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AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS KNOCKHARDY PUBLISHING.

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Presentation on theme: "AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS KNOCKHARDY PUBLISHING."— Presentation transcript:

1 AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS KNOCKHARDY PUBLISHING

2 INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard KNOCKHARDY PUBLISHING THE CHEMISTRY OF ALCOHOLS

3 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

4 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

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

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 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.

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. 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°

8 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

9 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

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 just van der Waals’ forces ethanolC 2 H 5 OH van der Waals’ forces + hydrogen bonding

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 just van der Waals’ forces ethanolC 2 H 5 OH van 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 methylbutan-2-ol (CH 3 ) 3 COH 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... 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 Alternative Method Pass vapour over a heated alumina (aluminium oxide) catalyst

15 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...

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 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°

17 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

18 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

19 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

20 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

21 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

22 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

23 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

24 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 +

25 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

26 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

27 INFRA-RED SPECTROSCOPY IDENTIFYING ALCOHOLS USING INFRA RED SPECTROSCOPY DifferentiationCompoundO-HC=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

28 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

29 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 ——> 2 C 2 H 5 OH

30 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 ——> 2 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

31 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

32 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.

33 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

34 You need to go over the relevant topic(s) again Click on the button to return to the menu

35 WELL DONE! Try some past paper questions

36 © JONATHAN HOPTON & KNOCKHARDY PUBLISHING THE THE END AN INTRODUCTION TO THE CHEMISTRY OF ALCOHOLS


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