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Topic: Alcohols: identification of 1/2/3 alcohols, dehydration mechanism· Organic Compounds Containing Oxygen

Major products A and B formed in the following reaction sequence, are[scheme: 2-methylcyclohexanol (OH and H_3C on adjacent ring carbons) PBr_3 A (major) [Δ]alc.\ KOH B (major), see figure]

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

[scheme: 2-methylcyclohexanol (OH and \(H_3C\) on adjacent ring carbons) \(\xrightarrow{PBr_3}\) A (major) \(\xrightarrow[\Delta]{alc.\ KOH}\) B (major), see figure]

reaction scheme with PBr3 then alc. KOH and the four A/B product-pair options

1. A = [structure: ring with OH, \(H_3C\) and Br], B = [structure: 2-methylcyclohexanone]

2. A = [structure: 1-bromo-2-methylcyclohexane], B = [structure: 1-methylcyclohexene (double bond adjacent to methyl carbon)]

3. A = [structure: 1-bromo-2-methylcyclohexane], B = [structure: 3-methylcyclohexene (double bond away from methyl carbon)]

4. A = [structure: ring with OH, \(H_3C\) and Br], B = [structure: 3-methylcyclohex-2-en-1-ol type, ring with OH, \(H_3C\) and a double bond]

1 answer

Best answer

Answer: 2. A = [structure: 1-bromo-2-methylcyclohexane], B = [structure: 1-methylcyclohexene (double bond adjacent to methyl carbon)]

Key idea: \(\mathrm{PBr_3}\) replaces OH by Br; alcoholic KOH causes elimination, with the more substituted alkene (Saytzeff) as the major product.

  1. Step 1 gives 1-bromo-2-methylcyclohexane (A).
  2. Step 2 removes H from the carbon bearing the methyl group (the more substituted, tertiary C-H), giving 1-methylcyclohexene with a trisubstituted double bond (B).

Option 1 and 4 keep OH or give a ketone, so they are wrong. Option 3 has the less substituted alkene. Option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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