Alkene Notes

Chemistry HYDROCARBONS ALKENE INTRODUCTION Alkenes are hydrocarbons with carbon-carbon double bonds, Alkenes are sometimes called olefins, a term derived from olefiant gas, meaning ‘oil forming gas’. Alkenes 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 largestvolume industrial organic compound, used to make polyethylene and a variety of other industrial and consumer chemicals. Alkenes polymeries 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 formula (G.F.) CnH2n (one double bond) (c) In Ethene C = C bonds length is 1.34 Å (d) Its bond energy is 146 kcal. mol–1 (e) The hybridization of (C = C) alchemic carbon is sp 2 (f) The 𝜋 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 1.34Å 1.10 Å 117.2° (b) H H C H 121.2 a H ° H H C H C C H 1 Chemistry Note: That angle a < b since repulsion due to 𝜋 electrons (double bond - single bond repulsion < single bond - single bond repulsion according to VSEPR theory. IUPAC Nomenclature alkenes and alkatrienes Table – 1 Sr. No. Compound Name 1. (CH3)2C = CH2 2-Methylpropane 2. CH3–CH = CH – CH2 – CH = CH2 Hexa-1, 4-diene 3. CH2 = CH – CH = CH2 Buta-1, 3-diene 4. CH3 – CH = C = CH – CH3 Penta-2, 3-diene 5. Cl 1 2 4 3 6 5 Type Alkene Isolated diene 6-Chlorohept-3-ene Conjugated diene Cumulated diene Alkene 4-methoxypenta-1, 3-diene Conjugated diene 3-Ethynylpenta-1, 4diene Isolated diene 3-Ethyl-4-methylhex3-ene Alkene 2, 3Dimethylcyclohex-1ene Cycloalkene Cumulated diene Alkene 7 CH3 – CH2 – CH = CH – CH2 – CH – CH3 6. 1 2 4 3 5 CH2 = CH – CH = C – CH3 OCH3 7. 1 2 4 3 5 CH2 = CH – CH = C CH3 OCH3 8. 6 CH3 C2H5 4 5 3 2 1 CH3 – CH2 – C = C – CH2 – CH3 9. 1 6 5 2 CH3 3 4 CH3 10. CH2=C=CH2 Propidine 11. CH2=C=O Ethenone 12. Methylene cyclopentane Alkene 13. Cyclopentolate Alkene 2 Chemistry 14. Ex. 3, 7, 11trimethyldodeca1, 6, 10 -triene Write IUPAC names of: CH3 Sol. Ex. Sol. (I) Isolated triene CH3 (a) (a) 2, 3-Dimethylcyclohexene (b) 1-(2-butenyl) cyclohex-1-ene (b) Give the structure for each of the following (a) 4-Methyl-1, 3-hexadiene, (b) 1-Isopropenylcyclopentene (a) (b) ISOMERISM IN ALKENES Alkenes show chain, Ring or functional, Position, Geometrical isomerism and optical isomerism. METHOD OF PREPARATION By Elimination Reaction E-1 Mechanism (a) It is called unimolecular elimination reaction (b) It takes place in two steps. Ist Step: Departure of leaving group from a molecule to form carbocation. C X C ⎯⎯⎯⎯⎯ C + ⊝ C +X IInd Step: Carbocation formed above loses a proton to the base and forms the C (c) (d) + C 𝐵𝑎𝑠𝑒 𝐹𝑎𝑠𝑡 C=C alkene. Ist step of E-1 mechanism is the rate determining step (R.D.S.) Since in R.D.S. only substrate undergoes covalency change i.e., it follows first order kinetics with respect to substrate Rate = K [substrate] 3 Chemistry (e) (f) (g) Since in R.D.S. carbocation is formed as an intermediate to the reactivity order of different substrate follows the stability of order of carbocation formed. Allylic > Tertiary > Secondary > Primary Loss of leaving group is the part of R.D.S. therefore the order of reactivity of alkyl halide in elimination reaction is R–I > R – Br > R – CI > R – F Above order is parallel to order of leaving group ability I >Br  >CI >F  E-1 mechanism does not show isotopic effect as the loss of hydrogen is not the part of R.D.S. (h) (i) =1 Since the ionization take place in step I. Therefore, polar solvent favors the E1 mechanism. E-1 mechanism involves carbocation as an intermediate so rearrangement of carbocation can take place if possible. More stable carbocation forms the major product in the reaction. E–2 Mechanism (a) It is known as bimolecular elimination mechanism. (b) It takes place in single step. (c) It involves the formation of transition state H C B. . . 𝐻 C Slow B− C=C L C=C L Transition State (d) Formation of transition state is the rate determination step. (e) Ratre of reacting depends on the concentration of substrate as well as of base i.e., it follows the second-order kinetics. R = [Substrate] [Base] (f) E-2 mechanism shows the isotopic effect as the loss of hydrogen is a part of R.D.S. = (g) E-2 mechanism shows the halogen effect as loss of leaving group is a part of R.D.S.  Reactivity of different alkyl halides follows the sequence Reactivity order of different substrate follows 3° > 2° > 1° 4 Chemistry (h) The rate of reaction increases with increasing strength and concentration of the base. (i) Non-polar solvent favors the E-2 mechanism. (j) Since E-2 mechanism involves the transition state rearrangement is not possible. E 1 CB Mechanism (a) In this mechanism, the reaction take place in two steps. (i) The first step involves rapid removal of a proton from the 𝛼-carbon (carbon adjacent to the carbon bearing halogen atom) forming a carbanion. (ii) The carbanion formed then loses the halides ion in the second-rate determining step.  O B ....H |  O Fast  R  CH  CH 2  R  C H  C H2  B  H  | | F F R CH CH2 Slow R CH = CH2+ F F (b) (c) The overall rate of the reaction is limited to slower second step & hence the rate of reaction depends only on the concentration of carbanion. Since carbanion is the conjugate base of the alkyl halide and rate of reaction depends on concentrating of carbanion hence the mechanism is designated as E1cB mechanism. Mechanism of -Eliminating 𝛼-elimination does not occur frequently and completes in two stages, the second of which is rate determining. for eq. CHCl3 + OH ⊝ CCl3 ⊝ CCl3 +H2O : CCl2 + Cl Di chlorocarbon Another example of 𝛼-elimination of reaction’s formation of diphenylacetylene from 2, 2-diphenyl vinyl bromide and soda lime. 5 Chemistry C6H5 H C=C C6H5 C6H5 – C Ξ C – C6H5 ⎯⎯⎯⎯ Br Mechanism C6H5 ⊝ C=C C6H5 C6H5 H C=C +NH −𝑁𝐻 ⊝ -Br Br C6H5 – C Ξ C – C6H5 Br C6H5 (II) ⊝ HYDROGENATION OF ALKYNE By partial reduction of Alkynes (a) By catalytic Hydrogenation of Alkynes in presence of poisoned catalyst (A) Syn Addition of Hydrogen Synthesis of cis-Alkenes: This is performed by (i) Landlar’s catalyst: Metallic palladium deposited on calcium carbonate conditioned with lead acetate and quinoline. (ii) P-2 catalyst (Ni2B nickel boride) General reaction R–CΞC–R H Pd/CaCO (Lindlar′s catalyst) quinoline R R C=C H H Mechanism of hydrogenation H–H+-C C-+H-H metal surface ( ) H–H+-C C-+H-H ⎯⎯⎯ ( ) ⎯⎯⎯ adsorption H C=C H H H desorption Steps: The reactant alkyne molecules and hydrogen molecules get adsorbed at the surface of metal catalyst. It is chemical adsorption (chemisorpti0on.) It this state, the reactants lie very close to each other and so the hydrogen atoms start forming bond with carbon. Two hydrogen atoms are added to two triply boned carbon atoms from the same side of 𝜋 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 in hiding further 6 Chemistry reduction to alkanes Quinoline therefore is called catalyst poison and such palladium is called deactivated catalyst or poisoned catalyst. Ex. CH3CH2C Ξ CCH2CH3 CH3CH2 H /Ni B(P− ) or H /Pd/CaCO (syn addition) CH2CH3 C=C H H (Z)-3-Hexene Cis-3-Hexene (97%) (b) Birch Reduction: (Anti addition of hydrogen: synthesis of trans-alkenes) R–C Ξ C–R Na/Li Liq. NH R H H H C=C R General reaction Mechanism: Reagents Na (or Li, K) + liq NH3  Na+ + e– (solvated electron)  R ̇ H R Θ C=C R–C Ξ R ⎯⎯⎯ C=C ⎯⎯⎯⎯⎯⎯⎯  R  R Na R NaNH2 + H H C=C H2N-H R N𝑎⊕ R H C=C (100%) (Trans alkene) Ex. CH3 – CH2 – C Ξ C – CH2 – CH3 / Θ (𝒷) ⎯⎯⎯⎯⎯⎯⎯⎯⎯ R H CH3CH2 C=C CH2CH3 trans Hex-3-ene H Note: This process of reduction is not eligible when terminal alkynes are taken (R–CCH) because terminal alkynes form sodium salt with Na metal. 7 Chemistry Ex. Identify the reagent for following synthesis. O O CH2 – C C – CH2CH3 ? ⎯ (A) Sol. H2/Landlar’s catalyst A Ex. cis − jasmone cis – Jasmone Identify the product in the following reaction: Sol. CH2 / CH2 – C CCH3 ⎯⎯⎯⎯⎯⎯⎯ (III) H H C=C CH3 FROM ALKYL HALIDES FROM MONO HALIDES: When mono halide reacts with alcoholic KOH or NaOH then respective alkenes are formed (i) H H H R – C – C – H + Alc. KOH H X ⎯⎯⎯⎯ R–C=C–H H Alkene ⊕ KOH + C2H5OH → C H OK + H O H H Θ R – C – C – H + C2H5O R – C = C – H +C2H5OH → H H H X (ii) CH3 – CH2 – CH – CH3 ⎯⎯⎯⎯⎯⎯⎯⎯ CH3 . Cl H C=C H Tran s CH3 This is in accordance with the say Zeff rule. 8 Chemistry Mechanism: E2 and E1 are possible. E2 Mechanism Cl Cl C–C Slow −OH H −Cl −H O C–C C=C H T.S. OH As molecularity of slowest step is two i.e., E2 mechanism. E1 Mechanism CH2 – H CH3 – C – Cl CH2 – H CH2 CH3-C+ CH3-C CH3 CH3 CH3 As molecularity of slowest step in one. That is why it is termed as E 1. Remember: Possibility of E1 increases with increasing stability of carbo cation. Similarly, possibility of E2 increases with decreasing stability of expected carbocation. From Dihalides From vicinal dihalides When vicinal dihalides are heated with Zn dust, alkene of same no. of carbon is obtained. H H R – C – C – H + Zn dust ∆ 300°C H H R – C = C – H + ZnX2 X X H H H H R – C – C – H + Zn dust → H – C = C – H + ZnCl2 Cl Cl From gem dihalides: When gem dihalides is heated with Na in ether then higher alkenes are formed. R–CH.X2 + 2Zn ∆ CH3 – CHCl2 + 2Zn ⎯⎯⎯⎯⎯⎯⎯ R – CH = CH – R CH3 – CH – CH – CH3 2 - butene 9 Chemistry Conclusion – If we take two different types of gems dihalides then we obtain three different types of alkenes. Note: The above react in is used in the formation of symmetrical alkenes only because if we take two different types of halides then mixture of alkenes is obtained so the yield of are individual alkene is reduced and is improper to separate each alkene from the mixture because the difference of boiling points in alkenes is very less. (IV) FROM ALCOHOLS: Alkenes can be prepared from monohydric alcohols or alkanols by the loss of H 2O and the reaction is known as dehydration reaction. β α -C – C- → -C=C- + H2O H H Alkene Alcohol The dehydration can be carried with Al2O3 or with mineral acid upon heating. (a) Dehydration with Al2O3: Ethene is prepared by heating ethanol with Al2O3 at 620 K. CH3 – CH2 – OH CH2 = CH2 + H2O Ethanol Ethene (b) Dehydration with mineral acid: Alcohols upon heating with conc. H2SO4 form alkenes and the reaction is called acidic dehydration. CH3–CH2 –OH % Ethanol CH2 = CH2 + H2O Ethene OH CH3 – CH – CH3 60% H SO 373 K Propan-2-ol Propene CH3 CH3–C–CH3 CH3 – CH = CH2 + H O CH3 30% H SO 363 K OH 2-Methylpropan-2-ol CH3 – C = CH2 + H2O 2-Methylpropene From the above reactions, it is clear that the order of acidic dehydration in different alcohols is Tertiary > Secondary > Primary Cycloalkenes can be prepared in the same way be the dehydration of cycloalkanols. 10 Chemistry Ex. OH / +H2O ⎯⎯⎯⎯⎯⎯⎯⎯⎯ Cyclohexanol Cyclohexene CH3 / ⎯⎯⎯⎯⎯⎯⎯⎯⎯ OH CH CH CH 3+ 3+ 2 (major) Regioselectivity of β elimination is governed by Zaitsev’s Rule. Mechanism of Reaction: The acidic dehydration of alcohol proceeds through the formation of a carbocation 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. : : ⊕ CH3 – CH2 – O – H + H ⊕ : ⇌ CH3 – CH2 – O – H Ethanol H (From acid) Protonated ethanol Step - II: Due to presence of positive change on electronegative oxygen, its electron accepting tendency increases. As a result, C – O bond becomes weak and cleaves as follows: ⊕ ⊕ : CH3 – CH2 – O – H ⎯⎯⎯ CH3 – CH2 + H2O Ethyl carbocation H This is a slow and is regarded as rate determining step. Step - III: Carbocation is unstable in nature and loses a H+ and changes into ethane in a fast step as follows: ⊕ H − CH − C H ⎯⎯⎯⎯ CH = CH + H Note: Dehydration of secondary and tart alcohol is best carried out by using dil. H 2SO4. Since alkenes produced from those alcohols have a tendency to form polymers under the influence of concentrated acid. 11 Chemistry Say Zeff 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-atoms. OH CH3 – CH2 – CH – CH3 H SO △ CH3 – CH = CH – CH3 + CH3 -CH2 – CH = CH2 Main Product 2-butene 80% 2-butanol CH3 – CH2 – CH2 – CH2-OH ⎯⎯⎯⎯⎯⎯ 1-butene 20% CH3 – CH = CH – CH3 + CH3CH2CH = CH2 2-butene 80% Main product 1-butanol 1-butene 20% Mechanism: Acid catalyzed dehydration of alkanols proceeds via the formation of more stable carbonium ion. .. ⊕ ⊕ CH CH CH − CH − 𝑂 − H + H → CH CH CH OH .. ⊕ ⊕ ⊕ CH CH CH CH OH − H + H → CH CH CH − CH + H O Primary Carbonium ion H H ⊕ CH3 –C–C–C–H H H H Rearrangement by 1,2 hybride ion shift ∘ 1 Carbonium H H H H H H H H 2∘ Carbonium more stable ⊕ CH3 –C–C–C–H H H ⊕ CH3–C–C–C–H Elimination of a proton CH3–CH=CH–CH3 2-butene (major product) CH3–CH2 – CH=CH2 1-butene (major product) (Say Zeff rule) Characteristic of dehydration of alcohol (1) Since the carbocation is formed s an intermediate in the R.D.S. of dehydration of alcohol The relative reactivity of alcohols decreases in the order. Tertiary > Secondary > Primary (2) Since the R.D.S. is reversible it is necessary to remove the H 2O molecule produced in the reaction. Therefore, in experimented condition conc. H 2SO4 is added in a regular interval. 12 Chemistry (3) For dehydration, different dehydrating agent like Al2O3, ThO2 ca also be employed. Rearrangement of carbocation can take place. –OH, is not a good leaving group. in alcohols so turn it into a good leaving group, it is converted into H2O by protonation. (4) (5) (6) H2O is a good leaving group. Rearrangement in Alcohol Dehydration (1) Some alcohols undergo dehydration to form alkene having carbon selector different from those of the straight alcohol. (2) This is due to rearrangement of carbocation formed in the reaction to more stable carbocation. CH3 CH3 𝐻 𝑆𝑂 H3C CH3 H3C CH3 –C – CH–CH3 CH3 –C–CH=CH2 + C=C + heat H3C CH3 H3C C – C – CH3 CH3OH CH3 CH3 (3%) (64%) (33%) (3) (4) An example of alcohol dehydration that is accompanied by rearrangement is. Percent yield of alkenes formed from rearranged carbocation is greater than the percent yield of alkene obtained from unrearranged carbocation. Rearrangement of carbocation can also lead to a change in ring size, as the following example shows (5) OH H3C CH-CH3 H , heat (−H O) H3C + CH-CH3 ⟶ H3C CH-CH3 + CH3 ≡ CH3 ⎯⎯⎯ H CH3 CH3 Regioselectivity in alcohol dehydration 1. In alcohols such as 2-methyl-2-butanol, dehydration can occur in two different directions to give alkenes that are constitution isomers. 2. More substituted alkene forms the major product and is called Say Zeff alkene or Zaitsev alkene. 3. Less substituted forms the minor product and is called Hofmann’s alkene. OH CH3 –C–CH2 –CH3 CH3 ⎯⎯⎯⎯⎯ 𝐶𝐻 CH3 𝐻 𝐶 + CH = CH = C C = C − CH H C CH3 (Minor) (Major) 13 Chemistry (V) FROM ESTERS When esters are heated in presence of liq. N2 and glass wool, then alkyl part of ester converts into respective alkene while alkenoate part of ester is converted into respective acid. H CH3 – CO – O Glass wool 450° liq. N CH2 – CH2 The reaction is called as ‘Pyrolysis of Ester.’ CH3 – COOH + CH2 = CH2 Mechanism: Pyrolysis of Ester is a type of E1 elimination which proceed via cyclic T.S. that is why the product becomes ‘cis. O H CH3 –C O CH2 CH2 Cyclic transition ⟶ CH2=CH–CH2–CH3+CH3–CH2–C–O–H O It is interesting to note that, in this case major produce is 1-alkene in the comparison to 2-alkene. O H O – C – CH3 CH3 – CH – CH – CH3 ⟶ CH3 – CH2 – CH = CH2 + CH3 – CH = CH – CH3 (Major) (Minor) Major product is obtained by removal of –H from less hindered carbon. (VI) KOLBE’S SYNTHESIS When aqueous solution of K or Na succinate is electrolysed, ethylene is released at anode. O –+ CH2 – C – O K –+ CH2 – C – O K O O ⟶ – CH2 – C – O.. – CH2 – C – O.. +2K+ O 14 Chemistry At Anode • CH2 • +CO2 CH2 CH2COO -2 CH2COO CH2COO• CH2 CH2COO• +2e- CH2 Unstable At Cathode +  2K 2K + 2e–   2KOH + H 2K + 2H2O  2 Note: If we use methyl succinic acid as reactant then propylene is formed. (VII) FROM QUATERNARY AMMONIUM HYDROXIDE CH3 H CH3 Θ CH3-N⊕ -CH2-CH2 ⎯⎯ CH3-N+CH2 = CH2 +H2O CH3 CH3 CH3 Ex. CH3-N ⊕ -CH-CH2-CH3 ⟶ CH-CH2CH3 CH3 ⊕ α Θ β CH3–N -CH-CH2-CH3 −OH −H O CH3 C⊕ H2 ∘ 1 𝑚𝑜𝑟𝑒 stable CH2 𝑚𝑎𝑗𝑜𝑟 product CH3 CH3 CH3 ⊖ CH3-N ⊕ -CH-CH-CH3 ⟶ CH=CH-CH3 ∘ CH3 CH3 CH3 minor product 2 less stable Ex. CH3 ⊕ α β CH3-N -CH2-CH2-CH3 ⊖ CH2 CH3 (a) (b) (c) Θ −OH −H O CH2=CH-CH3(minor) CH2=CH3(major) In this reaction β-hydrogen of tetra-alkyl ammonium ion is attached by a base and alkene is formed. In this reaction intermediate is carbanion. So, yield of product depends on stability of carbanion. In this reaction Hoffmann’s Rule is followed. 15 Chemistry (VIII) THE COPE ELIMINATION Tertiary amine oxide undergoes the elimination of a dialkylhydroxylamine when they heated. This reaction is called the Cope elimination. Reaction: CH3 CH3 △ CH3– CH2 – CH2 – N+ – CH3 ⎯⎯⎯ CH3 – CH = CH2 + N – CH3 O⊖ OH Mechanism: CH3 CH2 – N – CH3 CH3 – CH :O: H CH3 △ CH3-CH=CH2+ N – CH3 : ⎯⎯ OH3 (IX) WITTING REACTION The aldehydes and ketones are converted into alkenes by using a special class of compounds called phosphorus ylides, also called Witting reagents. The Triphenyl group of phosphorane has a strong tendency to pull oxygen atom of the aldehyde or ketone via a cyclic transition state forming an alkene. R” R” R” C=O ⊝ ⊕ R – C – pph3 C-O R” R-C-pph3 R’ R’’ R’ + Ph3P = 0 C=C R’ R’’ R’ (R, R',ylide R'' and R''' may be hydrogen or any alkyl group) Ex. Me ⊝ Methyl triphenyl Phosphonium salt : PH3P: + CH3-Br ⟶ [Ph3P ⊕ -CH3]Br ⊝ ⎯⎯⎯⎯⎯⎯ [Ph3P ⊕ -CH2] ⎯⎯⎯⎯⎯⎯⎯ Ylide C = CH Me Ph3P = O + Product alkene PHYSICAL PROPERTIES OF ALKENES / HYDROCARBON S.No. Physical properties Homologues series 1. Physical state C1 – C3 gases C4 – C20 liquids > C20: solids 2. Dipole moment (𝜇) - Isomers cis > trans 16 Chemistry 3. Polarity - 4. Melting point increases with M.W. 5. Boiling point increases with M.W. 6. Solubility 7. Stability Waals Practically insoluble in water but fairly soluble in nonpolar solvents like benzene petroleum ether, etc. cis > trans (Fro Cab = Cab type of alkenes) trans > cis (due to more packing capacity) cis > trans # branching decreases B.P. C | C-C=C trans Polarity increases, solubility in polar solvents increases. trans > cis (cis isomers has more Vander repulsion) CHEMICAL PROPERTIES OF ALKENES Alkenes are more reactive than alkane this is because – (a) The 𝜋 electrons of double bond are located much far from the carbon nuclei and are thus less firmly bound to them. (b) 𝜋 bond is weaker than 𝜎 bond and more easily 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 butanes may be compared, since all give the same products on combustion viz. 4CO2 + 4H2O 17 Chemistry Alkenes give the following type of reactions: (1) Addition reaction (2) Oxidation reaction (3) Ozonolysis (4) Substitution reaction (5) Polymerization Reaction (6) Isomerisation (1) Addition Reaction (a) REACTION WITH HYDROGEN – H H H H R-C=C-R+H2 R–C–C–R ⎯⎯⎯ H H Mechanism: The reaction takes place at the surface of Ni; therefore, the addition is cis addition. Ex. CH3 C=C CH3 D D (cis-alkene) CH3 +H2 H H D D CH3 (miso) CH3 C=C CH3 𝑁𝑖,△ (𝑐𝑖𝑠 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛) +H2 D D (Cis-alkene) CH3 𝑁𝑖, ∆ (cis addition) H H D D CH3 (miso) R R D R D (Cis-alkene) H Ni, ∆ (𝑐𝑖𝑠 − 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛) H D ≡ R D H H R R D D H (Meso-compound) 18 Chemistry R R D 𝐻 𝑁𝑖, ∆ (𝑐𝑖𝑠 − 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛) R D (trans-alkene) H D R D H D C=C D +H2 CH3 𝑁𝑖, ∆ (𝑐𝑖𝑠 𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛) D R H (Optically active) H CH3 R D ≡ CH3 H D (Trans-alkene) CH3 D + H H D D H CH3 CH3 ( ) Enantiomeric pair (i) Reduction of alkene via hydroboration Alkene can be converted in to alkane by hydroboration followed by proteolysis: H R-CH=CH2 BH2 ⎯⎯⎯⎯⎯ BH2 R-CH – CH2 4 memb. Cyclic T.s. This reaction is also represented as: ⟶ R-CH2 – CH2 / ⎯⎯⎯⎯⎯⎯⎯ 𝑅 − 𝐶𝐻 − 𝐶𝐻 / RCH = CH2 ⎯⎯⎯⎯⎯ (R-CH2-CH2)3B ⎯⎯⎯⎯⎯⎯ 3R = CH2 – CH3 + H3BO3 (ii) Alkene can be converted into alkane by hydroboration followed by treatment with AgNO3 + NaOH. CH3 CH3-CH2-CH2- C=CH2 CH3 B H AgNO / CH3 CH3-CH2-CH2- C-CH2-CH2-C-CH2-CH2-CH3 H H (b) HALOGENATION In presence of polar of medium alkene form vicinal dihalide with halogen. H H H H R–C=C–H+X–X ⎯⎯⎯⎯ R–C=C–H X X Vicinal dihalide 19 Chemistry Order of reactivity of halogens is: F2 > Cl2 > Br2 > I2 Mechanism: It is an electrophilic addition by molecular attack in which the addition takes place in trans manner. 𝛿 − 𝐵𝑟 𝛿 + 𝐵𝑟 C=C ⎯⎯⎯⎯ Br Br – C–C ⊕ C ⎯⎯⎯ ⎯⎯⎯ C Br H C=C + HX ⟶ alken e X cis CH3 CH3 Br H H Br Br H + H Br CH3 CH3 𝐵𝑟 𝐶𝐶𝑙 C=C H C–C Alkyl halide CH3 CH3 C–C H H H CH3 – CH=CH2 (c) ⎯⎯⎯ CH3-CH-CH2 REACTION WITH HX (HYDROHALOGENATION) Markownikoff’s Rule H X H CH3-CH=CH2 CH3 𝐵𝑟 𝐶𝐶𝑙 CH3 trans ⎯⎯ Iess stable CH3-CH-CH2 (Minor) H C || C H + CH3 – CH – CH2 ⎯⎯ H H CH3 Br Br CH3 miso When an unsaturated unsymmetrical hydrocarbon reacts with HX then halogen goes on that unsaturated carbon which has minimum number of hydrogen atom. Mechanism of the reaction is based on. 20 Chemistry Mechanism X ⊕ CH3-CH=CH2 ⎯⎯ CH3 – CH – CH3 CH3-CH-CH3 ⎯⎯ more stable (major) (i) Formation of Carbocation is Rate determining step. (ii) Rearrangement of carbocation to more stable carbocation can occur to form a more stable product. (iii) Regio chemistry – Markovnikov’s Addition of HX. Addition of HX on unsymmetrical alkenes (R – CH = CH 2) takes place according to Markownikaoff’s rule which states that, “the negative part of addendum is added on the carbon atom carrying lesser number of hydrogen atoms.” (iv) Reactivity order of different hydrogen haloed towards addition. HI > HBr > HCl > HF Anti Marchenko’s Principle / Kharasch Effect/ Peroxide Effect To understand anti Markovnikov’s pri0nciple let us consider the following reactions H H CH3-CH=CH2 + HBr ⟶ | | CH3 – C – C – H | | Br H Markownikoff’s rule H H CH3 – CH = CH2 + HBr ⎯⎯⎯⎯⎯⎯⎯⎯⎯ H H | | | | CH3 – C – C – H + CH3 – C – C – H | | | | Br H (minor) H Br (major) (Anti Markownikoff’s rule) It is bases on free radical mechanism. (a) Mechanism Concept – (i) R – O – O – R → RO + OR (ii) Peroxide RO + H – Br → ROH + xBr (iii) H H | | xBr + CH3 – C=C – H H H ⟶ | | | Br minor 20% H H | | | Br Major 80% 21 Chemistry HH | | CH3 –C–C–H+H Br • | Br (iv) ⟶ xBr + xBr (v) H H | | x CH3 –C–C–H+ Br | | H Br ⟶ Br2 Note: It is interesting to note that anti Markovnikov addition in the presence of peroxide is not applicable for HCl and HI (i) In the case of H – Cl, the step (ii) Cl Cl H CH2 – CH2 ⎯⎯⎯⎯⎯⎯ CH2 – CH2 + Cl is endothermic (as ∆ = + 12.6 KJ mol–1) In the case of H – I, the step I |   CH 2  CH 2  I  C H 2   C H 2 Is endothermic (as ∆H = + 46 KJ mol–1) But in the case of HBr both of the steps are exothermic, which results spontaneous reaction CCl4, CBrCl3 etc. can also be added to alkene in antiMarkovnikov manner. (d) REACTION WITH HYPOHALOUS ACID –   R  CH  CH 2  HO  C l  R  C H  C H 2 | | OH Cl (i) When chlorine water or bromine water is used. Cl R-CH ⊕ Cl CH2 ⎯⎯⎯⎯ R-CH-CH2 ⎯⎯⎯⎯ R-CH-CH2 : R-CH=CH2 Cl ∶ H +O-CH OH H 22 Chemistry (ii) When aq. solution of HOCl is added in the presence of strong acid. Cl Cl – O – H | ⊕ H-O-Cl | H + CH2-CH-R ⎯⎯⎯⎯ Cl Cl CH2 – CH – R ⎯⎯⎯ | H – O: + CH2 – CH-R | | H Zndust CH COOH (e) | | OH halohydrine C=C REACTION WITH H2SO4 – (a) With conc. H2SO4 R – CH=CH2+H2SO4 ⟶ R – CH – CH3 | O-SO3H Alkyl hydrogen Sulphate (b) With dil. H2SO4 𝐻 − 𝑆𝑂 𝐻 𝐻𝑂 R – CH=CH2 R – CH – CH2 | | OH OH Mechanism R-CH=CH2 ⎯⎯⎯⎯ + R-CH-CH3 ⎯⎯⎯⎯ R-CH-CH3 .. ∶ | H Classical carbocation | O+–H | ⎯⎯⎯⎯⎯ R-CH-CH3 | OH H The hydration of alkene is not stereoselective, as in the case of HCl addition. This fact can be explained on the basis of classical carbon cation formation. Since hydration proceeds via carbocation intermediate therefore rearrangement is always probable. 23 Chemistry CH3 | CH3-C-CH=CH2 ⎯⎯ | CH3 CH3 CH3 CH3 |+ | | CH3-C-CH-CH3 ⟶ CH3 – C – CH – CH3 ⎯⎯⎯⎯ ∶ | + | | CH3 H H CH CH | | | | CH3 – 3C – 3C – CH3 H-O+ H | H CH3CH3 ⎯⎯⎯⎯ | | CH3 – C – C – CH3 | | OH H Rearrangement, however, may be avoided by treatment of alkene by oxymercurationdemarcation method. OAc | Hg+ OAc | : : CH3CH=CH2+Hg – OAc ⟶ CH3CH – CH2 – CH2 ⎯⎯⎯⎯ CH3CHCH2–Hg–OAc +| :OH | H ↓ -:OAc CH3CHCH2 – Hg – OAc +AcO- | :OH + AcOH Sodium borohydride (NaBH4) converts the carbon-mercury bond into a carbonhydrogen bond. Because the reaction results in the loss of mercury, it is called demarcation. CH3CHCH2 – Hg – OAc OH (f) – CH3CHCH3 + Hg + AcO– OH ADDITION OF NITROSYL HALIDE – Alkene with nitrosyl bromide or nitrosyl chloride (Tilden reagent) react according to Markowinkoff’s rule to give alkene nitroso bromide and alkene nitroso chloride respectively. 24 Chemistry 𝛿⊖ 𝛿⊕ R – CH=CH2 + O=N – Cl ⟶ R – CH – CH2 𝛿⊕ | | Cl NO 𝛿⊖ R – CH=CH2 – O=N – Br ⟶ R – CH – CH2 𝛿⊕ 𝛿⊖ | | Br NO CH3 – CH – CH2 CH3 – CH=CH2 +O= N – Cl ⟶ | | Cl N=O Propylene nitroso chloride (g) ADDITION OF WATER (HYDRATION OF ALKENES): Propene and higher alkenes react with water in the 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 O ⎯⎯⎯⎯ CH − CH = CH | ⬚ Cl Propan-2-ol Propene OH (ii) | CH − C = CH + H O ⎯⎯ | CH − C − CH | CH3 CH 2-Methylpropene 2=Methylpropene-2-Ol Mechanism CH → CH = CH + H ( ⊕ ) ⎯⎯⎯⎯⎯⎯ CH − C H − CH Carbocation (2° ) ⊕ .. ( ) : CH − C H − CH + H − O − H ⎯⎯⎯⎯⎯⎯ CH − CH − CH .. | H–O–H − CH ⨁ ⎯⎯⎯⎯ CH − CH | O–H Propan-2-ol 25 Chemistry H+/H2O CH3 CH CH=CH2 CH3 OH CH3 C CH2 CH3 CH3 OH (i)BH3/THF CH3 CH CH CH2 (ii)H2O2/ –OH CH3 CH3 Cl CH3 C CH2 CH3 CH3 HCl without rearrangement anti-markownikoff's rule with rearrangement Markownikoff's rule OH CH CH CH2 CH3 H With out rearrangement markownikoff's rule HBr CH3 CH CH CH2 Peroxide CH3 H Br With out rearrangement anti-markownikoff's rule (ii) (AcO)2Hg/ CH3 H2 O (i) NaBH/NaoH (h) markownikoff's rule with rearrangement HYDROFORMYL OR OXO REACTION: Alkenes react with Carbon monoxide and hydrogen at 100 - 150°C temperature and high pressure (1200 atm) in the presence of Cobalt catalyst to produce an aldehyde. It does not follow 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–CH2–CH2–CHO Co./150° 𝐶 R-CH=CH2+CO+H2 High pressure R–CH–CH3 CHO (i) ALKENYLATION (ADDITION OF ALKENE) In presence of H2SO4 or H3PO4 at 80°C dimerization of isobutylene take place gives two isomers of octene. CH3 | 2CH3–C=CH2 CH3 , ° | CH3 | CH3 CH3 ⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH2=C–CH2–C–CH3+CH3–C=CH–C–CH3 | CH3 CH3 26 Chemistry Mechanism CH3 CH3 | CH3–C=CH2 + H ⊕ | ⟶ CH3–C–CH3 ⊕ CH3 CH3 CH3 | | | CH3 | CH3–C=CH2 + ⨁ C–CH3 ⟶ CH3=C–CH2–C–CH3 | ⊕ CH3 CH3 | | ⊕ CH3 –H CH3 CH3 | | CH2=C–CH2–C–CH3 | CH3–C=CH–C–CH3 | | CH3 (j) CH3 CH3 ADDITION OF CARBENE: The addition of carbene to alkene is always carried by diazomethane CH2N2 Carbene group obtained from diazomethane is added to alkene and give cycloalkanes. CH2 = CH2+CH2N2 △ ⎯⎯⎯⎯ CH2–CH2+N2 CH2 CH3 – CH=CH2+:CH2 ⟶ CH3–CH–CH2 CH2 Since: CH2 is an electrophilic (neutral) and there is more electron density on double bond so first attack of: CH2 will be at double bond. (k) ADDITION OF HCN: δ– δ– δ– δ CH3–CH=CH2+HCN (2) ⟶ CH3–CH–CH3 | CN Oxidation Reaction: Alkenes are easily oxidised by oxidising agents. Odidising agents attack on double bond and product formed during oxidation depends on oxidising agents. 27 Chemistry (a) ALKENE ON COMBUSTIION GIVES CO2 AND H2O CnH2n+ O2 ⟶ nCO2 + nH2O one mole of alkene requires (b) moles of O2 for complete combustion. WITH ACIDIC KMNO4 / HOT KMNO4 H H O || CH3 – CH–C–OH+CO2+H2O | R–C–O–H+CO +𝐻 𝑂 CH3 O [ ] CH3–CH–CH=CH ⎯⎯⎯ R –| C = C – H + [O] CH3 H H H–C=C–H ethene [ ] ⎯⎯⎯ 2CO + 2H O H H H–C–C=C–H H H [ ] ⎯⎯⎯ H H OH Propenes H H–C–C=O+CO2 + H2O Acetic acid H CH3 – C = C – CH3 [ ] ⎯⎯⎯ 2CH3 COOH 2–butene CH3 – C = CH2 | CH3 isobutylene (c) [ ] ⎯⎯⎯ CH3 – C=O+CO2+H2O | CH3 WITH ALKALINE KMNO4 / BAYER’S REAGENT H H H H | | R–C=C–H+[O]+H–OH . – R–C–C–H OH OH glycol 28 Chemistry Mechanism: Hydroxylation by alkaline KMnO4 (Bayer’s reagent) is cis addition and the mechanism is cyclic. H H O C + Mn C R O H H R O– O C ⟶ MN C O R ⎯⎯⎯⎯⎯⎯ O R (miso) Similarly H R O C + Os C R O H H R C C O O R H R H O ⎯⎯⎯⎯⎯⎯⎯ OH C OH H H R O O / Os C OH C OH ⎯⎯⎯⎯⎯⎯⎯ O O H R H R C R C O H / O C ⟶ R O Os C H R (d) O Os + O C ⟶ O H R O OH OH H H / O H R O– HYDROXYLATION BY OSO4 R H R O C O R R O O H C OH C OH ⎯⎯⎯⎯⎯⎯ H R H H C + OsO4 C H / Os C O R H C R O O ⟶ Os + C O H R ⎯⎯⎯ H HO OH H R (+) 29 Chemistry Ozonolysis This is the two-step reaction (i) Ozonide formation (ii) decomposition of ozonide (reductive hydrolysis) R C C O C=C O O I O (3) H O/H /Zn II O C + C || ZnO+ H O || R C + OsO4 C H HO ⎯⎯⎯ OH H R (+) Remember: Ozonolysis gives oxidative cleavage of alkene to form two carbonyl group for each >C=C<. Whether it is in acyclic or cyclic or in aromatic compound. e.g.: H R O O O O || || || || H–C–H+H–C–C–H+H–C–H 𝑶𝐳𝐨𝐧𝐨𝐥𝐲𝐬𝐢𝐬 (a) CH2 = CH – CH = CH2 ⎯⎯⎯⎯⎯⎯⎯⎯ (b) 𝐎𝐳𝐨𝐧𝐨𝐥𝐲𝐬𝐢𝐬 O + O ⎯⎯⎯⎯⎯⎯⎯⎯ O + O Trans Epoxidation by O2/Ag O (i) CH2 = CH2 + ½ O2 ⎯⎯ CH2 – CH2 O (ii) CH3 – CH = CH2 + ½ O2 ⎯⎯ CH3 – CH – CH2 30 Chemistry Epoxidation by per acid O C–O–O–H O (i) CH2 – CH2 CH2 = CH2 O (ii) O Hydroboration Alkene with borane hydride form an important compound called trialkyl borane. 3R – CH = CH2 + BH3 → (R – CH2–CH2)3B (a) Trialkyl borane is an important compound because it gives respective alkane on acidic hydrolysis. (R – CH2CH2)3B + 3H2O ⎯⎯⎯ 3R – CH2 – CH + B(OH)3 (b) It gives respective alcohol on alkaline hydrolysis (R – CH2CH2)3B + 3 H2O2 + 3NaOH → 3 R CH2CH2OH + 3NaCl + H3BO3 (c) It gives I° amine on alkaline hydrolysis with chloramine (R – CH2CH2)3B + 3NH2Cl + 3NaOH → 3R – CH2CH2NH2 +3NaCl + H3BO3 3R-CH2-CH2-CH2-R (R-CH2CH2)3B (d) (e) In the overall hydroboration-oxidation reaction, three moles of alkene read with one mole of BH3 to form three moles of alcohol. (f) Because carbocation intermediates are not formed in the reaction carbocation rearrangements do not occur. (g) Regio chemistry Anti-markownikoff’s Addition of water. (h) Alkyl boranes undergo coupling by means of silver nitrate in the presence of NaOH at 25°C. Higher alkanes are the products of this reaction. 31 Chemistry (4) Substitution Reaction (a) Except ethene other higher alkene having alkyl hydrogen when treated with chlorine or bromine ‘𝛼’ H is substituted. Allylic hydrogen atoms Allylic Carbon atoms H R – C – CH = CH2 H 𝟓𝟎𝟎° 𝑪 𝟔𝟎𝟎° 𝑪 CH3 – CH = CH2 + Cl2 ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ Cl - CH2 – CH = CH2 + HCl allyl chloride (3-chloropropene) ∆ ⎯ Cl – Cl (b) 2Cl-  CH2 – CH = CH2 Allylic bromination can be easily done by NBS (N-bromo succinimide) CH2-CO CH2-CO N-Br+H-CH2-CH=CH2 → NBS (5) N-H+Br-CH2-CH=CH2 CH2-CO CH2-CO Propene allyl bromide Polymerization (i) Two or more than two molecules of same compound unit with each other to form a long chain molecule with same empirical formula. This long chain molecules having repeating structural units called polymer, and the starting simple molecule as monomer and process is called addition polymerization. (ii) Molecular weight of polymer is simple multiple of monomer. (iii) The presence of oxygen initiates free radical mechanism. (iv) Addition polymerization can also be carried out by ionic mechanism by using Ziegler - Natta Catalysts (R3Al + TiZl4) 32 Chemistry Name of polymer 1. Polyvinyl chloride (PVC) Structure of monomer CH2 = CH – Cl 2. Polypetalous- F2C=CF2 ethylene or Te lon (PTFE) 3. Natural rubber vulcanized rubber which is strong 4. Orlon 5. Poly methyl methacrylate (PMMA) CH  CH  2 isoprene CH3 | C  CH3 acrylonitrile CH  2 CH | 3 C COOCH 3 Structure of Polymer Properties Uses Properties ( H C  H C  ) 2 | Pliable (easily moulded) Methyl methacrylate      CH  2        Used in handbag, raincoats, vinyl Cl looring, good electrical insulator for wires Flexible and For making non( F C  CF2  )n 2 inert to solvents, stick utensils boiling acids, coating even aquaregia stable up to 598K. Waxy and nonUsed as raw elastic material for making and elastic vulcanized rubber is used in making tyres hose, pipes etc. Used in making ( H C  H C  ) Fibrous 2 | Fabrics CN CH | C 3  | CO | OCH 3           n 33 Chemistry (6) Isomerisation Alkenes isomerase when heated at high temperature or at lower temperature in the presence of various catalysts as AlCl3. Ex. (i) (ii) CH3 – CH2 – CH2 – CH = CH2 𝐴𝑙𝐶𝑙 ⇌ CH3 – CH2 – CH = CH – CH3 CH3 CH3 AlCl | | 3    CH  CH CH  C  CH3 CH2 2 2  the mechanism proceeds through carbocation Uses (a) In plastic formation. (b) In oxy ethylene welding (c) As food preservatives and ripening fruits. (d) As general anaesthetic (C2H4 with8 10% O2) (e) In preparation of mustard gas CH2 CH2Cl CH2 → CH2 | + S2Cl2 | | | +S CH – S – CH 2 2 CH CH2+ 2 2,2’or (β, β′)trichloro diethyl-sulphide 2 (mustard gas) Functional Group (1) Bayer’s Reagent alk. dil. cold KMnO4 Reagent Observation Pink colour Disappears CH2=CH2+ H2O+O 4   (2) Br2/H2O Red colour decolorizes Br2+CH2= CH2 → C H 2  CH | 2 | BR Br C=C Reaction alk.KMnO Remarks Dihydroxylation CH 2  CH 2 | | OH OH Debromination White ppt. (3) O3 (ozone) C=O Compounds CH2=CH2+ O3 Zn/H O 2  Ozonolysis 34 Chemistry Dienes Dienes are the unsaturated hydrocarbons with carbon-carbon double bonds in the molecules. These are represented by the general formula C nH2n-2 which means that they are isomeric with alkynes (functional isomers). However, their properties 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, CH3 C H = C H − C H − CH = C H penta − 1,4 − diene C H = C H − C H − C H = CH 3 − Methylpenta − 1,4 − diene Conjugated dienes: In a conjugated diene, the two double bonds are present in the conjugated or alternate positions and are separated by a single bond. CH = CH − CH = CH Buta − 1,3 − diene CH = C − C H − CH = CH 2 − Methylpenta − 1,3 − diene Cumulate dienes: In this case, the two double in the molecules are present at adjacent positions. For example, CH = C = CH Buta − 1,2 − diene CH = C − CH − CH 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 justified on the basis of the theory of resonance. Penta-1, 3-diene (conjugated diene) is a hybrid of the following contributing structures. CH2=CH – CH=CH – CH3 ⊝ ⊕ ⊝ ⊕ ↔ CH2 – CH – CH = CH – CH3 ↔ CH2 – CH = CH – CH – CH3 The delocalization of 𝜋-electron charge because of resonance decreases the energy of the molecule or increases its stability. Penta-1, 4-diene (isolated diene) has only two contributing structures. ⊖ CH = C H − C H − CH = C H (I) ↔ ⊕ C H − C H − C H − CH = C H (II) 35 Chemistry Since the carbon atom C3 is not involved in any resonance, the contributing structures are less number are compared to the conjugated diene. The isolated diene is, therefore, less stable than a conjugated diene. 1. PROPETIES OF CONJUGATED DIENES The properties of the isolated dienes are similar to those of simple alkene but those of conjugated dienes are somewhat modified because of delocalization of the 𝜋electron charge. However, they also patriate in the addition reactions. The important chemical characteristics of the conjugated dienes are briefly discussed. Addition 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. Electrophilic Addition Reactions: The electrophilic addition is illustrated by the attack of halogen and halogen acid on Buta--1,3-diene, a conjugated diene. Addition of halogen: If one mole of halogen attacks per mole of the diene, two types of addition products are formed. There are 1, 2 and 1, 4 addition products. For example, Br | Br | 3,4-Dibromobut-1-ene (1,2-Addition) Br | Br | 1,4=Dibromobut-2-ene (1,4-Addition) 1, 2- addition is a normal addition in which one mole of halogen has been added to one of the double bonds. But 1, 4- addition is somewhat unexpected. Mechanism: The addition is electrophilic in nature and the halogen molecule (bromine) provides the electrophilic for the attack. Buta-1,3-diene Carbocation (2° ) 36 Chemistry The 2° carbocation get stablised by resonance as follows – The attack of ion on carbocation (I and II) Br ⊕ Br CH2=CH – CH – CH2 – Br (1,2-Addition) II CH2 – CH = CH – CH2 I Br Br 1,4addition (Main product) Addition of H - X CH2=CH – CH=CH2 ⎯⎯⎯⎯ ? Mechanism: The addition is electrophilic in nature as H + ion is the electrophilic The carbocation gets resonance stablised as follows: 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. CH 2  CH  CH  CH 3 | Br  Br   CH 2  CH  C H  CH 3  Br   C H 2  CH  CH  CH 3  1, 2  Addition product  CH 2  CH  CH  CH 3 | Br 1, 4  Addition product (2) Free Radical Addition Reaction: The addition to conjugated dienes can also proceed by free radical mechanism provided it is carried in the presence of a suitable regent which can help in forming a free radical. However, the addition also yields 1, 2 and 1, 4 addition products. The free radical is illustrated by the attack of 37 Chemistry bromodichloromethane (BrCCl3) on buta-1,3-diene in the presence of an organic peroxide such as benzoyl peroxide. 1 2 3 4 4 3 2 1 4 3 2 1 Benzyl CH2  CH  CH  CH2  BrCCl3   Cl3CCH2 CH  CH  CH2  Cl3CCH2  CH  CH  CH2 Peroxide | | Br Buta 1,3  diene Br Bromotri  chloromethane Mechanism: The mechanism of addition is free radical in nature which is initiated by benzyol free radical. It is explained in the following steps: Step – I Generation of free radical. C6 H5CO  O  O  COC6 H5 Homolysis    C6 H5  CO  O  C6 H5  BrCCl3   C 6 H 5  CO  O  O  COC6 H 5   C6 H5  CO 2  C6 H 5  Br  CCl3  Trichloromethyl free radical Step- II Attack of free radical on buta-1, 3-diene.  CCl3  CH 2  CH  CH  CH 2   Cl3C  CH 2  C H  CH  CH 2 Free radical (secondary) The free radical gets resonance stablised.  Cl3C  CH 2  C H  CH  CH 2 Step- III   Cl3C  CH 2  CH  CH  C H 2 Change of free radical into addition product. The free radical takes up 𝐵̇ 𝑟 from the attacking reagent to give the desired addition products. i.e., 1, 2 and 1, 4 addition products. 38 Chemistry 3. Cyclo-Addition Reaction (Diel Alder Reaction) Cyclo-addition reactions are one of the most important reactions of conjugated dienes. Cyclo-addition involves the combination between a conjugated diene (4𝜋electron system) and a compound containing a double bond (2 𝜋 -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 often termed as (4 + 2) cyclo-addition reaction because four 4 𝜋electron system adds to a two 2 𝜋 electron system. The addition products are called Diel Alder Adduct. For example. CH2 CH2 HC | HC + CH2 || CH2 → CH2 CH2 Buta-1,3-diene Cyclohexene (Diel Alder adduct) Ethene CH2 HC | HC CH2 | CH2 HC || HC CH2 + CH2 Buta-1,3-diene CH – CHO || CH2 Propenal (Acrolein) → CH – CHO | CH2 HC || HC CH2 3 - cyclohexene carbaldehyde 39