Aldehydes, Ketones, and Carboxylic Acids

sources:

• text: Standard textbook reference

These carbonyl compounds are among the most important functional groups in organic chemistry. Aldehydes have $-\text{CHO}$ at the end of a carbon chain; ketones have $>\text{C}=\text{O}$ in the middle.

• Aldehyde: RCHO (formaldehyde, acetaldehyde, benzaldehyde)
• Ketone: RCOR’ (acetone, acetophenone)
• Nucleophilic addition: the key reaction of carbonyl compounds
• Electrophilic nature: C=O is polar (C is $\delta^+$, O is $\delta^-$)
• Aldol condensation: aldehyde with $\alpha$-H undergoes self-addition
• Cannizzaro reaction: aldehydes without $\alpha$-H undergo disproportionation
• Oxidation: aldehydes oxidize efficiently (Tollens’, Fehling’s); ketones resist oxidation
• Reduction: to primary alcohols (aldehydes) or secondary alcohols (ketones)
• Haloform reaction: methyl ketones ($\text{CH}_3\text{CO}-$) give $\text{CHX}_3$ with X$_2$/NaOH

Problem: Write the product of the reaction of acetaldehyde with HCN.

Solution:

HCN adds across the C=O bond:

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

Product: 2-hydroxypropanenitrile (acetaldehyde cyanohydrin)

Mechanism: CN$^-$ attacks the electrophilic carbonyl carbon, then protonation of O$^-$ gives the product.

Common mistake: Writing the product as $\text{CH}_3\text{CH(OH)COOH}$. HCN adds as H and CN, not H and COOH.

Problem: Write the aldol condensation product of acetaldehyde.

Solution:

Step 1: Enolization (base removes $\alpha$-H): $\text{CH}_3\text{CHO} + \text{OH}^- \rightarrow \text{CH}_2=\text{CHO}^- + \text{H}_2\text{O}$

Step 2: Nucleophilic addition: $\text{CH}_2=\text{CHO}^- + \text{CH}_3\text{CHO} \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CHO}$

Step 3: Dehydration (on heating): $\text{CH}_3\text{CH(OH)CH}_2\text{CHO} \rightarrow \text{CH}_3\text{CH}=\text{CHCHO} + \text{H}_2\text{O}$

Product: but-2-enal (crotonaldehyde), an $\alpha,\beta$-unsaturated aldehyde.

Common mistake: Forgetting the dehydration step. Aldol condensation gives the unsaturated product upon heating.

Problem: How does Tollens’ test distinguish between acetaldehyde and acetone?

Solution:

Tollens’ reagent: ammoniacal silver nitrate $[\text{Ag(NH}_3)_2]^+\text{OH}^-$

Acetaldehyde: Oxidizes to acetate, reducing Ag$^+$ to metallic silver: $\text{CH}_3\text{CHO} + 2[\text{Ag(NH}_3)_2]^+ + 3\text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + 2\text{Ag} \downarrow + 4\text{NH}_3 + 2\text{H}_2\text{O}$

A silver mirror forms on the test tube wall (positive test).

Acetone: Does not react (ketones resist oxidation). No silver mirror (negative test).

Common mistake: Thinking all carbonyl compounds give a positive Tollens’ test. Only aldehydes (and $\alpha$-hydroxy ketones) react.

1. Write the product of propanal with 2,4-DNP reagent.
2. What is the product of the Cannizzaro reaction of benzaldehyde?
3. How would you distinguish between ethanol and acetaldehyde using chemical tests?

Aldehydes and ketones are found in formaldehyde (preservative), acetone (solvent), vanillin (flavoring), and many pharmaceuticals. Their reactivity makes them essential intermediates in organic synthesis.

• Aldehydes reduce Tollens’ and Fehling’s reagents; ketones do not
• Aldol condensation requires $\alpha$-hydrogen
• Cannizzaro reaction: no $\alpha$-H, concentrated base
• 2,4-DNP test: orange/red precipitate indicates C=O group
• Haloform test: methyl ketones give yellow CHI$_3$ precipitate with I$_2$/NaOH

The carbonyl carbon is an electrophile’s dream target: The C=O bond is like a tug-of-war where oxygen always wins — pulling electron density away from carbon, leaving it electron-deficient. Nucleophiles rush in to fill this electron gap, which is why nucleophilic addition is the signature reaction of aldehydes and ketones.

Why it matters: Aldehydes and ketones appear ineverything from formaldehyde preservatives to vanilla flavoring to drug molecules. The ability to distinguish aldehydes from ketones (Tollens’ test) and to build larger molecules (aldol condensation) are foundational organic chemistry skills.

The key insight: The presence or absence of an alpha-hydrogen determines whether a carbonyl compound undergoes aldol condensation or the Cannizzaro reaction — one H atom changes the entire reaction pathway.

Writing the aldol product as the beta-hydroxy aldehyde instead of the unsaturated product. Aldol condensation gives the alpha,beta-unsaturated aldehyde upon heating because dehydration occurs spontaneously. Students often stop at the beta-hydroxy stage and forget that the final product loses water to form a conjugated double bond.

Assuming Tollens’ test works for all carbonyl compounds. Only aldehydes (and alpha-hydroxy ketones) reduce Tollens’ reagent to give a silver mirror. Ketones like acetone do not react because they lack the hydrogen atom on the carbonyl carbon needed for oxidation. This is a key distinction between aldehydes and ketones.

Confusing the Cannizzaro reaction with the aldol reaction. The Cannizzaro reaction requires aldehydes that have no alpha-hydrogen atoms (such as benzaldehyde or formaldehyde). These undergo disproportionation in concentrated base, not self-addition. Only aldehydes with at least one alpha-hydrogen can undergo aldol condensation.

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