Using a 300 MHz NMR instrument: a. How many Hz downfield from TMS is a signal at 2.5 ppm? b. If a signal comes at 1200 Hz downfield from TMS, at what ppm does it occur? c. If two peaks are separated by 2 ppm, how many Hz does this correspond to?
The Correct Answer and Explanation is:
Correct Answers:
a. A signal at 2.5 ppm on a 300 MHz NMR instrument is:2.5 ppm×300 MHz=750 Hz2.5\ \text{ppm} \times 300\ \text{MHz} = 750\ \text{Hz}2.5 ppm×300 MHz=750 Hz
Answer: 750 Hz
b. A signal that is 1200 Hz downfield from TMS:1200 Hz300 MHz=4 ppm\frac{1200\ \text{Hz}}{300\ \text{MHz}} = 4\ \text{ppm}300 MHz1200 Hz=4 ppm
Answer: 4 ppm
c. A separation of 2 ppm on a 300 MHz instrument is:2 ppm×300 MHz=600 Hz2\ \text{ppm} \times 300\ \text{MHz} = 600\ \text{Hz}2 ppm×300 MHz=600 Hz
Answer: 600 Hz
Explanation:
Nuclear Magnetic Resonance (NMR) spectroscopy uses the interaction of atomic nuclei with magnetic fields to identify the environment of atoms in molecules. One key concept in NMR is the chemical shift, measured in parts per million (ppm), which indicates how far an NMR signal is from the reference signal of tetramethylsilane (TMS).
Chemical shift in ppm is independent of the strength of the magnetic field or the operating frequency of the instrument. However, the actual frequency difference in Hertz (Hz) does depend on the instrument’s magnetic field strength. This is because:Hz=ppm×spectrometer frequency (in MHz)\text{Hz} = \text{ppm} \times \text{spectrometer frequency (in MHz)}Hz=ppm×spectrometer frequency (in MHz)
For part a, multiplying the chemical shift (2.5 ppm) by the instrument frequency (300 MHz) gives 750 Hz. This means the signal appears 750 Hz downfield from TMS.
In part b, if a signal is 1200 Hz downfield, we divide that by the spectrometer frequency (300 MHz) to find the chemical shift, resulting in 4 ppm.
For part c, a difference of 2 ppm corresponds to a frequency difference of 600 Hz on a 300 MHz instrument. This shows that the same ppm separation results in a greater absolute frequency separation on higher-field instruments, improving resolution.
Understanding this conversion is essential for interpreting NMR spectra accurately, especially when analyzing peak splitting and comparing results from instruments with different field strengths.
