UPSC Combined Geo-Scientist, Mains · 2019 · Mains Paper II
Q1(p). Bromobutane with one NMR signal
Question
The ¹ H NMR spectrum of C₄H₉Br consists of a single line. Give its structure.
Answer in brief
The compound is tert-butyl bromide, 2-bromo-2-methylpropane, (CH₃)₃CBr.
Key points
- Its three methyl groups are symmetry-equivalent, and rapid methyl rotation makes all nine protons equivalent. They therefore give one signal integrating to 9H.
- The central carbon bears Br and no hydrogen, so there is no adjacent proton to cause ordinary vicinal splitting.
- The other constitutional isomers contain several inequivalent proton environments and cannot give a single ordinary 9H resonance. Resolved coupling to quadrupolar bromine is normally absent.
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.
CoversBromobutane with one NMR signalStructure elucidation
All parts of Question 1
- 1(a) Atomicity from heat capacities5 marks
- 1(b) Units of attraction parameter5 marks
- 1(c) Surface energy of cube and sphere5 marks
- 1(d) Atoms in a BCC sample5 marks
- 1(e) Volume independence of ideal-gas heat capacity5 marks
- 1(f) Entropy on doubling volume5 marks
- 1(g) Single-ion activity coefficients5 marks
- 1(h) Ammonia buffer5 marks
- 1(i) Pure-component vapour pressures5 marks
- 1(j) Zero-order half-life5 marks
- 1(k) Physical and chemical adsorption5 marks
- 1(l) Iron redox-electrode potential5 marks
- 1(m) Energy of one einstein5 marks
- 1(n) Bohr quantisation from matter waves5 marks
- 1(o) Spectral wavelength shifts5 marks
- 1(p) Bromobutane with one NMR signal5 marks