Question

Question 1 of 40 Incorrect X IncorrectX AH NMR signal is observed at 181.0 Hz (downfield of TMS) on a 125-MHz instrument. Map a) What is the observed chemical shift in parts per million (ppm)? Number 1.448 ppm b) What is the chemical shift, in parts per million (ppm), if the sample is analyzed with a 200.0-MHz instrument? Number 9.05 x 10 ppm c) Where will the proton signal appear, in hertz, if the sample is analyzed with a 200.0-MHz instrument? Number 0.905 Hz

Answer 1

a) Chemical shift in ppm on 125 MHz instrument = (shift from TMS in Hz / spectrometer frequency in MHz ) x 10⁶

= (181 Hz / 125 MHz ) = 1.448 ppm.

b) Here, we need to compare two instrumental frequency and abserved frequencies as

observed frequency in Hz / first instrumental frequency in MHz = unknown observed frequency in Hz (x) / second instrumental frequency in MHz.

So, 181 Hz / 120 MHz = x in Hz / 200 MHz

Unknown observed frequency in Hz = 181Hz x 200 MHz / 125 MHz = 289.6 Hz

Now chemical shift in ppm on 200 MHz instrument = 289.6Hz / 200 MHz = 1.448 ppm.

c) proton signal appears at 289.6 Hz from TMS on 200 MHz instrument.

What you learn from these questions? Simply , chemical shift value in ppm remains same irrespective of the spectrometer .