Sem - I Unit-III C) Aliphatic Hydrocarbons By Dr Pramod R Padole

C) Aliphatic Hydrocarbons:
a) Alkanes: Methods of formation: i) Wurtz reaction & ii) Corey-House reaction. Chemical reactions: i) Halogenation (With mechanism),
ii) Aromatisation.
b) Alkenes: Methods of formation (With mechanism): i) Dehydrohalogenation of alkyl halides (E1 & E2), ii) Dehydration of alcohols.
Chemical reactions: Electrophilic & free radical addition of HX and X2 (With mechanism).
c) Alkynes: Preparation from vicinal and germinal dihalides. Chemical reactions: Hydrogenation.
d) Alkadienes : Classification, 1,3-Butadiene: Preparation from cycolhexene, Reactions: Addition of H2, Br2 & HBr.

1. 1 B.Sc. First year Students Semester – I Unit- III C ) Aliphatic Hydrocarbons by Dr Pramod R Padol e

2. Our Inspiration 2

3. Organic Chemisrtry Unit-III A) Electronic Displacements: B) Reactive Intermediates: C) Aliphatic Hydrocarbons:

4. 3. Alkynes: Preparation from vicinal and germinal dihalides , Reaction- Hydrogenation 2. Alkenes: Methods of formation (With mechanism): i ) Dehydrohalogenation of alkyl halides (E1 & E2) ii) Dehydration of alcohols, Reactions: Electrophilic & free radical addition of HX and X 2 (with mechanism) Alkanes : Methods of formation: i ) Wurtz reaction and ii) Corey-House reaction, Reactoins : i ) Halogenatiom (With mechanism), ii) Aromatisation . Alkadienes : Classification,1,3-Butadiene- Preparation from cyclohexene , Reactions- Addition of H 2 , Br 2 and HBr . C) Aliphatic Hydrocarbons:

5. C) Aliphatic Hydrocarbons: Organic compounds containing carbon and hydrogen only are called hydrocarbons.

6. Do you know? What are Aliphatic Hydrocarbons?

7. What are Aliphatic Hydrocarbons? Aliphatic hydrocarbons are the organic molecules containing Carbon (C) and Hydrogen (H) atoms in their structure; in straight chains, branched chains or non-aromatic rings. Aliphatic hydrocarbons can be categorized into three main groups; alkanes ,  alkenes and alkynes . Two C=C bonds ALKADIENES or Dienes : X

8. [email protected] By Dr Pramod R Padole C) Aliphatic Hydrocarbons: Alkanes C-H & C-C bond Alkenes C=C bond Alkynes one C≡C bond Alkadines Two C=C bonds First Second Third Fourth

9. Alkanes organic compounds containing carbon and hydrogen only Saturated Hydrocarbons

10. [email protected] By Dr Pramod R Padole i ) Alkanes : (Saturated Hydrocarbons) Alkanes : The open chain saturated organic compounds containing carbon and hydrogen only are called as alkanes . ( i.e., Characteristic of C-C & C-H linkage) They have the general formula C n H 2n+2 . Where, n = 1,2,3,…, etc. Examples: 1) CH 4 - Methane, 2) C 2 H 6 – Ethane, 3) C 3 H 8 – Propane, 4) C 4 H 10 – Butane,…… etc.

11. www.themegallery.com By Dr Pramod R Padole In general, alkanes are non-reactive (inert) compared to other classes of organic compounds. 1 They don’t have a functional group 2 There bonds are quite strong, i.e., C-C (strong bond). Large energy is needed/required to break the bond hence less reactive 3 These two bonds are almost non-polar and therefore, neither electrophilic nor nucleophilic substitution reaction can take place. Can’t react with electron-loving species or a proton loving species ( Nucleophilic ) Reasons Reasons Reasons

12. [email protected] By Dr Pramod R Padole i ) Alkanes : (Saturated Hydrocarbons) Alkanes contain strong C-C and C-H covalent bonds. Therefore, this class of hydrocarbons is relatively chemically inert. Hence they are also called as Paraffins (Latin, parum = little , affinis = affinity ; Chemical attraction). Sources: The most important source of alkanes is petroleum or the natural gas associated with it. Fuel gases obtained from coal, e.g. Coal gas, etc. contain these hydrocarbons in small amounts. Methane (called as Marsh gas ) is formed during the decay of plants or animal tissues. Butane is used as a fuel in lighters. It is also used in same camping stoves.

13. Why methane is called marsh gas? Methane (called as Marsh gas ) is formed during the decay of plants or animal tissues.

14. Methods of Preparation: Reactions that produce a particular functional group are called preparations .

15. [email protected] By Dr Pramod R Padole Methods of Preparation : i ) Wurtz Reaction or Synthesis: or Preparation of Higher symmetrical alkanes : Preparation Of Alkanes ii) Corey-House Reaction or Synthesis: or Preparation of Higher unsymmetrical alkanes : or Coupling of alkyl halides with organometallic compounds:

16. Wurtz Reaction: or Preparation of Higher symmetrical alkanes : The Wurtz reaction is a very useful reaction in the fields of organic chemistry and organometallic chemistry for the formation of alkanes . In this reaction, two different alkyl halides are coupled to yield a longer  alkane  chain with the help of sodium and dry ether solution. Unhydrous Diethyl ether  (dry ether) is an especially good solvent for the formation of Higher symmetrical alkanes using Wurtz Reaction  because  ethers  are non-acidic ( aprotic ).

17. i ) Wurtz Reaction or Synthesis: or Preparation of Higher symmetrical alkanes : Q.1) What happens when alkyl halide is reacted with sodium metal in presence of ether? Q.2) How will you convert : Alkyl halide to Higher alkanes . Q.3) Explain: Wurtz reaction with suitable example. ( W-17, 4 Mark) Q.4) What happen when, methyl bromide is reacted with sodium metal in presence of dry ether? ( W-13 & S-16, 2 Mark) Q.5) What happen when, ethyl bromide is reacted with sodium metal in presence of dry ether? ( W-14, 2 Mark) Q.6) How will you prepare the Butane from ethyl chloride? ( Wurtz Reaction) ( S-18, 2 Mark) Q.7) How will you convert: Methyl chloride to Propane? ( W-19, 2 Mark ) Wurtz Reaction: When alkyl halide (two molecules) is reacted or treated with sodium metal in presence of dry ether as a solvent; to form higher alkanes .

18. i ) Wurtz Reaction or Synthesis: or Preparation of Higher symmetrical alkanes : {Note that: This method is particularly suitable for the preparation of symmetrical alkanes .}

19. Mechanism of Wurtz Reaction:

20. ii ) Corey-House Reaction or Synthesis: or Preparation of Higher unsymmetrical alkanes : or Coupling of alkyl halides with organometallic compounds: alkyl halide is reacted with lithium metal in presence of solvated ether

21. Corey-House Reaction or Synthesis: (also called the Corey–Posner– Whitesides –House reaction and other permutations) Corey-House Reaction: This method was developed in late 1960s by E.J. Corey and Herbert House, called as Corey-House synthesis. The coupling reaction is a good synthetic way to join two alkyl groups together to produce higher alkanes . This versatile method is known as the Corey-House reaction. (also called the  Corey –Posner– Whitesides – House reaction  and other permutations) Corey Q.1) What happens when alkyl halide is reacted with lithium metal in presence of solvated ether? Q.2) How will you convert :- i ) Alkyl halide to Higher alkanes . Q.3) Explain: Corey-House reaction with suitable example. Q.4) How will you prepare the Propane from ethyl bromide? (Corey-House Reaction) (S-18, 2 Mark) Q.5) How will you convert: Methyl chloride to Propane ? (W-19, 2 Mark)

22. Corey-House Reaction or Synthesis: (also called the Corey–Posner– Whitesides –House reaction and other permutations) Corey-House Reaction: When alkyl halide is reacted or treated with lithium metal in presence of ether as a solvent; to form first alkyl lithium , further it is reacted with cuprous halide; to form lithium dialkyl cuprate (LiR 2 Cu). This is then treated with the second alkyl halide ; to form a new alkane as a final product along with organocopper compound (R-Cu) and a lithium halide ( LiX ). Corey Where, R’ -X should be a primary halide (second alkyl halide); The alkyl group R- in the organometallic may be primary, secondary, or tertiary. {Note that: This method is particularly suitable for the preparation of unsymmetrical alkanes .}

23. Corey-House Reaction or Synthesis: (also called the Corey–Posner– Whitesides –House reaction and other permutations) Corey-House Reaction: When alkyl halide is reacted or treated with lithium metal in presence of ether as a solvent; to form first alkyl lithium , further it is reacted with cuprous halide; to form lithium dialkyl cuprate (LiR 2 Cu). This is then treated with the second alkyl halide ; to form a new alkane as a final product along with organocopper compound (R-Cu) and a lithium halide ( LiX ). Corey

24. Do you know? Why is Corey House reaction a better method for preparing alkane than Wurtz reaction?

25. Q.1) Why is Corey House reaction a better method for preparing alkane than Wurtz reaction? From  Wurtz reaction we can  prepare   only even carbon number of  alkane  but from  Corey House reaction we can  prepare   odd and even carbon number  alkan e . So  Corey House reaction  is preferred over  Wurtz reaction. First four alkanes methane, ethane, propane and butane are gases. Next thirteen members (C 5 to C 17 ) are colourless liquids. Higher alkanes are wax like solids. 2) Alkanes are nonpolar compounds.

26. Chemical Reactions of Alkanes : Halogenation (Addition of Halogen): 2) Aromatistion : Formation of Aromatic compound : Chemical Reactions: a) Chlorination b) Reaction with Iodine in presence of HIO 3 : i ) Reaction with Chlorine: ii) Reaction with Excess of Chlorine:

27. [email protected] By Dr Pramod R Padole Replacement of hydrogen atoms from alkane by halogen atom is known as Halogenation i ) Reaction with Chlorine in presence of U.V. light or heat at 623-673 K : Chlorination Of Alkanes ii) Reaction with Excess of Chlorine in presence of U.V. light or heat at 623-673 K :

28. a) Chlorination: 1) Reaction with Chlorine: When alkane is treated or heated or reacted with chlorine in presence of UV light or at high temperature (623-673 K); to form haloalkane . Q.1) What happens when methane is treated with Cl 2 in presence of U.V. light or heat at 623-673 K? (W-06, W-11 & S-15, 2 Mark) Q.2) Compete the following reaction: (W-17, 2 Mark) Q.3) How will you prepare Ethyl chloride from Ethane? (S-19, 2 Mark)

29. a) Chlorination: 1) Reaction with Chlorine: When alkane is treated or heated or reacted with chlorine in presence of UV light or at high temperature (623-673 K); to form haloalkane . Q.1) What happens when methane is treated with Cl 2 in presence of U.V. light or heat at 623-673 K? (W-06, W-11 & S-15, 2 Mark) Q.2) Compete the following reaction: (W-17, 2 Mark) Q.3) How will you prepare Ethyl chloride from Ethane? (S-19, 2 Mark)

30. a) Chlorination: 1) Reaction with Chlorine: When alkane is treated or heated or reacted with chlorine in presence of UV light or at high temperature (623-673 K); to form haloalkane . Q.1) What happens when methane is treated with Cl 2 in presence of U.V. light or heat at 623-673 K? (W-06, W-11 & S-15, 2 Mark) Q.2) Compete the following reaction: (W-17, 2 Mark) Q.3) How will you prepare Ethyl chloride from Ethane? (S-19, 2 Mark)

31. Reaction with excess of Chlorine: excess of chlorine in presence of UV light or at high temperature (623-673 K) Reaction with excess of Chlorine :

32. By Dr Pramod R Padole Reaction with excess of Chlorine: When methane is treated or heated or reacted with excess of chlorine in presence of UV light or at high temperature (623-673 K); hydrogen atoms of methane can be replaced by chlorine atoms one by one ; to form different product of chloromethane .

33. b) Iodination (in presence of HIO 3 ): Q.1) What happens when- i ) Methane is treated with iodine in presence of HIO 3 (Oxidizing Agent) ( S-04, W-04, S-05 & W-06, 1-2 Mark) Q.2) Complete the reaction: CH 4 + I 2 HIO 3 ? ( W-09, 2 Mark) Q.3) How will you convert: Methane to iodomethane ? ( S-10, 1 Mark )

34. b) Iodination (in presence of HIO 3 ): When alkane is treated or reacted with iodine in presence of strong oxidizing agent, such as HIO 3 , HNO 3 or HgO , etc.; to form alkyl iodide.

35. By Dr Pramod R Padole Mechanism of Chlorination of Methane

36. Mechanism of Chlorination of Methane: When methane is treated or heated or reacted with chlorine in presence of UV light or at high temperature (623-673 K); to form chloro -methane or methyl chloride. Q.1) Discuss / Explain the mechanism of chlorination of Methane. (S-04, S-06, W-07, W-13, W-15 & S-17, 4 Mark) Q.2) Discuss the mechanism of the following reaction: (S-05, 4 Mark) CH 4 + Cl 2 U.V light CH 3 Cl + HCl Q.3) Discuss / Explain free radical mechanism of chlorination of Methane.(S-07, S-08, W-09 & W-10, 3 Mark)

37. Mechanism of Chlorination of Methane: 1) Homolytic fission can be done by using UV light or at high temperature or peroxide. 2) Generally, a covalent bond between two same atoms or two atoms having nearly same eletronegativity , undergo homolysis . 3) Generally, homolysis is favoured by following factors: a) Gaseous state of reactants. b) At high temperature or in presence of sunlight, UV light, etc. c) Non-polar solvents. d) In presence of peroxide.

38. Mechanism: (Free Radical Mechanism ) Chlorination of methane follows Free radical mechanism and involved three steps. Step-I) Chain Initiation: where a radical species is generated, generally by heat, light, or other catalytic process. When chlorine molecule is heated in presence of UV light or at high temp. (623-673 K), undergoes homolytic fission; to form chlorine free radicals.

39. Mechanism: (Free Radical Mechanism) Step-II) Chain Propogation : here one radical species interacts with another molecule to create another radical species. a) When chlorine free radical attacks on methane; to form methyl free radical.

40. Mechanism: (Free Radical Mechanism) Step-II) Chain Propogation : here one radical species interacts with another molecule to create another radical species. b) When methyl free radical is reacted with chlorine; to form methyl chloride & Cl free radical. In above reaction, regeneration of chlorine free radical is take place. This two reactions (a & b) repeated over & over again. So, this step is known as chain propogation step.

41. Mechanism: (Free Radical Mechanism) Step-III) Chain Termination: where two radical species interact and quench the radical reaction an form a stable product.  Chain reaction can be stop or end by the combination of two F.R. as shown below, The final stage of a radical reaction is the  termination reaction  which quenches the radical species present.  For the methane – chlorine reaction this is the formation of the chloromethane (CH 3 Cl).

42. Aromatization: Formation of Aromatic Compound: Cr 2 O 3 catalyst supported over Al 2 O 3 at 873 K (600 o C)

43. Aromatization: Formation of Aromatic compound: When vapours of alkane containing six or more (C=6,7,8,…,etc.) Carbon atoms is passed over Cr 2 O 3 catalyst supported over Al 2 O 3 at 873 K (600 o C); to form aromatic compound. This process is known as aromatization. Q.1) Explain the term:- Aromatisation of an alkane with a suitable example. (S-04, W-04, W-05, S-06, W-06, W-08, S-09, S-10, W-10, S-11 & S-15, 2 Mark) Q.2) How can you bring the following conversions? (S-07 & S-09, 2 Mark) i ) Benzene from n-Hexane . Q.3) How will you convert: n-hexane to benzene? (W-11 & S-18, 2 Mark) Q.4) Complete the following reaction: Q.5) Write short note on: Aromatisation of an alkane . (W-16, 2 Mark) Q.6) What happen when n-hexane heated with Cr 2 O 3 supported over alumina?(S-17, 2 Mark)

44. Aromatization: Formation of Aromatic compound: e.g . 1) When vapours of n-hexane is passed over Cr 2 O 3 catalyst supported over Al 2 O 3 at 873 K ( 600 o C); to form Benzene (& H 2 is set free).     e.g . 2) When vapours of n- heptane is passed over Cr 2 O 3 catalyst supported over Al 2 O 3 at 873 K (600 o C); to form Toluene (& H 2 is set free).

45. Aromatization in alkane can be brought by using  Cr 2 O 3 catalyst supported over Al 2 O 3 at    873 K (600 o C ) . This process is known as reforming . In this reaction, one hydrogen atom is removed from each terminal carbon atom of n-hexane to form cyclohexane . This is cyclization reaction.  This is followed by the removal of three hydrogen molecules to form benzene. This is aromatization reaction.

46. D. Con. H 2 SO 4 B. Fe 2 O 3 A. Al 2 O 3 Aromatization in alkane can be brought by using: C. Cr 2 O 3 /Al 2 O 3 at 873 K (600 o C)

47. ii) Alkenes: Alkenes are also called olefins because the lower members forms an oily products on treatment with chlorine or bromine. ( Latin word : Olefiant - oil forming) or ( Oleum = oil ; fiacre = to make).

48. Ethylene is the hormone that causes tomatoes and apples to ripen .

49. Do you know? ALKENES

50. ii) ALKENES: Q.1) Define Alkenes. Q.2) Aliphatic unsaturated hydrocarbon containing one carbon-carbon double bond (C = C bond) is called _____. (W-11 & S-15, ½ Mark) Defination : Aliphatic unsaturated hydrocarbons containing one carbon-carbon double bond (C = C bond) are called alkenes . (W-11, ½ Mark)   Or Aliphatic unsaturated hydrocarbon containing one carbon-carbon double bond (C = C bond) are called alkenes . (S-15, ½ Mark) They are represented by the general formula CnH 2 n   Where , n is the number of C-atoms & n = 2, 3, 4, etc.   e.g. 1) CH 2 = CH 2 2) CH 3 CH= CH 2 Ethylene Propylene   Alkenes are also called olefins because the lower members forms an oily products on treatment with chlorine or bromine. (Latin word : Olefiant - oil forming) or ( Oleum = oil ; fiacre = to make).

51. Q.1) Write the IUPAC / Common names. According to the common name system olefins are named as ' alkylenes ' . In these suffix “ ane ” of the corresponding alkane is replaced by “ ylene ” In the IUPAC system their names are obtained by replacing the suffix ' ane ' of an alkane by ' ene '.

52. By Dr Pramod R Padole Sources of Alkenes or Preparation of Alkenes: Dehydration of alcohols: a) By using conc. H 2 SO 4 (as a dehydrating agent): (Liquid Phase Dehydration) & b) By using Alumina, (Al 2 O 3 ): ( Vapour Phase Dehydration ) Alkenes Preparation: Dehydrohalogenation of Alkyl Halides:

53. Dehydration of Alcohols: Preparation of Alkenes form Alcohols: Dehydration of alcohols: a) By using conc. H 2 SO 4 : (as a dehydrating agent): (Liquid Phase Dehydration) Dehydration of Alcohol: Dehydration of alcohols: b) By using Alumina, (Al 2 O 3 ): ( Vapour Phase Dehydration)

54. By Dr Pramod R Padole Preparation of Alkenes form Alcohols or Dehydration of Alcohols: 1) Dehydration of alcohols: Dehydration means removal of water molecule from adjacent C-atoms (α , β carbon atoms) and produces alkenes. Dehydration of alcohol is a β-elimination reaction. Alcohols are dehydrated by conc. H 2 SO 4 or Alumina (Al 2 O 3 ) or H 3 PO 4 or P 2 O 5 . The decreasing order of reactivity of alcohols in dehydration is, Tertiary (3 o ) > Secondary (2 o ) > Primary (1 o ) Q.1) How are alkenes prepared from alcohols? Give mechanism. Q.2) Discuss the mechanism of dehydration of alcohols. Q.3) Explain the mechanism of dehydration of ethyl alcohol (ethanol). (S-12 & S-13, 4 Mark) Q.4) Give the preparation of ethylene form ethyl alcohol and discuss its mechanism. (S-14, 4 M) Q.5) How will you prepare 1-Propene from 1-Propanol? (S-18, 2 Mark)

55. Dehydration of Alcohols: a) By using conc. H 2 SO 4 (as a dehydrating agent): (Liquid Phase Dehydration) Primary Alcohol Secondary Alcohol 3 o Alcohols When alcohols on heating with conc. H 2 SO 4 (at different temp.) undergo dehydration; to form alkenes conc. H 2 SO 4 (95% H 2 SO 4 ) 423-453 K (150 o C-180 o C) 60% H 2 SO 4 373 K (100 o C) 20% H 2 SO 4 at 363 K (90 o C)

56. [email protected] Dr Pramod R Padole Preparation of Alkenes form Alcohols: a ) By using conc. H 2 SO 4 ( as a dehydrating agent ): ( Liquid Phase Dehydration ) When alcohols on heating with conc. H 2 SO 4 (at different temp.) undergo dehydration; to form alkenes.

57. [email protected] Dr Pramod R Padole Preparation of Alkenes form Alcohols: a ) By using conc. H 2 SO 4 ( as a dehydrating agent ): ( Liquid Phase Dehydration )

58. Preparation of Alkenes form Alcohols: From 2 o Alcohols: { Note that:- Certain alcohols on dehydration produce two isomeric alkenes.}

59. Preparation of Alkenes form Alcohols: From 3 o Alcohols: When tert -butyl alcohol is heated with 20% H 2 SO 4 at 363 K (90 o C); to form iso-butylene .

60. Preparation of Alkenes form Alcohols: From 2 o Alcohols: Q.1) Explain Saytzeff rule . (S-18, 2 Mark) Saytzeff's rule state that, in elimination reaction; hydrogen atom from β- carbon atoms is preferentially eliminated having less no. of H-atoms & to form major product is a more substituted alkene . (at the  University of Kazan , Russian chemist  Alexander Zaitsev  studied a variety of different elimination reactions ) This is because the more substituted alkene is more stable and it is formed more easily. For example: When sec-butyl alcohol is heated with 60% H 2 SO 4 at 373 K (100 o C); to form 2-butene as a major product by Saytzeff rule. Zaitsev's rule   (or  Saytzeff's rule ,  Saytzev's rule ) 

61. Mechanism of Dehydration of Alcohol: Q.1) Explain the mechanism of dehydration of ethyl alcohol (ethanol). (S-12 & S-13, 4 Mark) Q. 2) Give the preparation of ethylene form ethyl alcohol and discuss its mechanism . (S-14, 4 Mark)

62. Mechanism of Dehydration of ethyl alcohol : The mechanism of acid catalysed dehydration of alcohols is illustrated here by taking example of ethyl alcohol. Consider the reaction: The mechanism of dehydration involves following three steps. Step-1) Protonation of alcohol; to form protonated alcohol.

63. Mechanism of Dehydration of ethyl alcohol : Step-1 ) Protonation of alcohol; to form protonated alcohol. Step-2) Dissociation of protonated alcohol; to form carbocation .

64. Mechanism of Dehydration of ethyl alcohol : Step-3) Loss of proton from carbocation ; to form alkene . Ethyl carbocation -----*****-----

65. Dehydration of 1-Propanol ( n- propyl alcohol): Q.1) How will you prepare 1-Propene from 1-propanol?

66. Mechanism of Dehydration of n- propyl alcohol: Mechanism: The mechanism of dehydration involves following three steps. Step-1) Protonation of alcohol; to form protonated alcohol. Reaction:

67. Mechanism of Dehydration of n- propyl alcohol: Mechanism: Step-2) Dissociation of protonated alcohol; to form carbocation . Step-3) Loss of proton from carbocation ; to form alkene .

68. b) Dehydration of alcohols: By using Alumina, ( Al 2 O 3 ): ( Vapour Phase Dehydration)

69. Dehydration of Alcohols: b) Dehydration of alcohols: By using Alumina, (Al 2 O 3 ): ( Vapour Phase Dehydration) Primary Alcohol Secondary Alcohol 3 o Alcohols When alcohols can also be dehydrated by passing their vapours over heated alumina (Al 2 O 3 ) tube at 623-423 K; to form alkenes . Al 2 O 3 623 K (________ o C ) Al 2 O 3 523 K (_____ o C ) Al 2 O 3 at 423 K (_____ o C ) 350 250 1 50

70. Dehydrohalogenation reaction or elimination reaction or α-β elimination reaction. Dehydrohalogenation of Alkyl Halides Preparation of Alkenes from Alkyl Halides:

71. Dehydrohalogenation of Alkyl Halides: Defination : The reaction in which hydrogen atom and halogen atom are removed from adjacent carbon atoms (α , β carbon atoms); to form alkene is known as dehydrohalogenation reaction or elimination reaction or α-β elimination reaction or β elimination reaction . Alkyl halides undergo dehydrohalogenation when heated with alcoholic KOH or NaOH (alkali); to form alkenes. The decreasing order of reactivity of alkyl halides in dehydrohalogenation is, Tertiary (3 o ) > Secondary (2 o ) > Primary(1 o )

72. Dehydrohalogenation of Alkyl Halides: From 1 o Alkyl halide:- Q.1) How will you convert Ethyl bromide to ethylene. (S-11, 1 Mark) Q.2) What happens when- Ethyl bromide is heated with alcoholic KOH? (W-11 & W-12, 1 Mark) Q.3) What happen when, Ethyl bromide is heated with alcoholic KOH? (S-14, 2 Mark) Q.4) Ethyl bromide when treated with alcoholic KOH gives _____________. (W-14, ½ Mark) Q.5) Give the preparation of ethylene from ethyl bromide and discuss its mechanism. (S-16, 4 M) Q.6) Complete the following reaction: (S-17, 2 Mark)

73. Dehydrohalogenation of Alkyl Halides: From 2 o Alkyl halide: e.g. ( i ) From iso-propyl chloride: Q.1) Explain Saytzeff rule. (S-18, 2 Mark) Saytzeff's rule state that , “In elimination reaction; hydrogen atom from β- carbon atoms is preferentially eliminated from that C-atom having less no. of H-atoms & to form major product is a more substituted alkene ”. This is because the more substituted alkene is more stable and it is formed more easily.

74. [email protected] By Dr. Pramod R. Padole Mechanism of Dehydrohalogenation : E 1 Mechanism: ( Unimolecular Elimination reaction): Mechanism of Dehydrohalogenation : E 2 Mechanism: (Bimolecular Elimination reaction) There are two mechanisms of dehydrohalogenation of alkyl halides through,

75. What Do The E1 and E2 Reactions Have In Common? Here’s what each of these two reactions has in common: in both cases, we form a new C-C π bond, and break a C-H bond and a C–(leaving group) bond in both reactions, a species acts as a base to remove a proton, forming the new π bond both reactions  follow  Zaitsev’s rule  (where possible) both reactions are  favored by heat .

76. How Are The E1 and E2 Reactions Different? Now, let’s also look at how these two mechanisms are different. Let’s look at this handy dandy chart:

77. E 1 Mechanism: ( Unimolecular Elimination reaction ): Unimolecular Elimination  ( E 1 ) is a  reaction  in which the removal of an HX substituent results in the formation of a double bond . It is similar to ( i ) a  unimolecular   nucleophilic   substitution reaction  (S N 1) in various ways. One being the formation of a carbocation intermediate. (ii) Also , the only rate determining (slow) step is the dissociation of the leaving group to form a  carbocation , hence the name  unimolecular . 

78. E 1 Mechanism: ( Unimolecular Elimination reaction ): Tertiary alkyl halides undergo dehydrohalogenation in a solution of low base concentration by E 1 mechanism. For example, Q.1) Explain E 1 mechanism with suitable example. (W-16 & W-17, 4 Mark) Q.2) E 1 mechanism, _______ reactants involved in rate determining step. (a) 1 (b) 2 (c) 3 (d) 4 (S-18, ½ Mark) Q.3) Explain the E 1 mechanism. (S-19, 4 Mark)

79. E 1 Mechanism: ( Unimolecular Elimination reaction ): The mechanism of involves following two steps - Step-1) Dissociation of alkyl halide; to form carbocation and halide ion (bromide ion). This is slow step hence called as rate determining step ( only one reactant is involved ). Step-2) Loss of proton from carbocation ; to form alkene .

80. E 2 Mechanism: ( Bimolecular Elimination reaction) Q.1) Explain E 2 mechanism and state Saytzeff rule. ( S-18, 4 Mark) Q.2) Explain E 2 mechanism with example. ( W-19, 4 Mark) E 2   stands for   bimolecular elimination . The   reaction   involves a one-step   mechanism   in which carbon-hydrogen and carbon-halogen bonds break to form a double bond (C=C Pi bond). E 2   is a single step   elimination , with a single transition state. E2, bimolecular elimination, was proposed in the 1920s by British chemist Christopher Kelk Ingold . Unlike  E1  reactions, E2 reactions remove two subsituents with the addition of a strong base, resulting in an alkene .

81. E 2 Mechanism: ( Bimolecular Elimination reaction) Primary alkyl halides undergo dehydrohalogenation in a solution of high base concentration by E 2 mechanism . For example,

82. E 2 Mechanism: ( Bimolecular Elimination reaction) The mechanism involves only one step (called concerted mechanism). The removal of halide ion from α(alpha) carbon atom and abstraction of proton from β(beta) carbon atom by OH - take place simultaneously. Note:- Secondary alkyl halides undergo dehydrohalogenation by both E 1 and E2 mechanisms. A low concentration of base favours E 1 mechanism while its high concentration favours E 2 mechanism.

83. Chemical Reactions of Alkenes: Halogenation (Addition of Halogen): Hydrohalogenation ( Addition of Halogen acid, HX): Chemical Reactions: a) Chlorination b) Bromination : in presence of CCl 4 (inert / organic solvent) Symmetrical alkene Unsymmetrical alkene Geminal Dihalides (1,1-dihalides) Vicinal dihalides (1,2-dihalides) X

84. Addition of halogen atoms is known as Halogenation [email protected] By Dr Pramod R Padole i ) Reaction with Chlorine in presence of CCl 4 (inert / organic solvent) : Halogenation Of Alkanes ii) Reaction with Bromine in presence of CCl 4 (inert / organic solvent) :

85. 1) Halogenation (Addition of Halogens): Or Preparation of Vicinal dihalide (1,2-dihalide) Alkenes  undergo an addition  reaction with halogens ; the  halogen  atoms partially break the carbon-carbon double bond in the  alkene  to a single bond and add across it. ... Addition of halogens to Alkene : When alkene is treated or reacted with halogens (chlorine and bromine only) in presence of CCl 4 (inert /organic solvent); to form vicinal dihalides (or 1,2-dihalides).

86. a) Chlorination (Addition of Chlorine): e.g. i ) From ethylene: When ethylene ( ethene ) is treated or reacted with chlorine in presence of CCl 4 (inert / organic solvent) ; to form ethylene dichloride (or 1,2-dichloroethane). Q.1) How will you bring the following conversions? (S-09 & W-09, 1 Mark) i ) Ethylene dichloride from ethylene ( ethene ). Q.2) How will you bring the following conversions? (S-10, 1 Mark) i ) Propylene dichloride from propylene ( propene ). Q.3) Discuss the ( Electrophilic or Free Radical) mechanism of addition of chlorine to ethylene (or propylene). Q.4) Complete the following reaction: (S-15, 2 Mark)   Q.5) What happens when: Propane on treatment with chlorine in presence of U.V. light? (W-18, 2 Mark)

87. a) Chlorination (Addition of Chlorine): e.g. ii) From Propylene:- When propylene ( propene ) is treated or reacted with chlorine in presence of CCl 4 (Inert / organic solvent); to form propylene dichloride (or 1,2-dichloropropane). Q.1) How will you bring the following conversions? (S-09 & W-09, 1 Mark) i ) Ethylene dichloride from ethylene ( ethene ). Q.2) How will you bring the following conversions? (S-10, 1 Mark) i ) Propylene dichloride from propylene ( propene ). Q.3) Discuss the ( Electrophilic or Free Radical) mechanism of addition of chlorine to ethylene (or propylene). Q.4) Complete the following reaction: (S-15, 2 Mark)   Q.5) What happens when: Propane on treatment with chlorine in presence of U.V. light? (W-18, 2 Mark)

88. b) Bromination (Addition of Bromine)

89. https://www.footprints-science.co.uk/index.php?quiz=Reaction_of_alkenes_and_hydrogen

90. b) Bromination (Addition of Bromine) e.g. i ) From ethylene: When ethylene reacts with bromine in presence of CCl 4 (inert / organic solvent); to form ethylene dibromide (or 1,2-dibromoethane). For example, Q.1) How will you bring the following conversions? (S-10 & S-12, 1 Mark) Ethylene dibromide from ethylene ( ethene ). Q.2) What happens when ethylene is treated with - (S-04, W-13 & W-14, 2 Mark) Bromine in presence of Carbon tetrachloride (CCl 4 )? Q.3) Discuss the ( Electrophilic or Free Radical) mechanism of addition of bromine to ethylene (or propylene). Q.4) Complete the following reaction. (S-13, 2 Mark) Q.5) How will you convert: Ethylene into ethyl bromide? (S-16, 2 Mark)

91. b) Bromination (Addition of Bromine): e.g. ii) From Propylene:- When propylene ( propene ) is treated or reacted with bromine in presence of CCl 4 (Inert / organic solvent); to form propylene dibromide (or 1,2-dibromopropane). Q.1) How will you bring the following conversions? (S-10 & S-12, 1 Mark) Ethylene dibromide from ethylene ( ethene ). Q.2) What happens when ethylene is treated with - (S-04, W-13 & W-14, 2 Mark) Bromine in presence of Carbon tetrachloride (CCl 4 )? Q.3) Discuss the ( Electrophilic or Free Radical) mechanism of addition of bromine to ethylene (or propylene). Q.4) Complete the following reaction. (S-13, 2 Mark) Q.5) How will you convert: Ethylene into ethyl bromide? (S-16, 2 Mark)

92. Confirmatory Test: On adding to an alkene, the brown colour of bromine in CCl 4 disappears. Mechanism of halogenations: Halogenation of an alkene takes place by electrophilic addition mechanism as well as Free radical mechanism.

93. By Dr Pramod R Padole Mechanism of halogenations: Free Radical Addition Mechanism of Br 2 to Ethylene: Electrophilic (or Ionic) addition Mechanism of Br 2 / Cl 2 : Mechanism Of Halogenations

94. By Dr Pramod R Padole Electrophilic ( or Ionic) addition Mechanism of Br 2 / Cl 2 : It is an ionic reaction initiated by the electrophile (Br + ) released from Br 2 . The mechanism of addition of Br 2 / Cl 2 to the ethylene ( ethene ) reaction involves three steps .   Step-1) Formation of an electrophile ( Br + ) :- When Br 2 undergoes heterolytic fission (ionizes on interaction with π – electron cloud) ; to form Br + ion ( bromonium ion) as an electrophile & Br - ion(bromide ion) as a nucleophile . Q.1) Explain the mechanism of addition of bromine to ethylene. (S-12, 4 Mark)

95. By Dr Pramod R Padole Electrophilic ( or Ionic) addition Mechanism of Br 2 / Cl 2 : Step -2) Formation of Carbonium ion Or Attack of an electrophile ( Br + ) :- When Br + ion, as an electrophile , attacks on the C=C bond of the ethylene; to form carbonium ion Q.1) Explain the mechanism of addition of bromine to ethylene. (S-12, 4 Mark)

96. By Dr Pramod R Padole Electrophilic ( or Ionic) addition Mechanism of Br 2 / Cl 2 : Step-3 ) Formation of Product Or Attack of a nucleophile ( Br - ): When Br – ion, as a nucleophile , then attacks to the carbonium ion from opposite side(back side) ; to form ethylene dibromide (vicinal dibromide ). Q.1) Explain the mechanism of addition of bromine to ethylene. (S-12, 4 Mark) Above mechanism of addition of bromine to ethylene / propylene in another way as below,

97. Mechanism ( Electrophilic addition): Above mechanism of addition of bromine to ethylene / propylene in another way as below, Consider the addition of bromine to propene , it involves following steps:

98. Mechanism ( Electrophilic addition): Step -1 : Electrophilic attack forms bromonium ion and bromide ion . Step - 2 : The bromide ion attacks on bromonium ion from back side; to form 1 , 2-dibromopropane.

99. Free Radical Addition Mechanism of Br 2 to Ethylene :

100. Free Radical Addition Mechanism of Br 2 to Ethylene : In the presence of sun light (h ע ) / peroxide / heat at high temp., the addition of Br 2 to ethylene / propylene takes place; to from vicinal dihalide . Mechanism : It is a free radical addition reaction initiated by bromine free radical (Br . ). The mechanism involves three steps. Step-1) Chain Initiation: When bromine molecule undergoes homolytic fission; to form bromine free radicals (Br*).

101. Free Radical Addition Mechanism of Br 2 to Ethylene : Mechanism: Step-2) Chain Propagation:   a) When bromine free radical attacks on ethylene; to form the ethylene free radical. b ) When ethylene free radical attacks on Bromine molecule; to form ethylene dibromide & bromine free radical again. In step-2) The reactions (a) & (b) are repeated over and over again, called as propagation step.

102. Free Radical Addition Mechanism of Br 2 to Ethylene : Mechanism: Step -3) ChainTermination – The above chain reaction comes to an end (or stopped / terminated) by combination of two free radicals. (For Home work) Q.1) Discuss the Electrophilic or ionic mechanism of addition of chlorine to ethylene / propylene. -----*****----- Q.2) Explain the mechanism for photochemical addition of chlorine to ethylene / propylene. (S-19, 4 Mark) -----*****----- Q.3) Discuss the Free radical mechanism of addition of chlorine to ethylene / propylene. -----*****-----

103. ( Addition of Halogen acid, HX): Hydrohalogenation :

104. By Dr Pramod R Padole Hydrohalogenation : Unsymmetrical alkene e.g. CH 3 -CH=CH 2 Symmetrical alkene e.g. CH 2 =CH 2 H-X Addition of Halogen acid,

105. Symmetrical alkene ( e.g. CH 2 =CH 2 ): a) Reaction with HBr : When ethylene is treated or reacted with strong aq. solution of HBr ; to form ethyl bromide . b) Reaction with HCl : When ethylene is treated or reacted with strong aq. solution of HCl ; to form ethyl chloride. Q.1) What happens when ethylene is treated / reacted with strong aqueous solution of HBr . Q.2) What happens when ethylene is treated with HBr ? (W-11 & S-16, 2 Mark) Q.3) Complete the following reaction: (S-14 & W-17, 2 Mark) The decreasing order of ease of addition of hydrogen halide is, HI> HBr > HCl

106. Mechanism of HBr addition: HBr molecule is polarized because there is a electronegativity difference between hydrogen and bromine atoms. Also, electrons density around the double bond is higher. (due to higher electrons density), those places can be attracted by positive charges.

107. On Unsymmetrical alkene ( e.g. CH 3 -CH=CH 2 ): Hydrohalogenation

108. [email protected] By Dr Pramod R Padole Hydrohalogenation on Unsymmetrical alkene : Markownikoff’s Rule: or Markovnikov Rule: (Anti Peroxide Effect) Alkene Unsymmetrical Anti- Markownikoff’s Rule : (Peroxide Effect Or Kharasch effect): In presence of or in absence of Oxygen / organic peroxide

109. Markownikoff’s Rule (Anti Peroxide Effect): Markownikoff’s Rule or Markovnikov Rule: Q.1) Give / Explain the mechanism of addition of HBr to propylene in absence of peroxide. (W-04, W-05, S-07, W-07, S-10, S-11 & W-11, 3-4 Mark) Q.2) Discuss the mechanism of addition of HBr to propylene. (S-05, 3 Mark) Q.3) Explain / What is Markownikoff’s Rule? Explain with suitable example. (W-06 & S-08, 2 Mark), (S-09, 1 1/2 Mark) & (W-10 & W-12, 3 Mark) Q.4) Explain the mechanism of addition of HBr to propylene. (W-14 & S-17, 4 Mark) Q.5) Give / Explain the mechanism of addition of HBr to unsymmetrical alkene (i.e., propylene) in absence of peroxide. (W-15, 4 Mark) Q.6) Give statement of Markownikoff’s Rule. (S-18 & S-19, 1 Mark) Or Preparation of isopropyl bromide: Or Anti Peroxide Effect: According to Markownikoff’s rule, “When propylene is treated or reacted with HBr reagent in absence of oxygen or absence of any organic peroxide , the negative part of the reagent (Br - ) adds to that carbon atom of the C=C bond, which carries lesser number of hydrogen atoms ; to form isopropyl bromide or 2-bromopropane ”.

110. Mechanism: Electrophilic (or Ionic) addition of HBr to Propylene: It is an ionic reaction initiated by the electrophile (H ) released from HBr . The mechanism involves following steps. Step -1) Formation of electrophile : When HBr undergoes heterolytic fission ( Ionisation ) ; to form H + ion (as an electrophile ) and Br - ion (as a nucleophilie ) Step-2) Formation of Carbonium ion or Attack of electrophile : When electrophile , H + ion, attacks on the C=C bond of the propylene; to form the more stable secondary carbocation . Note :- Order of stability of carbonium ion is, Tertiary (3 o ) > Secondary (2 o ) > Primary(1 o )>CH 3

111. Mechanism: Electrophilic (or Ionic) addition of HBr to Propylene: It is an ionic reaction initiated by the electrophile (H ) released from HBr . The mechanism involves following steps. Step -3) Formation of Product or Attack of nucleophile : When bromide ion (Br - ion, as a nucleophilie ) attack on the more stable 2 o carbocation; to form isopropyl bromide or 2-bromopropane ”.

112. Anti- Markownikoff’s Rule: Peroxide Effect or Kharasch effect

113. Anti- Markownikoff’s Rule: (Peroxide Effect or Kharasch effect): Or Preparation of n-propyl bromide (1-bromopropane ): Or In presence of Oxygen / organic peroxide: Free Radical Addition of HBr to Propylene ( Peroxide effect/Anti Markownikoff's addition): According to Anti- Markownikoff’s rule or Peroxide effect, “When propylene is treated or reacted with HBr reagent in presence of oxygen or any organic peroxide , abnormal addition will be take place, i.e ., the negative part of the reagent (Br - ) adds to that carbon atom of the C=C bond, which carries more number of hydrogen atoms ; to form n-propyl bromide or 1-bromopropane ”. Q.1) Give / Explain the mechanism of addition of HBr to propylene (propene) in presence of peroxide, H 2 O 2 (organic peroxide). ( S-04, S-06, W-06, S-08, W-08, W-09, S-11, W-11, W-12, S-15 & W-19, 3-5 Mark) Q.2) Discuss / Explain the mechanism of addition of HBr to propylene. ( S-05, 3 Mark) Q.3) Explain / What is Anti- Markonikoff’s Rule or Peroxide effect? Explain with suitable example. ( W-05, 1 1/2 Mark) Q.4) What happens when propylene on reaction with HBr in presence of organic peroxide? ( S-11, 1 Mark) Q.5) Write short note on: Peroxide effect. ( W-16, 2 Mark) Q.6) Explain the mechanism of addition of HBr to unsymmetrical alkene (propylene) in presence of organic peroxide. ( W-18, 4 Mark) Q.7) Give statement of Anti Markownikoffs rule. ( S-19, 1 Mark)

114. Anti- Markownikoff’s Rule: Free Radical Mechanism: It is a free radical addition reaction initiated by bromine free radical (Br . ). The mechanism involves three steps. Step-1) Chain Initiation: When organic peroxide molecule undergoes homolytic fission; to form alkoxy free radicals which then attack on HBr ; to form bromine free radical.

115. Anti- Markownikoff’s Rule: Free Radical Mechanism: Step-2) Chain Propagation: a ) Attack of bromine free radical (Br . ) on propylene; to form the more stable secondary free radical. b ) When more stable secondary free radical reacts with HBr ; to form n-propyl bromide and bromine free radical. Note :- Order of stability of Free radical is, Tertiary (3 o ) > Secondary (2 o ) > Primary(1 o )> CH 3 In step: 2) , reactions (a) & (b) are repeated over and over again, called as propagation step .

116. Anti- Markownikoff’s Rule: Free Radical Mechanism: Step -3) ChainTermination : The above chain reaction comes to an end (or stopped / terminated) by combination of two free radicals . -----*****-----

117. iii) ALKYNES (Acetylenes): Aliphatic unsaturated hydrocarbon containing C≡C bond is called _______.

118. [email protected] By Dr Pramod R Padole ALKYNES (Acetylenes): Defination : Aliphatic unsaturated hydrocarbons containing one C≡C bond in their molecule are called alkynes . The first & the most important member of alkyne series is acetylene and hence they are known as acetylenes . The characteristic functional group is -C≡C- . They have general formula C n H 2n-2 . Where , n is the number of carbon atoms .   e.g. CH≡CH - Ethyne or Acetylene. Q.1) Define / What are Alkynes? Q.2) Aliphatic unsaturated hydrocarbon containing C≡C bond is called _____. (S-16, ½ Mark) Note: The combustion of acetylene provides tremendous amounts of heat, sufficient to cut metal using an oxyacetylene torch.

119. [email protected] By Dr Pramod R Padole Synthesis or Preparation of Acetylene: From Vicinal dihalide : Preparation Of Acetylene From Geminal dihalide :

120. Synthesis or Preparation of Acetylene: , 1) From Vicinal dihalide : From Ethylene dichloride: When ethylene dichloride (1,2-dichlorethane) is heated with alcoholic solution of KOH; to form vinyl chloride, which is further reacts with NaNH 2 ( Sodamide ); to form acetylene . Q.1) How will you synthesize acetylene from 1,2-dichloro ethane? Q.2) Give the preparation of acetylene from Vinyl Chloride. (S-06 & S-11, 1 Mark) Q.3) How will you synthesize acetylene from Vinyl Chloride? (W-11, 1 Mark) Q.4) Complete the following reaction: (W-13, 2 Mark)   Q.5) What is the action of alcoholic KOH on ethylene dibromide followed by sodium amide? (S-16, 2 Mark)

121. Synthesis or Preparation of Acetylene: , 1) From Vicinal dihalide : From Ethylene dibromide : When ethylene dibromide (1,2-dibromoethane) is heated with alcoholic solution of KOH; to form vinyl chloride, which is further reacts with NaNH 2 ( Sodalime ); to form acetylene . Q.1) How will you prepare: Acetylene from ethylene dibromide . (W-15 & S-18, 2 Mark) Q.2) How will you convert: Ethylene dibromide to Acetylene? (W-19, 2 Mark)

122. 2) From Geminal Dihalide : Ethylidene dichloride: 1,1-dichlorethane

123. 2) From Geminal dihalide : From Ethylidene dichloride: When ethylidene dichloride (1,1-dichlorethane) is heated with alcoholic solution of KOH; to form vinyl chloride, which is further reacts with NaNH 2 ( Sodalime ); to form acetylene . Q.1) How will you synthesize acetylene from 1,1-dichloro ethane(or 1,1-dibromo ethane)? Q.2) Give the preparation of acetylene from Vinyl Chloride. (S-06, 1 Mark) Q.3) Complete the following reaction: (W-13, 2 Mark)

124. 2) From Geminal dihalide : From Ethylidene dibromide : When ethylidene dibromide (1,1-dibromoethane) is heated with alcoholic solution of KOH; to form vinyl bromide, which is further reacts with NaNH 2 ( Sodalime ); to form acetylene . Q.1) What is the action of: alcoholic solution of KOH followed by sodium amide on ethylidene dibromide ? (W-14, 2 Mark)

125. Chemical Reaction of Acetylene: Hydrogenation (Addition of Hydrogen) or Catalytic Hydrogenation (Catalytic Reduction): Acetylene Chemical Reaction Hydrogenation (Addition of Hydrogen): Controlled Reduction using Lindlar’s catalyst (Pd-BaSO 4 in Quinoline medium) :

126. Hydrogenation (Addition of Hydrogen) or Catalytic Hydrogenation (Catalytic Reduction): When acetylene is treated with hydrogen in presence of a catalyst, Ni / Pt / Pd; to form ethene (ethylene), which is further reduced; to form ethane. Q.1) How will you prepare:-ii) Ethane from Acetylene? (W-16, 2 Mark) Q.2) What happens when : ii) Acetylene is reacted with H 2 / Pt ? (W-06, 1 Mark)

127. Hydrogenation (Addition of Hydrogen): Controlled Reduction using Lindlar’s catalyst (Pd-BaSO 4 in Quinoline medium) : When acetylene is reduced with hydrogen in presence of Palladium suspended in barium sulphate (Pd-BaSO 4 ) in Quinoline medium ( Lindlar’s catalyst); to form ethene (ethylene) as a product, in very good yield. This is called as controlled reduction. The reaction can be stopped at ethylene stage by using poisoned catalyst. Q.1) Give the preparation of ethylene from acetylene. Q.2) Complete the following reaction. (W-10, W-13, S-15 & S-17, 2 Mark)

128. iv) ALKADIENES or Dienes :

129. iv) ALKADIENES or Dienes : Alkenes containing two carbon-carbon double (C=C) bonds are called dienes or alkadienes . They have general formula C n H 2n-2 Where , n is the number of carbon atom. e.g. Q.1) Define / What are Alkadienes or Dienes ? (S-07, W-13, S-15, W-17 & W-18, 1-2 Mark)

130. By Dr Pramod R Padole Classification of Dienes or ALKADIENES: Or Types of Dienes or ALKADIENES: Conjugated Diene : Isolated or Non-Conjugate Diene : Cumulated Diene : Classification of Dienes : Dienes are mainly classified on the basis of carbon-carbon double bond.

131. Classification of Dienes or ALKADIENES: Or Types of Dienes or ALKADIENES: Classification of Dienes : Dienes are mainly classified on the basis of carbon-carbon double bond. 1 ) Conjugated Diene : The diene in which the double bonds are separated by one single bond is called as conjugated diene . Q.1) Define / Explain and give one examples of :- (W-05, S-09 & W-11, 3 Mark) i ) Alternate ( or Conjugated ) dienes & ii) Isolated dienes Q.2) How are they ( dienes ) classified? Give one example of each diene . (S-07, 2 Mark) Q.3) Define the following terms with suitable example. (W-08 & S-14, 3-4 Mark) i ) Conjugated dienes & ii) Cummulated dienes Q.4) What are alkadienes ? Give examples of isolated and cumulated dienes . (W-13, 4 Mark) Q.5) What are isolated dienes ? (W-14, 1 Mark) Q.6) Explain the term : Isolated diene with suitable examples. (W-14 & S-16, 2 Mark) Q.7) Define the terms: Cummulated dienes with suitable example. (W-16, 2 Mark) Q.8) What are Alkadienes ? Give their types with one example of each. (W-17 & W-18, 4 Mark)

132. Classification of Dienes or ALKADIENES: Or Types of Dienes or ALKADIENES: Classification of Dienes : 1 ) Conjugated Diene : The diene in which the double bonds are separated by one single bond is called as conjugated diene .

133. Classification of Dienes or ALKADIENES: Or Types of Dienes or ALKADIENES: Classification of Dienes : 2 ) Isolated or Non-Conjugated Diene : The diene in which the double bonds are separated by more than one single bond is called as isolated or non-conjugated diene .

134. Classification of Dienes or ALKADIENES: Or Types of Dienes or ALKADIENES: Classification of Dienes : 3) Cumulated Diene : The diene in which the double bonds are adjacent to each other is called as comulated diene . The important class of dienes is that of conjugated dienes . Their chemical properties are different from those of ordinary alkenes.

135. Preparation of 1,3-Butadiene:

136. Synthesis or Preparation of 1,3-Butadiene: From Cyclohexene : When vapours of cyclohexene is passing over heated Ni-Cr wire (Nickel-chrome alloy) at 873 K (600 o C); to form 1,3-Butadiene & ethylene. (By Retro Diels-Alder Reaction) Or When cyclohexene is heated at 873 K in presence of Nickel-chrome alloy (Ni-Cr); to form 1,3-Butadiene & ethylene. (By Retro Diels-Alder Reaction) Q.1) What happens when cyclohexene is heated at 873 K in presence of Nichrome alloy (Ni-Cr)? Q.2) How will you prepare: 1,3-Butadiene from cyclohexene ? (W-15, W-16 & S-19, 2 Mark) Q.3) How will you convert: Cyclohexene into 1,3-butadiene? (S-18, 2 Mark) Q.4) Cyclohexene is converted to 1,3-butadiene in presence of Ni-Cr alloy. (W-18, 2 Mark) 1,3-Butadiene is a colourless gas, bp - 4.5 o C. Reactions that produce a particular functional group are called preparations

137. Chemicals Reaction of 1,3-Butadiene: Addition of Hydrogen (Hydrogenation or Catalytic Hydrogenation): Chemical Reaction of 1,3-Butadiene: Addition of Halogen Acid (HX): Addition of Halogen (Cl 2 or Br 2 ): Halogenation a) Chlorination: b) Bromination : in presence of CCl 4 (inert / organic solvent)

138. Prepared by Dr Pramod R. Padole Addition of Hydrogen (Hydrogenation or Catalytic Hydrogenation): Addition of Hydrogen: When 1,3-butadiene is reacted or treated or heated with hydrogen in presence of Ni / Pt / Pd as catalyst, under pressure; to form a mixture of 1-butene (1,2-addition) & 2-butene (1,4-addition), resp.   Q.1) How will you convert : i ) 1,3-Butadiene to 2-butene and 1-Butene. (W-04, 1 Mark) Q.2) What happens when: ii) 1,3-Butadiene is treated with hydrogen? (S-05, 1 Mark) Q.3) What happens when 1,3-butandiene is treated with H 2 /Pt? (W-10, 1 Mark) Q.4) What is the action of H 2 in presence of nickel or platinum on 1,3 butadiene? (W-15, 2 Mark) Q.5) How will you convert: 1,3-Butadiene to 2-butene? (S-16, 2 Mark)

139. Prepared by Dr Pramod R Padole 2) Addition of Halogen (Cl 2 or Br 2 ): Halogenation ( Chlorination or Bromination ): a) Chlorination: When 1,3-butadiene reacts with Cl 2 (chlorine) in presence of CCl 4 (as a organic / inert solvent); to form a mixture of 3,4-dichloro-1-butene (1,2-addition) & 1,4- dichloro -2-butene (1,4-addition), resp. Q.1) What happens when: 1,3-Butadiene is treated with Cl 2 in presence of CCl 4 ? (S-10 & S-17, 1 Mark) Q.2) What is the action of Cl 2 in CCl 4 on 1,3-Butadiene? (W-15, 2 Mark) Q.3) What is the action of Cl 2 on 1,3-Butadiene in presence of CCl 4 ? (S-16, 2 Mark)

140. Prepared by Dr Pramod R Padole 2) Addition of Halogen (Cl 2 or Br 2 ): Halogenation (Chlorination or Bromination ): b) Bromination : When 1,3-butadiene reacts with Br 2 (bromine) in presence of CCl 4 (as a organic / inert solvent); to form a mixture of 3,4-dibromo-1-butene (1,2-addition) & 1,4- dibromo-2-butene (1,4-addition), resp. Q.1) What happens when: 1,3-Butadiene is treated with Br 2 in presence of CCl 4 ? (W-04, W-05, S-15 & W-18, 2 Mark)

141. Prepared by Dr Pramod R. Padole 3) Addition of Halogen Acid (HX): Addition of Halogen Acid (HX): When 1,3-butadiene reacts with halogen acid (HX = HCl or HBr ); to form a mixture of 3-halo-1-butene at low temp . (1,2-addition) & 1-halo-2-butene at high temp. (1,4-addition), resp.   Q.1) What happens when: 1,3-Butadiene is treated with HBr ? (S-06 & S-14, 2 Mark) Q.2) What is the action of: HBr on 1,3-butadiene? (W-14, 2 Mark)

142. www.themegallery.com Thank You ! Unit – III by Dr Pramod R. Padole Stay Home. Take Care