Distinguishing between aldehydes and ketones

1. Distinguishing between aldehydes and ketones

2. Adehydes and ketones can be structural isomers of each other. Aldehydes are produced by the oxidation of a primary alcohol and have the C=O on end carbon. Ketones are produced by the oxidation of a secondary alcohol and have the C=O on a carbon atom in the middle of the carbon chain. Aldehydes can be further oxidised to carboxylic acids, while ketones are not oxidised further.

3. The tests we use to distinguish between aldehydes and ketones all involve oxidising the aldehyde but not the ketone. • While acidified dichromate or permanganate will distinguish between aldehydes and ketones, they are strong oxidising agents which will also change colour in the presence of alcohol or other reagents. • Oxidising agents which oxidise aldehydes are: • Tollen’s reagent • Benedict solution • Fehling’s solution

4. The ‘silver mirror’. You’re more likely to get a mirror with a very clean test tube.

5. Tollen’s reagent Tollen’s reagent is [Ag(NH3)2]+ which, when reduced, forms Ag(s). It must be freshly prepared. Silver nitrate solution A few drops of NaOH to form a precipitate. Add ammonia solution till the precipitate dissolves.

6. Add a few drops of the aldehyde or ketone, shake, and warm gently. The ketone remains colourless, the aldehyde will react. If you are lucky you will get a ‘silver mirror’ as elemental silver forms on the inside of the test tube. Less spectacular, but just as valid is the formation of a grey or black precipitate, also of elemental silver. A grey precipitate of silver.

7. Tollens’ Test • Tollens reagent is a complex of Ag(NH3)2+ . • When heated with an aldehyde a redox reaction occurs producing a silver mirror on the inner surface of the test tube. • The aldehyde is oxidised to a carboxylic acid. • The reduction half-equation is • Ag+(aq) + eAg(s) • If Tollens’ reagent is heated with a ketone or an alcohol no reaction occurs.

8. Tollens’ Test The overall reaction is : RCHO + 2Ag(NH3)2OH RCOONH4 + 2Ag + H2O + 3NH3 Silver mirror

9. Task – write the balanced redox reactions for the oxidation of propan-1-ol to propanal using Cr2072- /H+ and give all colour changes. (X3) 3 CH3CH2CH2OH 3 CH3CH2COH + 2H+ + 6H+ + 6e- + 2e- Cr2072- + 14H+ + 6e- 2Cr3+ + 7H20 Full balanced redox equation 3CH3CH2CH2OH + Cr2072- + 8H+ 3CH3CH2COH + 2Cr3+ + 7H20

10. Benedict solution Benedict solution is an alkaline solution of Cu2+, complexed with citrate ions to keep it in solution. It is a mild oxidising agent which is reduced to Cu+. In the alkaline solution the Cu+ is in the form of Cu2O which is a brick-red precipitate which is a positive test for an aldehyde.

11. Take about 2 mL of Benedict solution in each of two test tubes. Add a few drops of aldehyde to one tube, and ketone to the other tube, and shake to mix.

12. Heat the mixture by putting the tubes in hot water. Shake several times to mix.

13. A reaction has occurred in the left hand (aldehyde) tube, but not in the right hand (ketone) tube.

14. If you wait long enough you will see the red-brown precipitate of Cu2O form.

15. Benedicts and Fehlings Aldehydes reduce the copper (II) ions in both Fehlings and Benedicts solution to form a reddish brown copper (I) oxide which is precipitated The reaction is : RCHO + 2Cu 2+ + 2OH- RCOO- + Cu2O + 3H+ Brick red precipitate

16. Fehling’s solution Like Benedict solution, Fehling’s contains alkaline Cu2+, but Fehling’s uses potassium tartrate to complex the copper. The mixture is freshly prepared: Pour a little Fehling’s A solution into each test tube. Add the ‘B’ solution until a precipitate forms.

17. Keep adding ‘B’ solution until the precipitate has redissolved and the solution is a clear, dark blue.

18. A reaction occurs in the aldehyde tube as Cu2O forms. No reaction occurs in the ketone tube.

19. Add a few drops of aldehyde and ketone to separate tubes, shake, and heat in a beaker of hot water.

20. Glucose is an aldehyde and will form a mirror with Tollens and will also give a positive test with Benedicts and Fehlings solutions Do you remember doing this test for sugars in yr10 ?

21. In all of these reactions (Tollen’s, Benedict and Fehling’s), the aldehyde is oxidised to the carboxylic acid while no reaction occurs to the ketone.

22. Task – write the balanced redox reactions for the oxidation of the oxidation of propan-2-ol to propanone using Mn04- /H+ and give all colour changes. (X5) 5 CH3CHOHCH3 5 CH3COCH3 + 10H+ + 2H+ + 10e- + 2e- + 10e- (X2) 2MnO4- + Mn04- + 8H+ 16H+ + 5e- 2Mn2+ Mn2+ + 4H20 + 8H20 Full balanced redox equation 5CH3CHOHCH3 + 6H+ + 2Mn04- 5CH3COCH3 + 2Mn2+ +8H20

23. Producing an Aldehyde from a Primary Alcohol When forming the aldehyde (ethanal) from ethanol the alcohol and dichromate must be added to the hot concentrated H2SO4 We use a distillation technique to collect a volatile product

24. Producing an Aldehyde from a Primary Alcohol using distillation Dropping funnel with K2Cr2O7 and ethanol Distillation flask with hot H2SO4 The alcohol/dichromate mixture is added to the acid and only the first 2-3 mls of distillate is collected – why?

25. The alcohol/dichromate mixture is added to the acid and only the first 2-3 mls of distillate is collected – why? To make sure that the immediate production of ethanal with a lower boiling point of 21 deg C was vapourised and collected quickly and was not allowed to be converted to ethanoic acid.

26. To make sure a reaction goes to completion one such as: • Primary alcohol is completely oxidised to form a carboxylic acid • A secondary alcohol is oxidised to form a ketone • You would use a reflux arrangement

27. Making Aspirin – using methylsalicylate methylsalicylate Acetic anhydride Carboxylic acid Aspirin

28. Carboxylic Acids O R C OH Named with -oic on the end They are all organic acids which are weak acids

29. Reactions of Carboxylic Acids Carboxylic acids undergo 4 types of substitution reactions (of the –OH)

30. Reactions of Carboxylic Acids 1. Forming acid chlorides from carboxylic acids – reagentsare PCl5,PCl3 or SOCl2 acid chlorides are named at the end by the –oyl group Functional group of the acid chloride O R C Cl

31. Reactions of Acid Chlorides 2. Forming esters from acid chloride – reagents are a primary alcohol Ester From acyl chloride O + HCl R C O R’ From alcohol CH3COCl + CH3OH CH3COOCH3 + HCl ethanoyl chloride + methanol methyl ethanoate + hydrogen chloride

32. Reactions of Acyl Chlorides 3. acyl chlorides form amides – reagent ammonia and heat Functional group of the amide O R C NH2 CH3COCl + 2NH3 CH3CONH2 + NH4Cl ethanoyl chloride + ammonia ethanamide + ammonium chloride

33. Reactions of Acyl Chlorides 4. Acyl chlorides forming N - substituted amides – reagent amine O N substituted amide R C NH R’ O O CH3 + NH2 CH3 + HCl CH3 C C NH CH3 Cl Ethanoyl chloride + aminomethane N – methyl ethanamide + hydrochloric acid

34. Reactions of Acyl Chlorides Acyl chlorides react with water to form acidic solutions O O CH3 CH3 O H C + HCl C O H Cl H ethanoyl chloride + water ethanoic acid + hydrogen chloride

35. Making Acyl Chlorides Carboxylic acids react with PCl3, PCl5 or SOCl2 (not HCl) to form Acyl chlorides by substituting the –OH for a Cl. O O PCl5 CH3 CH3 C C Cl OH Ethanoic acid + PCl5 ethanoyl chloride

36. Lysergic acid diethylamide (LSD)

37. ESTERS Carboxylic acids react with alcohols, in the presence of conc sulfuric acid as a catalyst, to form esters. The reagents are heated together to bring about a reaction. Any excess acid is neutralised by the addition of sodium carbonate. If ethanoic acid is reacted with methanol the ester, methyl ethanoate is formed. H3O+ + H2O + CH3OH heat

38. Hydrolysis of esters The hydrolysis of an ester in aqueous solution results in the break up of the ester and the formation of an alcohol and the carboxylic acid or carboxylate ion *(depending on the pH of the solution). Hydrolysis in acid produces the alcohol + carboxylic acid CH3CH2COOCH3 +  CH3CH2COOH + CH3OH H2O / H+ methyl propanoate propanoic acid methanol

39. Hydrolysis of esters Hydrolysis in NaOH soln gives alcohol + the sodium salt of the carboxylic acid. CH3CH2COOCH3 + NaOH CH3CH2COONa+ + CH3OH methyl propanoate methanol sodium propanoate

40. hydrolysing an ester (methyl salicylate) OH Methyl salicylate C O CH3 O Write the products if we hydrolyse it in acid ie H2O/H+

41. hydrolysing an ester (methyl salicylate) if we hydrolyse it in acid ie H2O/H+ the products are OH + CH3OH C O H O methanol Salicylic acid

42. Write the products if we hydrolyse it in alkaline conditions ie H2O/NaOH OH + CH3OH C O – Na + O methanol sodium salicylate

43. Turn to the last page in the booklet Ester Hydrolysis read the method (the ester is methyl salicylate) Change the method as follows: Weigh out 4.8 grams of NaOH place in boiling flask then add 20mls of water (careful! it may get very hot) Measure out and add 5mls of oil of wintergreen (methyl salycilate) to flask Place boiling chips in boiling flask and reflux carefully for 30 mins

44. Reflux – heating mixture without losing volatile substances

45. Distillation – using the deferring boiling points of substances to separate them

46. Write equations using structural formulae for the formation of the following esters a) ethyl methanoate from ethanol and methanoic acid b) butyl propanoate from butan-1-ol and propanoic acid

47. complete the following scheme giving all Formula H+ Ethanoyl chloride Formula? CH3COOH + C2H5OH CH3COCl ethanamide CH3CONH2 NH3 / ethanol C2H5OH Reagents? salt + ethanol CH3COO- + C2H5OH OH- Ethyl ethanoate Formula? CH3COOC2H5 C2H5OH + Conc H2SO4 Reagents? Ethanoic acid CH3COOH N-ethylethanamide CH3CONH2C2H5 C2H5NH2 / ethanol

48. Fats and oils Fats and oils (lipids) are all triesters made from glycerol (propane-1,2,3-triol) and three long chain carboxylic acids (fatty acids) as shown below. Glycerol is an example of a”triol” which has three -OH groups present. Each of these can form an ester link with a different carboxylic acid, for example the fat called stearin. + R1COOH + R3COOH + R2COOH  + 3H2O 3 fatty acids Fat or oil Glycerol

49. Soap The three ester links present in these molecules can be broken (or hydrolysed) by heating with sodium hydroxide solution. This releases the original glycerol molecule plus the sodium salts of the long chain fatty acids which are soaps. This “saponification” process is shown in the diagram below. + R1COONa+ + R2COONa+ + R3COONa+ Heat + 3NaOH 3 soap molecules

50. Soaps work because the tail of the molecule is a long non-polar hydrocarbon chain (from the fatty acid) which readily dissolves grease and dirt (as “like dissolves like”). Then the ionic carboxylate ion readily dissolves in water (which is also polar) and is able to carry away the grease with it in the rinse water. non-polar hydrocarbon tail (dissolves grease) polar carboxylate head (dissolves in water)