Q1(b). Benzyltrimethylammonium salt with sodium amide at low and at high temperature
Discuss the product(s) formation when the above quaternary ammonium salt (benzyltrimethylammonium halide, PhCH₂N⁺(CH₃)₃ X⁻) is treated with sodium amide at low temperature and at high temperature. [structure of the salt given in the paper]
Answer in brief
Low temperature gives the Sommelet–Hauser rearrangement to N,N-dimethyl-2-methylbenzylamine; high temperature gives the Stevens rearrangement to N,N-dimethyl-1-phenylethylamine.
Key points
- NH₂⁻ removes a proton next to N⁺: a nitrogen ylide.
- Low temperature: methyl ylide undergoes a [2,3]-sigmatropic shift onto the ortho carbon.
- A hydrogen shift restores the aromatic ring: ortho-methyl benzylamine (Sommelet–Hauser).
- High temperature: the benzylic ylide undergoes a [1,2]-shift of CH₃ (Stevens).
- Concerted [1,2]-shift is symmetry-forbidden: caged radical pair, higher barrier.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Topic notes: molecular rearrangements, with every Civil Services, IFoS and Geo-Scientist PYQ →
CoversSommelet–Hauser rearrangementStevens rearrangementnitrogen ylide[2,3]-sigmatropic shiftsodium amidequaternary ammonium salt
All parts of Question 1
- 1(a) Origin of the CN group in RCOO⁻ + BrCN10 marks
- 1(b) Benzyltrimethylammonium salt with sodium amide at low and at high temperature10 marks
- 1(c) 2-Cyclohexylpropan-2-ol with BF₃·OEt₂: products A and B10 marks
- 1(d) Two aldehydes towards a nucleophile in the presence of BF₃·OEt₂10 marks
- 1(e) Two transformations by electrocyclic relays10 marks