Aldehydes and ketones are the most important carbonyl compounds studied at the NEET level, and the carbonyl group (C=O) is the most versatile functional group in organic chemistry. Understanding this single functional group unlocks reactions ranging from simple nucleophilic addition to complex multi-step named reactions.

Structure and Electronic Character

The carbonyl carbon is sp2-hybridized, meaning it forms three sigma bonds and one pi bond with oxygen. This makes the carbonyl carbon and its directly attached atoms coplanar (trigonal, 120° bond angles). Because oxygen is highly electronegative, the C=O bond is strongly polar: carbon carries a partial positive charge ($C^{delta+}$) and oxygen carries a partial negative charge ($O^{delta-}$). This makes the carbonyl carbon electrophilic — the defining feature that drives all nucleophilic addition reactions. The reactivity order in nucleophilic addition is: HCHO (formaldehyde, no alkyl groups, maximum electrophilicity, no steric hindrance) > other aldehydes (one alkyl group) > ketones (two alkyl groups). Alkyl groups both donate electrons (+I effect, reducing $C^{delta+}$) and create steric obstruction, both of which oppose nucleophilic attack.

Nucleophilic Addition Mechanism

The general mechanism has three steps: (1) the nucleophile (Nu-) attacks the electrophilic carbonyl carbon — this is the rate-determining step; (2) the C=O pi bond breaks, both electrons move to oxygen, and the carbon becomes sp3 (tetrahedral), giving an alkoxide intermediate; (3) the alkoxide is protonated to give the final addition product. This mechanism applies to all additions: HCN addition (gives cyanohydrin), Grignard addition (gives alcohols — primary from HCHO, secondary from RCHO, tertiary from R2CO), and hydride reductions (NaBH4, LiAlH4).

Reactions with NH3 Derivatives (Condensation)

Four important reagents react with both aldehydes and ketones via condensation (addition followed by elimination of water, forming a C=N bond): hydroxylamine (NH2OH) gives oximes (C=NOH); phenylhydrazine (C6H5NHNH2) gives phenylhydrazones (C=NNHC6H5); 2,4-dinitrophenylhydrazine (2,4-DNP) gives a characteristic orange or yellow precipitate used to detect any C=O group; semicarbazide (H2NCONHNH2) gives semicarbazones (C=NNHCONH2). These crystalline derivatives are used to characterize specific aldehydes and ketones.

Distinction Tests: Tollens', Fehling's, Iodoform

The 2,4-DNP test is positive for ALL carbonyl compounds (both aldehydes and ketones) and serves as the first step in carbonyl identification. Tollens' reagent (ammoniacal AgNO3) gives a silver mirror ONLY with aldehydes (RCHO + 2[Ag(NH3)2]+ + 2OH- → RCOO- + 2Ag + 4NH3 + H2O). Ketones do not react. Fehling's solution (alkaline Cu2+ tartrate) gives a brick-red precipitate of Cu2O ONLY with aldehydes; ketones do not react. Iodoform test $\frac{I2}{NaOH}$ gives a yellow CHI3 precipitate with methyl ketones (CH3CO-R), acetaldehyde (CH3CHO), and compounds that can be oxidized to these: ethanol is oxidized to acetaldehyde and isopropanol is oxidized to acetone in the reaction conditions.

Reduction Methods

Three fundamentally different reduction methods exist. NaBH4 (in methanol) and LiAlH4 (in dry ether) reduce C=O to C-OH (alcohol) — oxygen is retained as a hydroxyl group. These are the mild reductions. Clemmensen reduction (Zn-Hg amalgam / concentrated HCl, acidic, reflux) and Wolff-Kishner reduction (NH2NH2 / KOH in ethylene glycol, heat, basic) both completely remove the carbonyl oxygen, converting C=O to CH2 (methylene). The choice between Clemmensen and Wolff-Kishner depends on the stability of other functional groups: use Clemmensen for acid-tolerant substrates, Wolff-Kishner for base-tolerant/acid-sensitive substrates.

Aldol Condensation

Aldol condensation requires at least one alpha-hydrogen (H on the carbon adjacent to C=O). With dilute NaOH, the base abstracts an alpha-H to form an enolate (nucleophile), which attacks another carbonyl molecule, forming a new C-C bond. The product is a beta-hydroxy carbonyl compound (the "aldol"). Heating eliminates water to give an alpha,beta-unsaturated carbonyl compound. Classic example: 2 CH3CHO → CH3CH(OH)CH2CHO (3-hydroxybutanal) → CH3CH=CHCHO (2-butenal/crotonaldehyde) + H2O.

Cannizzaro Reaction

Aldehydes WITHOUT alpha-hydrogen undergo disproportionation (self-redox) with concentrated NaOH. One molecule is oxidized to the carboxylate salt and one is reduced to the alcohol. Key substrates: HCHO (2HCHO → HCOONa + CH3OH), benzaldehyde (2C6H5CHO → C6H5COONa + C6H5CH2OH), and pivaldehyde ((CH3)3CCHO). The critical exam point: if alpha-H is present, aldol occurs; if no alpha-H, Cannizzaro occurs.

Haloform Reaction

Methyl ketones (CH3COR) react with I2 + NaOH in three stages: (1) trihalogenation of the CH3 group: CH3CO- → CI3CO-; (2) cleavage of C-C bond by OH-: CI3CO-R + OH- → CHI3 + RCOO-. Yellow CHI3 (iodoform) precipitates. Stoichiometry: 3 I2 + 3 NaOH per methyl ketone. The test is also positive for ethanol (oxidized to CH3CHO) and isopropanol (oxidized to acetone).

Summary Table for NEET

The essential one-line rules: (i) 2,4-DNP = C=O detection (both); (ii) Tollens'/Fehling's = aldehyde only; (iii) NaBH4 = alcohol (C-OH); (iv) Clemmensen (acidic) or Wolff-Kishner (basic) = CH2; (v) Alpha-H → Aldol; No alpha-H → Cannizzaro; (vi) CH3CO- → iodoform positive; (vii) HCHO > aldehydes > ketones in reactivity.