JEE Advanced level
A JEE Advanced-level example on Alcohols, Phenols and Ethers
Question. Compound A (C₆H₆O) gives a violet color with neutral FeCl₃ solution. When A is treated with CHCl₃ and aqueous NaOH followed by acidification, compound B (C₇H₆O₂) is obtained as the major ortho-product. When A is treated with NaOH, then CO₂ at 125°C under 4-7 atm pressure, followed by acidification, compound C (C₇H₆O₃) is formed. (i) Identify compounds A, B, and C with their structures. (ii) A tertiary alcohol D with molecular formula C₄H₁₀O is subjected to Lucas test. What is D and what will be observed?
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- A gives violet color with FeCl₃, so A contains a phenolic -OH group. With formula C₆H₆O, A is phenol (C₆H₅OH).
- Phenol + CHCl₃ + NaOH is the Reimer-Tiemann reaction. It introduces a -CHO group at the ortho position. B is salicylaldehyde (2-hydroxybenzaldehyde, o-HOC₆H₄CHO).
- Phenol + NaOH + CO₂ under heat and pressure is the Kolbe-Schmitt reaction. It introduces a -COOH group at the ortho position. C is salicylic acid (2-hydroxybenzoic acid, o-HOC₆H₄COOH).
- Tertiary alcohol with formula C₄H₁₀O is tert-butyl alcohol or 2-methylpropan-2-ol, (CH₃)₃COH.
- Lucas reagent is ZnCl₂ + concentrated HCl. Tertiary alcohols react immediately at room temperature giving a cloudy solution (turbidity) due to formation of insoluble tert-butyl chloride.
Answer: (i) A = Phenol (C₆H₅OH), B = Salicylaldehyde (2-hydroxybenzaldehyde), C = Salicylic acid (2-hydroxybenzoic acid). (ii) D = tert-butyl alcohol (2-methylpropan-2-ol). Lucas test shows immediate turbidity at room temperature.
This example was solved twice independently and both solutions agree.