Calculate the resulting first-order ¹H NMR multiplet from several groups of equivalent neighbouring protons (number n, coupling constant J) – including a short-hand name (e.g. "dt" for doublet of triplets) and a stick spectrum with relative intensities.
This calculator assumes first-order coupling (J much smaller than the chemical-shift difference to the coupling partners). With J values of a similar magnitude to the shift difference (Δδ), roofing effects and more complex patterns occur that are not modelled here.
In ¹H NMR spectroscopy, a proton couples with magnetically non-equivalent neighbouring protons through chemical bonds. Each group of equivalent neighbouring protons splits the signal into multiple lines – the result is the signal's characteristic coupling pattern (multiplet).
If a proton couples simultaneously with several different groups of neighbours (with different coupling constants J), the individual splittings overlap into a more complex pattern, which can be predicted by successive convolution (the tree-diagram method).
n+1 rule
Number of lines = n + 1 (per group of equivalent neighbours)
Number of equivalent neighbours (n)
The number of magnetically equivalent protons in this neighbouring group that couple with the proton under consideration (e.g. n=3 for a neighbouring CH₃ group).
Coupling constant (J)
The spacing between lines within a splitting pattern, in Hz. Typical values: ³J(H,H) in open-chain systems approx. 6–8 Hz, in alkenes (cis) approx. 6–12 Hz, (trans) approx. 12–18 Hz.
Pascal's triangle
The relative intensities within a single (n+1)-line system correspond to the binomial coefficients of the n-th row of Pascal's triangle (e.g. 1:3:3:1 for a quartet, n=3).
A proton couples with one neighbouring proton (J₁ = 8 Hz) and another, chemically distinct neighbouring proton (J₂ = 4 Hz). What pattern results?
Given
Group 1: n=1, J=8 Hz; Group 2: n=1, J=4 Hz
Solution
Convolving the two doublets gives a doublet of doublets (dd) with 4 equally intense lines at offsets −6, −2, +2, and +6 Hz relative to the centre of the signal.
A common mistake is treating chemically distinct neighbouring groups that happen to have a similar J as a single group instead of entering them separately – the convolution automatically accounts for lines coinciding due to equal J (see the note in the result). A second mistake: the n+1 rule only applies to first-order coupling with magnetically equivalent neighbours, not to strong coupling (Δδ on the order of J).
The best approach
First identify all chemically distinct neighbouring groups individually (don't count all neighbours together if they have different J values), enter each group separately, and finally check the short-hand name against the pattern you expect from the spectrum.