Question
Map the major reactions of aldehydes and ketones via nucleophilic addition. What determines the reactivity order, and how do different nucleophiles give different products?
(JEE Main, NEET, CBSE 12 — carbonyl reactions dominate organic chemistry questions)
Solution — Step by Step
The carbonyl group () has a polarised double bond — carbon is electrophilic () and oxygen is nucleophilic (). Nucleophiles attack the carbon, breaking the pi bond, and oxygen accepts the electrons.
Aldehydes are more reactive than ketones because: (a) less steric hindrance at carbonyl carbon, and (b) fewer electron-donating alkyl groups (less stabilisation of the electrophilic carbon).
| Nucleophile | Product | Reaction Name |
|---|---|---|
| HCN | Cyanohydrin (R-CH(OH)(CN)) | Cyanohydrin formation |
| NaHSO3 | Bisulphite addition product | Bisulphite addition |
| RMgX (Grignard) | Alcohol (after hydrolysis) | Grignard reaction |
| NH2OH | Oxime (C=N-OH) | Oximation |
| NH2-NH2 | Hydrazone (C=N-NH2) | Hydrazone formation |
| 2,4-DNP | 2,4-dinitrophenylhydrazone | Confirmatory test for C=O |
| NH2-NH-CO-NH2 | Semicarbazone | Semicarbazone formation |
Aldol condensation: Two aldehyde molecules combine in the presence of dilute NaOH — one molecule acts as nucleophile (alpha carbon), the other as electrophile (carbonyl carbon). Product: beta-hydroxy aldehyde.
Cannizzaro reaction: Non-enolisable aldehydes (like HCHO, C6H5CHO) undergo disproportionation with strong NaOH — one molecule is oxidised to acid, the other is reduced to alcohol.
Clemmensen reduction: using (acidic conditions).
Wolff-Kishner reduction: using (basic conditions).
graph TD
A["Carbonyl C=O"] --> B["HCN: Cyanohydrin"]
A --> C["RMgX: Alcohol"]
A --> D["NH2OH: Oxime"]
A --> E["2,4-DNP: Hydrazone"]
A --> F["NaOH dil: Aldol"]
A --> G["NaOH conc: Cannizzaro"]
A --> H["Zn-Hg/HCl: Clemmensen"]
A --> I["NH2NH2/KOH: Wolff-Kishner"]Why This Works
All these reactions share the same first step — nucleophilic attack on the electrophilic carbonyl carbon. The product depends on which nucleophile attacks and whether the reaction stops at addition or continues to elimination (forming C=N derivatives with amines).
The reactivity order of carbonyls: HCHO > other aldehydes > ketones. Formaldehyde is the most reactive because it has no alkyl groups (zero steric hindrance, zero electron donation). This order is tested directly in JEE.
Alternative Method
For JEE, remember the Clemmensen vs Wolff-Kishner choice: both reduce C=O to CH2, but Clemmensen uses acidic conditions (Zn-Hg/HCl) and Wolff-Kishner uses basic conditions (NH2NH2/KOH). Choose based on other functional groups present — if the molecule has an acid-sensitive group, use Wolff-Kishner, and vice versa.
Common Mistake
The biggest error: applying Cannizzaro reaction to aldehydes that have alpha-hydrogens. Cannizzaro works only on aldehydes with no alpha-hydrogen (HCHO, benzaldehyde, trimethylacetaldehyde). If alpha-hydrogens are present, the aldehyde undergoes aldol condensation instead. JEE frequently gives an aldehyde and asks “what happens with concentrated NaOH?” — check for alpha-hydrogens first.
Also, the 2,4-DNP test gives a yellow/orange precipitate with aldehydes and ketones — it is a confirmatory test for the carbonyl group. It does NOT distinguish between aldehydes and ketones. For that distinction, use Tollen’s test (silver mirror — aldehydes only) or Fehling’s test (red precipitate — aldehydes only).