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

Before you start, three words you must have crisp (from the parent note):

Chemical shift $δ$ — WHERE a peak sits, in ppm, relative to TMS at $0$ . Low $δ$ = shielded (electron-rich); high $δ$ = deshielded (electron-poor).
Integration — the AREA under a peak, proportional to the NUMBER of equivalent protons. Only the ratio matters.
Multiplicity ( $n + 1$ rule) — a proton with $n$ equivalent neighbours on adjacent carbons splits into $n + 1$ lines, from spin–spin coupling, with intensities read off Pascal's triangle.

Level 1 — Recognition

Exercise 1.1

A molecule has three signals whose integrals measure $9.0$ , $6.0$ , $3.0$ (in arbitrary height units). What is the smallest whole-number proton ratio?

Recall Solution

WHAT we do: integrals give relative areas only, so we divide every number by the smallest to expose the simplest whole-number ratio. WHY: NMR cannot report absolute counts; only ratios are reliable, so we normalise. $9.0 : 6.0 : 3.0 \div 3.0 3 : 2 : 1$ Answer: $3 : 2 : 1$ . (Six protons could sit inside those groups too — e.g. $6 : 4 : 2$ — but the simplest ratio is $3 : 2 : 1$ .)

Exercise 1.2

Match each proton to "shielded (low $δ$ )" or "deshielded (high $δ$ )": (a) $- C H_{3}$ of an alkane, (b) aromatic $A r - H$ , (c) a carboxylic acid $- C O O H$ .

Recall Solution

Rule: electron-poor surroundings pull electron density away, so the proton feels more of $B_{0}$ → deshielded → high $δ$ . See the ruler figure below.

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

(a) $- C H_{3}$ alkane: only C and H neighbours, electron-rich → shielded, $δ \approx 0.9$ .
(b) $A r - H$ : ring current strongly deshields → deshielded, $δ \approx 7$ .
(c) $- C O O H$ : extreme electron withdrawal + H-bonding → very deshielded, $δ \approx 11$ .

Level 2 — Application

Exercise 2.1

A proton has $4$ equivalent neighbouring protons. State (a) the multiplicity name, (b) the intensity pattern.

Recall Solution

WHAT/WHY: $n = 4$ equivalent neighbours → by the $n + 1$ rule the signal splits into $n + 1$ lines. The relative intensities count how many spin combinations give each net field — that is row $n$ of Pascal's triangle, the binomials $(k 4)$ . $n + 1 = 4 + 1 = 5 \Rightarrow quintet$ $(0 4) : (1 4) : (2 4) : (3 4) : (4 4) = 1 : 4 : 6 : 4 : 1$ Sum $= 2^{4} = 16$ (the row total).

Exercise 2.2

Predict the full ¹H NMR (shift ballpark, multiplicity, integral) of chloroethane, $C H_{3} C H_{2} C l$ .

Recall Solution

Two distinct proton environments: $C H_{3}$ and $C H_{2}$ .

$C H_{3}$ (3H): its only vicinal neighbours are the 2 protons of $C H_{2}$ → $n = 2$ → triplet, integral $3$ . Environment: plain alkyl → $δ \approx 1.5$ .
$C H_{2}$ (2H): vicinal neighbours are the 3 protons of $C H_{3}$ → $n = 3$ → quartet, integral $2$ . Attached to electronegative Cl (Electronegativity) → deshielded → $δ \approx 3.5$ . Spectrum: triplet (3H, ~1.5 ppm) + quartet (2H, ~3.5 ppm) — the classic ethyl signature.

Level 3 — Analysis

Exercise 3.1

A compound of formula $C_{3} H_{6} O$ shows: a singlet (6H, $δ \approx 2.1$ ) and nothing else. Identify it and explain the singlet.

Recall Solution

Degree of unsaturation: $\frac{2 ( 3 ) + 2 - 6}{2} = 1$ , so one ring or double bond — with an O this hints at a C=O. A single 6H singlet means all six protons are equivalent and have no vicinal H neighbours. Two equivalent $C H_{3}$ groups, each attached to a carbon bearing no H, fits: acetone, $(C H_{3})_{2} C = O$ .

Why singlet: the carbonyl carbon between the two methyls carries no protons, so $n = 0$ → $0 + 1 = 1$ line.
Why $δ \approx 2.1$ : the methyls sit next to a C=O (mild deshielding), landing in the 2.0–2.5 window.

Exercise 3.2

Isopropyl chloride $(C H_{3})_{2} C H C l$ : predict multiplicities and the integral ratio.

Recall Solution

Two environments: the six equivalent methyl H's and the single methine (CH) H.

$(C H_{3})_{2}$ (6H): each methyl's vicinal neighbour is the 1 methine H → $n = 1$ → doublet. $δ \approx 1.5$ .
$C H$ (1H): its vicinal neighbours are the 6 methyl H's (all equivalent) → $n = 6$ → septet (7 lines, $1 : 6 : 15 : 20 : 15 : 6 : 1$ ). Attached to Cl → deshielded → $δ \approx 4.1$ .
Integral ratio $= 6 : 1$ . Note the two-way symmetry of coupling: the methyls see 1 neighbour, the CH sees 6.

Level 4 — Synthesis

Exercise 4.1

An unknown $C_{2} H_{4} C l_{2}$ gives one singlet only. Which isomer is it — 1,1-dichloroethane or 1,2-dichloroethane — and why?

Recall Solution

1,1-dichloroethane, $C H C l_{2} C H_{3}$ : two environments ( $C H C l_{2}$ and $C H_{3}$ ), so it would show a quartet (1H) + doublet (3H) — NOT a singlet.
1,2-dichloroethane, $C l C H_{2} C H_{2} C l$ : all 4 protons are equivalent by symmetry (each $C H_{2}$ is identical). Equivalent protons don't split each other → one singlet. Answer: 1,2-dichloroethane, singlet (4H) at $δ \approx 3.7$ (deshielded by Cl).

Exercise 4.2

Design the ¹H NMR fingerprint of ethyl acetate, $C H_{3} - C O - O - C H_{2} - C H_{3}$ : list every signal with shift, multiplicity, and integral.

Recall Solution

Three distinct environments (the acetyl $C H_{3}$ is isolated; the ethyl $C H_{2}$ and $C H_{3}$ couple to each other):

Group	Neighbours ( $n$ )	Multiplicity	Integral	$δ$ (ppm)	Why the shift
$C H_{3} - C O -$ (acetyl)	0	singlet	3	~2.0	next to C=O, no vicinal H
$- O - C H_{2} -$	3 (of ethyl $C H_{3}$ )	quartet	2	~4.1	attached to O → strongly deshielded
$- C H_{2} - C H_{3}$	2 (of $C H_{2}$ )	triplet	3	~1.3	plain alkyl
Fingerprint: singlet(3H, 2.0) + quartet(2H, 4.1) + triplet(3H, 1.3). The isolated acetyl singlet is the tell-tale; the quartet+triplet is the ethyl pair.

Level 5 — Mastery

Exercise 5.1

A neutral compound $C_{8} H_{8} O$ gives exactly: a singlet (3H, $δ \approx 2.6$ ), and aromatic signals integrating to 5H near $δ \approx 7.5$ – $8.0$ . Deduce the structure and justify every signal.

Recall Solution

Step 1 — degrees of unsaturation: $\frac{2 ( 8 ) + 2 - 8}{2} = 5$ . Five degrees = strong hint of a benzene ring (4: three C=C + one ring) plus one more (a C=O). Step 2 — read the integrals: total observed H $= 3 + 5 = 8$ , matching $C_{8} H_{8} O$ . The 5H aromatic block means a mono-substituted benzene ( $C_{6} H_{5} -$ , five ring H's). Step 3 — the 3H singlet at 2.6: three equivalent H, no vicinal neighbours ( $n = 0$ → singlet), mildly deshielded → a $C H_{3}$ attached to a carbon bearing no H. Next to C=O fits the shift. Assemble: $C_{6} H_{5} - + - C O - + - C H_{3} = C_{6} H_{5} C O C H_{3}$ = acetophenone.

Aromatic 5H, $δ \approx 7.5$ – $8.0$ : ring current deshields; the ortho H's are pushed further downfield by the carbonyl.
$C H_{3}$ 3H singlet, $δ \approx 2.6$ : isolated methyl adjacent to C=O.

Exercise 5.2

Two isomers $C_{3} H_{8} O$ : 1-propanol $C H_{3} C H_{2} C H_{2} O H$ and 2-propanol $(C H_{3})_{2} C H O H$ . Using multiplicity and integration ALONE (ignore the exchangeable OH), decide which spectrum belongs to which.

Recall Solution

Ignore OH (broad, exchangeable — see the trap below). 1-propanol $C H_{3} C H_{2} C H_{2} O H$ — three carbon environments:

$C H_{3}$ (3H): 2 vicinal H → triplet.
middle $C H_{2}$ (2H): 3 + 2 = 5 non-equivalent vicinal H → a complex multiplet (sextet-like).
$O - C H_{2}$ (2H): 2 vicinal H → triplet, deshielded ( $δ \approx 3.6$ ). Carbon-H integral pattern (excluding OH) $= 3 : 2 : 2$ . 2-propanol $(C H_{3})_{2} C H O H$ — two carbon environments:
$(C H_{3})_{2}$ (6H): 1 vicinal H (the CH) → doublet.
$C H$ (1H): 6 vicinal H → septet, deshielded ( $δ \approx 4.0$ ). Carbon-H integral pattern (excluding OH) $= 6 : 1$ . Decision: a spectrum with a 6H doublet + 1H septet is 2-propanol; a spectrum with two triplets flanking a central multiplet (3:2:2) is 1-propanol.

Active Recall

Recall Rapid-fire (cover the answers)

Integral heights $12 : 8 : 4$ simplest ratio? ::: $3 : 2 : 1$ Proton with 6 equivalent neighbours → lines? ::: 7 (septet), $1 : 6 : 15 : 20 : 15 : 6 : 1$ Acetone shows how many ¹H signals? ::: one singlet (6H) Ethyl acetate acetyl $C H_{3}$ multiplicity? ::: singlet Which $C_{2} H_{4} C l_{2}$ isomer is a lone singlet? ::: 1,2-dichloroethane How to make an OH peak disappear? ::: shake with $D_{2} O$ (H/D exchange) 6H doublet + 1H septet identifies which alcohol? ::: 2-propanol

Recall One-glance cheat map

WHERE (ppm) = environment · AREA = how many H · LINES = $n + 1$ neighbours · intensities = Pascal's triangle row.