Exercise practice: Aldehydes, Ketones and Carboxylic Acids

4.1 Exercise 8.1 and 8.2: terms and names

Cyanohydrin is the product of hydrogen cyanide addition to a carbonyl compound, for example CH3CH(OH)CN from ethanal. Acetal is the product formed when an aldehyde reacts with two molecules of an alcohol in dry HCl. Semicarbazone is the condensation product with semicarbazide, and oxime is the product with hydroxylamine. Aldol is the beta hydroxy aldehyde or ketone formed from two carbonyl molecules with dilute base. Hemiacetal has one alkoxy group and one OH on the same carbon. Ketal is the acetal of a ketone. Imine and Schiff base have a carbon nitrogen double bond from condensation with ammonia or a primary amine. A 2,4-DNP derivative is the orange or red precipitate with 2,4-dinitrophenylhydrazine, used as a test.

8.2(i) CH3CH(CH3)CH2CH2CHO is 3-methylbutanal; the CHO carbon is C1, so the methyl sits at C3. 8.2(ii) CH3CH2COCH(C2H5)CH2CH2Cl has the ketone at C3 and is named 1 chloro 3 ethyl 5 methylhexan-3-one after choosing the chain that gives the carbonyl the lowest locant. 8.2(iii) CH3CH=CHCHO is but-2-enal. 8.2(iv) CH3COCH2COCH3 is pentane-2,4-dione. 8.2(vi) (CH3)3CCH2COOH is 3,3-dimethylbutanoic acid.

4.2 Exercise 8.4 and 8.5: derivatives

8.4(i) CH3CO(CH2)4CH3 is heptan-2-one; the common name is methyl pentyl ketone. 8.4(ii) the compound with a bromine and a methyl branch is named after the aldehyde end, so the CHO carbon is C1 and the name is 3 bromo 5 methylhexanal. 8.4(vi) PhCOPh is diphenylmethanone, common name benzophenone.

A 2,4-DNP derivative forms from any aldehyde or ketone by condensation with 2,4-dinitrophenylhydrazine, losing water. Semicarbazone forms from the same carbonyl with semicarbazide, oxime with hydroxylamine and hydrazone with hydrazine. Acetal forms only from an aldehyde with an alcohol and dry HCl. The structure of each derivative keeps the carbon skeleton and replaces the C=O oxygen by the nitrogen group.

4.3 Exercise 8.6 and 8.7: distinguishing tests

8.6 To tell propanal from propanone, use Tollens reagent. Propanal gives a silver mirror and propanone does not. Fehling solution also works, giving a red precipitate with propanal only. Iodoform does not help here because neither is a methyl ketone.

8.7 To tell ethanal from propanal, the iodoform test is enough: ethanal has a methyl group next to the carbonyl and gives a yellow precipitate, while propanal does not. Another route is oxidation: ethanal gives ethanoic acid and propanal gives propanoic acid, which can be identified by a derivative or by the acid test.

4.4 Exercise 8.8 and 8.9: reactivity and acidity orders

8.8 Arrange in increasing reactivity towards nucleophilic addition. The rule is steric crowding first, then electron release. Methanal is most reactive, then ethanal, then propanone, then the aromatic aldehyde such as benzaldehyde is lower than the aliphatic ones because of conjugation. Between butanone and benzaldehyde the ketone has two alkyl groups but no ring conjugation, so benzaldehyde usually comes lower than an aliphatic ketone of similar size.

8.9 Carboxylic acids in increasing acid strength: propanoic acid, then ethanoic acid, then methanoic acid, then chloroethanoic acid, then trichloroethanoic acid. Alkyl groups weaken the acid, and the chlorine groups withdraw electron density and stabilise the carboxylate.

4.5 Exercise 8.10 and 8.11: named reactions

8.10 Write the reactions of the following. (a) Benzaldehyde with concentrated NaOH gives the Cannizzaro products, benzyl alcohol and sodium benzoate. (b) Propanone with dilute NaOH gives the aldol product, 4 hydroxy 4 methylpentan-2-one, and on heating it gives 4 methylpent-3-en-2-one. (c) Ethanal with sodium hydroxide gives aldol and then crotonaldehyde on heating.

8.11 Complete the reactions. (a) An acid chloride with H2 over palladium on barium sulphate gives the aldehyde, the Rosenmund reaction. (b) An alkene with ozone followed by zinc and water is cleaved to two carbonyl compounds. (c) A ketone with zinc amalgam and concentrated HCl gives the hydrocarbon, the Clemmensen reduction.

4.6 Exercise 8.12 and 8.13: conversions

Ethanol to propanone. Oxidise ethanol to ethanoic acid, convert the acid to its calcium salt and heat it to get propanone; the salt decomposes to the ketone and calcium carbonate. Ethanal to propan-2-ol. Reduce ethanal with sodium borohydride, or add methylmagnesium iodide followed by hydrolysis. Benzaldehyde to acetophenone. Convert benzaldehyde to the nitrile through the oxime and dehydration, then react the nitrile with methylmagnesium iodide and hydrolyse. Benzene to m-nitrobenzaldehyde. Nitrate benzene to nitrobenzene, use a Friedel Crafts or oxidation route to the substituted benzaldehyde, and control the orientation with the meta directing nitro group.

4.7 Exercise 8.14 and 8.15: mechanism and dicarboxylic acids

Mechanism of nucleophilic addition. Step 1: the nucleophile attacks the carbonyl carbon, the pi bond breaks and a tetrahedral alkoxide forms. Step 2: protonation gives the addition product. In acidic medium the carbonyl oxygen is first protonated, which makes the carbon more electrophilic and speeds the attack.

Dicarboxylic acids. Ethanedioic acid, propanedioic acid and butanedioic acid are the first three. On heating, ethanedioic acid decarboxylates to methanoic acid and carbon dioxide, propanedioic acid gives ethanoic acid and carbon dioxide, and butanedioic acid cyclises by losing water to give the cyclic anhydride.