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Organic Chemistry

Organic Chemistry. P 267. Hydrocarbon. Hydro: hydrogen Carbon Additional element: Nitrogen, Oxygen, Sulphur Atomic number of carbon = 6 Valence electrons of carbon = 4 Each carbon atom has 4 bond in compound Ex: alkanes

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Organic Chemistry

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  1. Organic Chemistry P 267

  2. Hydrocarbon • Hydro: hydrogen • Carbon • Additional element: Nitrogen, Oxygen, Sulphur Atomic number of carbon = 6 Valence electrons of carbon = 4 Each carbon atom has 4 bond in compound Ex: alkanes Alkanes: hydrocarbon that consist of hydrogen and carbon, (single bond)

  3. CH4, methane C2H6, ethane C3H8, propane C4H10, butane C5H12, pentane

  4. C6H14, hexane C7H16, heptane C8H18, octane C9H20, nonane C10H22, decane

  5. Structure Types of Propane Stick and ball model Displayed Formula 3 dimensional Formula H H Structural Formula H H CH3CH2CH3 H H H H Molecular Formula Skeletal Formula C3H8

  6. ALCOHOL (-OH) PROPANE PROPANOL

  7. ALCANOIC ACID BUTANOIC ACID BUTANE

  8. ALDEHID/ALCANAL PENTANAL PENTANE

  9. Ketones Butan-2-one

  10. Alkylamine R-NH2 CH3CH2CH2NH2 Propylamine

  11. ESTHER (Alkyl Alkanoate) propanoate methyl

  12. Hydrocarbon with Branch 6 5 4 3 2 1 -Find the longest carbon chain, give the name - Find branch and give the name to the Branch: Carbon:Alkyl; 1 C atom: methyl 2 C atom: ethyl 3 C atom: propyl Halogen: F(Floro), Cl(Chloro), Br(Bromo), I(Iodo) - Give the number, branch number is the smallest number Same branch: 2 branchs: di 3 branchs: tri 4 branchs: tetra Hexane 2- Methyl

  13. the branches names arranged alphabetically 3-bromo 9-chloro 4-methyldecan-3-ol

  14. 6-chloro 4-ethyl octan-3-one

  15. Alcanal, and alcanoic numbering begin from C attached to the function 5 4 3 2 1 4- methyl Pentanal

  16. 4 6 1 2 3 5 7 8 9 • Same branch: • 2 branchs: di • 3 branchs: tri • 4 branchs: tetra 4,6 -diethyl 2,4,7 -trimethyl nonane

  17. Alcanal, and alcanoic numbering begin from C attached to the function 6-ethyl 3-floro 4,7-dimethyl octanoic acid Remember that the branches names arranged alphabetically

  18. Task • Copy the table 15.2 page 270 in to your book

  19. Determination of element mass in compound Compound: AmBn Mass of element A = Mass of element B = Find mass of carbon in 88 gr CO2 Mass of carbon = = = 24 gr

  20. Determination of empirical formulae of organic compound 100,00 g amino acid(CpHqOrNs) react with oxygen and produced 117,10 g of carbon dioxide and 59,92 g of water. In kjeldahl a second of amino acid produced 22,64 g ammonia, determine the empirical formula of amino acid Answer - write the information mass of amino acid = 100 gr mass of carbon dioxide (CO2) = 117,10 gr mass of water(H2O)= 59,92 g mass of ammonia (NH3) = 22,64

  21. Find mass of carbon in carbon dioxide = 31,94 g • Find mass of hydrogen in water = 6,66 g • Find mass of nitrogen in ammonia = 18,64 g • Find mass of oxygen in amino acid =mass of amino acid – carbon mass – hydrogen mass - nitrogen mass = 100 – 31,94 – 6,66 – 18,64 = 42,76 gr

  22. Compare mass of the elements C : H : O : N 31,94 : 6,66 : 42,7 : 18,64 • Compare mol of the elements 31,94/12 : 6,66/1 : 42,76/16 : 18,64/14 2,67 : 6,66 : 2,67 : 1,33 • Divide by the smallest mole 2 : 5 : 2 : 1 • Determine the empirical formula C2H5O2N

  23. Reactions Subtitution H H H H H C C Br + OH- H OH + Br- C C  H H H H

  24. Addition Double bondsingle bond Triple bonddouble or single bond Elimination single bondDouble bond double Triple bond + H2O H H H H H H H H C C C C H H OH OH C C C C  H H H H H H H H H2O + 

  25. Hydrolysis Reaction involve breaking covalent bonds by reaction with water H H H H H C C Br + H2O H OH + HBr C C  H H H H

  26. Breaking Bonds in Different Ways Homolytic fission Each atom break to free radical (atom with unpaired electron  + Heterolytic fission Each atom break to ion (cation and anion) - + + 

  27. Another example Homolytic fission CH3Br  CH3+ Br Heterolytic fission CH3Br  CH3++ Br- + Carbocation

  28. Cycloalkane cyclohexane cyclobutane cyclopentane cyclopropane Methyl cyclohexane

  29. Alkylamine (R-NH2) CH3NH2 methylamine CH3CH2NH2 ethylamine CH3CH2CH2NH2 propylamine

  30. Alkenes C=C ethene CH2=CH2 CH2=CH-CH3 CH2=CH-CH2-CH3 CH2=CH-CH2-CH2-CH3 CH3-CH=CH-CH2-CH3 Double bond number is the smallest number in the longest carbon chain propene butene pentene pent-2-ene

  31. Isomerism Structural isomer Different structural formula same molecular formula CH3-CH2-CH2-CH3 Butane, C4H10 Methyl propane, C4H10

  32. Functional isomerism • Alcohol and eter • Ethanol: CH3-CH2-OH, molecular formula: C2H6O • Ether : CH3-O-CH3, molecular formula: C2H6O • Aldehid and ketones • pentanal: ,molecular formula:C5H10O • Penta-2-one C5H10O • Alkanoic acid and esthers • Propanoic acid • Methyl ethanoate Prove it !

  33. Stereo isomer Geometric isomer (cis-trans isomerism) But-2ene: CH3-CH=CH-CH3 Cis-but-2-ene trans-but-2-ene

  34. The star sign indicate chiral center. Chiral center is the carbon atom which carries four different groups Optical isomerism use 3 dimensional formula Occurs in the chiral molecules Butan-2-ol

  35. Reactions Routes Write the reactions routes to convert: • Propene to propanol • pass propene and steam over heated phosphoric acid catalyst under pressure b. Propene to propanal • pass propene and steam over heated phosphoric acid catalyst under pressure untillpropanol forms • Distill on propanol to acidified potassium dichromate

  36. Draw the structure of 5,7 dibromo, 2,6 dimethyl octan-3-one Solution steps Draw the longest chain Give the number Put the function Put the branches C O 1 2 3 4 5 6 7 8 C-C-C-C-C-C-C-C C Br Br

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