¹H NMR — chemical shift, multiplicity (n + 1 rule), integration; common ranges: Visual walkthrough

Prerequisites we lean on: Spin-spin coupling (J), Pascal's triangle, and the parent ¹H NMR topic note.

Step 1 — A proton is a tiny bar magnet with two choices

WHAT. Before any splitting, we need the raw object. A hydrogen nucleus (one proton) behaves like a microscopic bar magnet — it has a property called spin. In a big external magnetic field $B_{0}$ this little magnet can only sit in two ways.

WHY these two and not a smooth range? Because nuclear spin is quantised — the proton is a "spin-½" particle, which is physics-speak for "exactly two allowed orientations, no in-between." We label them:

==spin-up ( $↑$ )==: aligned with $B_{0}$ → slightly lower energy (comfortable).
==spin-down ( $↓$ )==: aligned against $B_{0}$ → slightly higher energy (strained).

The energy gap between them is $Δ E = fixed constants \frac{γ h}{2 π} \cdot field the proton actually feels B_{local}$

$γ$ — the gyromagnetic ratio, a fixed number for a proton (how strongly its magnet responds).
$h$ — Planck's constant, another fixed number.
$B_{local}$ — the only thing that changes from proton to proton. This is the star of the whole show.

PICTURE. A magnet in a field, two allowed orientations, two energy rungs.

Step 2 — One proton, no neighbours = one peak (a singlet)

WHAT. Take an isolated proton whose neighbouring carbons carry no hydrogens. It feels exactly one value of $B_{local}$ , absorbs one radio frequency, and gives one line: a singlet.

WHY start here? This is our "zero neighbours" baseline, $n = 0$ . The $n + 1$ rule predicts $0 + 1 = 1$ line — we want to see that one line before we perturb it.

$ν = \frac{γ}{2 π} B_{local} ⟹ single frequency ⟹ single peak$

$ν$ — the frequency the proton absorbs.
one $B_{local}$ value $\to$ one $ν$ $\to$ one vertical line on the spectrum.

PICTURE. A clean single spike sitting at its chemical shift $δ$ .

Step 3 — Bring in ONE neighbour: the peak splits into two (a doublet)

WHAT. Now put a single hydrogen on the adjacent carbon. That neighbour is itself a tiny magnet (Step 1), so it too points $↑$ or $↓$ — and its field adds to or subtracts from the field our proton feels.

WHY does that split the line? Because our proton no longer feels one $B_{local}$ . Across the millions of molecules in the tube:

in half of them the neighbour is $↑$ → our proton feels $B_{local} + \frac{1}{2} J^{'}$ → absorbs at a slightly higher frequency.
in the other half the neighbour is $↓$ → our proton feels $B_{local} - \frac{1}{2} J^{'}$ → absorbs slightly lower.

Two sub-populations → two lines of equal height → a doublet. The gap between them is the ==coupling constant $J$ == (see Spin-spin coupling (J)) — a fixed splitting measured in Hz that does not change with the spectrometer.

$1 neighbour \to {↑, ↓} \to 2 equally likely fields \to 2 lines, ratio 1 : 1$

PICTURE. The single line of Step 2 tears into two even lines, split by $J$ .

Step 4 — TWO equivalent neighbours: the 1 : 2 : 1 triplet is born

WHAT. Two equivalent H's on the adjacent carbon. Each independently points $↑$ or $↓$ , so we list all combinations and add up their field contributions.

WHY enumerate combinations? Because the field our proton feels depends on the net tilt of the two neighbours. Different combinations can give the same net field — and when they do, those molecules pile onto the same line, making it taller.

Neighbour states	Net field shift	Line
$↑↑$	$+ 1$ unit	high-frequency line
$↑↓$ or $↓↑$	$0$ (they cancel)	centre line
$↓↓$	$- 1$ unit	low-frequency line

The top and bottom rows have 1 way each.
The middle field ( $0$ ) is reached 2 ways ( $↑↓$ and $↓↑$ ) → twice as tall.

Result: 3 lines in ratio $1 : 2 : 1$ — a triplet. Three distinct net fields = $2 + 1 = 3$ lines. There is the $n + 1$ rule appearing for $n = 2$ .

${↑↑} ∣ {↑↓, ↓↑} ∣ {↓↓} \Rightarrow 1 : 2 : 1$

PICTURE. The four spin arrangements sorted into three field bins, with the middle bin twice as full.

Step 5 — The pattern is exactly Pascal's triangle (and why)

WHAT. Generalise. With $n$ equivalent spin-½ neighbours, the number of ways to get $k$ of them pointing $↑$ is the binomial coefficient $(k n) = \frac{n !}{k ! ( n - k )!}, k = 0, 1, \dots, n$

$n$ — how many neighbours.
$k$ — how many of them chose $↑$ .
$(k n)$ — the count of arrangements giving that particular net tilt = the height of one line.

WHY is this Pascal's triangle? Because $(k n)$ is the entry in row $n$ of Pascal's triangle — the same "add the two numbers above you" object. Each new neighbour is like adding one more coin flip; the counts propagate exactly like Pascal's rule. So:

$number of lines = values k can take: 0 \dots n n + 1, total intensity = \sum_{k = 0}^{n} (k n) = 2^{n}$

$n + 1$ — because $k$ runs over $n + 1$ values → the $n + 1$ rule, derived.
$2^{n}$ — total number of spin arrangements ( $2$ choices, $n$ neighbours) = the row sum of Pascal's triangle.

$n$	lines	Pascal row
0	1 (singlet)	1
1	2 (doublet)	1 1
2	3 (triplet)	1 2 1
3	4 (quartet)	1 3 3 1
4	5 (quintet)	1 4 6 4 1

PICTURE. Pascal's triangle drawn as a stack of NMR peaks — each row is a multiplet.

Step 6 — Edge case: why equivalent protons DON'T split each other

WHAT. The three H's of a lone $CH_{3}$ never split each other, even though each is a magnet with a neighbour. A CH₃ on an otherwise-H-free carbon is a singlet, not a quartet-of-itself.

WHY? Splitting is visible only when a spin flip changes the transition energy you observe. For chemically equivalent protons, every spin flip shifts all their energy levels by the same amount — the up-shift and down-shift line up and cancel in the observed transition. Formally, there is no net difference in $B_{local}$ to report, so the coupling is real but silent.

$equivalent partners \Rightarrow Δ (observed frequency) = 0 \Rightarrow no splitting$

PICTURE. Two equivalent protons: their coupling shifts both levels identically, the observed line stays single.

Step 7 — Edge case: zero neighbours and the "many neighbours" limit

WHAT. Two boundary checks so no scenario surprises you.

$n = 0$ (no vicinal H): $0 + 1 = 1$ → a singlet. Matches Step 2. ✔
$n$ large: the $1 : n : (2 n) : \dots$ pattern gets very lopsided — a tall middle, tiny wings. The outermost lines ( $1$ at each end) become so short relative to the centre they may vanish into noise, so a septet ( $n = 6$ , e.g. the CH of isopropyl) shows a strong centre and faint edges.

WHY show the limit? So that when you see a "5-line-looking" septet in real data, you know the two faint outer lines are there — the maths says $1 : 6 : 15 : 20 : 15 : 6 : 1$ — they're just dwarfed by the $20$ in the middle.

$n = 6 : 1 : 6 : 15 : 20 : 15 : 6 : 1 (sum = 2^{6} = 64)$

PICTURE. Doublet, triplet, quartet, septet side by side — watch the wings shrink as $n$ grows.

Step 8 — Putting all three readouts together (ethanol)

WHAT. Combine chemical shift (Step-1 field idea) + integration + the splitting we just derived, on ethanol $CH_{3} CH_{2} OH$ .

WHY ethanol? It exercises every piece: one triplet, one quartet, one exchange-broadened singlet.

Group	Neighbours ( $n$ )	Multiplicity ( $n + 1$ )	Integral	$δ$ / ppm
$CH_{3}$	2 (the CH₂)	triplet $1 : 2 : 1$	3	~1.2
$CH_{2}$	3 (the CH₃)	quartet $1 : 3 : 3 : 1$	2	~3.7 (near O → deshielded, cf. Electronegativity)
$O - H$	fast exchange	singlet (broad)	1	variable

CH₃ sees the $2$ CH₂ protons → Step 4 → triplet.
CH₂ sees the $3$ CH₃ protons → Step 5, $n = 3$ → quartet.
OH: fast proton exchange averages away the coupling → Step 6's "silence" for a different reason → broad singlet.

PICTURE. The full ethanol spectrum, each peak labelled with why it has that shape.

The one-picture summary

Recall Feynman retelling — say it back in plain words

Every hydrogen is a minuscule magnet that can only face two ways in a strong field. On its own it feels one field and gives one spike. Now a hydrogen on the next-door carbon is also a magnet, and its choice — face up or face down — nudges the field our hydrogen feels. One neighbour has two choices, so our spike splits into two even lines. Two neighbours give four choices, but two of them (up-down, down-up) cancel to the same field, so the middle line is double: that's the 1-2-1 triplet. Keep adding neighbours and the "how many ways can I get this net tilt" count is exactly Pascal's triangle — which instantly tells you there are $n + 1$ lines and their heights sum to $2^{n}$ . The only twist: neighbours that are identical to our hydrogen shift everything equally, so their coupling never shows up — same-neighbour means no splitting. Read shift for where, integral for how many, and this line-count for how many neighbours, and you've decoded the molecule.

Recall Rapid self-test

$n = 3$ gives how many lines and what pattern? ::: 4 lines, $1 : 3 : 3 : 1$ (quartet) Why do equivalent protons not split each other? ::: their coupling shifts all levels equally → no observed frequency difference What is the total intensity of a multiplet from $n$ neighbours? ::: $2^{n}$ Why is a lone CH₃ a singlet? ::: $n = 0$ vicinal H (its own 3 H are equivalent, so silent) → $0 + 1 = 1$ A septet's outer lines are faint — are they really there? ::: yes, ratio $1 : 6 : 15 : 20 : 15 : 6 : 1$ , wings dwarfed by centre

Related: ¹³C NMR · Aromaticity & ring current · Tetramethylsilane (TMS) · Integration in IR/MS.