Q7(a). Pinacol rearrangement; Wolff–Kishner beside a ketal
Identify the major products in the following reactions and discuss their formation:
(i) [structure] → (H⁺)
(ii) [structure] → (1) N₂H₄, 2) KOH, Δ)
Answer in brief
(i) 1,1,1-triphenylpropan-2-one: Ph₂C⁺ forms and phenyl migrates. (ii) 6-butyl-1,4-dioxaspiro[4.5]decane: only the unprotected ketone is reduced to CH₂; the ketal survives the basic conditions (and can later be removed with aqueous acid to give 2-butylcyclohexanone).
Key points
- Which OH leaves. Protonation and loss of water can occur at C-1 (the Ph₂C(OH) end) or at C-2 (the PhC(CH₃)(OH) end). Loss from C-1 gives the diphenyl-substituted cation Ph₂C⁺, stabilised by two phenyl rings; this more stable…
- Which group migrates. C-2 carries Ph, CH₃ and OH. Phenyl has the higher migratory aptitude (it migrates through a bridged phenonium-like transition state) and moves to C-1 with its bonding pair.
- The positive charge now sits on C-2, stabilised by the oxygen lone pair (oxocarbenium ion); loss of H⁺ gives the ketone.
- 1,1,1-triphenylpropan-2-one, Ph₃C-CO-CH₃ (C₂₁H₁₈O). (Methyl migration would give 1,2,2-triphenylpropan-1-one, PhCO-CPh₂CH₃, a minor product at most.)
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
(i) Pinacol rearrangement of 1,1,2-triphenylpropane-1,2-diol
Scheme: The more stable cation forms; the group with the higher migratory aptitude (phenyl) moves.
- Which OH leaves. Protonation and loss of water can occur at C-1 (the Ph₂C(OH) end) or at C-2 (the PhC(CH₃)(OH) end). Loss from C-1 gives the diphenyl-substituted cation Ph₂C⁺, stabilised by two phenyl rings; this more stable carbocation forms (rate-determining ionisation).
- Which group migrates. C-2 carries Ph, CH₃ and OH. Phenyl has the higher migratory aptitude (it migrates through a bridged phenonium-like transition state) and moves to C-1 with its bonding pair.
- The positive charge now sits on C-2, stabilised by the oxygen lone pair (oxocarbenium ion); loss of H⁺ gives the ketone.
- Product: 1,1,1-triphenylpropan-2-one, Ph₃C-CO-CH₃ (C₂₁H₁₈O). (Methyl migration would give 1,2,2-triphenylpropan-1-one, PhCO-CPh₂CH₃, a minor product at most.)
(ii) Wolff–Kishner reduction with the ketal untouched
Scheme: C=O of the side chain becomes CH₂; the ethylene ketal survives strong base.
- The substrate is a monoprotected diketone: the ring C=O is masked as an ethylene ketal (1,3-dioxolane), and the side chain carries an unprotected ketone (2-oxobutyl).
- Hydrazine condenses with the unprotected ketone to give the hydrazone. KOH at high temperature (Huang-Minlon: in hot diethylene glycol) removes N–H protons; the diimide anion loses N₂ and the carbanion is protonated: C=O → CH₂.
- Ketals are stable to base, so the protected ring carbonyl is not touched. Product: 6-butyl-1,4-dioxaspiro[4.5]decane (C₁₂H₂₂O₂), the ethylene ketal of 2-butylcyclohexanone.
- Clemmensen reduction (Zn(Hg)/HCl) would not do this job: the acid would hydrolyse the ketal and reduce both carbonyl groups.
Conclusion. (i) 1,1,1-triphenylpropan-2-one: Ph₂C⁺ forms and phenyl migrates. (ii) 6-butyl-1,4-dioxaspiro[4.5]decane: only the unprotected ketone is reduced to CH₂; the ketal survives the basic conditions (and can later be removed with aqueous acid to give 2-butylcyclohexanone).
Application. Pinacol rearrangements build quaternary carbon centres (pinacolone, spiro ketones); ketal protection with base-mediated reduction is a standard strategy in steroid synthesis, where one carbonyl must be reduced beside another.
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All parts of Question 7
- 7(a) Pinacol rearrangement; Wolff–Kishner beside a ketal15 marks
- 7(b) Electrocyclic reactions: hexatriene, pentadiene, and ethene15 marks
- 7(c) Synthetic rubbers: Buna-S and neoprene10 marks
- 7(d) 1H NMR of ethyl (E)-but-2-enoate10 marks