1. 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 THE CHEMISTRY OF ALCOHOLS

2. 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 THE CHEMISTRY OF ALCOHOLS

3. 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

4. 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.

5. 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°

6. 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

7. 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

8. 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

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. 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

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. 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

11. Boiling temperature comparision Alcohol  london forces and Hydrogen bonding Halagenoalkane  C-F bonds stronger than C-C Alkenes/alkanes  only london forces Boiling Temp. No. of carbon atoms

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

13. 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

14. 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 Note This is potentially an extremely dangerous preparation because of the close proximity of the very hot concentrated sulphuric acid and the sodium hydroxide solution.

15. ELIMINATION OF WATER (DEHYDRATION) Reagent/catalyst Aluminium Oxide ( Al 2 O 3 ) Conditions Heat Product alkene Equation e.g. C 2 H 5 OH(l) ——> CH 2 = CH 2 (g) + H 2 O(l)  This is a simple way of making gaseous alkenes like ethene. If ethanol vapour is passed over heated aluminium oxide powder, the ethanol is essentially cracked to give ethene and water vapour.

16. 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...

17. 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°

18. OXIDATION OF PRIMARY ALCOHOLS Primary alcohols are easily oxidised to aldehydes e.g. CH 3 CH 2 OH(l) CH 3 CHO(l) + H 2 O(l) ethanol ethanal It is essential to distil off the aldehyde before it gets oxidised to the (Carboxilic) acid CH 3 CHO CH 3 COOH ethanal ethanoic acid Practical details 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 K 2 Cr 2 O 7 + DIL. H 2 SO 4 Heat under distillation Note : K 2 Cr 2 O 7  Oxidisong agent K 2 Cr 2 O 7 + DIL. H 2 SO 4

19. OXIDATION OF SECONDARY ALCOHOLS Secondary alcohols are easily oxidised to ketones CH 3 CHOHCH 3 (l) CH 3 COCH 3 (l) + H 2 O(l) propan-2-ol propanone  Prolonged treatment with oxidising agent with secondary alcohol does not produce carboxylic acid OXIDATION OF TERTIARY ALCOHOLS Tertiary alcohols are resistant to normal oxidation K 2 Cr 2 O 7 + DIL. H 2 SO 4 Heat under reflux (CH 3 ) 3 COH NO REACTION K 2 Cr 2 O 7 + DIL. H 2 SO 4 Heat under reflux/distillation

21. OXIDATION OF PRIMARY ALCOHOLS Aldehyde has a lower boiling point so distils off before being oxidised further PRIMARY ALCOHOLS OXIDATION TO ALDEHYDES DISTILLATION

22. SECONDARY ALCOHOLS OXIDATION TO CARBOXYLIC ACIDS REFLUX Aldehyde condenses back into the mixture and gets oxidised to the acid OXIDATION OF SECONDARY ALCOHOLS Reflux condenser prevents volatile compounds like ethanal, ethanol and ethanoic acid vapours from leaving the flask Note : volatile compounds  Chemical compounds that transition to gas at low temperatures

23. 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.

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. 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°

25. OTHER REACTIONS OF ALCOHOLS  Reactions involving Oxygen  Reactions involving Sodium  Reactions involving hydrogen halides

26. OTHER REACTIONS OF ALCOHOLS OXYGEN 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 2CH 3 CH 2 OH(l) + 2Na(s) 2CH 3 CH 2 O¯ Na + + H 2 (g) sodium ethoxide room temperature Ignite/spark Alkoxides are white, ionic crystalline solids e.g. CH 3 CH 2 O¯ Na + * Observation : Bubbles will form, Sodium dissolves, White solid formed 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.

28. The general reaction looks like this:  Reactions involving hydrogen halides ROH + HX RX + H 2 O

29. CHLORINATION OF ALCOHOLS When PCl 5 is added to dry alcohol, clods of hydrogen cloride fumes are produced CH 3 CH 2 OH + PCl 5 CH 3 CH 2 Cl + POCl 3 + HCl (g) Hydrogen chloride test Hydrogen chloride gas forms a white smoke with ammonia. BROMINATION OF ALCOHOLS C 2 H 5 OH + HBr C 2 H 5 Br + H 2 O Dry conditions Room temp NaBr / KBr + 50% CONC. H 2 SO 4 Heat under reflux 3C 2 H 5 OH + PBr 3 3C 2 H 5 Br + H 3 PO 3 Moist red Phosperous + Br 2 KBr + H 2 SO 4 ---> KHSO 4 + HBr Heat under reflux

30. IODINATION OF ALCOHOLS 3C 2 H 5 OH + PI 3 3C 2 H 5 I + H 3 PO 3 Moist red Phosperous + I 2 In this case the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H 3 PO 4, and the iodoalkane is distilled off. The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide which reacts with the alcohol. Phosphoric(V) acid is used instead of concentrated sulphuric acid because sulphuric acid oxidises iodide ions to iodine and produces hardly any hydrogen iodide. A similar thing happens to some extent with bromide ions in the preparation of bromoalkanes, but not enough to get in the way of the main reaction. There is no reason why you couldn't use phosphoric(V) acid in the bromide case instead of sulphuric acid if you wanted to. C 2 H 5 OH + HI C 2 H 5 I + H 2 O NaI / KI + CONC. H 3 PO 4 H 3 PO 4 + KI ----> KH 2 PO 4 + HI Heat under reflux

31. SECONDARY ALCOHOLS WITH HALOGENS + PCl 5 Butane-2-ol + PCl 5  2-cloro-butane

32. Tertiary alcohols react reasonably rapidly with concentrated hydrochloric acid, but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance. A tertiary alcohol reacts if it is shaken with with concentrated hydrochloric acid at room temperature. A tertiary halogenoalkane (haloalkane or alkyl halide) is formed TERTIARY ALCOHOLS WITH HALOGENS

33. Haloalkanes can be prepared from the vigorous reaction between cold alcohols and phosphorus(III) halides SUMMARY

34. QUESTION TIME

35.  Propanoic acid may be prepared by oxidizing propan-1-ol in acidic conditions. X + H2SO4 CH3CH2CH2OH CH3CH2COOH Procedure 1. Pour 10 cm of distilled water into a boiling tube and add 12 g of oxidizing agent X. Shake the mixture and leave X to dissolve. 2. Pour 3 cm of propan-1-ol into a round-bottom flask and add 10 cm of distilled water and a few anti-bumping granules. Set up the apparatus for heating under reflux. 3. Add 4 cm of concentrated sulfuric acid, drop by drop, to the propan-1-ol. While the mixture is still warm, add the solution of oxidizing agent X, drop by drop. The energy released from the reaction should cause the mixture to boil without external heating. 4. When all of the solution of X has been added, use a low Bunsen burner flame to keep the mixture boiling for 10 minutes, not allowing any vapour to escape. 5. Distil the mixture in the flask using the apparatus shown below. Collect 5–6 cm of distillate, which is an aqueous solution of propanoic acid.

36. (a)Suggest, by name or formula, a suitable oxidising agent Potassium dichromate ((VI))/ K2Cr2O7 Sodium dichromate ((VI))/ Na2Cr2O7 (b) What colour change does X undergo when it oxidizes propan-1-ol? From Orange to green (c) Draw a labelled diagram showing the apparatus for heating under reflux. (d) Give two reasons why the escape of vapour in step 4 should be prevented. Reason 1 Yield would be reduced/reactants and or products would be lost complete oxidation could not occur Reason 2 Vapour is flammable/toxic/ hazardous/harmful/ acidic/irritant

37. (e) How does the reflux apparatus prevent escape of vapour? Mixture being heated returns to the flask The vapour is (cooled and) condensed The water in the condenser is cold (and flowing) (f) Some water can be removed from the distillate in step 5 by adding a solid drying agent. The solution of propanoic acid can then be decanted leaving the drying agent behind. (i)Suggest a suitable solid drying agent. (Anhydrous) calcium chloride/ (Anhydrous) magnesium sulfate/ (Anhydrous) sodium sulfate Silica gel

38. (ii) Suggest why removing excess solid drying agent by decanting, rather than filtering through filter paper, improves the yield. Some product is absorbed BY/ INTO filter paper (g) In a larger scale preparation of propanoic acid, 10.0 g of propan-1-ol was used. (i)Calculate the maximum mass of propanoic acid which could be formed from 10.0 g of propan-1-ol. Propan-1-ol Molar mass = 60.1 Propanoic acid Molar mass =74.1 Mol propanol = (10/60.1) = 0.166/ 0.17 = (mol propanoic acid) Mass propanoic acid = (0.166 x 74.1) = 12.32945 = 12.33/12.3 (g) If 0.17 mol then 12.597/12.6 (g) 8.1066 (g)

39. (ii) After purification, 6.0 cm of dry propanoic acid was obtained. Calculate the percentage yield in the preparation. The density of propanoic acid is 0.99 g cm. Mass propanoic acid = 6 x 0.99 = 5.94 (g) % yield = (5.94/ 12.33)x100 = 48.17% = (h) In another experiment, the same reaction mixture (propan-1-ol, X and concentrated sulfuric acid) was heated in the apparatus shown in step 5. Identify the main organic product which would be collected and explain why propanoic acid is not produced. Product Propanal/CH 3 CH 2 CHO Explanation Product removed as formed/ Incomplete oxidation

40. This question is about the alcohol, propan-1-ol. (a)Give two observations when propan-1-ol reacts with a small piece of sodium. Observation 1 Sodium dissolves/disappears/gets smaller Observation 2 Bubbles/effervescence/fizzes (b) A student investigated the rate of reaction of propan-1-ol with sodium. Suggest one suitable measurement which could be made to determine the rate of this reaction. Measure volume of gas in fixed time/measure time to collect a volume of gas/measure time for sodium to dissolve

41. (c) A small amount of phosphorus(V) chloride (phosphorus pentachloride), PCl5, is added to propan-1-ol in a test tube. (i)Describe the appearance of the fumes at the mouth of the test tube. misty (fumes) white (fumes) (ii) An open bottle of concentrated ammonia is held near the mouth of the tube. Describe what would be seen at the mouth of the test tube. White smoke

42. 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

43. 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

44. 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

45. 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

46. INDUSTRIAL PREPARATION OF ALCOHOLS HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditionscatalyst - phosphoric acid (H 3 PO 4 ) high temperature and pressure EquationC 2 H 4 + H 2 O ——> C 2 H 5 OH

47. INDUSTRIAL PREPARATION OF ALCOHOLS HYDRATION OF ETHENE Reagent(s) ETHENE - from cracking of fractions from distilled crude oil Conditionscatalyst - phosphoric acid (H 3 PO 4 ) 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

48. 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

49. 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

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