alkenes and alkynes


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alkenes and alkynes

  1. 1. 1) Occurrence of carbon :- i) Carbon is found in the atmosphere, inside the earth’s crust and in all living organisms. ii) Carbon is present in fuels like wood, coal, charcoal, coke, petroleum, natural gas, biogas, marsh gas etc.iii) Carbon is present in compounds like carbonates, hydrogen carbonates etc.iv) Carbon is found in the free state as diamond, graphite, fullerenes etc.
  2. 2. 2) Bonding in carbon – Covalent bond :- The atomic number of carbon is 6, its electronic arrangement is 2,4, ithas 4 valence electrons. It can attain stability by gaining 4 electrons,losing 4 electrons or sharing 4 electrons with other atoms. It does not gain 4 electrons because it is difficult for the 6 protons tohold 10 electrons. It does not lose 4 electrons because it needs a large amount ofenergy to lose 4 electrons. So it shares 4 electrons with other atoms to attain stability resulting inthe formation of covalent bonds. Since carbon atom needs 4 electrons to attain stability, its valency is 4and it is tetravalent. X I _ _ X C X C X I
  3. 3. 3) Formation of covalent bonds :- Covalent bond is chemical bond formed by the sharing of electronsbetween atoms. The sharing of one pair of electrons results in the formation of singlecovalent bond, sharing of two pairs of electrons results in the formationof double covalent bond and sharing of three pairs of electrons resultsin the formation of triple covalent bond.Eg :- Formation of single covalent bond in Hydrogen molecule - H2The atomic number of hydrogen is 1, its electronic arrangement is 1, ithas 1 valence electron. It needs 1 electron more to attain stability. Sotwo hydrogen atoms share 1 pair of electrons resulting in the formationof a single covalent bond in hydrogen molecule H2. Hx + x H H XX H H–H H2
  4. 4. Formation of double covalent bond in oxygen molecule - O2 The atomic number of oxygen is 8, its EC is 2,6, it has 6 VE, it needs 2electrons more to attain stability. So two oxygen atoms share twopairs of electrons resulting in the formation of a double covalent bondin oxygen molecule O2 X X XX XX XX X X XX X X XX XX XX XX XX O + O O O O=O O2Formation of triple covalent bond in Nitrogen molecule - N2 The atomic number of nitrogen is 7, its EC is 2,5, it has 5 VE, it needs3 electrons more to attain stability. So two nitrogen atoms share threepairs of electrons resulting in the formation of a triple covalent bond innitrogen molecule N2 X X X X X X XX X X XX X X X XX X X X N + N N N NΞN N2
  5. 5. 4) Electron dot structures :- Methane molecule – CH4 Ethane molecule – C2H6 H X H X H X X X X X X X X X X H C H H C C H H H H H H H I I I
  6. 6. 5) Formation of a very large number of carbon compounds :- Carbon forms a very large number of compounds. The number ofcarbon compounds is more than three million. It is more than thenumber of compounds formed by all other elements. This is because :- i) Carbon atom can form bonds with other carbon atoms to form long chains, branched chains and closed rings. This property is called catenation.ii) Since the valency of carbon is 4, it can form bonds with other carbon atoms or with atoms of other elements like hydrogen, oxygen, nitrogen, halogens etc. I _ _ C I I I I I I I I I I C _ _ _ _ C–C–C–C–C–C C–C–C–C C C I I I I I I I I I I C C _ _ C C I Long chain Branched chain Closed ring
  7. 7. 6) Hydrocarbons, Saturated and Unsaturated hydrocarbons :- i) Hydrocarbons :- are compounds containing carbon and hydrogen atoms. ii) Saturated hydrocarbons :- are hydrocarbons having all single covalent bonds between the carbon atoms. Eg : Alkanes :- have all single covalent bonds between the carbon atoms and their names end with – ane. H I Methane – CH4 H–C–H I H H H I I Ethane – C2H6 H–C–C–H I I H H
  8. 8. iii) Unsaturated hydrocarbons :- are hydrocarbons having a double or triple covalent bond between two carbon atoms. Eg : Alkenes and Alkynes. Alkenes :- have a double covalent bond between two carbon atoms. and their names end with – ene. H H H H I I I I Ethene - C2H4 C=C Propene – C3H6 H–C=C– C–H I I I I H H H H Alkynes :- have a triple covalent bond between two carbon atoms and their names end with – yne. Ethyne – Ethyne – C2H2 H–CΞC–H H I Propyne - C3H4 H–C ΞC–C–H I H
  9. 9. 7) Isomerism :- Carbon compounds having the same molecular formula but differentstructural formulae are called isomers. This property is called isomerism. Eg:- Butane – C4H10 has 2 isomers. They are Normal butane and Iso butane. H H H H H H H I I I I I I I H–C–C–C–C–H H–C–C–C–H Iso butane I I I I I I H H H H H H H– C –H Normal butane I H Pentane – C5H12 has 3 isomers. They are Normal pentane, Iso pentane and Neo pentane. Neo pentane Iso pentane H H I I H– C–H Normal pentane H–C–H H H I I H H H H H H H H H–C– C– C–H I I I I I I I I I I H–C–C–C–C–C–H H–C–C–C–C–H H H I I I I I I I I I H–C–H H H H H H H H H H I H
  10. 10. 8) Functional groups :- An atom or a group of atoms which decides the properties of acarbon compound is called a functional group. i) Halide ( Halo group) :- - Cl, - Br, etc. ( Names end with – ane ) Eg :- CH3Cl – Chloro methane, C2H5Br – Bromo ethaneii) Alcohol :- - OH ( Names end with – ol ) Eg :- CH3OH – Methanol, C2H5OH – Ethanol Hiii) Aldehyde :- - CHO -C ( Names end with – al ) O Eg :- HCHO – Methanal, CH3CHO – Ethanal O IIiv) Carboxylic acid :- - COOH - C - OH (Names end with – oicacid ) Eg :- HCOOH – Methanoic acid, CH3COOH – Ethanoic acid v) Ketone :- - CO - C - (Names end with – one ) II O
  11. 11. 9) Homologus series :- Homologus series is a group of carbon compounds having similarstructures, similar chemical properties and whose successive membersdiffer by a – CH2 group. Eg :- Alkanes, Alkenes, Alkynes etc. Alkanes :- have general molecular formula CnH2n+2 . Their names endwith – ane and the members are as follows :- Methane - CH4 Ethane - C2H6 Propane - C3H8 Butane - C4H10 Pentane - C5H12 H IMethane :– CH4 H – C – H I H H H H H H I I I I IEthane :– C2H6 H–C–C–H Propane – C3H8 H–C–C–C–H I I I I I H H H H H
  12. 12. Alkenes :- Alkenes have general molecular formula CnH2n . Their names endwith – ene and the members are as follows :- Ethene - C2H4 Propene - C3H6 Butene - C4H8 Pentene - C5H10 H H I IEthene :- C2H4 C=C I I H H H H H H H H H I I I I I I IPropene :- C3H6 H–C=C–C–H Butene :- C4H8 H–C=C–C–C–H I I I H H H
  13. 13. Alkynes :- Alkynes have general molecular formula CnH 2n – 2 .Their names endwith – yne and the members are as follows :- Ethyne - C2H2 Propyne - C3H4 Butyne - C4H 6 Ethyne :- C2H2 H–C C–H H I Propyne :- C3H4 H–C C–C–H I H H H I I Butyne :- C H H–C C–C–C–H
  14. 14. Saturated and UnsaturatedCompounds Saturated compounds (alkanes) have the maximum number of hydrogen atoms attached to each carbon atom Unsaturated compounds have fewer hydrogen atoms attached to the carbon chain than alkanes Unsaturated compounds contain double or triple bonds
  15. 15. 3 Classes of UnsaturatedHydrocarbons1st CLASSAlkenes – contains one or more C-C doublebonds H2C=CH2 ethene (ethylene)2nd CLASSAlkynes – contains one or more C-C triplebonds HC≡CH ethyne (acetylene)
  16. 16. Unsaturated Hydrocarbons3rd CLASS Arenes- aromatic hydrocarbons H Benzene CH-C C-HH-C C-H C H(not chemically reactive under any of theconditions described)Arenes are found in proteins, nucleic acids, andpharmaceuticals like aspirin
  17. 17. AlkenesCarbon-carbon double bondsNames end in –en-e H2C=CH2 ethene (ethylene) H2C=CH-CH3 propene (propylene) cyclohexene
  18. 18. Shapes of AlkenesVSEPR predicts 120o for bond angles inethene and propene H2C=CH2 H2C=CH-CH3 ∠ 121.7° ∠ 124.7°The actual ∠ for these molecules are closeto the predicted; however, in other alkenesthe predicted angles will have a largerdeviation from that predicted in the VESPRmodel b/c there is limited rotation around adouble bond
  19. 19. AlkynesCarbon-carbon triple bondsNames end in -yne HC≡CH ethyne(acetylene) HC≡C-CH3 propyne
  20. 20. Shapes of AlkynesVSEPR predicts 180o for bond angles inethyne H C≡C H ethyne ∠ 180°
  21. 21. Naming AlkenesThe double bond takes precedence over substituents in numbering the parent chain.2. Use the infix “en” for all alkenes and cycloalkenes3. Use the suffix “e” for all alkenes and cycloalkenesChange the infix “an” to the corresponding alkane to “en” • butane butene • propane propene • octane octene
  22. 22. Naming Alkenes1. For open chain alkenes, identify theparent chain as the longest sequence ofcarbons that includes the double bond. CH2 CH3CH2CCH2CH2CH2CH3There is a longer chain of 7C but it does not include the double bond.
  23. 23. Naming Alkenes4. For an open chain alkenes, number the parent chainfor whichever end gives the lower number to the firstcarbon of double bonds.These rules give precedence to the location of the doublebond over the location of the first substituent on theparent chain. CH3 CH3CHCH2CH=CH2 double bond is at position 1 4-methyl-1-pentene
  24. 24. Naming Alkenes5. For cycloalkenes always give position 1 to oneof the two carbons at the double bond CH3 3-methylcyclohexene
  25. 25. Naming Alkenes6. Place the # that locates the 1st carbon of the doublebond as a prefix, and separate this number from thename by a hyphen1 2 3 4CH2=CHCH2CH3 1-buteneCH3CH=CHCH3 2-butene*Remember to separate # from numbers by commas,but use hyphens to connect a number to a word
  26. 26. Naming Alkenes7. When a compound has two double bonds, it isnamed as a diene with 2 numbers in the name tospecify the locations of the double bonds. 6 CH3 5 1 CH2=CCH=CH2 4 22-methyl-1,3-butadiene 3 1,4 - cyclohexadieneThis pattern can be easily extended to trienes,tetraenes, etc.
  27. 27. Learning Check HA2 Write the IUPAC name for each of the following unsaturated compounds: CH3 CH3A. CH3C=CHCH3 B.
  28. 28. Solutions HA2 Write the IUPAC name for each of the following unsaturated compounds: CH3 CH3A. CH3C=CHCH3 B.
  29. 29. Isomers of Alkenes3rd Geometric isomerism•No free rotation at the double bond in a ring•Have identical constitution including the location of thedouble bond but differ in geometry•Differ only in the direction taken by their end of chainmethyl group•Common at the molecular level , particularly in ediblefats, oils and in related compounds that make up most ofa cell membrane
  30. 30. Alkenes• Alkene Nomenclature – Cis isomer: • two groups (on adjacent carbons) on the same side of the C = C double bond – Trans isomer: • two groups (on adjacent carbons) on opposite sides of the C = C double bond
  31. 31. Isomers of AlkenesAlkenes can exist as isomers in 3 waysConstitutional isomers:1st different carbon skeletons CH3CH2=CHCH2CH3 CH2=CCH3 1-butene 2-methylpropene2nd H atoms attached differently to the skeleton CH3 CH3CH=CHCH3 CH2=CCH3 2-butene 2-methylpropene
  32. 32. Alkenes• Alkene Nomenclature• Different geometric isomers are possible for many alkenes. – Compounds that have the same molecular formula and the same groups bonded to each other, but different spatial arrangements of the groups • cis isomer • trans isomer
  33. 33. Geometric Isomers•When there are two identical groups at one endof a double bond, geometric isomers are notpossible•Cyclic compounds can also have geometricisomers•This cis-trans isomerism is found in the manycyclic structures of carbohydrates.•Geometric differences alone make mostcarbohydrates unusuable in human nutrition.
  34. 34. AlkeneCH3 CH3 CH3 H C=C C=CH H H CH3 trans-2-butene cis-2-butene
  35. 35. Geometric IsomersDouble bond is fixedCis/trans Isomers are possibleCH3 CH3 CH3 H C=C C=C H H H CH3(bp 3.7°C) (bp 0.9°C)
  36. 36. Naming Alkenes – Name all other substituents in a manner similar to the alkanes. – Use a prefix to indicate the geometric isomer present, if necessary.
  37. 37. Learning CheckDraw the structures for thefollowing compounds:cis-6-methyl-3-heptene
  38. 38. Alkynes• Alkynes: – unsaturated hydrocarbons that contain a C C triple bond• Alkyne Nomenclature: – Identify the longest continuous chain containing the triple bond – To find the base name, change the infix of the corresponding alkane from “an” to “yn”
  39. 39. Alkynes• Alkyne Nomenclature: – Use a number to designate the position of the triple bond • number from the end of the chain closest to the triple bond – just like with alkenes – Name substituents like you do with alkanes and alkenes
  40. 40. Learning Check HA3Write the IUPAC name for each of thefollowing unsaturated compounds: CH3CH2C≡CCH3
  41. 41. Solutions HA3 Write the IUPAC name for each of the following unsaturated compounds:• CH3CH2C≡CCH3 2-pentyne
  42. 42. Alkynes Name the following compounds: CH3CH2C CCHCH3 CH2CH3 CH3CH2C C Cl
  43. 43. AlkynesDraw the following alkynes.4-chloro-2-pentyne3-propyl-1-hexyne
  44. 44. Physical Properties• alkenes and alkynes are nonpolar compounds• the only attractive forces between their molecules are London dispersion forces• their physical properties are similar to those of alkanes with the same carbon skeletons• alkenes and alkynes are insoluble in water but soluble in one another and in nonpolar organic liquids• alkenes and alkynes that are liquid or solid at room temperature have densities less than 1 g/ mL; they float on water
  45. 45. Reactions of Alkenes &Alkynes• More reactive than alkanes or aromatics, why?• Generally undergo addition reactions• Presence of easily accessible π electrons• Unsaturated: can fit more atoms around the carbons
  46. 46. Reaction of AlkenesAdditions reactions of the double bondThe new double bond is broken and inits place single bonds are formed to the newatoms or groups of atomsThe double bond becomes a single bond H H H HH– C= C–H + X-Y H–C–C–H XY
  47. 47. Hydrogenation Adds a hydrogen atom to each carbon atom of a double bond H H H H NiH–C=C–H + H2 H–C–C–H H H ethene ethaneReacts with H2 in the presence of transitionmetal catalyst (Pd, Pt, Ru, Ni)
  48. 48. Products of Hydrogenation Adding H2 to vegetable oils produces compounds with higher melting points Margarines Soft margarines Shortenings (solid)
  49. 49. Trans FatsIn vegetable oils, the unsaturated fats usually contain cis double bonds.During hydrogenation, some cis double bonds are converted to trans double bonds (more stable) causing a change in the fatty acid structureIf a label states “partially” or “fully hydrogenated”, the fats contain trans fatty acids.
  50. 50. Learning Check HA4What is the product of adding H2(Ni catalyst) to 1-butene?
  51. 51. Solution HA4What is the product of adding H2(Ni catalyst) to 1-butene? NiCH2=CHCH2CH3 + H2 CH3CH2CH2CH3
  52. 52. Learning Check HA5 Write the product of the following addition reactions: CH3CH=CHCH3 + H2 + Br2
  53. 53. Solution HA5Write the product of the following additionreactions:CH3CH=CHCH3 + H2 CH3CH2CH2CH3 + Br2 Br Br
  54. 54. Addition of Bromine• Br2 (in CCl4) is added to an unknown liquid• The unknown is saturated b/c Br2 does not lose its red color.• The unknown was unsaturated. The deep red color of Br2 is decolorized as it reacts with the double bond.
  55. 55. Orientation of Addition• Both alkene & reagent are symmetric: one possible product• One is symmetric and the other is asymmetric: one possible product• Both alkene & reagent are asymmetric: two possible products
  56. 56. Markovinkov’s Rule• When an unsymmetrical reactant of the type X-Y adds to an unsymmetrical alkene, the carbon with the greater number of hydrogens gets more H• Used to predict the product of many alkene addition reactions however it does not explain Why???
  57. 57. Markovinkov’s Rule Try This !!!
  58. 58. Addition of Hydrogen HalidesAdds a H atom and Cl to each carbon atom of a double bond H H H HH–C=C–H + HCl H–C–C– H H Cl ethene chloroethane
  59. 59. Addition of HydrogenHalidesMarkovnikov’s rule – when unsymmetrical reagent adds to an unsymmetrical carbon, the carbon with the greater # of hydrogens gets more H ClCH3C=CH2 + HCl CH3CCH3 CH3 CH3
  60. 60. Learning CheckCH3CH=CH2 +HI + HBr =CH2 + HBr
  61. 61. Question 3• What is the major product of the following reaction? + HBr
  62. 62. Addition of Water(Hydration)• Water does not react with an alkene in the absence of an acid catalyst• Water is a weak donor of H+ b/c it holds it protons too strongly
  63. 63. Addition of H2O• Addition of water is called hydration – hydration is acid catalyzed, most commonly by H2SO4 – hydration follows Markovnikov’s rule; H adds to the less substituted carbon and OH adds to the more substituted carbon OH H H 2 SO4CH3 CH=CH 2 + H2 O CH3 CH-CH2 Propene 2-Propanol
  64. 64. CH3 CH3 H2 SO4 CH3 C=CH2 + H2 O CH3 C-CH2 HO H2-M ethylpropene 2-Methyl-2-propanol
  65. 65. Alkene Addition
  66. 66. 10) Chemical properties of Carbon compounds :-a) Combustion :- Carbon compounds burn in oxygen to form water, carbon dioxide, heat and light. Eg :- C + O2 CO2 + heat + light CH4 + 2O2 2H2O + CO2 + heat + light C2H5OH + 3O2 3H2O + 2CO2 heat + lightb) Oxidation :- Carbon compounds like alcohols are oxidised to carboxylic acids on heating with oxidising agents like alkaline Potassium permanganate – KMnO4 or acidic potassium dichromate - K2Cr2O7 . Eg:- Alcohols are oxidised to Carboxylic acids alkaline KMnO4 + heat C2H5OH CH3COOH Ethanol acidic K2Cr2O7 + heat Ethanoic acid
  67. 67. c) Addition reaction :- Unsaturated hydrocarbons undergo addition reaction with hydrogen in thepresence of nickel or palladium as catalyst to form saturated hydrocarbons. Eg:- Ethene undergoes addition reaction with hydrogen to form ethane in thepresence of nickel or palladium as catalyst. Ni or Pd catalyst C2H4 + H2 C2H6 H H H H I I Ni or Pd catalyst I I C = C + H2 H–C–C–H I I I I H H H H The addition of hydrogen to unsaturated hydrocarbons to form saturatedhydrocarbons is called hydrogenation. Hydrogenation is used to convertunsaturated oils and fats to saturated oils and fats.d) Substitution reaction :- Saturated hydrocarbons undergo substitution reaction with halogens toform substitution products. Eg :- Methane undergoes substitution reaction with chlorine in the presenceof sunlight to form substitution products. CH4 + Cl2 CH3Cl + HCl CH3Cl + Cl2 CH2Cl2 + HCl CH2Cl2 + Cl2 CHCI3 + HCl CHCI3 + Cl2 CCl4 + HCl
  68. 68. 11) Some important carbon compounds :- a) ETHANOL :- C2H5OH - Ethyl alcohol Properties :- i) Ethanol is a colourless liquid with a pleasant smell and burning taste. ii) It is soluble in water.iii) Ethanol reacts with sodium to form sodium ethoxide and hydrogen. 2C2H5OH + 2Na 2C2H5ONa + H2iv) Ethanol reacts with hot conc. H2SO4 to form ethene and water. Conc. H2SO4 is a dehydrating agent and removes water from ethanol. conc. H2SO4 C2H5OH C2H4 + H2O Uses :- i) Ethanol is used for making alcoholic drinks. ii) It is used as a solvent.iii) It is used for making medicines like tincture iodine, cough syrups, tonics etc.
  69. 69. b) ETHANOIC ACID :- CH3COOH – Acetic acid Properties :- i) Ethanoic acid is a colourless liquid with a pungent smell and sour taste. ii) It is soluble in water.iii) A solution of 5% to 8% ethanoic acid in water is called Vinegar.iv) Esterification :- Ethanoic acid reacts with ethanol to form the ester ethyl ethanoate in the presence of conc. H2SO4. conc.H2SO4 CH3COOH + C2H5OH CH3COOC2H5 + H2O The reaction between carboxylic acid and alcohol to form an ester is called esterification. v) Saponification :- When an ester reacts with sodium hydroxide solution, the sodium salt of the carboxylic acid and the parent alcohol are formed. This reaction is called saponification. Eg :-Ethyl ethanoate reacts with sodium hydroxide to form sodium acetate and ethanol. CH3COOC2H5 + NaOH CH3COONa + C2H5OH vi) Ethanoic acid reacts with bases to form salt and water. CH3COOH + NaOH CH3COONa + H2Ovii) Ethanoic acid reacts with carbonates and hydrogen carbonates to form salt, water and carbon dioxide. 2CH3COOH + Na2CO3 2CH3COONa + H2O + CO2 CH3COOH + NaHCO3 CH3COONa + H2O + CO2
  70. 70. 12) Soaps and detergents :-a) Soaps :- Soaps are long chain sodium or potassium salts of carboxylicacids. Eg:- Sodium stearate – C17H35COONa Structure of soap molecule :- A soap molecule has two parts. A longhydrocarbon part which is hydrophobic (water repelling) and soluble in oil andgrease and a short ionic part which is hydrophyllic (water attracting) andinsoluble in oil and grease. + COO Na Hydrocarbon part Ionic part (Water repelling) (Water attracting) Cleansing action of soap :- When soap is dissolved in water it formsspherical structures called micelles. In each micelle the soap molecules arearranged radially such that the HC part is towards the centre and the ionic partis towards the outside. The HC part dissolves the dirt, oil and grease and formsan emulsion at the centre of the micelles which can be washed away by water.
  71. 71. b) Detergents :- Detergents are long chain sodium salts of sulphonic acids. Soaps do not wash well with hard water because it forms insolubleprecipitates of calcium and magnesium salts in hard water. Detergents wash well with hard water because it does not form insolubleprecipitates of calcium and magnesium salts in hard water.c) Differences between soaps and detergents :- Soaps Detergents i) Soaps are sodium salts of Detergents are sodium salts of fatty acids. sulphonic acids. ii) Soaps clean well in soft water but Detergents clean well with both do not clean well in hard water. hard and soft water.iii) Soaps do not clean as well as Detergents clean better than soaps. detergents.iv) Soaps are biodegradable and Some detergents are non biodegradable do not cause pollution. and cause pollution.