NMR Spin-Spin Splitting: The n+1 Rule, Pascal Triangle and J

By Aniket Bhardwaj · 6 November 2026 · Updated 6 November 2026 · IIT-JAM / GATE / CSIR-NET Spectroscopy

Chemical shift tells you where a proton signal appears. Spin–spin splitting tells you how many neighbouring protons it has. In IIT-JAM, GATE, CSIR-NET and CUET-PG, a typical NMR question gives you a formula and a few signals with their shape, and asks for the structure. If you read the splitting pattern correctly, the answer follows quickly. This article explains the n + 1 rule, intensity ratios, the coupling constant J, and what to do when a proton has two different sets of neighbours.

An ethyl CH₂ quartet has 4 lines with intensities 1:3:3:1 at δ 2.373, 2.391, 2.409 and 2.427 ppm, with adjacent lines separated by J = 7.2 Hz. — Three equivalent neighbours split CH₂ into a 1:3:3:1 quartet; adjacent lines are separated by J.

The n + 1 rule

Number of lines = n + 1
n = number of equivalent neighbouring protons, on atoms next to the one carrying the observed proton (three bonds away, H–C–C–H).

The intensities of the lines follow Pascal's triangle. Each row is made by adding the two numbers above it:

n	Lines	Name	Intensity ratio
0	1	singlet (s)	1
1	2	doublet (d)	1 : 1
2	3	triplet (t)	1 : 2 : 1
3	4	quartet (q)	1 : 3 : 3 : 1
4	5	quintet	1 : 4 : 6 : 4 : 1
5	6	sextet	1 : 5 : 10 : 10 : 5 : 1
6	7	septet	1 : 6 : 15 : 20 : 15 : 6 : 1

The reason is simple. Each neighbouring proton can have its spin aligned with the field or against it. One neighbour gives two arrangements, two neighbours give four arrangements, grouped as 1 + 2 + 1, and so on. The number of ways to reach each total gives the line intensities. The total intensity of a multiplet is proportional to the number of protons that produce it, and the lines are centred on the chemical shift of those protons.

Important rules and limits

Protons that are equivalent do not split each other. The three H of a CH₃ group show one signal, not a pattern from their mutual coupling.
Normal splitting needs three bonds between the protons (H–C–C–H). Protons on the two sides of an oxygen, as in CH₃–O–CH₂–, are four bonds apart (H–C–O–C–H) and usually do not split each other.
O–H and N–H protons usually appear as broad singlets because of fast exchange, and generally do not split their neighbours. Check the conditions given in the question.
The simple n + 1 rule works when the coupling constants to all the neighbours are similar, and when the shift difference between the coupled protons (in Hz) is much larger than J (a first-order spectrum).

The coupling constant J

The distance between neighbouring lines in a multiplet is the coupling constant J. It is measured in hertz, not ppm, and it does not change with the spectrometer field. Chemical shift in Hz, on the other hand, does change with the field:

Shift in Hz = δ (ppm) × spectrometer frequency (MHz)

So a given J takes up fewer ppm on a stronger magnet, and the multiplets become better separated. This is why higher-field instruments give clearer spectra.

Worked examples

Example 1: ethyl group, line positions. In an ethyl group CH₃CH₂–X at 400 MHz, the CH₂ appears at δ 2.40 and the CH₃ at δ 1.20, with J = 7.2 Hz. Describe both signals.

CH₂ has 3 equivalent neighbours (the CH₃), so n + 1 = 4: a quartet, intensity 1 : 3 : 3 : 1.
CH₃ has 2 equivalent neighbours (the CH₂), so n + 1 = 3: a triplet, intensity 1 : 2 : 1.
Frequencies: 2.40 ppm × 400 MHz = 960 Hz; 1.20 ppm × 400 = 480 Hz.
Quartet lines sit at −1.5 J, −0.5 J, +0.5 J, +1.5 J from the centre, which is −10.8, −3.6, +3.6, +10.8 Hz. In ppm (divide by 400): δ 2.373, 2.391, 2.409, 2.427.
Triplet lines sit at −J, 0, +J, i.e. −7.2, 0, +7.2 Hz, or δ 1.182, 1.200, 1.218.
The same J (7.2 Hz) appears in both multiplets, which proves the two groups are coupled to each other. Integration gives the ratio CH₂ : CH₃ = 2 : 3.

Example 2: identify an isomer. A compound C₃H₇Cl shows a doublet (6H), a septet (1H) and nothing else. Find the structure.

A 6H doublet means six equivalent protons with one neighbour (n = 1), which fits two CH₃ groups on a CH. The septet (n + 1 = 7, so n = 6) means one proton with six equivalent neighbours, the same CH. So the compound is (CH₃)₂CHCl, 2-chloropropane.
Its isomer 1-chloropropane would show a triplet (3H), a sextet (2H, five neighbours) and a triplet (2H), so the two isomers are easy to tell apart by splitting alone.

Example 3: two different neighbours. In CH₃CH₂CH₂Cl, the middle CH₂ is next to a CH₃ (3 H) and another CH₂ (2 H). If the two J values are nearly equal, the n + 1 rule gives (3 + 2) + 1 = 6 lines, a sextet.

If the J values are different, say J_a = 7.2 Hz to the CH₃ and J_b = 6.5 Hz to the CH₂, the pattern is a quartet of triplets: 4 × 3 = 12 lines. Count it as (3 + 1) × (2 + 1). In most textbook problems the J values are taken as equal, so the sextet is the usual answer; always read the question to see which is wanted.

Example 4: counting lines. A proton has 1 neighbour of one type and 2 neighbours of another type, with clearly different J. How many lines?

(1 + 1) × (2 + 1) = 2 × 3 = 6 lines, called a doublet of triplets (dt). The order in the name goes from the largest J to the smallest.

Common mistakes

Counting the protons on the same carbon. Splitting comes from the neighbours, not from the group itself.
Saying that J is given in ppm, or that J changes when the magnet is changed.
Forgetting that equivalent protons (like those in a CH₃) do not split each other.
Expecting splitting from O–H protons in an exchanging solvent.
Using n + 1 with n = number of lines. The rule uses the number of neighbouring protons.
Forgetting that the intensity ratio must match Pascal's triangle; a quartet is 1 : 3 : 3 : 1, not 1 : 2 : 2 : 1.

Quick summary

Fragment	Signal of the first group	Signal of the second group
CH₃–CH₂–	CH₃: triplet (3H)	CH₂: quartet (2H)
(CH₃)₂CH–	CH₃: doublet (6H)	CH: septet (1H)
CH₃–CH<	CH₃: doublet (3H)	CH: quartet (1H)
–CH₂–CH₂– (different ends)	each: triplet (2H)	each: triplet (2H)
CH₃ on a carbon with no H neighbour	singlet (3H)	—