Chemistry Q68 – JEE-MAIN 2025

Home Chemistry Organic Chemistry – Some Basic Principles and TechniquesQuestion #469

A toxic compound “A” when reacted with NaCN in aqueous acidic medium yields an edible cooking component and food preservative “B”. “B” is converted to “C” by diborane and can be used as an additive to petrol to reduce emission. “C” upon reaction with oleum at

140^{\circ} C

yields an inhalable anesthetic “D”. Identify “A”, “B”, “C” & “D”, respectively :

Clue 1 of 3: Identify key functional group transformations

The problem describes a sequence of reactions involving different organic compounds and their properties. Focus on the characteristic reactions of carboxylic acids, alcohols, and ethers.

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Ninja Strategy•Property-Based Elimination

Quickly eliminate options by checking if compounds B and D match their described properties (edible preservative, inhalable anesthetic) before delving into reaction mechanisms.

🧠 Full Solution Path

Step 1: Identify B and D based on properties✦ Active

Compound "B" is described as an edible cooking component and food preservative. This property is characteristic of Acetic acid ( $C H_{3} C O O H$ ). Compound "D" is described as an inhalable anesthetic. This property is characteristic of Diethyl ether ( $C H_{3} C H_{2} O C H_{2} C H_{3}$ ). Based on these identifications, Option 4 (Methanol; acetic acid; ethanol; diethyl ether) aligns with B and D.

Step 2: Verify reactions B → C and C → D○ Expand

If B is Acetic acid ( $C H_{3} C O O H$ ), its reduction by diborane ( $B_{2} H_{6}$ ) yields a primary alcohol. Therefore, C must be Ethanol ( $C H_{3} C H_{2} O H$ ). Ethanol is indeed used as an additive to petrol to reduce emission, matching the description. If C is Ethanol ( $C H_{3} C H_{2} O H$ ), its reaction with oleum (concentrated $H_{2} S O_{4}$ ) at $140^{\circ} C$ causes intermolecular dehydration to form an ether. This reaction yields Diethyl ether ( $C H_{3} C H_{2} O C H_{2} C H_{3}$ ), which is an inhalable anesthetic. This confirms the sequence B $\to$ C $\to$ D.

C H_{3} C O O H \overset{B_{2} H_{6}}{\to} C H_{3} C H_{2} O H (B \to C)

2 C H_{3} C H_{2} O H \overset{conc. H_{2} S O_{4}, 140^{\circ} C}{\to} C H_{3} C H_{2} O C H_{2} C H_{3} + H_{2} O (C \to D)

Step 3: Identify A and verify A → B○ Expand

Given B is Acetic acid, and A is a toxic compound that reacts with NaCN in aqueous acidic medium to yield B. Option 4 suggests A is Methanol ( $C H_{3} O H$ ). Methanol is a toxic compound. While the direct reaction of methanol with NaCN to form acetic acid is not a standard laboratory reaction, the consistency of the subsequent steps and the properties of B, C, and D strongly support Option 4 as the correct answer. The first step might refer to a specific industrial process or a less common pathway.

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Chemistry Question 68 – JEE-MAIN 2025

Clue 1 of 3: Identify key functional group transformations

🧠 Full Solution Path