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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 Alkanes: hydrocarbon that consist of hydrogen and carbon, (single bond) CH4, methane C2H6, ethane C3H8, propane C4H10, butane C5H12, pentane C6H14, hexane C7H16, heptane C8H18, octane C9H20, nonane C10H22, decane Structure Types of Propane Stick and ball model Displayed Formula 3 dimensional Formula H Structural Formula CH3CH2CH3 Molecular Formula C3H8 H H H H H H H Skeletal Formula ALCOHOL (-OH) PROPANE PROPANOL ALCANOIC ACID BUTANE BUTANOIC ACID ALDEHID/ALCANAL PENTANE PENTANAL KETONES Butan-2-one Alkylamine R-NH2 CH3CH2CH2NH2 Propylamine ESTHER (Alkyl Alkanoate) methyl propanoate 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: Same branch: Carbon:Alkyl; 2 branchs: di 1 C atom: methyl 3 branchs: tri 4 branchs: tetra 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 2- Methyl Hexane the branches names arranged alphabetically 3-bromo 9-chloro 4-methyldecan-3-ol 6-chloro 4-ethyl octan-3-one Alcanal, and alcanoic numbering begin from C attached to the function 5 4 3 2 1 4- methyl Pentanal 1 • • • • 2 3 4 5 6 7 8 9 Same branch: 2 branchs: di 3 branchs: tri 4 branchs: tetra 4,6 -diethyl 2,4,7 -trimethyl nonane 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 Task • Copy the table 15.2 page 270 in to your book 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 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 • 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 • 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 Reactions Subtitution H H H H H C C Br + OH- H C C OH + BrH H H H Addition Double bondsingle bond Triple bonddouble or single bond H H H C C H + H2O H C C OH H H H H Elimination single bondDouble bond double Triple bond H H H C C OH H C C H + H2O H H H H Hydrolysis Reaction involve breaking covalent bonds by reaction with water H H H H H C C Br + H2O H C C OH + HBr H H H H 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) ++ - Another example Homolytic fission CH3Br CH3 + Br Heterolytic fission CH3Br CH3+ + Br- + Carbocation Cycloalkane cyclopropane cyclobutane Methyl cyclohexane cyclopentan e cyclohexane Alkylamine (R-NH2) CH3NH2 methylamine CH3CH2NH2 ethylamine CH3CH2CH2NH2 propylamine Alkenes C=C ethene CH2=CH2 propene CH2=CH-CH3 butene CH2=CH-CH2-CH3 pentene CH2=CH-CH2-CH2-CH3 CH3-CH=CH-CH2-CH3 pent-2-ene Double bond number is the smallest number in the longest carbon chain Isomerism Structural isomer Different structural formula same molecular formula CH3-CH2-CH2-CH3 Butane, C4H10 Methyl propane, C4H10 Functional isomerism - Alcohol and eter - Ethanol: CH3-CH2-OH, molecular formula: C2H6O - Ether : CH3-O-CH3, molecular formula: C2H6O - Aldehid and ketones - pentanal: - Penta-2-one ,molecular formula:C5H10O C5H10O - Alkanoic acid and esthers - Propanoic acid - Methyl ethanoate Prove it ! Stereo isomer Geometric isomer (cis-trans isomerism) But-2ene: CH3-CH=CH-CH3 Cis-but-2-ene trans-but-2-ene Optical isomerism use 3 dimensional formula Occurs in the chiral molecules The star sign indicate chiral center. Chiral center is the carbon atom which carries four different groups Butan-2-ol Reactions Routes Write the reactions routes to convert: a. 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 untill propanol forms – Distill on propanol to acidified potassium dichromate 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-CC-C C B B