Analytical Techniques: Question 5
Syllabus 37.4
In the proton () NMR spectrum of a compound, one signal is a quartet integrating for 2H, and a second signal is a triplet integrating for 3H. No other signals are present, and there is no evidence of an exchangeable O–H or N–H proton.
Which structural fragment is consistent with this pattern?
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Worked solution
Applying the n + 1 rule
The n + 1 rule states that a proton’s signal is split into lines by equivalent protons on the adjacent carbon. Testing an isolated ethyl group, (flanked, on the far side, by an atom or group with no protons of its own. Such as an oxygen, a carbonyl carbon, or a halogen):
- The protons are adjacent to the group (3 equivalent H), so : a quartet, integrating for 2H.
- The protons are adjacent to the group (2 equivalent H), so : a triplet, integrating for 3H.
This is exactly the pattern described. A 2H quartet paired with a 3H triplet, and no other signals, because the and are coupled only to each other.
Why the other options are wrong
- B: an isopropyl group, , has two equivalent groups (6H total) split into a doublet by the single adjacent proton, and the proton (1H) split into a multiplet (a septet, from 6 equivalent neighbouring protons), a doublet + multiplet, not a quartet + triplet.
- C: a propyl group, , has three distinct proton environments (terminal , middle , and the attached to the rest of the molecule), giving three signals rather than two.
- D: an isolated group with no neighbouring protons on any adjacent carbon gives a single singlet, with no quartet involved at all.
Final answer
A, an isolated ethyl group, .