Aldehydes, Ketones and Carboxylic AcidsNEET Chemistry · Class 12 · NCERT Chapter 8

Two factors work together. (1) Steric effect: in an aldehyde (R-CHO), the carbonyl carbon has one H and one alkyl group attached. A nucleophile can approach relatively easily. In a ketone (R-CO-R'), both substituents are bulky alkyl groups, which physically block the nucleophile from approaching the carbonyl carbon. (2) Electronic effect: alkyl groups donate electrons by induction (+I effect), pushing electron density onto the carbonyl carbon. This reduces the partial positive charge on C, making it less electrophilic and less attractive to nucleophiles. Ketones have two electron-donating alkyl groups; aldehydes have only one. The result: formaldehyde (HCHO) is most reactive (H + H), then aliphatic aldehydes (H + R), then aliphatic ketones (R + R), then aryl ketones (Ar + R, ArCO-R) least reactive among common carbonyls.

Tollens' test: uses Tollens' reagent (ammoniacal silver nitrate, Ag(NH3)2+). The aldehyde reduces Ag+ to metallic silver, which deposits as a silver mirror on the test tube. Both aliphatic AND aromatic aldehydes give a positive Tollens' test. Ketones do NOT react. Fehling's test: uses Fehling's solution (Fehling A = CuSO4, Fehling B = sodium potassium tartrate in NaOH). The aldehyde reduces Cu2+ (deep blue) to Cu+ as brick-red Cu2O precipitate. Only ALIPHATIC aldehydes (formaldehyde, acetaldehyde, propanal, etc.) give a positive Fehling test. Aromatic aldehydes (benzaldehyde, cinnamaldehyde) do NOT give Fehling's test — they are weaker reducing agents and cannot reduce the Cu2+ complex. Ketones give no positive Fehling test. Key distinction: Fehling distinguishes aliphatic from aromatic aldehydes; Tollens does not (both aryl and alkyl react).

Aldol condensation occurs when an aldehyde or ketone with at least one alpha-H (H on the carbon adjacent to C=O) reacts under dilute NaOH (base catalyst) or dilute acid catalyst. Step 1: base removes an alpha-H to form an enolate anion. Step 2: the enolate attacks the C=O carbon of another molecule nucleophilically. Step 3: protonation gives a beta-hydroxy aldehyde or ketone (the "aldol product" — named from having both ALDehyde and alcohOL groups). On gentle heating, the beta-hydroxy carbonyl undergoes dehydration to give an alpha,beta-unsaturated carbonyl compound (conjugated enone or enal). The reaction is very important in synthesis. If one of the two components has no alpha-H (e.g., benzaldehyde, formaldehyde), a crossed-aldol condensation gives only one product because only the alpha-H-containing partner can act as the nucleophile.

The Cannizzaro reaction is a disproportionation reaction — one molecule is oxidised while another is reduced. It occurs with aldehydes that have NO alpha-H, in the presence of concentrated NaOH. Examples: formaldehyde (HCHO), benzaldehyde (C6H5CHO), and trimethylacetaldehyde (pivaldehyde). Mechanism: NaOH adds to one aldehyde to form an alkoxide. This alkoxide transfers a hydride (H-) to the carbonyl carbon of a second aldehyde molecule (intermolecular hydride transfer). Products: one molecule is oxidised to carboxylate (salt), the other is reduced to alcohol. Example: 2 HCHO + NaOH → HCOONa + CH3OH. Why does it not happen with aldehydes that have alpha-H? Because those aldehydes preferentially undergo aldol condensation, which is much faster than Cannizzaro for alpha-H-bearing aldehydes.

Both are carboxylic acids. Their acidity depends on how stable the carboxylate anion (RCOO-) is after losing a proton. Formic acid (HCOOH, pKa 3.74) has no alkyl group — the carboxylate HCOO- has only H attached to the carbonyl carbon. No electron donation occurs. Acetic acid (CH3COOH, pKa 4.74) has a methyl group (-CH3) that donates electrons to the ring by induction (+I effect). This electron donation increases the electron density on the carboxylate oxygen, destabilising the negative charge slightly, making it slightly harder to form and making acetic acid weaker. In general: as alkyl chain length increases, acidity decreases (more electron donation). Formic acid is the most acidic among simple aliphatic carboxylic acids because it has the least electron donation.

The Hell-Volhard-Zelinsky (HVZ) reaction selectively halogenates the alpha position of a carboxylic acid. Reagents: Cl2 (or Br2) + a small amount of red phosphorus (catalyst). The red P reacts with Cl2 to form PCl3, which converts the carboxylic acid to an acyl chloride (more reactive toward enolisation). The acyl chloride enolises at the alpha carbon. Cl2 (or Br2) then halogenates the alpha carbon. The acyl chloride intermediate is hydrolysed back to give the alpha-halo carboxylic acid. Product: R-CHCl-COOH (2-chloroalkanoic acid) or R-CHBr-COOH. The reaction is important because it installs a halogen at the alpha position, which can then be displaced by -OH, -NH2, -CN, etc. to make alpha-substituted carboxylic acids.

Aldehydes, Ketones and Carboxylic Acids is one of the highest-weightage organic chapters in NEET Class 12 Chemistry. You can expect 3-6 questions every year. The most tested topics are: (1) Identification tests — Tollens vs Fehling vs Schiff, which aldehyde reacts with which. Know the exceptions (benzaldehyde: Tollens yes, Fehling no). (2) Aldol condensation — what product forms, conditions, and whether crossed-aldol is clean or gives a mixture. (3) Cannizzaro reaction — recognise aldehydes with no alpha-H; know products are one acid + one alcohol. (4) Acidity order — formic acid most acidic among simple acids; EWG on alpha-C increases acidity; remember the chloroacetic acid series. (5) Named reactions — Clemmensen, Wolff-Kishner, Rosenmund, HVZ. Know conditions and products precisely.