NEET 2024 Chemistry Question 86 Solution
Question
Major products A and B formed in the following reaction sequence, are: 2-methylcyclohexan-1-ol → (PBr₃) A (major) → (alc. KOH, Δ) B (major)
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Major products A and B formed in the following reaction sequence, are: 2-methylcyclohexan-1-ol → (PBr₃) A (major) → (alc. KOH, Δ) B (major)
PBr3 substitutes -OH with -Br to give 1-bromo-2-methylcyclohexane (A). alc. KOH/Δ eliminates HBr following Saytzeff’s rule to form the MORE SUBSTITUTED alkene: 1-methylcyclohexene (B)!
Reaction of secondary alcohols with phosphorus tribromide (PBr3) cleanly converts the -OH into a bromide leaving group without carbocation rearrangement, yielding 1-bromo-2-methylcyclohexane:
Alcoholic potassium hydroxide (alc. KOH) is a strong dehydrohalogenating reagent. Upon heating, the ethoxide/hydroxide base abstracts a proton from a β-carbon concurrent with the expulsion of the bromide ion in a concerted anti-coplanar E2 pathway:
The base can abstract a proton from either: - The tertiary β-carbon (C-2): Abstraction of the single tertiary hydrogen yields 1-methylcyclohex-1-ene (a trisubstituted alkene with endocyclic double bond). - The secondary β-carbon (C-6): Abstraction yields 3-methylcyclohex-1-ene (a disubstituted alkene). By Zaitsev’s rule, the more substituted alkene possesses more hyperconjugative α-hydrogens and is thermodynamically more stable, predominating as the major product B:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
PBr3 replaces -OH, it does not add to the ring while retaining -OH.
alc. KOH does not oxidize alcohols to ketones.
Accurate substitution by PBr3 followed by Saytzeff E2 elimination to the more substituted alkene.
Forms the minor Hofmann-type disubstituted alkene instead of the major Saytzeff product.
With unhindered bases like alcoholic KOH, elimination strictly obeys Saytzeff’s rule yielding the more substituted alkene (Option 3). Option (4) gives the less substituted alkene (Hofmann product), which only forms with bulky bases like t-BuOK.
"In dehydrohalogenation of secondary alkyl halides with alcoholic KOH, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms."
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