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

The scenario matrix

Every ¹H NMR problem is really a combination of a few case-types. Here is the complete list — every worked example below is tagged with the cell(s) it covers.

Cell	Case class	What makes it tricky	Covered by
A	Simple ethyl / alkyl splitting	plain $n + 1$ , no traps	Ex 1
B	Isolated singlet (no vicinal H)	$n = 0$ → 1 line	Ex 2
C	Equivalent protons don't split	symmetry "degenerate" case	Ex 2, Ex 3
D	Two different neighbour sets	which H couples to which	Ex 4
E	Integration-only counting	ratios, tert-butyl vs CHO trap	Ex 5
F	Electronegativity → $δ$ ordering	rank shifts by environment	Ex 6
G	Aromatic ring current (limiting high $δ$ )	why $\sim 7$ ppm	Ex 7
H	Exchangeable OH/NH (degenerate splitting)	broad singlet, D₂O trick	Ex 8
I	Real-world / structure-from-data	reverse-engineer a molecule	Ex 9
J	Exam twist (symmetry hides atoms)	fewer signals than carbons	Ex 10

The sign/quadrant analogue in NMR is the direction of the shift (upfield vs downfield) and the degenerate cases (equivalent protons, $n = 0$ , fast exchange). We cover all of them.

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

Ex 1 — Bromoethane $C H_{3} C H_{2} B r$ (Cell A)

Worked example Predict the full spectrum of

$C H_{3} C H_{2} B r$ . Forecast: How many signals? What multiplicity and integration for each? Which one is more downfield?

Steps

Identify the distinct proton environments. There are two: the three $C H_{3}$ protons (all equivalent to each other) and the two $C H_{2}$ protons. Why this step? Signals = number of chemically different proton sets. Equivalent H's share one peak.
Apply $n + 1$ to $C H_{3}$ . Its vicinal neighbours are the 2 $C H_{2}$ protons → $n = 2$ → $2 + 1 = 3$ lines → triplet. Why this step? The rule counts protons on adjacent carbons only.
Apply $n + 1$ to $C H_{2}$ . Its neighbours are the 3 $C H_{3}$ protons → $n = 3$ → quartet. Why this step? Same rule from the other side.
Assign shifts. $C H_{2}$ is attached to Br (electronegative → deshielded) so it sits downfield ( $δ \approx 3.4$ ); $C H_{3}$ is only next to carbon ( $δ \approx 1.7$ ). Why this step? See Electronegativity — pulling electron density away lowers shielding, raising $δ$ .
Integration: $C H_{3} : C H_{2} = 3 : 2$ .

Verify: Total H predicted $= 3 + 2 = 5$ . Formula $C_{2} H_{5} B r$ has 5 H ✓. Multiplicities $3 + 4 =$ lines fit Pascal's triangle rows $121$ and $1331$ ✓.

Ex 2 — Neopentane vs acetone (Cells B, C)

Worked example Why does neopentane

$C (C H_{3})_{4}$ give a single peak, and 2,2-dimethylpropane's spectrum look "empty"? Forecast: How many signals, and what multiplicity?

Steps

Draw neopentane: a central carbon with no H, bonded to four $C H_{3}$ groups. Why this step? Splitting needs vicinal H's on the adjacent carbon — the central carbon has none.
All four $C H_{3}$ groups are equivalent by symmetry → one signal, integration = all 12 H. Why this step? Symmetry-equivalent protons collapse into one peak (Cell C).
Multiplicity: neighbours = 0 (central C carries no H) → $n = 0$ → $0 + 1 = 1$ line → singlet. Why this step? This is the degenerate $n = 0$ case — the analogue of "zero input."
Shift: only C,H neighbours → $δ \approx 0.9$ (alkane range).

Verify: $C (C H_{3})_{4} = C_{5} H_{12}$ , so 12 H in one singlet. $12$ H, $1$ signal ✓. The "empty" look = one small- $δ$ singlet, no splitting.

Ex 3 — Why equivalent protons never split each other (Cell C)

Ex 4 — 1,1,2-trichloroethane $C H C l_{2} C H_{2} C l$ (Cell D)

Worked example Assign both signals of

$C H C l_{2} C H_{2} C l$ . Forecast: Which proton is a doublet and which a triplet? Which is more downfield?

Steps

Two environments: the lone $C H C l_{2}$ proton and the two $C H_{2} C l$ protons. Why this step? Different neighbours + different substituents → two peaks.
$C H_{2} C l$ : vicinal neighbour is the single $C H C l_{2}$ proton → $n = 1$ → doublet. Why this step? $1 + 1 = 2$ ; one neighbour with two spin states gives two lines.
$C H C l_{2}$ : vicinal neighbours are the two $C H_{2}$ protons → $n = 2$ → triplet. Why this step? $2 + 1 = 3$ ; two equivalent spins give populations $1 : 2 : 1$ .
Shifts: $C H C l_{2}$ carries two Cl → very deshielded, $δ \approx 5.8$ . $C H_{2} C l$ carries one Cl, $δ \approx 3.9$ . Why this step? More electronegative atoms on the same carbon ⇒ higher $δ$ .

Verify: Total H $= 1 + 2 = 3$ ; $C_{2} H_{3} C l_{3}$ has 3 H ✓. Doublet integrates 2, triplet integrates 1, so 2:1 ratio ✓.

Ex 5 — Integration trap: tert-butyl acetate (Cell E)

Worked example

$C H_{3} - C (= O) - O - C (C H_{3})_{3}$ gives peaks with raw integral heights 27 : 9. Assign them. Forecast: The tall peak has more protons — is it the more downfield one?

Steps

Reduce the ratio: divide $27 : 9$ by 9 → 3 : 1... but wait, count H's: acetyl $C H_{3}$ has 3 H; $C (C H_{3})_{3}$ has 9 H. Why this step? The true H ratio is $3 : 9 = 1 : 3$ , so height $27$ = the 9-H group and $9$ = the 3-H group. Integration reports area = number of H, not position.
So the taller peak (integral 27) is the $C (C H_{3})_{3}$ , and it sits upfield at $δ \approx 1.4$ . Why this step? This kills the "tall = downfield" myth. Height tracks H-count only.
The acetyl $C H_{3}$ (integral 9, meaning 3 H) is next to $C = O$ , $δ \approx 2.0$ . Why this step? Being adjacent to a carbonyl is mild deshielding.
Both are singlets — no vicinal H on either side.

Verify: H ratio: tert-butyl $9$ vs acetyl $3 = 3 : 1$ . Given heights $27 : 9 = 3 : 1$ ✓ (the 9-H group is 3× taller). See Integration in IR/MS for the general "area = amount" idea.

Ex 6 — Rank the shifts by electronegativity (Cell F)

Ex 7 — Aromatic ring current: benzene (Cell G, limiting high $δ$ )

Worked example Why is benzene's single proton signal way out at

$δ \approx 7.3$ , far downfield of a normal vinyl $= C H$ ( $δ \approx 5.5$ )? Forecast: Ring current — does it shield or deshield the ring H's?

Steps

In benzene the $π$ electrons circulate around the ring like current in a loop, driven by $B_{0}$ (see Aromaticity & ring current). Why this step? A moving charge makes its own magnetic field — this is the extra effect beyond plain electronegativity.
This ring current creates a field that, outside the ring (where the H's sit), adds to $B_{0}$ . Why this step? The induced field reinforces $B_{0}$ at the ring-edge protons → extra deshielding → higher $δ$ .
Result: aromatic H's land at $δ \approx 6.5$ – $8.0$ , higher than isolated $π$ (vinyl) protons.

Verify: Benzene: 6 equivalent H → one singlet, integration 6, at $δ \approx 7.3$ ✓. This is the "limiting" high- $δ$ case short of aldehyde/acid.

Ex 8 — Exchangeable OH: ethanol revisited (Cell H)

Worked example Ethanol

$C H_{3} C H_{2} O H$ shows CH₃ triplet, CH₂ quartet, and OH... what, and what happens with D₂O? Forecast: Does the OH split the CH₂ into extra lines? What does adding D₂O do?

Steps

The OH proton undergoes fast chemical exchange — it hops between molecules faster than the NMR timescale. Why this step? Fast exchange averages out the coupling, so OH shows as a broad singlet and doesn't cleanly split its neighbour.
So CH₂ stays a clean quartet (coupling only to CH₃'s 3 H, $n = 3$ ), not a more complex pattern. Why this step? This is the degenerate "no-coupling" case for exchangeable protons.
Add D₂O: the OH proton swaps for deuterium (D), which NMR doesn't detect at ¹H frequency → the OH peak vanishes. Why this step? A diagnostic trick to identify OH/NH protons.

Verify: Ethanol summary: triplet (3H) + quartet (2H) + singlet (1H) = 6 H total; $C_{2} H_{6} O$ has 6 H ✓. After D₂O: 5 H visible ✓.

Ex 9 — Structure from data (Cell I, real-world)

Worked example A compound

$C_{3} H_{6} O$ shows: singlet, integral 6, $δ \approx 2.1$ , and nothing else. What is it? Forecast: One signal, 6 H, near a carbonyl-ish shift — guess the structure.

Steps

Count H accounted for: 6 H in one signal, but formula says 6 H total → all H are equivalent. Why this step? One signal for all H demands high symmetry.
Molecular formula $C_{3} H_{6} O$ : degrees of unsaturation $= \frac{2 ( 3 ) + 2 - 6}{2} = 1$ → one ring or one double bond. Why this step? The formula tells us a $C = O$ or ring is present; $δ \approx 2.1$ and the missing 6 H on a carbonyl points to a $C = O$ .
Two equivalent $C H_{3}$ groups flanking a $C = O$ = acetone, $(C H_{3})_{2} C = O$ . Why this step? Both methyls are equivalent (singlet, 6 H), next to carbonyl ( $δ \approx 2.1$ ), no vicinal H → no splitting.

Verify: Acetone $C_{3} H_{6} O$ : 6 H, all equivalent, singlet at $δ \approx 2.1$ ✓. Degrees of unsaturation $= 1$ (one C=O) ✓.

Ex 10 — Exam twist: symmetry hides carbons (Cell J)

Worked example

p-xylene (1,4-dimethylbenzene) has 8 carbons. How many ¹H signals? Give multiplicities and integrations. Forecast: Eight carbons — but how many different kinds of proton?

Steps

Identify proton types. The two $C H_{3}$ groups are equivalent by symmetry → one methyl signal (6 H). The four aromatic H's are all equivalent (para-symmetry) → one aromatic signal (4 H). Why this step? Molecular symmetry collapses many atoms into few signals — the classic exam trap.
Multiplicity of $C H_{3}$ : attached to the aromatic ring carbon (no H on it) → $n = 0$ → singlet, $δ \approx 2.3$ . Why this step? Benzylic methyls are next to a carbon bearing no H.
Multiplicity of Ar–H: all four are equivalent → equivalent protons don't split each other → singlet, $δ \approx 7.0$ . Why this step? Same degenerate rule as Ex 3, now on the ring.

Verify: Two signals: $C H_{3}$ (6 H, singlet, 2.3) and Ar–H (4 H, singlet, 7.0). Total H $= 6 + 4 = 10$ ; $C_{8} H_{10}$ has 10 H ✓. Ratio $6 : 4 = 3 : 2$ ✓. Compare ¹³C NMR, which would show more carbon environments.

Active Recall

Recall Cover the answers

Neopentane $C (C H_{3})_{4}$ : how many signals? ::: One singlet, 12 H (Cell B/C).
$C H C l_{2} C H_{2} C l$ : which proton is the triplet? ::: The $C H C l_{2}$ proton ( $n = 2$ neighbours) at $δ \approx 5.8$ .
Tert-butyl (9 H) at 1.4 vs aldehyde (1 H) at 9.7 — which peak is taller? ::: The tert-butyl (area = H count, not position).
Why is benzene at $δ \approx 7.3$ ? ::: Ring current adds to $B_{0}$ at ring-edge H's → extra deshielding.
What does D₂O do to an OH peak? ::: Exchanges H for D → the OH signal disappears.
$C_{3} H_{6} O$ , one singlet 6 H at 2.1 — what is it? ::: Acetone.