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NEET UG Haloalkanes, Alcohols and Ethers — practice questions — page 3

Explore 61 free Haloalkanes, Alcohols and Ethers questions in NEET UG Chemistry. Open each question or exercise to read its content, reference answers and available explanation.

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Haloalkanes, Alcohols and Ethers questions and exercises with answers

Showing questions 61–61 of 61.

Worked practice questions

The KOLBE reaction converts phenol (in the form of sodium phenoxide) into salicylic acid. The conditions and key reagent are

  1. $\mathrm{CO_2}$ under pressure
  2. $\mathrm{CHCl_3}$ + NaOH
  3. $\mathrm{Br_2}$ in $\mathrm{CS_2}$
  4. $\mathrm{H_2SO_4}$ at $443\,\text{K}$
Show answer and solution

Correct answer: A. $\mathrm{CO_2}$ under pressure

1. Your syllabus: KOLBE reaction (also Kolbe-Schmitt) introduces a carboxylic acid group onto the phenoxide. 2. PROCEDURE: sodium phenoxide is heated with $\mathrm{CO_2}$ under PRESSURE (~125°C, ~4 atm), then ACIDIFIED. 3. MECHANISM: the phenoxide ion (activated nucleophile, with negative charge delocalised to ortho/para) attacks CO$_2$ at the ortho position. After workup with acid, the product is ORTHO-hydroxybenzoic acid (SALICYLIC ACID). 4. PRODUCT: salicylic acid is the precursor of ASPIRIN (acetylsalicylic acid) — historically important industrial chemistry. 5. Option B describes the REIMER-TIEMANN reaction (gives aldehyde, not acid). Option C is mild bromination (gives bromophenol). Option D is dehydration of alcohol, unrelated. _Source: syllabus Chemistry Part 2, Ch 7, §7.7 (Kolbe reaction), p. 18–19._

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