Discussion
In the 1H-NMR spectrum of the following molecule, the signal of proton Ha appears as
(The molecule is C6H5CD2Ha: a benzene ring bearing a CD2Ha group.)
Answer: C
The signal of Ha is a quintet.
Ha is on the same carbon as two deuterium atoms. Deuterium has nuclear spin I = 1, so a group of n equivalent deuterons splits a neighbouring proton into \(2nI + 1\) lines:
\[ 2(2)(1) + 1 = 5 \]
The quintet has intensities 1:2:3:2:1 and a small spacing, since an H-D coupling is about one-sixth of the corresponding H-H coupling.
A triplet is what the n + 1 rule for spin-1/2 nuclei would give for two neighbours, a common slip here. A singlet would apply only if there were no coupled nuclei, and a quartet would need three spin-1/2 neighbours.
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