In 400 MHz 1H MNR spectrum of an organic compound exhibited a doublet. The two lines of the doublet are at δ = 2.35 and 2.38 ppm. The coupling constant (J) value is:a)3Hzb)6Hzc)9Hzd)12 HzCorrect answer is option 'D'. Can you explain this answer?

Question

In 400 MHz 1H MNR spectrum of an organic compound exhibited a doublet. The two lines of the doublet are at &delta; = 2.35 and 2.38 ppm. The coupling constant (J) value is:a)3Hzb)6Hzc)9Hzd)12 HzCorrect answer is option 'D'. Can you explain this answer?

Answer

Chemical shifts of 3.5 ppm and 3.7 ppm, and have equal intensities. The coupling constant (J value) between the two protons responsible for the doublet is:

To determine the coupling constant, we need to use the formula:

J = Δν / (n-1)

Where J is the coupling constant, Δν is the distance between the two peaks in Hz, and n is the number of peaks in the multiplet (in this case, n = 2).

First, we need to convert the chemical shift difference (Δδ) between the two lines of the doublet from ppm to Hz:

Δδ = (3.7 ppm - 3.5 ppm) = 0.2 ppm
Δν = Δδ x ν = 0.2 ppm x 400 MHz = 80 Hz

Now we can plug these values into the formula to get the coupling constant:

J = Δν / (n-1) = 80 Hz / (2-1) = 80 Hz

Therefore, the coupling constant (J value) between the two protons responsible for the doublet is 80 Hz.

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