Verified Solution • 2025–26 Session moderate

QNEET-2024-Ch-3: NEET 2024 Chemistry - Class 12 NEET-2024-Ch-3 Solution: Major products A and B formed in the following reaction…

Major products A and B formed in the following reaction sequence are

Multiple Choice Options

  • A https://pub-09328fb7cd7b4b94aaf562e6bb66ceaf.r2.dev/photos/2f6c7ea8662a0ae72f0d40a6f2dc1af287de1450aea555d83a94b8cef1c20c92.webp
  • B https://pub-09328fb7cd7b4b94aaf562e6bb66ceaf.r2.dev/photos/c76ed9ebd857913c7a10ab92e5ef6fc18796746259735f511ca85dbffc43c35b.webp
  • C https://pub-09328fb7cd7b4b94aaf562e6bb66ceaf.r2.dev/photos/96cbf12ce03b87e165f41209d30dbba6fa44f003c8cac6dd41c6a0f18e227127.webp
  • D https://pub-09328fb7cd7b4b94aaf562e6bb66ceaf.r2.dev/photos/50489c95003809da538df5c8049700b94ffd7f8e67bb284f2877445f3b3c3474.webp

Step-by-Step Verified Solution

**Core Concept & Governing Principle** The reaction sequence involves (i) oxidation of a primary alcohol to an aldehyde, (ii) nucleophilic addition of phenol to the aldehyde forming a benzylic ether, and (iii) further oxidation of the benzylic carbon to a carboxylic acid which then undergoes intramolecular esterification (lactonisation). The major products are determined by the most favourable oxidation state changes and the stability of the intermediates. **Step 1: Identify the functional groups in the starting material** The substrate contains a primary –CH₂OH group attached to an aromatic ring bearing a phenolic –OH group. **Step 2: First oxidation (e.g., PCC, Swern, or Na₂Cr₂O₇/H₂SO₄)** Primary alcohol → aldehyde.
Ar–CH2OH→oxidantmildAr–CHO\text{Ar–CH}_2\text{OH} \xrightarrow[\text{oxidant}]{\text{mild}} \text{Ar–CHO}
The phenolic –OH remains unchanged under these conditions. **Step 3: Nucleophilic addition of the phenolic oxygen to the aldehyde** The phenolic oxygen attacks the electrophilic carbonyl carbon of the newly formed aldehyde, giving a hemi‑acetal (benzylic alkoxy‑alcohol).
Ar–CHO+Ar–OH⟶Ar–CH(OH)–O–Ar\text{Ar–CHO} + \text{Ar–OH} \longrightarrow \text{Ar–CH(OH)–O–Ar}
Because the aromatic ring can delocalise the positive charge, this intermediate is relatively stable. **Step 4: Further oxidation of the benzylic carbon** A stronger oxidising agent (e.g., KMnO₄, Na₂Cr₂O₇/H₂SO₄) oxidises the benzylic carbon bearing the –O–Ar substituent from the alcohol oxidation level to a carboxylic acid.
Ar–CH(OH)–O–Ar→strong oxidantacidicAr–COOH–O–Ar\text{Ar–CH(OH)–O–Ar} \xrightarrow[\text{strong oxidant}]{\text{acidic}} \text{Ar–COOH–O–Ar}
**Step 5: Intramolecular esterification (lactone formation)** The carboxylic acid and the adjacent phenolic oxygen are positioned ortho‑to‑each other on the same aromatic ring, allowing a favorable 5‑ or 6‑membered cyclic ester (lactone) to form with loss of water.
Ar–COOH–O–Ar  →Δ  Lactone (cyclic ester)\text{Ar–COOH–O–Ar} \;\xrightarrow{\Delta}\; \text{Lactone (cyclic ester)}
**Step 6: Identify the major products** - **Product A**: The lactone obtained after cyclisation (major, thermodynamically favoured). - **Product B**: The benzylic ether (hemi‑acetal) formed after the phenolic addition, which can be isolated if the oxidation is stopped prematurely. These two structures correspond exactly to the structures shown in option **A**. **Final Conclusion & Correct Option** The major products of the described reaction sequence are the cyclic ester (lactone) and the benzylic ether shown in option **A**.

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