UPSC Civil Services (Mains), Chemistry optional · 2021 · Paper II
Q1(a). Origin of the CN group in RCOO⁻ + BrCN
Question
In the reaction R–COO⁻ + Br–CN → R–CN + Br⁻ + CO₂↑, what is the origin of the –CN group in the product? Explain by using isotopic labelling technique.
Answer in brief
The nitrile carbon is the original carboxyl carbon and only the nitrogen comes from BrCN; R–¹⁴COO⁻ gives radioactive R–¹⁴CN and inactive CO₂.
Key points
- Tracer: prepare R–¹⁴COO⁻; treat with unlabelled BrCN at 250–300 °C.
- Nitrile radioactive, CO₂ inactive: the carboxyl carbon stays on R.
- CN not transferred as a unit: simple replacement of CO₂ ruled out.
- Path: acyl cyanate, then acyl isocyanate (isolated), then a four-membered ring.
- Ring splits to R–C≡N and CO₂: CO₂ carbon is the BrCN carbon.
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.
Coversisotopic labelling¹⁴C tracercyanogen bromideacyl isocyanatenitrile from carboxylate
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