Q7(c). Chloromethylation, then Sommelet–Hauser rearrangement
Complete the following reactions and give the mechanism for the conversion of the step-2 product into the step-3 product: benzene + HCHO, HCl (step 1) → ?; (i) (CH₃)₂NH, (ii) CH₃Cl (step 2) → ?; NaNH₂ (step 3) → ?
Answer in brief
Step 1: benzyl chloride. Step 2: benzyltrimethylammonium chloride. Step 3: (2-methylbenzyl)dimethylamine by the Sommelet–Hauser rearrangement.
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CoversSommelet–Hauser rearrangementchloromethylationnitrogen ylide[2,3]-sigmatropic shiftStevens rearrangementquaternary ammonium salt
All parts of Question 7
- 7(a)(i) Schmidt reaction of a cyclohexanone with an alkyl azide5 marks
- 7(a)(ii) Robinson annulation of 2-methylcyclohexane-1,3-dione5 marks
- 7(b) Oxepin: 6π electrocyclic ring closure, then Diels–Alder addition5 marks
- 7(c) Chloromethylation, then Sommelet–Hauser rearrangement5 marks
- 7(d)(i) Iodolactonisation of a cyclopentenylacetic acid5 marks
- 7(d)(ii) Epoxidation, then acid-catalysed ring expansion to a ketone5 marks