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AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS

AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS. THE CHEMISTRY OF ALCOHOLS. CONTENTS Structure of alcohols Nomenclature Isomerism Physical properties Chemical properties of alcohols Identification using infra-red spectroscopy Industrial preparation and uses of ethanol

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AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS

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  1. AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS

  2. THE CHEMISTRY OF ALCOHOLS • CONTENTS • Structure of alcohols • Nomenclature • Isomerism • Physical properties • Chemical properties of alcohols • Identification using infra-red spectroscopy • Industrial preparation and uses of ethanol • Revision check list

  3. THE CHEMISTRY OF ALCOHOLS • Before you start it would be helpful to… • Recall the definition of a covalent bond • Recall the difference types of physical bonding • Be able to balance simple equations • Be able to write out structures for simple organic molecules • Understand the IUPAC nomenclature rules for simple organic compounds • Recall the chemical properties of alkanes and alkenes

  4. CLASSIFICATION OF ALCOHOLS Aliphatic• general formula CnH2n+1OH - provided there are no rings • the OH replaces an H in a basic hydrocarbon skeleton

  5. CLASSIFICATION OF ALCOHOLS Aliphatic• general formula CnH2n+1OH - provided there are no rings • the OH replaces an H in a basic hydrocarbon skeleton Aromatic • in aromatic alcohols (or phenols) the OH is attached directly to the ring • an OH on a side chain of a ring behaves as a typical aliphatic alcohol The first two compounds are classified as aromatic alcohols (phenols) because the OH group is attached directly to the ring.

  6. CLASSIFICATION OF ALCOHOLS Aliphatic• general formula CnH2n+1OH - provided there are no rings • the OH replaces an H in a basic hydrocarbon skeleton Aromatic • in aromatic alcohols (or phenols) the OH is attached directly to the ring • an OH on a side chain of a ring behaves as a typical aliphatic alcohol The first two compounds are classified as aromatic alcohols (phenols) because the OH group is attached directly to the ring. Structural differences • alcohols are classified according to the environment of the OH group • chemical behaviour, eg oxidation, often depends on the structural type PRIMARY 1° SECONDARY 2° TERTIARY 3°

  7. NAMING ALCOHOLS Alcohols are named according to standard IUPAC rules • select the longest chain of C atoms containing the O-H group; • remove the e and add ol after the basic name • number the chain starting from the end nearer the O-H group • the number is placed after the an and before the ol ... e.g butan-2-ol • as in alkanes, prefix with alkyl substituents • side chain positions are based on the number allocated to the O-H group e.g. CH3 - CH(CH3) - CH2 - CH2 - CH(OH) - CH3 is called 5-methylhexan-2-ol

  8. STRUCTURAL ISOMERISM IN ALCOHOLS Different structures are possible due to... A Different positions for the OH group and B Branching of the carbon chain butan-1-ol butan-2-ol 2-methylpropan-2-ol 2-methylpropan-1-ol

  9. BOILING POINTS OF ALCOHOLS Increases with molecular size due to increased van der Waals’ forces. Alcohols have higher boiling points than similar molecular mass alkanes This is due to the added presence of inter-molecular hydrogen bonding. More energy is required to separate the molecules. Mr bp / °C propane C3H8 44 -42 just van der Waals’ forces ethanol C2H5OH 46 +78 van der Waals’ forces+ hydrogen bonding

  10. BOILING POINTS OF ALCOHOLS Increases with molecular size due to increased van der Waals’ forces. Alcohols have higher boiling points than similar molecular mass alkanes This is due to the added presence of inter-molecular hydrogen bonding. More energy is required to separate the molecules. Mr bp / °C propane C3H8 44 -42 just van der Waals’ forces ethanol C2H5OH 46 +78 van der Waals’ forces+ hydrogen bonding Boiling point is higher for “straight” chain isomers. bp / °C butan-1-ol CH3CH2CH2CH2OH 118 butan-2-ol CH3CH2CH(OH)CH3 100 2-methylpropan-2-ol (CH3)3COH 83 Greater branching = lower inter-molecular forces

  11. BOILING POINTS OF ALCOHOLS Increases with molecular size due to increased van der Waals’ forces. Alcohols have higher boiling points than similar molecular mass alkanes This is due to the added presence of inter-molecular hydrogen bonding. More energy is required to separate the molecules. Mr bp / °C propane C3H8 44 -42 just van der Waals’ forces ethanol C2H5OH 46 +78 van der Waals’ forces+ hydrogen bonding Boiling point is higher for “straight” chain isomers. bp / °C butan-1-ol CH3CH2CH2CH2OH 118 butan-2-ol CH3CH2CH(OH)CH3 100 2-methylpropan-2-ol (CH3)3COH 83 Greater branching = lower inter-molecular forces

  12. SOLVENT PROPERTIES OF ALCOHOLS SolubilityLow molecular mass alcohols are miscible with water Due to hydrogen bonding between the two molecules Heavier alcohols are less miscible Solvent propertiesAlcohols are themselves very good solvents They dissolve a large number of organic molecules

  13. CHEMICAL PROPERTIES OF ALCOHOLS The OXYGEN ATOM HAS TWO LONE PAIRS; this makes alcohols... BASESLewis bases are lone pair donors Bronsted-Lowry bases are proton acceptors The alcohol uses one of its lone pairs to form a co-ordinate bond NUCLEOPHILESAlcohols can use the lone pair to attack electron deficient centres

  14. ELIMINATION OF WATER (DEHYDRATION) Reagent/catalystconc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4) Conditions reflux at 180°C Product alkene Equation e.g. C2H5OH(l) ——> CH2 = CH2(g) + H2O(l) Mechanism Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation Step 3 loss of a proton (H+) to give the alkene Alternative Method Pass vapour over a heated alumina (aluminium oxide) catalyst

  15. ELIMINATION OF WATER (DEHYDRATION) MECHANISM Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation Step 3 loss of a proton (H+) to give the alkene Note 1There must be an H on a carbon atom adjacent the carbon with the OH Note 2 Alcohols with the OH in the middle of a chain can have two ways of losing water. In Step 3 of the mechanism, a proton can be lost from either side of the carbocation. This gives a mixture of alkenes from unsymmetrical alcohols...

  16. OXIDATION OF ALCOHOLS All alcohols can be oxidised depending on the conditions Oxidation is used to differentiate between primary, secondary and tertiary alcohols The usual reagent is acidified potassium dichromate(VI) PrimaryEasily oxidisedto aldehydes and then to carboxylic acids. SecondaryEasily oxidised to ketones TertiaryNot oxidised under normal conditions. They do break down with very vigorous oxidation PRIMARY 1° SECONDARY 2° TERTIARY 3°

  17. OXIDATION OF PRIMARY ALCOHOLS Primary alcohols are easily oxidised to aldehydes e.g. CH3CH2OH(l) + [O] ——> CH3CHO(l) + H2O(l) ethanol ethanal it is essential to distil off the aldehyde before it gets oxidised to the acid CH3CHO(l) + [O] ——> CH3COOH(l) ethanal ethanoic acid

  18. OXIDATION OF PRIMARY ALCOHOLS • Primary alcohols are easily oxidised to aldehydes • e.g. CH3CH2OH(l) + [O] ——> CH3CHO(l) + H2O(l) • ethanol ethanal • it is essential to distil off the aldehyde before it gets oxidised to the acid • CH3CHO(l) + [O] ——> CH3COOH(l) • ethanal ethanoic acid • Practical details • the alcohol is dripped into a warm solution of acidified K2Cr2O7 • aldehydes have low boiling points - no hydrogen bonding - they distil off immediately • if it didn’t distil off it would be oxidised to the equivalent carboxylic acid • to oxidise an alcohol straight to the acid, reflux the mixture • compound formula intermolecular bonding boiling point • ETHANOL C2H5OH HYDROGEN BONDING 78°C • ETHANAL CH3CHO DIPOLE-DIPOLE 23°C • ETHANOIC ACID CH3COOH HYDROGEN BONDING 118°C

  19. OXIDATION OF PRIMARY ALCOHOLS Controlling the products e.g. CH3CH2OH(l) + [O] ——> CH3CHO(l) + H2O(l) then CH3CHO(l) + [O] ——> CH3COOH(l) OXIDATION TO ALDEHYDES DISTILLATION OXIDATION TO CARBOXYLIC ACIDS REFLUX Aldehyde has a lower boiling point so distils off before being oxidised further Aldehyde condenses back into the mixture and gets oxidised to the acid

  20. OXIDATION OF SECONDARY ALCOHOLS Secondary alcohols are easily oxidised to ketones e.g. CH3CHOHCH3(l) + [O] ——> CH3COCH3(l) + H2O(l) propan-2-ol propanone The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol.

  21. OXIDATION OF SECONDARY ALCOHOLS Secondary alcohols are easily oxidised to ketones e.g. CH3CHOHCH3(l) + [O] ——> CH3COCH3(l) + H2O(l) propan-2-ol propanone The alcohol is refluxed with acidified K2Cr2O7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol. OXIDATION OF TERTIARY ALCOHOLS Tertiary alcohols are resistant to normal oxidation

  22. OXIDATION OF ALCOHOLS Why 1° and 2°alcohols are easily oxidised and 3° alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms.

  23. H H R C O + [O]R C O + H2O H H OXIDATION OF ALCOHOLS Why 1° and 2°alcohols are easily oxidised and 3° alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms. 1°

  24. H H R C O + [O]R C O + H2O H H H H R C O + [O]R C O + H2O RR OXIDATION OF ALCOHOLS Why 1° and 2°alcohols are easily oxidised and 3° alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms. 1° 2°

  25. H H R C O + [O]R C O + H2O H H H H R C O + [O]R C O + H2O RR OXIDATION OF ALCOHOLS Why 1° and 2°alcohols are easily oxidised and 3° alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms. 1° 2° This is possible in 1° and 2° alcoholsbut not in 3° alcohols.

  26. 1° 2° H H R C O + [O]R C O + H2O H H H H R C O + [O]R C O + H2O RR R H R C O + [O] R 3° OXIDATION OF ALCOHOLS Why 1° and 2°alcohols are easily oxidised and 3° alcohols are not For oxidation to take place easily you must have two hydrogen atoms on adjacent C and O atoms. This is possible in 1° and 2°alcoholsbut not in 3° alcohols.

  27. ESTERIFICATION OF ALCOHOLS Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 ) Conditions reflux Product ester Equatione.g. CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l) ethanol ethanoic acid ethyl ethanoate NotesConcentratedH2SO4 is a dehydrating agent - it removes water causing the equilibrium to move to the right and increases the yield

  28. ESTERIFICATION OF ALCOHOLS Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 ) Conditions reflux Product ester Equation e.g. CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l) ethanol ethanoic acid ethyl ethanoate Notes Concentrated H2SO4 is a dehydrating agent - it removes water causing the equilibrium to move to the right and increases the yield Uses of esters Esters are fairly unreactive but that doesn’t make them useless Used as flavourings Naming esters Named from the alcohol and carboxylic acid which made them... CH3OH + CH3COOHCH3COOCH3+ H2O fromethanoic acidCH3COOCH3from methanol METHYLETHANOATE

  29. OTHER REACTIONS OF ALCOHOLS OXYGEN Alcohols make useful fuels C2H5OH(l) + 3O2(g) ———> 2CO2(g) + 3H2O(l) Advantages have high enthalpies of combustion do not contain sulphur so there is less pollution can be obtained from renewable resources

  30. OTHER REACTIONS OF ALCOHOLS OXYGEN Alcohols make useful fuels C2H5OH(l) + 3O2(g) ———> 2CO2(g) + 3H2O(l) Advantages have high enthalpies of combustion do not contain sulphur so there is less pollution can be obtained from renewable resources SODIUM Conditionsroom temperature Product sodium alkoxide and hydrogen Equation2CH3CH2OH(l) + 2Na(s) ——> 2CH3CH2O¯ Na + + H2(g) sodium ethoxide Notes alcohols are organic chemistry’s equivalent of water water reacts with sodium to produce hydrogen and so do alcohols the reaction is slower with alcohols than with water. Alkoxides are white, ionic crystalline solids e.g. CH3CH2O¯ Na+

  31. BROMINATION OF ALCOHOLS Reagent(s)conc. hydrobromic acid HBr(aq) or sodium (or potassium) bromide and concentrated sulphuric acid Conditions reflux Product haloalkane EquationC2H5OH(l) + conc. HBr(aq) ———> C2H5Br(l) + H2O(l) Mechanism The mechanism starts off similar to that involving dehydration (protonation of the alcohol and loss of water) but the carbocation (carbonium ion) is attacked by a nucleophilic bromide ion in step 3 Step 1 protonation of the alcohol using a lone pair on oxygen Step 2 loss of a water molecule to generate a carbocation (carbonium ion) Step 3 a bromide ion behaves as a nucleophile and attacks the carbocation

  32. INFRA-RED SPECTROSCOPY Chemical bonds vibrate at different frequencies. When infra red (IR) radiation is passed through a liquid sample of an organic molecule, some frequencies are absorbed. These correspond to the frequencies of the vibrating bonds. Most spectra are very complex due to the large number of bonds present and each molecule produces a unique spectrum. However the presence of certain absorptions can be used to identify functional groups. BOND COMPOUND ABSORBANCE RANGE O-H alcohols broad 3200 cm-1 to 3600 cm-1 O-H carboxylic acids medium to broad 2500 cm-1 to 3500 cm-1 C=O ketones, aldehydes strong and sharp 1600 cm-1 to 1750 cm-1 esters and acids

  33. INFRA-RED SPECTROSCOPY IDENTIFYING ALCOHOLS USING INFRA RED SPECTROSCOPY DifferentiationCompound O-H C=O ALCOHOLYESNO ALDEHYDE / KETONENOYES CARBOXYLIC ACIDYESYES ESTERNOYES ALCOHOL CARBOXYLIC ACID ESTER BUTAN-1-OL PROPANOIC ACID ETHYL ETHANOATE O-H absorption O-H + C=O absorption C=O absorption

  34. INDUSTRIAL PREPARATION OF ALCOHOLS FERMENTATION Reagent(s) GLUCOSE - produced by the hydrolysis of starch Conditions yeast warm, but no higher than 37°C EquationC6H12O6 ——> 2 C2H5OH + 2 CO2 Advantages LOW ENERGY PROCESS USES RENEWABLE RESOURCES - PLANT MATERIAL SIMPLE EQUIPMENT Disadvantages SLOW PRODUCES IMPURE ETHANOL BATCH PROCESS

  35. INDUSTRIAL PREPARATION OF ALCOHOLS HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditions catalyst - phosphoric acid high temperature and pressure EquationC2H4+ H2O ——> C2H5OH

  36. INDUSTRIAL PREPARATION OF ALCOHOLS HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditions catalyst - phosphoric acid high temperature and pressure Equation C2H4 + H2O ——> C2H5OH Advantages FAST PURE ETHANOL PRODUCED CONTINUOUS PROCESS Disadvantages HIGH ENERGY PROCESS EXPENSIVE PLANT REQUIRED USES NON-RENEWABLE FOSSIL FUELS TO MAKE ETHENE Uses of ethanol ALCOHOLIC DRINKS SOLVENT - industrial alcohol / methylated spirits FUEL - petrol substitute in countries with limited oil reserves

  37. USES OF ALCOHOLS ETHANOL DRINKS SOLVENT industrial alcohol / methylated spirits (methanol is added) FUEL used as a petrol substitute in countries with limited oil reserves METHANOL PETROL ADDITIVE improves combustion properties of unleaded petrol SOLVENT RAW MATERIAL used as a feedstock for important industrial processes FUEL Health warning Methanol is highly toxic

  38. LABORATORY PREPARATION OF ALCOHOLS from haloalkanes - reflux with aqueous sodium or potassium hydroxide from aldehydes - reduction with sodium tetrahydridoborate(III) - NaBH4 from alkenes - acid catalysed hydration using concentrated sulphuric acid Details of the reactions may be found in other sections.

  39. REVISION CHECK What should you be able to do? Recallandexplain the physical properties of alcohols Recall the different structural types of alcohols Recall the Lewis base properties of alcohols Recallandexplain the chemical reactions of alcohols Write balanced equations representing any reactions in the section Understand how oxidation is affected by structure Recall how conditions and apparatus influence the products of oxidation Explain how infrared spectroscopy can be used to differentiate between functional groups

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