Organic Chemistry Lecture 4 PDF

CHEM 0901 ORGANIC CHEMISTRY LECTURE 4 Lecture Outline  Reactions of alkanes  Combustion analysis of hydrocarbons  Free radical halogenation  Reaction of methane with chlorine  Reactions of higher alkanes (product ratio/ radical stabili...

CHEM 0901 ORGANIC CHEMISTRY LECTURE 4 Lecture Outline  Reactions of alkanes  Combustion analysis of hydrocarbons  Free radical halogenation  Reaction of methane with chlorine  Reactions of higher alkanes (product ratio/ radical stability)  Cycloalkanes  Double bond equivalents: concept and calculations Reactions of alkanes  Bonds are non-polar, covalent so alkanes are relatively inert  Alkanes do not react with most acids, bases or reducing agents Reactions of alkanes  Alkanes are used as fuel  Heat is evolved in the process  This is an oxidation reaction  Combustion is the process of burning and is accompanied by the release of light and heat. CH4 + 2O2 CO2 + 2H2O + heat Reactions of alkanes  Combustion analysis  Gay-Lussac Law  Volumes of gases taking part in chemical reaction bear a simple numerical relationship to one another if all measurements are done at STP.  Avogadro Law  Equal volumes of all gases under the same conditions of temp and pressure contain the same number of molecules CxHy + x + y/4 O2 x CO2 + y/2 H2O Combustion analysis of hydrocarbons  A known volume of the hydrocarbon is reacted with a known volume of oxygen. The volume occupied by the products and unreacted oxygen is measured after cooling.  What are the products?  This method can be used to determine the molecular weight of an alkane  Problem question  Twenty mL of a gaseous hydrocarbon was combusted with 200 ml oxygen. After cooling, the residual gases occupied 160 mL. Eighty ml of the mixture was found to be carbon dioxide. Determine the molecular formula of the hydrocarbon. CxHy + x + y/4 O2 x CO2 + y/2 H2O Past Paper Question  1997  A gaseous hydrocarbon X of volume 15 cm3 was mixed with oxygen to give a total volume of 145 cm3. After combustion and cooling, the residual gas occupied 100 cm3. One fourth of the gas remained after it was passed through aqueous sodium hydroxide. Determine the molecular formula of the hydrocarbon. (All measurements were made at STP) (10 marks)  Determination of molecular formulae using percent composition  The percent composition (the percent by mass) of the elements in a sample may be obtained experimentally and is used to determine the empirical and molecular formula of a compound.  The empirical formula is a chemical formula that indicates the relative proportions of the elements in a molecule rather than the actual number of atoms. Percent composition of an element = n x molar mass of element X 100% molar mass of compound n = no. of moles of element/ mole of compound  Determination of molecular formulae using percent composition Example from Past Paper, December 2000  Liquid hydrocarbon G is composed of 12.27% hydrogen. Determine the molecular formula of this compound. Mass Convert to grams Moles of Divide by the Mole ratios percent and divide by each element smallest number of elements molar mass of moles Convert to whole numbers Empirical formula Free radical halogenation  Reaction of methane with chlorine  Initiation  Propagation  Termination CH3 CH3CH2 (CH3)2CH (CH3)3C methyl 1o 2o 3o Increasing stability  Multiple substitution reactions H H Monohalogenation H C H + Cl2 H C Cl + H Cl H H H H Dihalogenation H C Cl + Cl2 H C Cl + H Cl H Cl H H Trihalogenation H C Cl + Cl2 Cl C Cl + H Cl Cl Cl H Cl Tetrahalogenation Cl C Cl + Cl2 Cl C Cl + H Cl Cl Cl  Reactions with Alkanes Free radical halogenation of alkanes Br h R CH CH3 + Br2 R C CH3 + R CH CH2Br or  CH3 CH3 CH3 o major (3 ) minor (1o) F2 – explosive Cl2/Br2 – diffused light or heat I2 - unreactive  MECHANISM INITIATION..Br Br Br Br +. R R PROPAGATION. C Br + H C CH3 H Br + CH3 CH3 CH3 R +. R Br Br C CH3 Br C CH3 + Br. CH3 CH3 TERMINATION..Br Br + Br Br. R R H3C C +.Br H3C C Br CH3 CH3 R R.+. R R C CH3 H3C C C CH3 H3C C CH3 CH3 CH3 CH3  Reaction selectivity Bromine is more selective than either chlorine or fluorine CH 3 CH 3 CH 3 H2 Cl2, hv H2 H2 H3C C C CH 3 H3C C C CH 2Cl + H3C C C CH 3 H H Cl CH 3 CH 3 H2 H ClH2C C C CH 3 + H3C C C CH 3 H Cl H CH3 CH3 H CH3 H2 Br2, light H2 H3C C C CH3 H3C C C CH3 + H3C C C CH2Br H Br H H high yield low yield  Free Radicals in Diseases Thought to contribute to various human diseases Most age-related diseases can be explained through reactions of free radicals with biological substances. – cancer [carcinogenesis] – Strokes – Alzheimer's disease – Rheumatoid arthritis – Parkinson’s disease – Development of Cataracts Past Paper Question July/ August 2004  2-Methylpropane when heated with bromine in the presence of light produces 2-bromo-2-methylpropane (>99%) and 1-bromo-2- methylpropane(