NCERT Solutions: Class 12 Chemistry Chapter 8 - Aldehydes, Ketones and Carboxylic Acids

Examples: CH3CHO + HCN → CH3CH(OH)CN, a cyanohydrin. CH3CHO + 2C2H5OH/H+ → CH3CH(OC2H5)2, an acetal. Aldehyde/ketone + NH2NHCONH2 → semicarbazone. 2CH3CHO/dilute base → CH3CH(OH)CH2CHO, an aldol. CH3CHO + CH3OH ⇌ CH3CH(OH)OCH3, a hemiacetal. CH3CHO + NH2OH → CH3CH=NOH, an oxime. Ketone + glycol/H+ → ketal. Aldehyde/ketone + NH3 or primary amine gives imine/Schiff base. Carbonyl compound + 2,4-dinitrophenylhydrazine gives the 2,4-DNP hydrazone.

Number aldehydes and carboxylic acids from the functional carbon. For ketones, choose the longest chain containing the carbonyl group and give the carbonyl carbon the lowest locant; then add substituents alphabetically with locants.

Construct each parent chain or benzene derivative from the name, then place the aldehyde, ketone or carboxylic acid group at its suffix locant. Prefixes such as p-nitro, p-methyl, bromo, chloro and hydroxy are then placed at the indicated positions.

Hydrazones and semicarbazones replace the carbonyl oxygen by =NNHAr or =NNHCONH2. Oximes have =NOH. Acetals/ketals replace C=O by two –OR groups; cyclic ethylene ketals use –OCH2CH2O–. A hemiacetal has one –OH and one –OR on the former carbonyl carbon.

PhMgBr adds phenyl to the aldehyde carbonyl and hydrolysis gives a secondary alcohol. Tollens' reagent oxidises aldehyde to acid. Semicarbazide condenses at the carbonyl group. Excess ethanol in acid gives an acetal. Clemmensen conditions reduce the aldehyde group to –CH3.

Aldol condensation requires at least one alpha hydrogen in an aldehyde or ketone. Cannizzaro reaction is shown by aldehydes without alpha hydrogen in concentrated alkali. Ketones without alpha hydrogen and alcohols do not fit either reaction. Cannizzaro products are the corresponding alcohol and carboxylate salt; aldol products are beta-hydroxy carbonyl compounds that can dehydrate on heating.

2CH3CHO →(dilute NaOH) CH3CH(OH)CH2CHO; reduction gives CH3CH(OH)CH2CH2OH, butane-1,3-diol. Heating the aldol gives CH3CH=CHCHO, but-2-enal. Oxidation of but-2-enal gives CH3CH=CHCOOH, but-2-enoic acid.

Formation of 2,4-DNP derivative and reduction of Tollens' reagent show an aldehyde group. Cannizzaro reaction shows absence of alpha hydrogen at the aldehyde carbon, consistent with an aromatic aldehyde. Vigorous oxidation gives 1,2-benzenedicarboxylic acid, so the side chain and –CHO are ortho. Formula C9H10O fits o-C2H5C6H4CHO.

CH3CH2CH2COOCH2CH2CH2CH3 + H2O/H+ → CH3CH2CH2COOH + CH3CH2CH2CH2OH. CH3CH2CH2CH2OH + 2[O] → CH3CH2CH2COOH + H2O. CH3CH2CH2CH2OH →(H2SO4, heat) CH2=CHCH2CH3 + H2O.

Nucleophilic addition to carbonyl increases when steric hindrance and electron donation decrease; aldehydes are more reactive than ketones. Acid strength rises with electron-withdrawing groups and with proximity of the withdrawing group to –COOH; electron-donating groups lower acidity.

(i) Propanal gives Tollens' silver mirror; propanone does not. (ii) Acetophenone gives positive iodoform test; benzophenone does not. (iii) Benzoic acid gives brisk effervescence with NaHCO3; phenol does not. (iv) Benzoic acid gives NaHCO3 effervescence; ethyl benzoate does not. (v) Pentan-2-one gives iodoform test; pentan-3-one does not. (vi) Benzaldehyde gives Tollens' test; acetophenone does not. (vii) Ethanal gives iodoform test; propanal does not.

(i) Acetylation: introduction of an acetyl group using acetyl chloride or acetic anhydride, e.g. alcohol + (CH3CO)2O → acetate ester. (ii) Cannizzaro: 2HCHO + NaOH → CH3OH + HCOONa. (iii) Cross aldol: CH3CHO + C6H5CHO/dilute NaOH → cinnamaldehyde after dehydration. (iv) Decarboxylation: CH3COONa + NaOH/CaO, heat → CH4 + Na2CO3.

(i) Cyanide attacks the carbonyl carbon; methyl groups near the carbonyl block approach in 2,2,6-trimethylcyclohexanone. (ii) The –NH2 attached directly to carbonyl in semicarbazide has its lone pair delocalised with C=O and is less nucleophilic; the terminal –NH2 reacts. (iii) By Le Chatelier's principle, removing water or ester shifts the equilibrium toward ester formation.

Oxygen percentage = 100 - 69.77 - 11.63 = 18.60. Moles: C = 69.77/12 = 5.81, H = 11.63/1 = 11.63, O = 18.60/16 = 1.16. Ratio ≈ C5H10O; empirical mass = 86, equal to molecular mass, so formula is C5H10O. It is not an aldehyde, gives bisulphite addition and positive iodoform test, so it is a methyl ketone. Oxidation to ethanoic and propanoic acids fits CH3COCH2CH2CH3.

In phenoxide ion, some resonance structures place negative charge on ring carbon atoms, which are less electronegative than oxygen, and the resonance forms are not equivalent. In carboxylate ion, the two resonance structures are equivalent and the negative charge is shared by two oxygen atoms. Greater conjugate-base stabilisation makes carboxylic acids stronger acids than phenol.