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UPSC Civil Services (Mains)

Chemistry optional

  • 2 papers
  • 250 marks each
  • 3 h per paper
  • 5 of 8 questions, Q1 and Q5 compulsory

Paper II covers organic chemistry and spectroscopy.

UPSC Civil Services (Mains), Chemistry optional · 2018 · Paper II

Q5(d). Chemical shift is independent of the spectrometer frequency

  • 10 marks
  • Answered
Question

A signal has been reported to occur at 120 Hz downfield from TMS in an NMR spectrometer with a 60 MHz operating frequency.
(i) What is its chemical shift?
(ii) What would its chemical shift be in an instrument operating at 100 MHz?
(iii) How many hertz downfield from TMS would the signal be in a 100 MHz spectrometer?

Answer in brief

(i) δ = 2.00 ppm; (ii) δ = 2.00 ppm (independent of the instrument); (iii) Δν = 200 Hz downfield from TMS at 100 MHz.

Key points

  • At 60 MHz: δ = (120 Hz)/(60 MHz) = 2.00 ppm
  • δ is a ratio of a frequency difference to the operating frequency; both scale with B₀, so δ is the same on every instrument: at 100 MHz, δ = 2.00 ppm.
  • At 100 MHz: Δν = δ × ν₀ = 2.00 × 100 = 200 Hz
  • Since ν = γ B₀(1-σ)/2π, both the signal–TMS separation and ν₀ are proportional to B₀; their ratio depends only on the shielding constant σ, a property of the nucleus in its molecule.

The full model answer

Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.

Introduction. The separation of a signal from TMS in hertz grows in proportion to the field (and so to the operating frequency). Dividing by the operating frequency gives the chemical shift δ, a field-independent number in parts per million:

δ (ppm) = (Δν (Hz))/(ν₀ (MHz))

  1. At 60 MHz: δ = (120 Hz)/(60 MHz) = 2.00 ppm
  2. δ is a ratio of a frequency difference to the operating frequency; both scale with B₀, so δ is the same on every instrument: at 100 MHz, δ = 2.00 ppm.
  3. At 100 MHz: Δν = δ × ν₀ = 2.00 × 100 = 200 Hz

Figure: The signal stays at δ 2.00; its distance in Hz scales with ν₀.

  1. Since ν = γ B₀(1-σ)/2π, both the signal–TMS separation and ν₀ are proportional to B₀; their ratio depends only on the shielding constant σ, a property of the nucleus in its molecule.
  2. Coupling constants J (in Hz) do not change with the field; so at 100 MHz multiplets occupy a smaller δ range and overlap less: higher-field instruments give spectra closer to first order.

Conclusion. (i) δ = 2.00 ppm; (ii) δ = 2.00 ppm (independent of the instrument); (iii) Δν = 200 Hz downfield from TMS at 100 MHz.

Application. Reporting shifts in δ lets spectra recorded at 60, 300 or 600 MHz be compared directly; a 2.0 ppm signal is typical of CH₃CO⁻ or CH₃-C=C protons.

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Model answer to Q5(d), 2018 Paper II: solution page 18
Solution page 18 of 24