Q3(c). Simmons–Smith; SOCl₂ on crotyl alcohol; Beckmann rearrangement
Predict the major products in the following reactions and write the mechanism:
(i) [structure] → (CH₂I₂, Zn-Cu)
(ii) H₃C-CH=CH-CH₂OH → (SOCl₂)
(iii) C₆H₅-C(CH₃)=O → (1) NH₂OH, 2) PCl₅)
Answer in brief
(i) 1,1,2,2-tetramethylcyclopropane (Simmons–Smith, syn, concerted). (ii) 3-chlorobut-1-ene (SNi' with allylic shift; 1-chlorobut-2-ene with pyridine). (iii) acetanilide (Beckmann: phenyl anti to OH migrates).
Key points
- The zinc–copper couple inserts into a C–I bond: CH₂I₂ + Zn → ICH₂ZnI, a carbenoid (it behaves like singlet CH₂ but is not a carbene itself).
- The carbenoid transfers CH₂ to the alkene in a concerted, three-centred (“butterfly”) transition state; ZnI₂ is released. Both new C–C bonds form on the same face: stereospecific syn addition.
- The electron-rich tetrasubstituted alkene reacts well; product: 1,1,2,2-tetramethylcyclopropane (C₇H₁₄). No C–H insertion products form, unlike with uncomplexed CH₂.
- The OH attacks S of SOCl₂; loss of HCl gives the alkyl chlorosulfite ROS(O)Cl.
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) Simmons–Smith cyclopropanation
Scheme: The zinc carbenoid delivers CH₂ to both carbons at once.
- The zinc–copper couple inserts into a C–I bond: CH₂I₂ + Zn → ICH₂ZnI, a carbenoid (it behaves like singlet CH₂ but is not a carbene itself).
- The carbenoid transfers CH₂ to the alkene in a concerted, three-centred (“butterfly”) transition state; ZnI₂ is released. Both new C–C bonds form on the same face: stereospecific syn addition.
- The electron-rich tetrasubstituted alkene reacts well; product: 1,1,2,2-tetramethylcyclopropane (C₇H₁₄). No C–H insertion products form, unlike with uncomplexed CH₂.
(ii) Thionyl chloride and crotyl alcohol: SNi'
Scheme: Cl is delivered intramolecularly to C-3 while the double bond shifts to C1=C2 (allylic rearrangement).
- The OH attacks S of SOCl₂; loss of HCl gives the alkyl chlorosulfite ROS(O)Cl.
- In the absence of a base the chlorosulfite reacts intramolecularly: the Cl on sulfur is delivered to the far end of the allyl unit (C-3) through a six-membered cyclic transition state, while the C1–O bond breaks and SO₂ leaves (SNi').
- Major product: 3-chlorobut-1-ene, CH₃CHClCH=CH₂, formed with allylic rearrangement. This SNi' course has been shown for both but-2-en-1-ol and but-3-en-2-ol, which each give the rearranged chloride.
- If pyridine is added, uncombined Cl⁻ attacks C-1 of the chlorosulfite (SN2) and the unrearranged 1-chlorobut-2-ene is the main product.
(iii) Oxime formation and Beckmann rearrangement
Scheme: The group anti to the leaving oxygen migrates from C to N.
- Hydroxylamine adds to C=O and water is lost: acetophenone oxime. The (E)-oxime, with OH anti to the bulky phenyl, is the main isomer.
- PCl₅ converts the oxime OH into a good leaving group (-OPCl₄).
- As the leaving group departs, the group anti-periplanar to it, the phenyl, migrates from C to N in one concerted step (Beckmann rearrangement), giving a nitrilium ion CH₃C#N+Ph. The migrating group keeps its configuration.
- Water adds to the nitrilium carbon and the imidic acid tautomerises: product acetanilide, C₆H₅NHCOCH₃ (N-phenylacetamide).
Conclusion. (i) 1,1,2,2-tetramethylcyclopropane (Simmons–Smith, syn, concerted). (ii) 3-chlorobut-1-ene (SNi' with allylic shift; 1-chlorobut-2-ene with pyridine). (iii) acetanilide (Beckmann: phenyl anti to OH migrates).
Application. Simmons–Smith reactions make cyclopropane rings of drugs and pyrethroids; the Beckmann rearrangement of cyclohexanone oxime is the industrial route to caprolactam for nylon-6; acetaminophen (paracetamol) has been made by a Beckmann route from 4-hydroxyacetophenone oxime.
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All parts of Question 3
- 3(a) Toluene to m-bromotoluene: an amino group as a temporary director10 marks
- 3(b) MCPBA epoxidation of cis- and trans-but-2-ene10 marks
- 3(c) Simmons–Smith; SOCl₂ on crotyl alcohol; Beckmann rearrangement30 marks