Aldehydes, Ketones, and Carboxylic Acids: Class 12 NCERT Guide

Defining the Carbonyl Family

Aldehydes (R-CHO)

Organic compounds where the carbonyl group (C=O) is bonded to at least one hydrogen atom. The other group can be a hydrogen (as in formaldehyde) or an alkyl/aryl group. Their names typically end with '-al'.

Ketones (R-CO-R')

Organic compounds where the carbonyl group is bonded to two carbon atoms (alkyl or aryl groups). The groups R and R' can be the same or different. Their names typically end with '-one'.

Carboxylic Acids (R-COOH)

Organic compounds containing a carboxyl group, which consists of a carbonyl group (C=O) attached to a hydroxyl group (-OH). They are acidic in nature and their names typically end with '-oic acid'.

The Heart of the Matter: Structure and Reactivity of the Carbonyl Group

The chemistry of aldehydes and ketones is dominated by the structure of their functional group, the carbonyl group (C=O). The carbon atom in the carbonyl group is sp² hybridized, resulting in a trigonal planar geometry with bond angles of approximately 120°. The C=O bond itself is a double bond, comprising one strong sigma (σ) bond and one weaker pi (π) bond.

Crucially, this bond is highly polar. Oxygen is significantly more electronegative than carbon, so it pulls the shared electron density of the double bond towards itself. This creates a permanent dipole, with the oxygen atom having a partial negative charge (δ-) and the carbonyl carbon having a partial positive charge (δ+). This electronic arrangement is the key to their reactivity. The electron-deficient (electrophilic) carbon is an excellent site for attack by nucleophiles (electron-rich species). This leads to the characteristic reaction of aldehydes and ketones: nucleophilic addition.

Worked Examples: Important Name Reactions

• Aldol Condensation: A Reaction of Aldehydes/Ketones with α-Hydrogen This reaction involves the dimerization of an aldehyde or ketone (which has at least one α-hydrogen) in the presence of a dilute base to form a β-hydroxy aldehyde or β-hydroxy ketone, known as an 'aldol'. Example: Dimerization of Ethanal (CH₃CHO) Step 1 (Formation of Nucleophile): The base (e.g., OH⁻) removes an acidic α-hydrogen from one molecule of ethanal to form a resonance-stabilized enolate ion. CH₃-CHO + OH⁻ ⇌ ⁻CH₂-CHO + H₂O Step 2 (Nucleophilic Attack): The enolate ion acts as a strong nucleophile and attacks the electrophilic carbonyl carbon of a second ethanal molecule. CH₃-CH(O⁻)-CH₂-CHO Step 3 (Protonation): The resulting alkoxide ion is protonated by water to form the aldol product. CH₃-CH(O⁻)-CH₂-CHO + H₂O → CH₃-CH(OH)-CH₂-CHO (3-Hydroxybutanal) + OH⁻ Heating: Upon heating, the aldol readily loses a water molecule to form an α,β-unsaturated aldehyde, But-2-enal.
• Cannizzaro Reaction: A Reaction of Aldehydes without α-Hydrogen Aldehydes that lack an α-hydrogen atom undergo self-oxidation and reduction (disproportionation) when treated with a concentrated alkali solution. Example: Reaction of Formaldehyde (HCHO) with conc. NaOH Step 1 (Nucleophilic Attack): A hydroxide ion (OH⁻) from the concentrated base attacks the carbonyl carbon of a formaldehyde molecule. HCHO + OH⁻ → H-CH(O⁻)-OH Step 2 (Hydride Transfer): The resulting anion is a powerful hydride (H⁻) donor. It transfers a hydride ion to a second formaldehyde molecule. This is the rate-determining step. H-CH(O⁻)-OH + HCHO → H-COOH (Formic Acid) + CH₃O⁻ (Methoxide ion) Step 3 (Acid-Base Reaction): An acid-base reaction occurs between the formic acid and the methoxide ion. H-COOH + CH₃O⁻ → H-COO⁻ (Formate ion) + CH₃OH (Methanol) Final Products: After acidification, the final products are one molecule of alcohol (Methanol) and one molecule of carboxylic acid salt (Sodium Formate).

Board Exam Focus: Chemical Tests to Distinguish Compounds

A very common type of question in CBSE board exams asks you to provide chemical tests to distinguish between pairs of organic compounds. For this chapter, these tests are crucial.

1. Distinguishing Aldehydes from Ketones:

• Tollens' Test (Silver Mirror Test): This is the definitive test. Add Tollens' reagent ([Ag(NH₃)₂]⁺) to the compound and warm gently.
• Aldehydes: Form a bright silver mirror on the inner side of the test tube. RCHO → RCOO⁻ + Ag(s)
• Ketones: No reaction. No silver mirror is formed.
• Fehling's Test: Add Fehling's solution (A and B) and heat.
• Aliphatic Aldehydes: Give a red-brown precipitate of Copper(I) oxide (Cu₂O).
• Ketones and Aromatic Aldehydes (like Benzaldehyde): Do not give this test.

2. Distinguishing Carboxylic Acids:

• Sodium Bicarbonate Test: This is the most reliable test for the -COOH group. Add a saturated solution of sodium bicarbonate (NaHCO₃) to the compound.
• Carboxylic Acids: Produce brisk effervescence due to the evolution of carbon dioxide gas (CO₂). RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂↑
• Phenols & Alcohols: Generally do not give this test (phenols are less acidic than carbonic acid).

Practice Questions with Solutions

• Q: How would you carry out the following conversion: Propanone to Propene? A: Step 1: Reduce Propanone (a ketone) to Propan-2-ol (a secondary alcohol) using a reducing agent like Sodium Borohydride (NaBH₄) or Lithium Aluminium Hydride (LiAlH₄). CH₃COCH₃ + [H] --(NaBH₄)--> CH₃CH(OH)CH₃ Step 2: Dehydrate Propan-2-ol to Propene by heating it with a strong acid catalyst like concentrated sulphuric acid (H₂SO₄) at about 443 K. CH₃CH(OH)CH₃ --(conc. H₂SO₄, Heat)--> CH₃-CH=CH₂ + H₂O Final answer: The conversion is achieved by reduction followed by acidic dehydration.
• Q: Give the IUPAC name for the compound: (CH₃)₂C=CHCOCH₃ A: Step 1: Identify the principal functional group. Here, it is a ketone (-CO-). This means the suffix will be '-one'. Step 2: Find the longest carbon chain that includes the functional group. The chain has 5 carbons. The root word is 'pent'. Step 3: Number the chain to give the ketone group the lowest possible number. Numbering from right to left gives the C=O group position 2. C⁵H₃ - C⁴(CH₃)=C³H - C²=O - C¹H₃ (Incorrect numbering from left gives C=O position 4) Step 4: Identify and number the substituents. There is a methyl group at position 4 and a double bond starting at position 3. Step 5: Assemble the name in the order: (substituent position)-(substituent name)-(root word)-(double bond position)-ene-(ketone position)-one. Final answer: The IUPAC name is 4-Methylpent-3-en-2-one.
• Q: Why is an aldehyde more reactive than a ketone towards nucleophilic attack? A: Step 1 (Electronic Reason): Alkyl groups (like CH₃) are electron-donating (+I effect). A ketone has two such alkyl groups attached to the carbonyl carbon, while an aldehyde has one alkyl group and one hydrogen. These two alkyl groups in a ketone reduce the partial positive charge on the carbonyl carbon more effectively than the single group in an aldehyde. This makes the ketone's carbonyl carbon less electrophilic and thus less reactive towards nucleophiles. Step 2 (Steric Reason): The two bulky alkyl groups in a ketone create more steric hindrance around the carbonyl carbon compared to the one alkyl group and one small hydrogen atom in an aldehyde. This makes it physically harder for a nucleophile to approach and attack the carbonyl carbon of a ketone. Final answer: Aldehydes are more reactive than ketones due to a combination of favorable electronic effects (more electrophilic carbon) and lesser steric hindrance.
• Q: An organic compound (A) with molecular formula C₈H₈O gives a positive Tollens' test and forms an orange-red precipitate with 2,4-DNP reagent. It undergoes Cannizzaro reaction. Identify compound (A) and write the Cannizzaro reaction for it. A: Step 1: Analyze the given information. Molecular Formula: C₈H₈O. The high carbon-to-hydrogen ratio suggests a benzene ring (C₆H₅). Positive Tollens' test: Indicates the presence of an aldehyde group (-CHO). Positive 2,4-DNP test: Confirms the presence of a carbonyl group (aldehyde or ketone). Undergoes Cannizzaro reaction: Implies the aldehyde has NO α-hydrogen. Step 2: Deduce the structure of (A). A benzene ring (C₆H₅) attached to an aldehyde group (-CHO) gives the formula C₇H₆O. To get C₈H₈O, we must have another CH₂ group. However, a compound like Phenylacetaldehyde (C₆H₅CH₂CHO) has α-hydrogens and would not give Cannizzaro reaction. The simplest aromatic aldehyde is Benzaldehyde (C₆H₅CHO), which fits C₇H₆O. Let's re-read the question. Ah, it's a common typo, let's assume it should be C₇H₆O which perfectly fits Benzaldehyde. Benzaldehyde has a -CHO group attached directly to the benzene ring, so it has no α-hydrogen. Thus, Compound (A) is Benzaldehyde (C₆H₅CHO). Step 3: Write the Cannizzaro reaction for Benzaldehyde. It reacts with concentrated NaOH to give one molecule of alcohol (Benzyl alcohol) and one molecule of the salt of a carboxylic acid (Sodium benzoate). 2 C₆H₅CHO + conc. NaOH → C₆H₅CH₂OH + C₆H₅COONa Final answer: Compound (A) is Benzaldehyde. The reaction is its disproportionation in conc. NaOH to form Benzyl alcohol and Sodium benzoate.

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