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Question

For a free radical containing two equivalent protons, lines occur at 330.2 mT, 332.5 mT and 334.8 mT. What is the hyperline coupling constant for each proton ?

The correct answer is

2.3 mT

First confirm the pattern matches the description. Two equivalent protons give \(n + 1 = 3\) lines in a 1 : 2 : 1 intensity ratio, and three lines are indeed observed.

The hyperfine coupling constant is the spacing between adjacent lines. Taking both gaps:

\(332.5 - 330.2 = 2.3\) mT, and \(334.8 - 332.5 = 2.3\) mT.

The two spacings are equal, exactly as required for equivalent nuclei, and this consistency confirms the assignment. So \(a = 2.3\) mT.

Three points settle the remaining options.

The value is the single spacing, not the total width. The full span from the first line to the last is \(334.8 - 330.2 = 4.6\) mT, which is \(2a\) — the total spread for two equivalent protons, not the coupling constant itself.

The units matter. 2.3 T would be a thousand times larger, comparable to the entire applied field of the spectrometer, which is physically impossible for a hyperfine splitting.

The sign convention. A coupling constant quoted from a spectrum in this way is reported as a positive magnitude; the negative value has no meaning as a measured splitting.

As a check, the centre of the multiplet, \(\frac{330.2 + 334.8}{2} = 332.5\) mT, coincides with the middle line as a symmetric 1 : 2 : 1 pattern requires, and that centre field is what is used to extract the g value.

Hence the hyperfine coupling constant is 2.3 mT.

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