Distinguishing between aldehydes and ketones

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.

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

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

Tollen’s reagent Tollen’s reagent is [Ag(NH 3 ) 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.

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.

Tollens’ Test Tollens reagent is a complex of Ag(NH 3 ) 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.

Tollens’ Test The overall reaction is : RCHO + 2Ag(NH 3 ) 2 OH RCOONH 4 + 2Ag + H 2 O + 3NH 3 Silver mirror

Task – write the balanced redox reactions for the oxidation of propan-1-ol to propanal using Cr /H+ and give all colour changes. CH 3 CH 2 CH 2 OH Cr CH 3 CH 2 COH + 2H + + 2e- 2Cr H H + + 6e- (X3)33+ 6H + + 6e- 3CH 3 CH 2 CH 2 OH + Cr Cr H H + 3CH 3 CH 2 COH Full balanced redox equation

Benedict solution Benedict solution is an alkaline solution of Cu 2+, 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 Cu 2 O which is a brick-red precipitate which is a positive test for an aldehyde.

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.

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

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

If you wait long enough you will see the red-brown precipitate of Cu 2 O form.

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 OH- RCOO- + Cu 2 O + 3H+ Brick red precipitate

Fehling’s solution Like Benedict solution, Fehling’s contains alkaline Cu 2+, 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.

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

A reaction occurs in the aldehyde tube as Cu 2 O forms. No reaction occurs in the ketone tube.

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

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 ?

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.

Task – write the balanced redox reactions for the oxidation of the oxidation of propan-2-ol to propanone using Mn0 4 - /H+ and give all colour changes. CH 3 CHOHCH 3 Mn0 4 - CH 3 COCH 3 + 2H + + 2e- Mn H H + + 5e- (X5)55+ 10H e- 5CH 3 CHOHCH 3 + 2Mn Mn 2+ +8H H + 5CH 3 COCH 3 Full balanced redox equation (X2) 2MnO H e- + 8H 2 02Mn 2+

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 H 2 SO 4 We use a distillation technique to collect a volatile product

Producing an Aldehyde from a Primary Alcohol using distillation Dropping funnel with K 2 Cr 2 O 7 and ethanol Distillation flask with hot H 2 SO 4 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. The alcohol/dichromate mixture is added to the acid and only the first 2-3 mls of distillate is collected – why?

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

Making Aspirin – using methylsalicylate methylsalicylate Acetic anhydride Aspirin Carboxylic acid

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

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

Reactions of Carboxylic Acids 1. Forming acid chlorides from carboxylic acids – reagents are PCl 5,PCl 3 or SOCl 2 acid chlorides are named at the end by the –oyl group Functional group of the acid chloride O Cl C R

Reactions of Acid Chlorides 2. Forming esters from acid chloride – reagents are a primary alcohol O O C R R’ From acyl chloride From alcohol + HCl CH 3 COCl + CH 3 OH CH 3 COOCH 3 + HCl ethanoyl chloride + methanol methyl ethanoate + hydrogen chloride Ester

Reactions of Acyl Chlorides 3. acyl chlorides form amides – reagent ammonia and heat O NH 2 C R Functional group of the amide CH 3 COCl + 2NH 3 CH 3 CONH 2 + NH 4 Cl ethanoyl chloride + ammonia ethanamide + ammonium chloride

4. Acyl chlorides forming N - substituted amides – reagent amine O NH C R N substituted amide Reactions of Acyl Chlorides R’ O Cl C CH 3 + NH 2 CH 3 O C + HCl CH 3 NH Ethanoyl chloride + aminomethane N – methyl ethanamide + hydrochloric acid

Acyl chlorides react with water to form acidic solutions Reactions of Acyl Chlorides O Cl C CH 3 H H O O C HO + HCl ethanoyl chloride + water ethanoic acid + hydrogen chloride

Carboxylic acids react with PCl 3, PCl 5 or SOCl 2 (not HCl) to form Acyl chlorides by substituting the –OH for a Cl. Making Acyl Chlorides O OH C CH 3 O C Cl Ethanoic acid + PCl 5 ethanoyl chloride PCl 5

Lysergic acid diethylamide (LSD)

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. + CH 3 OH H3O+H3O+ heat + H 2 O

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 CH 3 CH 2 COOCH 3 +  CH 3 CH 2 COOH + CH 3 OH H 2 O / H + methyl propanoate propanoic acid methanol

Hydrolysis of esters Hydrolysis in NaOH soln gives alcohol + the sodium salt of the carboxylic acid. CH 3 CH 2 COOCH 3 + NaOH  CH 3 CH 2 COO  Na + + CH 3 OH methyl propanoate methanol sodium propanoate

O CH 3 C O OH hydrolysing an ester (methyl salicylate) Methyl salicylate Write the products if we hydrolyse it in acid ie H 2 O/H+

O HC O hydrolysing an ester (methyl salicylate) Salicylic acid if we hydrolyse it in acid ie H 2 O/H+ the products are + CH 3 OH methanol

O – Na + C O OH Write the products if we hydrolyse it in alkaline conditions ie H 2 O/NaOH sodium salicylate + CH 3 OH methanol

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

Reflux – heating mixture without losing volatile substances

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

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

complete the following scheme giving all Formula Ethyl ethanoate Formula? Ethanoyl chloride Formula? Ethanoic acid CH 3 COOH CH 3 COCl CH 3 COOC 2 H 5 Reagents? C 2 H 5 OH + Conc H 2 SO 4 Reagents? H+ CH 3 COOH + C 2 H 5 OH OH- C 2 H 5 NH 2 / ethanol salt + ethanol CH 3 COO - + C 2 H 5 OH ethanamide CH 3 CONH 2 NH 3 / ethanol N-ethylethanamide CH 3 CONH 2 C 2 H 5

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. + R 1 COOH + R 3 COOH + R 2 COOH  Glycerol 3 fatty acids + 3H 2 O Fat or oil

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. + R 1 COO  Na + + R 2 COO  Na + + R 3 COO  Na + Heat 3 soap molecules + 3NaOH

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)

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bestchoice.net.nz is an excellent revision website run by the Auckland University- To get on to this site you must use the following details – go to new user on main page register using the following ID: 919 Password: compound Go on and try the organic section – lasts years class went nuts on this site Log on using your own username and the supplied password and ID