1. Polymer Revolution Organic Functional Group

2. Learning Outcomes (d) recognise and write formulae for alkenes and use systematic nomenclature to name and interpret the names of alkenes; (e) recognise members of the following homologous series: aldehydes, ketones, carboxylic acids; (f) recall the difference between primary, secondary and tertiary alcohols from their structures and identify examples of them;

4. Answer Aldehyde

5. Starter Activity

6. CLASSIFICATION OF ALCOHOLS Aliphatic general formula C n H 2n+1 OH - provided there are no rings the OH replaces an H in a basic hydrocarbon skeleton

7. CLASSIFICATION OF ALCOHOLS Aliphatic general formula C n H 2n+1 OH - 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.

8. CLASSIFICATION OF ALCOHOLS Aliphatic general formula C n H 2n+1 OH - 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°

9. 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. CH 3 - CH(CH 3 ) - CH 2 - CH 2 - CH(OH) - CH 3 is called 5-methylhexan-2-ol NAMING ALCOHOLS

10. 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-olbutan-2-ol 2-methylpropan-1-ol2-methylpropan-2-ol

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. M r bp / °C propaneC 3 H 8 44 -42just van der Waals’ forces ethanolC 2 H 5 OH 46 +78van der Waals’ forces + hydrogen bonding

12. 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. M r bp / °C propaneC 3 H 8 44 -42just van der Waals’ forces ethanolC 2 H 5 OH 46 +78van der Waals’ forces + hydrogen bonding Boiling point is higher for “straight” chain isomers. bp / °C butan-1-ol CH 3 CH 2 CH 2 CH 2 OH118 butan-2-ol CH 3 CH 2 CH(OH)CH 3 100 2-methylpropan-2-ol (CH 3 ) 3 COH 83 Greater branching = lower inter-molecular forces

13. 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. M r bp / °C propaneC 3 H 8 44 -42just van der Waals’ forces ethanolC 2 H 5 OH 46 +78van der Waals’ forces + hydrogen bonding Boiling point is higher for “straight” chain isomers. bp / °C butan-1-ol CH 3 CH 2 CH 2 CH 2 OH118 butan-2-ol CH 3 CH 2 CH(OH)CH 3 100 2-methylpropan-2-ol (CH 3 ) 3 COH 83 Greater branching = lower inter-molecular forces

14. 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 Show the relevant lone pair(s) when drawing hydrogen bonding

15. CHEMICAL PROPERTIES OF ALCOHOLS The OXYGEN ATOM HAS TWO LONE PAIRS; this makes alcohols... BASES Lewis 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 NUCLEOPHILES Alcohols can use the lone pair to attack electron deficient centres

16. ELIMINATION OF WATER (DEHYDRATION) Reagent/catalyst conc. sulphuric acid (H 2 SO 4 ) or conc. phosphoric acid (H 3 PO 4 ) Conditions reflux at 180°C Product alkene Equation e.g. C 2 H 5 OH(l) ——> CH 2 = CH 2 (g) + H 2 O(l) Mechanism Step 1protonation of the alcohol using a lone pair on oxygen Step 2loss of a water molecule to generate a carbocation Step 3loss of a proton (H + ) to give the alkene Alternative Method Pass vapour over a heated alumina (aluminium oxide) catalyst

17. ELIMINATION OF WATER (DEHYDRATION) MECHANISM Step 1protonation of the alcohol using a lone pair on oxygen Step 2loss of a water molecule to generate a carbocation Step 3loss 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 2Alcohols 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...

18. 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 oxidised to 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°

19. OXIDATION OF PRIMARY ALCOHOLS Primary alcohols are easily oxidised to aldehydes e.g. CH 3 CH 2 OH(l) + [O] ——> CH 3 CHO(l) + H 2 O(l) ethanol ethanal it is essential to distil off the aldehyde before it gets oxidised to the acid CH 3 CHO(l) + [O] ——> CH 3 COOH(l) ethanal ethanoic acid

20. OXIDATION OF PRIMARY ALCOHOLS Primary alcohols are easily oxidised to aldehydes e.g. CH 3 CH 2 OH(l) + [O] ——> CH 3 CHO(l) + H 2 O(l) ethanol ethanal it is essential to distil off the aldehyde before it gets oxidised to the acid CH 3 CHO(l) + [O] ——> CH 3 COOH(l) ethanal ethanoic acid Practical details the alcohol is dripped into a warm solution of acidified K 2 Cr 2 O 7 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 formulaintermolecular bondingboiling point ETHANOL C 2 H 5 OHHYDROGEN BONDING 78°C ETHANAL CH 3 CHODIPOLE-DIPOLE 23°C ETHANOIC ACID CH 3 COOHHYDROGEN BONDING 118°C

21. OXIDATION OF PRIMARY ALCOHOLS Controlling the products e.g. CH 3 CH 2 OH(l) + [O] ——> CH 3 CHO(l) + H 2 O(l) then CH 3 CHO(l) + [O] ——> CH 3 COOH(l) Aldehyde has a lower boiling point so distils off before being oxidised further OXIDATION TO ALDEHYDES DISTILLATION OXIDATION TO CARBOXYLIC ACIDS REFLUX Aldehyde condenses back into the mixture and gets oxidised to the acid

22. OXIDATION OF SECONDARY ALCOHOLS Secondary alcohols are easily oxidised to ketones e.g. CH 3 CHOHCH 3 (l) + [O] ——> CH 3 COCH 3 (l) + H 2 O(l) propan-2-ol propanone The alcohol is refluxed with acidified K 2 Cr 2 O 7. 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.

23. OXIDATION OF SECONDARY ALCOHOLS Secondary alcohols are easily oxidised to ketones e.g. CH 3 CHOHCH 3 (l) + [O] ——> CH 3 COCH 3 (l) + H 2 O(l) propan-2-ol propanone The alcohol is refluxed with acidified K 2 Cr 2 O 7. 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

24. 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.

25. 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. H H R C O + [O] R C O + H 2 O HH 1°1°

26. 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. H H R C O + [O] R C O + H 2 O HH R C O + [O] R C O + H 2 O RR 1°1° 2°2°

27. 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. H H R C O + [O] R C O + H 2 O HH R C O + [O] R C O + H 2 O RR This is possible in 1° and 2° alcohols but not in 3° alcohols. 1°1° 2°2°

28. 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. H H R C O + [O] R C O + H 2 O HH R C O + [O] R C O + H 2 O RR R H R C O + [O] R This is possible in 1° and 2° alcohols but not in 3° alcohols. 1°1° 2°2° 3°3°

30. Answer:

31. ESTERIFICATION OF ALCOHOLS Reagent(s)carboxylic acid + strong acid catalyst (e.g conc. H 2 SO 4 ) Conditionsreflux Productester Equation e.g.CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) ethanol ethanoic acid ethyl ethanoate NotesConcentrated H 2 SO 4 is a dehydrating agent - it removes water causing the equilibrium to move to the right and increases the yield

32. ESTERIFICATION OF ALCOHOLS Reagent(s)carboxylic acid + strong acid catalyst (e.g conc. H 2 SO 4 ) Conditionsreflux Productester Equation e.g.CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) ethanol ethanoic acid ethyl ethanoate NotesConcentrated H 2 SO 4 is a dehydrating agent - it removes water causing the equilibrium to move to the right and increases the yield Uses of estersEsters are fairly unreactive but that doesn’t make them useless Used as flavourings Naming estersNamed from the alcohol and carboxylic acid which made them... CH 3 OH + CH 3 COOH CH 3 COOCH 3 + H 2 O from ethanoic acid CH 3 COOCH 3 from methanol METHYL ETHANOATE

33. OTHER REACTIONS OF ALCOHOLS OXYGENAlcohols make useful fuels C 2 H 5 OH(l) + 3O 2 (g) ———> 2CO 2 (g) + 3H 2 O(l) Advantages have high enthalpies of combustion do not contain sulphur so there is less pollution can be obtained from renewable resources

34. OTHER REACTIONS OF ALCOHOLS OXYGENAlcohols make useful fuels C 2 H 5 OH(l) + 3O 2 (g) ———> 2CO 2 (g) + 3H 2 O(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 Productsodium alkoxide and hydrogen Equation2CH 3 CH 2 OH(l) + 2Na(s) ——> 2CH 3 CH 2 O¯ Na + + H 2 (g) sodium ethoxide Notesalcohols 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. CH 3 CH 2 O¯ Na +

35. BROMINATION OF ALCOHOLS Reagent(s)conc. hydrobromic acid HBr(aq) or sodium (or potassium) bromide and concentrated sulphuric acid Conditionsreflux Producthaloalkane EquationC 2 H 5 OH(l) + conc. HBr(aq) ———> C 2 H 5 Br(l) + H 2 O(l) MechanismThe 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 1protonation of the alcohol using a lone pair on oxygen Step 2loss of a water molecule to generate a carbocation (carbonium ion) Step 3a bromide ion behaves as a nucleophile and attacks the carbocation

36. 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

37. INFRA-RED SPECTROSCOPY IDENTIFYING ALCOHOLS USING INFRA RED SPECTROSCOPY DifferentiationCompoundO-HC=O ALCOHOLYESNO ALDEHYDE / KETONENOYES CARBOXYLIC ACIDYESYES ESTERNOYES ALCOHOL ALDEHYDE CARBOXYLIC ACID PROPAN-1-OL PROPANAL PROPANOIC ACID O-H absorption C=O absorption O-H + C=O absorption

39. Answer:

40. INDUSTRIAL PREPARATION OF ALCOHOLS FERMENTATION Reagent(s)GLUCOSE - produced by the hydrolysis of starch Conditionsyeast warm, but no higher than 37°C EquationC 6 H 12 O 6 ——> 2 C 2 H 5 OH + 2 CO 2

41. INDUSTRIAL PREPARATION OF ALCOHOLS FERMENTATION Reagent(s)GLUCOSE - produced by the hydrolysis of starch Conditionsyeast warm, but no higher than 37°C EquationC 6 H 12 O 6 ——> 2 C 2 H 5 OH + 2 CO 2 AdvantagesLOW ENERGY PROCESS USES RENEWABLE RESOURCES - PLANT MATERIAL SIMPLE EQUIPMENT DisadvantagesSLOW PRODUCES IMPURE ETHANOL BATCH PROCESS

42. INDUSTRIAL PREPARATION OF ALCOHOLS HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditionscatalyst - phosphoric acid high temperature and pressure EquationC 2 H 4 + H 2 O ——> C 2 H 5 OH

43. INDUSTRIAL PREPARATION OF ALCOHOLS HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditionscatalyst - phosphoric acid high temperature and pressure EquationC 2 H 4 + H 2 O ——> C 2 H 5 OH AdvantagesFAST PURE ETHANOL PRODUCED CONTINUOUS PROCESS DisadvantagesHIGH ENERGY PROCESS EXPENSIVE PLANT REQUIRED USES NON-RENEWABLE FOSSIL FUELS TO MAKE ETHENE Uses of ethanolALCOHOLIC DRINKS SOLVENT - industrial alcohol / methylated spirits FUEL - petrol substitute in countries with limited oil reserves

44. 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

45. LABORATORY PREPARATION OF ALCOHOLS from haloalkanes- reflux with aqueous sodium or potassium hydroxide from aldehydes - reduction with sodium tetrahydridoborate(III) - NaBH 4 from alkenes- acid catalysed hydration using concentrated sulphuric acid Details of the reactions may be found in other sections.

46. REVISION CHECK What should you be able to do? Recall and explain the physical properties of alcohols Recall the different structural types of alcohols Recall the Lewis base properties of alcohols Recall and explain 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 CAN YOU DO ALL OF THESE? YES NO