Identify the incorrect option from the following:
Alcohols, Phenols & Ethers · Class 12 · JEE Main Previous Year Question
Identify the incorrect option from the following:
- a
Option (a)
- b✓
Option (b)
- c
Option (c)
- d
Option (d)
Option (b)
Step 1: Evaluate each option
(a) Allyl bromide + KOH(aq) → Allyl alcohol: Aqueous KOH promotes SN2 (nucleophilic substitution). Allyl bromide is a primary allylic halide, highly reactive toward SN2. Product = allyl alcohol. Correct ✓
(b) sec-Butyl bromide + KOH(alc) → Alcohol: KOH in alcoholic (ethanolic) solution is a condition specifically designed for E2 elimination, NOT for substitution. The product should be an alkene (but-2-ene or but-1-ene), not an alcohol.
If the option claims KOH(alc) gives an alcohol, that is INCORRECT.
- KOH(aq) → substitution → alcohol.
- KOH(alc) → elimination → alkene.
This is the incorrect statement ✓
(c) Chlorobenzene + CH3COCl/AlCl3 → Acetophenone: Friedel-Crafts acylation: activated by Lewis acid (AlCl3). Product = p-chloroacetophenone (aromatic ketone). Correct ✓
(d) Chlorobenzene + NaOH (623K) → Phenol: Dow process: SNAr under forcing conditions gives phenol. Correct ✓
Answer: Option (b) is incorrect
Key Points to Remember:
- KOH(aq) → SN1/SN2 → alcohol.
- KOH(alc) → E2 → alkene.
- This is one of the most frequently tested distinctions in JEE organic chemistry.
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