Aldehyde reactions map — oxidation, reduction, addition, condensation routes

Question

Starting from acetaldehyde (CH₃CHO), show the products formed by: (a) Tollens’ test, (b) reduction with NaBH₄, (c) aldol condensation, and (d) reaction with HCN. Write mechanisms where relevant.

(JEE Main / NEET pattern)

Solution — Step by Step

flowchart TD
    A["RCHO\n(Aldehyde)"] -->|"Tollens' reagent\n(Ag⁺/NH₃)"| B["RCOO⁻ + Ag mirror\n(Oxidation)"]
    A -->|"NaBH₄ or LiAlH₄"| C["RCH₂OH\n(Reduction)"]
    A -->|"HCN"| D["RCH(OH)CN\n(Cyanohydrin)"]
    A -->|"dil. NaOH\n(Aldol)"| E["β-hydroxy aldehyde\n(Aldol product)"]
    A -->|"NH₂OH"| F["RCH=NOH\n(Oxime)"]
    A -->|"2,4-DNP"| G["Orange ppt\n(2,4-DNP test)"]
    E -->|"Heat"| H["α,β-unsaturated aldehyde\n(Crotonaldehyde)"]

\text{CH}_3\text{CHO} + 2[\text{Ag(NH}_3)_2]^+ + 2\text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + 2\text{Ag} \downarrow + 3\text{NH}_3 + \text{H}_2\text{O}

The silver mirror on the test tube wall is the positive result. Aldehydes are oxidised to carboxylate ions; ketones do not react (this distinguishes them).

\text{CH}_3\text{CHO} \xrightarrow{\text{NaBH}_4} \text{CH}_3\text{CH}_2\text{OH}

NaBH₄ delivers hydride ( $\text{H}^-$ ) to the electrophilic carbonyl carbon. The result is a primary alcohol from an aldehyde (or a secondary alcohol from a ketone). NaBH₄ is selective — it reduces only the carbonyl and leaves C=C double bonds intact. LiAlH₄ is more powerful and reduces almost everything.

Two molecules of acetaldehyde react in dilute NaOH:

2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}

This beta-hydroxy aldehyde (3-hydroxybutanal) is the aldol product. On heating, it loses water to form crotonaldehyde ( $\text{CH}_3\text{CH=CHCHO}$ ), an alpha-beta unsaturated aldehyde.

The mechanism: NaOH generates the enolate from one molecule, which attacks the carbonyl of the second molecule (nucleophilic addition).

\text{CH}_3\text{CHO} + \text{HCN} \rightarrow \text{CH}_3\text{CH(OH)CN}

CN⁻ (nucleophile) attacks the carbonyl carbon, then protonation gives the cyanohydrin. This is a nucleophilic addition — the characteristic reaction of the carbonyl group.

The cyanohydrin can be hydrolysed to a hydroxy acid, making this useful in synthesis.

Why This Works

The carbonyl group ( $\text{C=O}$ ) has a partial positive charge on carbon (due to oxygen’s electronegativity). This makes aldehydes excellent electrophiles that react with a wide range of nucleophiles: $\text{H}^-$ (reduction), $\text{CN}^-$ (cyanohydrin), enolates (aldol), $\text{NH}_2\text{OH}$ (oxime), etc.

Oxidation works because aldehydes have an H on the carbonyl carbon that can be removed. Ketones lack this H, so they resist mild oxidation — this is why Tollens’ and Fehling’s tests work specifically for aldehydes.

Alternative Method — Identifying Aldehydes Systematically

Use a decision tree: positive 2,4-DNP test → carbonyl compound (could be aldehyde or ketone). Then positive Tollens’/Fehling’s → aldehyde confirmed. Positive iodoform → CH₃CO- group present.

In JEE, the most tested aldehyde reactions are: aldol condensation (including crossed aldol), Cannizzaro reaction (for aldehydes without alpha-H), and nucleophilic addition mechanism. For NEET, focus on the named tests — Tollens’, Fehling’s, 2,4-DNP, and iodoform.

Common Mistake

Students confuse which reagent is selective and which is not. NaBH₄ reduces only C=O (selective). LiAlH₄ reduces C=O, COOH, COOR, and even some C=C in conjugated systems (non-selective). If a question says “selective reduction of aldehyde in the presence of a double bond,” the answer is NaBH₄, not LiAlH₄. Getting this wrong means you pick the wrong product entirely.

Question

Solution — Step by Step

Why This Works

Alternative Method — Identifying Aldehydes Systematically

Common Mistake

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