Aliphatic hydrocarbon alkanes

Chemistry B.Sc.-I Sem-Ist
Sant Gadge Baba Amravati University, Amravati,
Late Ku. Durga K. Banmeru Science College, Lonar

1. 1 Late Ku. Durga K. Banmeru Science College , LONAR DIST. BULDANA (Maharashtra), India . Section –II Unit-3 C) Aliphatic Hydrocarbon “ALKANES” B. Sc. Ist year Sem-Ist Subject :- Chemistry

2. 2 Dr. Suryakant B. Borul (M.Sc., M.Phil. , Ph.D.) Head Of Department Department of Chemistry Late Ku. Durga K. Banmeru Science College, Lonar Teacher Profile

3. C) Aliphatic Hydrocarbons Aliphatic hydrocabons are the main class of organic compounds. Which involved basic or parent compounds of organic. I) Alkanes II) Alkenes III) Alkynes IV) Alkadienes

4. Q.- What are alkanes ? Ans - “The aliphatic saturated hydrocarbon containing carbon-carbon single bond they have general formula CnH2n+2 are called as alkanes .” “ALKANES” Some members of alkanes as follows- CH4 -Methane CH3-CH3 -Ethane CH3-CH2-CH3 -Propane C4H10 n-Butane

5. Methods of Formation of Alkanes Wurtz Synthesis- Q. Write a note on Wurtz synthesis. Ans - Alkyl halide reacts with sodium metal in presence of dry ether then it gives higher alkanes , this reaction is called as Wurtz reaction.

6. ii) When ethyl bromide reacts with sodium metal in presence of dry ether then it gives n-butane. Examples- i ) When methyl bromide reacts with sodium metal in presence of dry ether then it gives ethane.

7. 2. Corey House Alkane Synthesis- Q. Write a note on Corey House Alkane Synthesis. Ans - When alkyl halide reacts with lithium metal in presence of ether then it gives alkyl lithium, which on treated with CuI (copper iodide) to from lithium dialkyl cuprate , then alkyl halide react with dialkyl cuprate to from alkane .

8. Q. How will you prepare propane from ethyl bromide? Ans - When ethyl bromide reacts with lithium metal in presence of ether then it gives ethyl lithium, which on treated with CuI (copper iodide) to from lithium diethyl cuprate , then methyl bromide react with lithium diethyl cuprate to from Propane.

9. Chemical Reactions of Alkanes 1. Halogenation - “The process in which replacement of hydrogen atom from alkane by halogen is called as halogenation .” Chlorination- “When alkanes react with chlorine in presence of UV light or mild sunlight or at high temp (300-400oC) it gives mixture of haloalkane i.e. chloroalkane .” Ex.- Methane react with Cl2 in UV light then it gives methyl chloride, methylene chloride, trchloromethane carbon tetrachloride.

10. If excess of methane is used then it gives methyl chloride is major product.

11. Q. Explain the mechanism of chlorination of methane. Ans - When methane reacts with chlorine in presence of ultra violet light then it gives methyl chloride. Step-1: Chain Initiation- Chlorine molecule undergoes homolytic bond fission to give chlorine radicals Free radical mechanism involved following steps-

12. Step-2: Chain Propagation- Chlorine free radical attacks on methane to form HCl and methyl free radical. Step-3: Chain Termination- Chain propagation steps when any two free radical combine.

13. Q. What is action of Chlorine in presence of UV light on 1) Ethane 2) Propane. Ans - 1) Ethane with chlorine- When ethane reacts with chlorine in presence of ultra violet light then it gives ethyl chloride. 2) Propane with chlorine- When propane reacts with chlorine in presence of ultra violet light then it gives ethyl chloride.

14. 2. Aromatization- “The process in which formation of aromatic ring from alkenes is called as Aromatization of Alkane .” Q. How will you convert 1) n-Hexane to Benzene 2) n- Heptane to Toluene. Ans - 1) n-Hexane to Benzene – When n-hexane heated with Cr 2 O 3 -Al 2 O 3 at 600oC and10 atmospheric pressure then it undergoes aromatization to form benzene.

15. 2 ) n- Heptane to Toluene – When n- heptane heated with Pt-Al 2 O 3 at 600oC and10 atmospheric pressure then it undergoes aromatization to form toluene.

16. Q.- What are alkenes? Ans - “The aliphatic unsaturated hydrocarbon containing carbon-carbon double bond they have general formula CnH2n are called as alkenes or olefins.” “ALKENES” Some members of alkenes as follows- C 2 H 4 - Ethene / Ethylene C3H6 - Propene / Propylene C4H10- Butene / Butylene CH 2 =CH 2 CH 3 -CH=CH 2 CH 3 -CH=CH-CH 3 They have general formula CnH2n, where n=number of carbon atom They have Functional group -C=C- Ethylene linkage It is also called as Olefins.

17. Methods of Formation of Alkenes I. Dehydrohalogenation of Alkyl halides Or Elimination Reaction Q. Write a note on Elimination Reaction . Ans - When alkyl halide reacts with alcoholic KOH or NaOH to form alkene . Two types of Elimination reactions E1 reaction- E2 reaction-

18. E1 stand for; 1- stand for Unimolecular , E- stand for Elimination reaction Mechanism of Elimination reactions E 1 Reaction Mechanism- Q. Explain E1 Reaction Mechanism with example. Ans - When 2-Bromo 2-methyl Propane (t- butyl bromide) reacts with alcoholic NaOH to form 2-methyl propane. Rate of reaction (R) = K

19. Mechanism:- Step-I – Formation of Carbocation - In this step breaking C-Br bond take place and this is slow step hence it is rate determine step (RDS). Step-II – Elimination of Proton- In this step breaking C-H bond and formation of carbon carbon double bond take place. This is fast step. Rate of this reaction depends on only one reactant molecule hence it is unimolecular i.e. E1

20. E2 stand for; 2- stand for bimolecular, E- stand for Elimination reaction 2. E 2 Reaction Mechanism- Q. Explain E2 Reaction Mechanism with example. Ans - When 1-Bromo Propane (n- Propyl bromide) reacts with sodium ethoxide in presence of alcoholic NaOH it gives Propane and ethanol.

21. Rate of Reaction- Rate of reaction depends on two molecules hence it is biomolecular reaction. Mechanism:- Rate This is one step mechanism hence breaking of C-Br, C-H bond and making of carbon-carbon double bond take place simultaneously.

22. Saytzeff rule - If dehydrohalogenation of alkyl halides gives more than one alkenes, then the most highly substituted alkene is major product.

23. II. Dehydration of Alcohols- When alcohol treated with strong acid like H 2 SO 4 or p-toluene Sulphonic acid then it undergoes dehydration to form alkenes. Mechanism- Step-I- Protonation of Alcohol- Step-II- Loss of H+ (proton) & H 2 O to form Alkene -

24. Alkenes- “The aliphatic unsaturated hydrocarbon containing carbon-carbon double bond they have general formula CnH2n are called as alkenes or olefins.” “Reactions of Alkenes” C4H8- Butene / Butylene CH3-CH=CH-CH3, or CH3-CH2-CH=CH2 C5H10- Pentene/ Pentylene CH3-CH2-CH2-CH=CH2 or CH3-CH2-CH=CH-CH3 From this example it is clear that there are Two Types of Alkenes Symmetrical Alkenes- Unsymmetrical Alkenes-

25. Symmetrical Alkenes- “The alkenes which have same or equal number of atoms on the both sides of double bonds such alkenes are called as symmetrical alkenes” Ex- C4H8- Butene - CH3-CH=CH-CH3 2-Butene C6H12- Butene - CH3-CH2-CH=CH-CH2-CH3 3-Hexene Unsymmetrical Alkenes- “The alkenes which have different or un-equal number of atoms on the both sides of double bonds such alkenes are called as unsymmetrical alkenes” Ex- C4H8- Butene CH3--CH2-CH=CH2 1-Butene C5H10- Pentene CH3-CH2-CH2-CH=CH2 or 1-Pentene CH3-CH2-CH=CH-CH3 2 - Pentene

26. Reactions of Alkenes 1. Hydrohalogenation or Addition of HX (Halo acid)- Addition of HX in Symmetrical alkenes- The symmetrical alkenes react with HX then it gives alkyl halides” Ex- i ) When Ethene react with HBr then it gives Ethyl bromide. ii) When 2-butene react with HBr then it gives 2-bromo butane.

27. B) Addition of HX in Unsymmetrical alkenes- The unsymmetrical alkenes react with HX then it gives mixture of alkyl halides” Ex- i ) Propene react with HBr then it gives 1-bromo propene and 2-bromo propene . Here mixture of 1-bromo propene and 2-bromo propene obtained but which is major product. According to Markownikoff addition rule 2- bromo propene is the major product.

28. Markownikoff’s Addition Rule Q.- State and Explain the Markownikoff’s Addition Rule Ans - Statement- “When an addition of unsymmetrical reagent (HX) in an unsymmetrical alkenes then the negative part of reagent (X-) get attached to the double bonded carbon atom which carries less number of hydrogen atom”. Ex- i ) When Propene react with HBr then it gives 2-bromo propene . Here Br- ion get attached to middle double bonded carbon atom of propene because according to M.K. rule which carries less number of hydrogen atom.

29. Example ii) When 1-butene react with HBr then it gives 2-bromo butene .

30. Q.-Explain the Mechanism of addition of HBr to Propylene. Ans - Propene react with HBr then it gives 2-bromo propene . Mechanism- Mechanism of addition of HBr in Propylene in three steps are as follows- Step-I:- Formation of Electrophile - In this HBr undergoes heterolytic fission.

31. Step-II:- Formation of Carbocation - Attack of Proton on propylene to from primary (less stable) carbocation and secondary (more stable) carbocation . Step-III:- Formation of Product- Attack of Br- ion on more stable carbocation to from 2- bromo Propylene ( iso propyl bromide).

32. Peroxide Effect or Anti- Markownikoff’s Addition Rule Q.- Explain the Peroxide effect or Anti- Markownikoff’s Addition Rule Ans - “When an addition of unsymmetrical reagent (HX) in an unsymmetrical alkenes in presence of peroxide then the negative part of reagent (X-) get attached to the double bonded carbon atom which carries more number of hydrogen atom” is called as Peroxide effect or Anti Markownikoff’s addition rule. For Example – When Propylene or Propene react with HBr in presence of peroxide (R-O-O-R) then it gives n- propyl brmode or 1-bromo propene .

33. Q.-Explain the Mechanism of addition of HBr to Propylene in presence of peroxide. Ans - When Propylene or Propene react with HBr in presence of peroxide (R-O-O-R) then it gives n- propyl brmode or 1-bromo propene . Mechanism- Mechanism of addition of HBr to Propylene in presence of peroxide occur in three steps, these are as follows- Step-I:- Chain Initiation- Step-II:- Chain Propagation- Step-III:- Chain Termination-

34. Step-I:- Chain Initiation- In this step peroxide undergoes homolysis to form free radicals of alkoxy group, which attack on HBr to from alcohol and bromine free radical.

35. Step-II:- Chain Propagation- In this step bromine free radical attack on propylene to form iso-propyl bromide primary (less stable) and n- propyl bromide secondary free radical (more stable). Then n- propyl bromide secondary free radical attack on HBr to form n- propyl bromide and bromine free radical.

36. Step-III:- Chain Termination- In this step bromine free radicals attack one another to form stable bromine molecule.

37. 2. Addition of Halogen ( Halogenation ) or Alkenes- When alkenes reacts with halogens (chlorine & Bromine) then it gives vicinal dihalidess . Q.- What is action of i ) Br 2 in CCl 4 ii) Cl 2 in CCl 4 on ethylene. Ans - i ) Action of Br2 in CCl4 on ethylene / ethene - ii) Action of Cl2 in CCl4 on ethylene / ethene -

38. Electrophilic Addition Mechanism Mechanism of Halogenation of Alkenes Q- Explain electrophilic addition mechanism of bromine to propylene. Mechanism- Step- I- Electrophilic attack forms bromonium ion and bromide ion. Ans -

39. Step- II- Attack of Bromide on bromonium ion-

40. Free Radical Addition Mechanism Mechanism of Halogenation of Alkenes Q- Explain Free radical addition mechanism of bromine to propylene. Ans - Stewart in 1935 proposed this mechanism, When ethene react with chlorine in presence of UV light then it undergoes homolysis to form 1, 2 dichloroethane . Step-I- Chain Initiation- In this step chlorine undergoes homolysis in presence of UV light then it gives chlorine free radicals. Mechanism-

41. Step-II- Chain Propagation- In this step chlorine free radical attack on ethylene to form ethylene chloride free radical first then it react with chlorine molecule to form ethylene dichloride. Step-II- Chain Termination- In this step chlorine free radical combine with chloride free radical to form chlorine molecule.

42. Peroxide Effect or Anti- Markownikoff’s Addition Rule Q.- Explain the Peroxide effect or Anti- Markownikoff’s Addition Rule Ans - “When an addition of unsymmetrical reagent (HX) in an unsymmetrical alkenes in presence of peroxide then the negative part of reagent (X-) get attached to the double bonded carbon atom which carries more number of hydrogen atom” is called as Peroxide effect or Anti Markownikoff’s addition rule. For Example – When Propylene or Propene react with HBr in presence of peroxide (R-O-O-R) then it gives n- propyl brmode or 1-bromo propene .

43. Q.-Explain the Mechanism of addition of HBr to Propylene in presence of peroxide. Ans - When Propylene or Propene react with HBr in presence of peroxide (R-O-O-R) then it gives n- propyl brmode or 1-bromo propene . Mechanism- Mechanism of addition of HBr to Propylene in presence of peroxide occur in three steps, these are as follows- Step-I:- Chain Initiation- Step-II:- Chain Propagation- Step-III:- Chain Termination-

44. Step-I:- Chain Initiation- In this step peroxide undergoes homolysis to form free radicals of alkoxy group, which attack on HBr to from alcohol and bromine free radical.

45. Step-II:- Chain Propagation- In this step bromine free radical attack on propylene to form iso-propyl bromide primary (less stable) and n- propyl bromide secondary free radical (more stable). Then n- propyl bromide secondary free radical attack on HBr to form n- propyl bromide and bromine free radical.

46. Step-III:- Chain Termination- In this step bromine free radicals attack one another to form stable bromine molecule.

47. 2. Addition of Halogen ( Halogenation ) or Alkenes- When alkenes reacts with halogens (chlorine & Bromine) then it gives vicinal dihalidess . Q.- What is action of i ) Br 2 in CCl 4 ii) Cl 2 in CCl 4 on ethylene. Ans - i ) Action of Br2 in CCl4 on ethylene / ethene - ii) Action of Cl2 in CCl4 on ethylene / ethene -

48. Electrophilic Addition Mechanism Mechanism of Halogenation of Alkenes Q- Explain electrophilic addition mechanism of bromine to propylene. Mechanism- Step- I- Electrophilic attack forms bromonium ion and bromide ion. Ans -

49. Step- II- Attack of Bromide on bromonium ion-

50. Free Radical Addition Mechanism Mechanism of Halogenation of Alkenes Q- Explain Free radical addition mechanism of bromine to propylene. Ans - Stewart in 1935 proposed this mechanism, When ethene react with chlorine in presence of UV light then it undergoes homolysis to form 1, 2 dichloroethane . Step-I- Chain Initiation- In this step chlorine undergoes homolysis in presence of UV light then it gives chlorine free radicals. Mechanism-

51. Step-II- Chain Propagation- In this step chlorine free radical attack on ethylene to form ethylene chloride free radical first then it react with chlorine molecule to form ethylene dichloride. Step-II- Chain Termination- In this step chlorine free radical combine with chloride free radical to form chlorine molecule.

52. 2. Addition of Halogen ( Halogenation ) or Alkenes- When alkenes reacts with halogens (chlorine & Bromine) then it gives vicinal dihalidess . Q.- What is action of i ) Br 2 in CCl 4 ii) Cl 2 in CCl 4 on ethylene. Ans - i ) Action of Br2 in CCl4 on ethylene / ethene - ii) Action of Cl2 in CCl4 on ethylene / ethene -

53. Electrophilic Addition Mechanism Mechanism of Halogenation of Alkenes Q- Explain electrophilic addition mechanism of bromine to propylene. Mechanism- Step- I- Electrophilic attack forms bromonium ion and bromide ion. Ans -

54. Step- II- Attack of Bromide on bromonium ion-

55. Free Radical Addition Mechanism Mechanism of Halogenation of Alkenes Q- Explain Free radical addition mechanism of bromine to propylene. Ans - Stewart in 1935 proposed this mechanism, When ethene react with chlorine in presence of UV light then it undergoes homolysis to form 1, 2 dichloroethane . Step-I- Chain Initiation- In this step chlorine undergoes homolysis in presence of UV light then it gives chlorine free radicals. Mechanism-

56. Step-II- Chain Propagation- In this step chlorine free radical attack on ethylene to form ethylene chloride free radical first then it react with chlorine molecule to form ethylene dichloride. Step-II- Chain Termination- In this step chlorine free radical combine with chloride free radical to form chlorine molecule.

57. Q.- What are alkynes? Ans - “The aliphatic unsaturated hydrocarbon containing carbon-carbon triple bond they have general formula CnH2n-2 are called as alkynes or acetylenes.” “ALKYNES” Some members of alkynes as follows- They have general formula CnH2n-2, where n=number of carbon atom They have Functional group acetylene linkage It is also called as Acetylenes.

58. Methods of Formation of Alkynes Dehydrohalogenation of Vicinal dihalides - When vicinal dihalides reacts with alcoholic KOH or NaOH then it gives vinyl halide which on react with sodium amide it gives acetylene. General reaction-

59. Q. How will you convert ethylene dibromide to Acetylene. Ans - When ethylene dibromide first reacts with alcoholic KOH or NaOH then it gives vinyl bromide, which on react with sodium amide it gives acetylene.

60. II. Dehydrohalogenation of Geminal dihalides - When geminal dihalides reacts with alcoholic KOH or NaOH then it gives vinyl halide which on react with sodium amide it gives acetylene. General reaction-

61. Q. How will you convert ethylidene dibromide to Acetylene. Ans - When ethylidene dibromide first reacts with alcoholic KOH or NaOH then it gives vinyl bromide, which on react with sodium amide it gives acetylene.

62. Reactions of Alkynes Hydrogenation of Alkynes- A. Catalytic hydrogenation- When ethyne reacts with hydrogen in presence of catalyst as Ni or Pt or Pd then it first gives ethylene and finally ethane. B. Controlled hydrogenation- When ethyne reacts with hydrogen in Pd-BaSO4 in quinoline then it gives ethylene. Palladium in Barium Sulphate in quinoline medium (Pd-BaSO4 in quinoline ) is called as “ Lindlar’s Catalyst”

63. Q.- What are alkadienes ? Ans - “The aliphatic unsaturated hydrocarbon containing carbon-carbon two double bond they have general formula CnH2n-2 are called as alkadienes .” “ALKADIENES” Some members of alkynes as follows- They have general formula CnH2n-2, where n=number of carbon atom They have Functional group - diene -

64. Q.- How alkadienes are classified? Ans - On the basis of position of double bond alkadienes are classified into three classes. Conjugated dienes - Isolated dienes - Cumulated dienes - Conjugated dienes - The alkadienes in which two double bonds are separated by one single bond are called as conjugated dienes . ex.-

65. 2. Isolated dienes - The alkadienes in which two double bonds are separated by more than one single bond are called as isolated dienes . ex.- 3. Cumulated dienes - The alkadienes in which two double bonds are adjacent to each other are called as cumulated dienes . ex.-

66. Synthesis of 1, 3 Butadiene A) From Cyclohexene - Q. How will you prepared 1, 3 butadiene from cyclohexene ? Ans - When cyclohexene heated with 873k in presence of Ni-Cr alloy then it gives 1,3 butadiene.

67. Reactions of 1, 3 Butadiene 1) Addition of Hydrogen- Q. What is the action of H 2 in presence of Ni/Pt on 1, 3 butadiene. Ans - When 1, 3 butadiene react with hydrogen gas in presence of Ni or Pt, then it undergoes 1, 2 addition & 1, 4 addition then it gives 1-butene & 2-butene.

68. 2) Addition of Halogen- Q. What is the action of chlorine in presence of CCl4 on 1, 3 butadiene. Ans - When 1, 3 butadiene react with halogen gas in presence of CCl4 then it undergoes 1, 2 addition & 1, 4 addition then it gives 3, 4 dichloro-1-butene & 1, 4 dichloro-2-butene .

69. 3) Addition of Halogen acid (HX)- Q. What is the action of HBr on 1, 3 butadiene. Ans - When 1, 3 butadiene react with halogen acid then it undergoes 1, 2 addition & 1, 4 addition then it gives 3-dibromo-1-butene & 1-bromo-2-butene .

70. Thank you