C – X  C=C + IX It involves an of halogen atomsantielimination Remarks : (i) Both are E2 elimination. (ii) Both are stereospecific antielimination. Example : CH3–CHBr–CHBr–CH3 NaI Acetone  CH3–CH=CH–CH3 E x . Identify the product in the following reactions : (a) CH3 H H Br Br CH3 NaI / acetone (b) CH3 H H Br Br CH3 NaI / acetone S o l . (a) H CH3 HH3C ; (b) H CH3 H CH3 ( B ) From gem dihal ide : Higher alkene obtained CH3CH 2 2X + 2Zn + X CHCH3  CH3—CH CH—CH3 + ZnX2 ( 3 ) By Pyrolysis of e ster : CH3 O C O CH2 CH R H 400 500 C  CH3COOH + CH2 CHR Hoffmann's Rule : Less subst ituted or less stable alkene is major product. Example : CH3 O C O CH CH2 CH3 CH3    CH3 CH2 COOH + CH CH2 CH + CH3 CH CH3 CH3 (Major) (Minor) In the react ion to form an alkene a -hydrogen from alkyl ester is at tracted by oxygen atom of keto group & Hoffmann's alkene will be the major product. ( 4 ) By Pyrolysis of tetra alkyl ammonium ion : CH3 CH3 N CH3  CH2 CH2 H OH  CH3 CH3 N CH3 + CH2 CH + HO2 2 Example : CH3 CH3 N CH3  CH CH2 CH3   CH3 —OH  —HO2 CH3 CH3 N CH3 CH CH2 CH2 CH3   CH CH2 CH3 CH2 1° more stable major product CH3 CH3 N CH3 CH CH CH3 CH3  CH CH CH3 CH3 2° less stable minor product   E x . Identify the product in the following reaction : CH–C CCH2 3 3Na / NH S o l . C = C CH2 H CH3 H ( 9 ) Witt ig Reaction : The aldehydes and ketones are converted into alkenes by using a special class of compounds called phosphorus y l ides, also cal led Wit t ig reagents. The Triphenyl group of phosphorane has a strong tendency to pul l oxygen atom of the aldehyde or ketone via a cyclic transit ion state forming an alkene. R" R" R" R" R" R" C=O C=C C – O R–C–PPh3 R' ylide   R–C – PPh3 R'  R' R' + PhP = O3 (R, R', R" and R''' may be hydrogen or any alkyl group) e.g. PhP: + CH – Br [PhP –CH] Br 3 3 3 3   PhP –CH] PhP=O +3 2 3 Bu–Li .. Me Me C = O Me Me C = CH2 Product alkene YlideMethyltriphenyl phosphorium salt  Phys ical Proper t ie s of Alkenes / Hydrocarbons : Physical proper t ie s Homologus ser ies I s o m e r s 1. Physical state C1 – C3 gases C4 – C20 liquids > C20 : solids 2. Dipolemoment () cis > trans 3. Polarity _ cis > trans (for Cab=Cab type of alkenes) 4. Melting point increases with M.W. trans > cis (due to more packing capacity 5. Boiling point increases with M.W. cis > trans # branching decreases B.P. C C–C=C < C – C = C – C Polarity increases, boiling point increases 6. Solubility Practically insoluble in cis > trans water but fairly soluble Polarity increases, solubility in polar in nonpolar solvents solvents increases. like benzene petroleum ether, etc. 7. Stability trans > cis (cis isomers has more Vander Waals repulsion  Chemical Proper t ie s : 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 less firmly bound to them. (b)  bond is weaker than bond and more easi ly broken. The reactivity order for alkenes - CH2 CH2 > R—CH CH2 > R2C CH2  RCH CHR > R2C CHR > R2C CR2 (Trans < Cis) The reactivity order of alkenes has been delt in terms of heat of hydrogenation of alkene, more is the heat 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. 4CO2+4H2O Alkenes give the fol low ing t ype of react ions : (a) Addition reaction (b) Oxidation reaction. (c) Substitution reaction. (d) Polymerization Reaction. (e) Isomerisation A lk ene Hea t o f com bu s t ion (k J/m ol ) Hea t o f hyd rog ena t ion (k ca l/m ol ) 1-Butene 2719 30.3 Isobutene 2703 27.2 Cis-2-butene 2712 28.6 trans-2-butene 2707 27.6 ( A ) Addit ion Reaction : [ A 1 ] free radical addit ion :  Addition of H2 : R—CH CH2+H2 Ni,Pt or Pd R—CH2—CH3+ Heat of Hydrogenation. N o t e : (a) React ion is exothermic, It is cal led heat of hydrogenat ion. (b) 2 1 1 Stability of alkene heat of hydrogenation reactivity of alkene with H   (c) The process is used to obtain vegetable (saturated fats) ghee from hydrogenation of oil. ( B ) [ A 2 ] Electroph i l ic addit ion react ions : Because of the presence of >C C< bond in molecules, alkenes generally take part in the addition rea c t i o n s . C C + AB  C C A B Alkene Attacking molecule Addition product From mechanism point of view, the addit ion in alkenes is general ly electrophi lic in nature which means that attacking reagent which carries the initial attack is an electrophile (E+). This is quite expected also as there is high electron density in the double bond. The mechanism proceeds in two steps. 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 simultaneously undergoes electromeric effect and the attack by the electrophi le is accomplished in slow step (also cal led rate determining step) to form a carbocat ion intermediate. C C (Slow ) Rate determining step (RDS)  Addition product Step II : The nucleophile (: –Nu ) released in the slow step combines with the carbocation to give the desired addition product in the fast step. 1 . Addit ion of Halogen : It is a electrophilic addition reaction. R CH CH X CH + X R2 2  CH2 X (Vicinal halides) N o t e : (a) Reactivity order of halogen is : Cl2 > Br2 > I2 (b) Addit ion of F2 is exothermic react ion so it is difficult to control. (c) The addition of Br2 on alkenes provides a useful test for unsaturation in molecule.The brown colour of the bromine being rapidly discharged. Thus decolarization of 5% Br2 in CCl4 by a compound suggest unsaturat ion in it. Colourless dibromo compound is formed. (d) I2 reacts slowly with alkenes to form Vicinal di-iodides which are unstable and eliminated I2 molecule very readi ly to give original alkene due to large size of Iodine they overlap. CH3—CH CH2 + I2  CH CH 2 I I CH 3 Unstable  Mechanism : It is interesting to note that product which is mainly formed as a result of addit ion is trans in nature whereas the cis isomer is obtained in relatively smaller proportions. Since carbocat ion intermediate is planar (sp2 hybridised), both cis and trans addit ion products must be formed almost in equal proportions. The trans product can be justified in case a cyclic ion is formed by the initial electrophile attack. CH2 = CH2 + Br—Br + – ( )Slow  CH2 Br CH 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. All polar reagents of the general structure  Y  Z (such as  H  X,  H  OH  H  SOH,3  X  OH) add on unsymmetrical unsaturated compound in accordance with Markownikoff's rules. Such additions are cal led normal Markownikoff's rule, where as addit ions in the opposite manner are reffered to as abnormal or ant imarkownikoff 's addit ions.  ANTI MARKONIFF'S RULE OR PEROXIDE EFFECT OR KHARASCH RULE : (i) In the presence of oxygen of peroxides the addition of HBr on unsaturated unsymmetrical compound takes place contrary to Markownikoff's rule. This is called peroxide effect and is due to the difference in the mechanism of the addition. (ii) In the normal Markownikoff's addit ion the mechanism is ionic. (iii) In the presence of peroxide the addition of HBr takes place via free radicals. CH3 CH CH2 CH3 CH Br CH3 Isopropyl bromide Markownikoff's addition. CH3 n–Propyl bromide Anti Markownikoff's addition HBr R O O R HBr CH—CH—Br2 2  Mechanism : ( i ) Chain ini t iat ion - (a) R—O—O—R  2RO• (b) HBr + RO•  ROH + Br• ( i i ) Chain propagat ion CH3 CH CH + Br2 CH3 CH Br CHBr2 2° free radical more stable 1° free radical less stable . . HBr CHCHCHBr + Br3 2 2 . (major) CH3 CH CH2 . HBr Br CHCHCH + Br3 3 . ( i i i ) Chain terminat ion : R R   R—R R Br   R—Br Br Br   Br—Br E 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 obtained from peroxides, but the addit ion of iodine atom on alkene is endothermic as compared to Br atom therefore iodine atoms so formed combine with each other to yield iodine. 3 . Addit ion of Hypohalous acid (or X2/H2O, or HOX) : It is a electrophilic addition and follows Markownikoff's rule. CH2 CH2 Cl CH + Cl2Cl Cl + HC2   Slow  – Carbocation CH2 CH2 + H Cl   .. O.. H (Fast) CH2 CH2 Cl OH H..  –H CH2 Cl CH2 OH – Ethylene chlorohydrin In the fast step, there is competition between –Cl ion and H2O molecule to act as nucleophile but H2O is a better nucleophile. Re activity order is HOCl > HOBr > HOI 4 . Addit ion of H2SO4: Alkene react with conc. H2SO4 to produce alkyl hydrogen sulphate. Which gives alcohols on hydrolyses.This reaction used to seprate alkene from a mixture of alkane and alkene. CH+ HOSOH2 3 CH3 OSOH3 CHCH3 CH  CH3 HO2 CH3 OH CH CH + HSO3 2 4 Isopropyl alcohol CH2 CH + HSO2 2 4 CHCHHSO3 2 4  HO2 CH + HSO2 4 2 4 CH OH + HSO2 5 2 4 Ethyl hydrogen sulphate give ethylene when heated 430-440K while ethanol is obtained on boiling it with water. 5 . Addit ion of water (Hydration of alkenes) : Propene and higher alkenes react with water in the presence of acid to form alcohol. This react ion is known as the hydrat ion react ion . Intermediate in this react ion is carbo cat ion, so rearrangement wil l take place. (i) CH3—CH=CH + H O2 2 CH—CH=CH3 3 OH Propene H + Propan-2-ol (ii) CH3 C CH + HO2 2CH3 2–Methylpropene H  CH3 C CH3CH3 OH 2-Methylpropan-2-ol  Mechanism : CH CH + H2 +CH3 (Slow ) CH CH 3CH3  Carbocation (2°) CH3 CH CH + H3  O .. .. H ( )Fast  H O H CH3 CH CH3 ..  –H  O H CH3 CH CH3 Propan-2-ol 6 . Addit ion of NOCl (Ti lden reagant) : CH CH + NOCl2CH3 Cl CH3    CH NO CH Propylene nitrosochloride 7 . Hydroborat ion : It obeys markoni'koff's rule.Diborane readi ly reacts with alkenes giv ing tr ialkyl boranes. The reaction is called hydroboration. CH CH + BH2 2 62R    2(R CH2 CH) BH2 2    CH CH2R CH2 CH) B2 3(R  CHR CH2 CH2 CH)2 22(R BH + – + – Trialkylborane BH3 does not exist or stable as monomer so a solvent THF (tetra hydro furane) is used. Example : CH HCH + B23CH3  H H     THF (CH 3—CH 2—CH 2) 3B BHR2 also can be taken. Example : CH CH + BHR2 2CH3    CH2CH3 CH2 BR2 3CH3 HO/H2 CH2(CH3 CH2 CH) B2 3 CH + HBO3 3 3 Propane CH3 H O /OH2 2 CH2 CH2 Propanol OH (1° alcohal) CH3 Cl CH2 CH2 (1° amine) NH + NaCl + HBO2 3 3 Propanamine NH2 NaOH Tripropyl Borane +  8 . Oxymercurat ion – demercurat ion : Mercuric acetate in tetrahydro furan (THF) is treated with an alkene.The addition product on reduction with sodium Boro hydride in aqueous NaOH solution gives alcohol. It follows the markonikoff's rule. CH3—CH CH 2  OH CH3 CH CH3 (i) (AcO)2 Hg/H2O (Mercuric acetate) or (CH3COO)2 Hg/H2O (ii) NaBH4/NaOH  Mechanism : CH3 COO CH3 COO Hg H o2  CH3—COO — + CH3—COOHg+ (Electrophile) Ex : 90 mL of oxygen is required for complete combustion of unsatuarated 20 mL gaseous hydrocarbon, hydrocarbon is ? S o l . Fol lowing two formulae can be used for solut ion of the above asked quest ion. 2 Volume of Hydrocarbon 2 Volume of O 3n  (for Alkene) 2 Volume of Hydrocarbon 2 Volume of O 3n 1   (for Alkyne) By putting the values in above formulae we can find the hydrocarbon for which n is natural number. 20 2 90 3n  n = 3 So hydrocarbon is Propene [C3H6]. E x . How many mole of oxygen is required for complete combustion of 1 mole of Alkene. S o l . n 2n 2 2 22C H 3nO 2nCO 2nH O   keeping in mind, the above equation.  for 2 mole of Alkene, 3n mole of O2 is required for combustion.  for 1 mole of Alkene, 3n 2 mole of O2 is required for combustion. = 1.5n mole of O2 E x . 30 mL mixture of ethylene and Butylene is burnt in presence of oxygen then 150 mL of oxygen is required, what is the volume of Ethylene & Butylene in mixture. S o l . Let the volume of C2H4 = x mL So volume of Butylene = (30–x) mL For C2H4 2 4 2 2C H 3O 2C 2H O    from equation  for 1 volume C2H4, 3 volume of O2 is required.  for x mL vol. of C2H4, 3x ml volume of O2 is required. For C4H8 C4H8 + 6O2  4CO2 + 4H2O  for 1 volume C4H8, 6 volume of O2 is required.  for (30–x) mL " " , 6 (30–x) mL of O2 is required. Total volume of O2 = 3x + 6 (30–x) mL = 150 mL(Given) x = 10  Volume of C2H4 in mixture is 10 mL  Volume of C2H4 in mixture is 20 mL ( 2 ) Ozonolysis : (A test for unsaturation in molecule) (i) The addition of ozone on the double bonds and subsequent a reductive hydrolysis of the ozonide formed is termed as ozonolysis. (ii) When ozone is passed through an alkene in an inert solvent, it adds across the double bond to form an ozonide. Ozonides are explosive compound they are not isolated. (iii) On warming with Zn and H2O, ozonides cleave at the site of the double bond, the products are carbonyl compound (aldehyde or ketone) depending on the nature of the alkene. Example : CH3 C CH CH3 CH3 Ozonolysis CH3 C O + CHCHO3 CH3 (iv) Ozonolysis of alkenes helps in locating the position of double bond in an alkene. It can be achieved by joining together the carbon atoms of the two carbonyl compounds formed as the products of ozonolysis with double bond. Example : CH3 C H O + O C H CH3  CH3 CH CH CH3 1 2 3 4 Ethanal But-2-ene Example : H C H O + O C H CH2  CH2 CH CH2 CH3 1 2 3 4 CH3 Methanal Propanal But-1-ene Example : CH3 C CH3 O + O C  CH3 C C CH3CH3 CH3 CH3CH3 Propanone 2,3–Dimethyl but–2–ene It may be noted that reaction with bromine water or Baeyer's reagent detects the presence of double bond (or unsaturation) in an alkene while ozonolysis helps in locating the position of the double bond. In an reduction of ozonide by LiAlH4 or NaBH4 gives corresponding alcohols. R' CH O CH R" O O  LiAlH4 R'CHOH + R"CHOH2 2 (Alcohols) ( 3 ) Hydroxylat ion : Oxidat ion of carbon-carbon double bond to OH C C OH is known as hydroxylation. ( a ) Oxidation by Baeyer's reagent (A test for unsaturation) : Alkenes on passing through dilute alkaline 1% cold KMnO4 (i.e., Baeyer's reagent) decolourise the pink colour of KMnO4 and gives brown ppt MnO2 and glycol. C C + HO + [O]2 4 OH KMnO –  OH C OH C Glycol (cis–addition) ( b ) By OsO4 : R CH  R CH + OsO4 R CH R CH O O Os O O  HO2 R CH R CH OH OH + HOsO2 4 cis–addition ( c ) By peracid : O H C O O H> C C < + HCOOH >C C< O 2H O HO >C H C< OH trans glycol ( 4 ) Epoxidation : (a) Alkenes reacts with oxygen in the presence of Ag catalyst at 250°–400° C to form epoxide. CH2 CH2 2 1 O 2  Ag  O CH2 CH2 2H O OH OH CH2 CH2 (anti addition) (b) Pri leschiaev react ion: When an alkene is treated with perbezoic acid an epoxide is formed. Such an epoxidation is known as Prileschiave reactions. RCH CH2 + C6H5COOOH R CH2 CH O + CHCOOH6 5 Epoxide Emmons have found that perbenzoic oxy trifluoroacetic acid (CF3COO2H) is a very good reagent for epoxidation and hydroxylation. Anti addition Syn addition cis alkene Meso compound Racemic mixture Anti addition Syn addition trans alkene Meso compound Racemic mixture 2 2 4 2 2 2 2 2 Syn addition on alkene H ,O ,Baeyer ' s reagent,OSO / H O Anti addition on alkene X ,HOX,RCOOH / H O, Ag O / H O   Example : C C HCH3 CH3 H Br2 Anti addition D2 Syn addition H Br Br H CH3 CH3 H CH3 CH3 D H D cis–alkene + Racemic mixture H Br Br H CH3 CH3 Meso compound (i) C C HCH3 CH3H Baeyer's Reagent RCOOOH/H O2 Anti addition OH H H OH CH3 CH3 H CH3 CH3 OH H OH trans-alkene + Racemic mixture OH H H OH CH3 CH3 Meso compound Syn addition (ii)  Uses : (a) In plast ic format ion. (b) In oxy ethylene welding (c) As food preservatives and ripening fruits. (d) As general anaesthetic (C2H4 with 10% O2) (e) In preparation of mustard gas S +S CH2 CH2 + SCl 2 2 + CH 2 CHCl 2 CHCl2 CH2 CH2 CH2 2,2' or (,  ') dichloro diethyl-sulphide (mustard gas)  Laboratory test of alkene : Fu nc tiona l Grou p Reag en t Obser va t ion Reacti on Rem ark s (1 ) Bayer's Reagent alk . dil. cold KMnO4 Pink colour disappears Dihydroxylation (2) Br2/H 2O Red colour decolourises Dibrominaiton (3) O3 (ozone) Ozonolysis CH =CH +H O+O2 2 2 CH –CH2 2 alk. KMnO4 OH OH Br +CH =CH2 2 2 CH –CH2 2 Br Br White ppt. CH =CH +O2 2 3 2HCHO Zn/H O2 C = C C=O Compounds : DIENES : Dienes are the unsaturated hydrocarbons with carbon-carbon double bonds in their molecules. These are represented by the general formula n 2n 2C H  which means that they are isomeric with alkynes (functional isomers). However, their proper ties are quite different from those of alkynes. Depending upon 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 are separated by more than one single bond. For example, CH 2 CH2CH2 CH 1 2 3 CH 4 5 CH3 CH CH2CH2 CH 1 2 3 CH 4 5 Penta–1,4–diene 3–Methylpenta–1, 4–diene Cunjugated dienes : In a conjugated diene, the two double bonds are present in the conjugated or alternate posit ion and are separated by a single bond. CH CH2 CH 1 2 3 CH 2 4 CH3 CH CH3CH2 C 1 2 3 CH 4 5 Buta-1,3-diene 2-Methylpenta-1,3-diene Commulate dienes : In this case, the two double bonds in the molecules are present at adjacent positions. For example, CH 2CH2 C 1 2 3 CH CH3CH2 C 1 2 3 4 Propa-1,2–diene Buta–1, 2–diene  Compar ison of relat ive stabi l i t ie s of isolated and conjugated dienes : Resonance Theory : The relative stabili ties of the two types dienes can also be justifid on the basis of the theory of resonance. Penta-1,3-diene (conjugated diene) is a hybrid of the following contributing structures. CH CH3CH2 CH CH  CH CH2 CH   CH CH3  CH CH2 CH   CH CH3 The 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 2 CH2CH2 CH 1 2 3 CH 4 5  CH 2 CH2CH2 CH 1 2 3 CH 4 5  (I) (II) Since the carbon atom C3 is not involved in any resonance, the contributing structures are less in number are compared to the conjugated diene. The isolated diene is, therefore, less stable than a conjugated diene.  Proper t ie s of Conjugated Dienes : The properties of the isolated dienes are similar to those of simple alkene but those of conjugated dienes 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 dienes are briefly discussed. 1 . Addit ion Reaction : Conjugated or 1,3–dienes take part in the addition reactions which can proceed by electrophilic as well as free radical mechanism depending upon the nature of the attacking reagent and the reaction conditions. ( A ) Electrophi lic Addit ion React ions : The electrophilic addit ion is i l lustrated by the attack of halogen and halogen acid on buta–1,3–diene, a conjugated diene. ( a ) Addit ion of halogen : If one mole of halogen attacks per mole of the diene, two types of addit ion products are formed. There are 1, 2 and 1, 4 addit ion products. For example, CH2 CH2 CH CH CH CH2 CH CH2 1 2 3 4 Br2 1 2 Br 3 Br 4 3,4–Dibromobut–1–ene (1,2–Addition) CH2 CH CH CH2 1 2 Br 3 Br 4 1,4–Dibromobut–2–ene (1,4–Addition) Buta–1, 3 diene 1, 2–addit ion is a normal addit ion in which one mole of halogen has been added to one of the double bond. But 1, 4-addition is somewhat unexpected.  Mechanism : The addition is electrophilic in nature and the halogen molecule (bromine) provides the electrophilie for the attack. CH BrCH2 CH CH + Br2   CH CH2 CH CHBr + 2 Br   Buta-1, 3-diene Carbocation (2°) The 2° carbocation get stablised by resonance as follows – CH CH2 CH CH2  Br  CH CH2 CH CH2  Br I II The attack of Br  ion on carbocation ( I and II ) CH2 CH CH CH2 Br Br (1,2–Addition) CH2 CH CH CH2 Br  Br Br CH2 CH CH CH2 Br  CH2 CH CH CH2 Br  1,4–addition (Main product) ( b ) Addit ion of H – X : CH2 CH—CH CH2 HBr ?  Mechanism : The addition is electrophilic in nature as H ion is the electrophile. CH BrCH2 CH CH + H2   CH CH2 CH CH+ 3 Br  – Carbocation (2°) The carbocation gets resonance stabilised as follows : CH CH2 CH CH3  CH 2 CH  CH3CH (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 + CH2 CH CH CH3  –  Br CH2 CH CH CH3 1,2–Addition product Br + CH2 CH CH CH3  –  Br CH2 CH CH CH3 1,4–Addition product ( B ) Free Radical Addit ion React ion : The addit ion to conjugated dienes can also proceed by free radical mechanism provided it is carried in the presence of a suitable reagent which can help in forming a free radical. However, the addit ion also yields 1,2 and 1,4 addit ion products. The free radical addition is illustratated by the attack of bromotrichloromethane (BrCCl3) on buta–1,3–diene in the presence of an organic peroxide such as benzoyl peroxide. CH2 CH CH CH + BrCCl2 3 1 2 3 4 Benzoyl peroxide  Br Cl CCHCH3 2 CH CH + Cl CCH2 3 2 CH CH CH2 4 3 2 1 4 3 2 1 Buta–1,3–diene Bromotri- chloromethane