The ¹H Chemical Shift Table

Introduction

Every chemist interpreting proton NMR spectra relies on a chemical shift table: a compact summary showing where each kind of proton usually absorbs. Exam papers provide one on the data sheet; professionals carry the key values in their heads. The table is not a random list. Its order follows directly from the ideas of shielding, electronegativity and anisotropy, so understanding the reasons makes it much easier to use sensibly. This page presents a working table, explains its structure, and shows how to use it to narrow down proton environments in an unknown compound.

Core explanation

A working ¹H chemical shift table. Typical ranges (δ in ppm relative to TMS) are:

Proton environment Example Typical δ --- --- --- R–CH₃, R–CH₂–R (alkyl) CH₃ of propane 0.9–1.7 C=C–CH (allylic) CH₃ of propene 1.6–2.2 R–C≡C–H (terminal alkyne) ethyne 1.8–3.1 CH₃–C=O, –CH₂–C=O CH₃ of propanone 2.0–2.7 Ar–CH (benzylic) CH₃ of methylbenzene 2.3–3.0 R₂N–CH CH₃ of methylamine 2.3–3.0 halogen–CH CH₂ of bromoethane 2.1–4.3 RO–CH (alcohol, ether) CH₃ of methanol 3.3–4.0 RCOO–CH (ester, O side) OCH₃ of methyl ethanoate 3.7–4.2 C=C–H (alkene) ethene 4.5–6.5 Ar–H (aromatic) benzene 6.5–8.0 R–CHO (aldehyde) ethanal 9.0–10.0 R–COOH (carboxylic acid) ethanoic acid 10–12 (broad) R–OH (alcohol) ethanol 0.5–5 (variable) Ar–OH (phenol) phenol 4–7 (variable) R–NH₂, R₂NH (amine) ethylamine 1–5 (variable) R–CONH– (amide N–H) ethanamide 5–8.5 (variable)

Reading the table as three zones.

- δ 0–3: protons on sp³ carbon not directly attached to a heteroatom. Simple alkyl groups are the most shielded. Neighbouring C=O, C=C or aromatic rings deshield slightly, giving the 2–3 ppm band. - δ 3–4.5: protons on carbon bonded to O, N or halogen. Inductive withdrawal by the electronegative atom deshields them. - δ 4.5–12: protons on sp² carbon or on acidic groups. Anisotropy of π systems moves alkene, aromatic and aldehyde protons far downfield; carboxylic acid O–H protons are strongly deshielded by hydrogen bonding and the carbonyl group.

Exchangeable protons are unreliable. O–H and N–H protons are shown with wide ranges because their δ depends on concentration, solvent, temperature and the extent of hydrogen bonding. They often give broad signals and do not usually split, or get split by, neighbouring C–H protons. Their identity is confirmed by exchanging them for deuterium, which makes the signal disappear.

Overlap and judgement. Some ranges overlap: an alcohol O–H might appear anywhere from 0.5 to 5, and a CH₂ next to bromine may sit close to one next to oxygen. The table gives probabilities, not certainties. Chemical shift is always combined with integration, splitting patterns, the molecular formula and evidence from IR and mass spectrometry. Multiple substituents also add, so a CH₂ between two oxygens or next to both an aromatic ring and oxygen appears higher than either range alone suggests.

Step-by-step reasoning

To use the table on an unknown signal:

1. Note the δ value and decide which zone it lies in. 2. List every environment in the table whose range includes that value. 3. Eliminate options that conflict with the molecular formula or other spectra, for example no oxygen means no RO–CH. 4. Check integration and splitting for consistency before assigning.

Visual explanation

Picture a δ axis from 12 to 0 coloured in bands: red at 10–12 for COOH, orange at 9–10 for CHO, purple at 6.5–8 for aromatic, green at 4.5–6.5 for alkenes, blue at 3.3–4.2 for O–CH, and grey at 0.9–3 for alkyl and near-functional-group protons, with a hatched band showing where variable O–H and N–H signals may fall.

Real-world analogy

A shift table is like a postcode map. A postcode tells you the district a house is in, which narrows the search dramatically, but you still need the street name and number (integration and splitting) to pinpoint the exact address.

Real-world example

Ethanol, CH₃CH₂OH, gives signals at about δ 1.2 (CH₃, alkyl range), δ 3.7 (CH₂, RO–CH range) and a variable O–H signal, often near δ 2–3 in CDCl₃. Every signal lies in the range the table predicts.

Why?

Why do carboxylic acid protons appear even further downfield than aldehyde protons? The O–H proton is attached directly to electronegative oxygen, the adjacent carbonyl withdraws further electron density, and the acids form strongly hydrogen-bonded dimers, which deshield the proton still more.

Common misconception

"A signal at δ 2.1 must be a CH₃ next to C=O." Several environments fall near 2.1, including allylic, alkyne, iodoalkane protons and possibly an O–H. The table suggests candidates; other evidence must be used to decide.

Worked example

Question: A compound C₃H₆O₂ shows ¹H signals at δ 11.7 (1H, broad), 2.4 (2H) and 1.2 (3H). Use the table to identify it.

Reasoning: δ 11.7, broad, 1H fits a COOH proton. δ 2.4 (2H) fits CH₂ next to C=O. δ 1.2 (3H) fits an alkyl CH₃. The formula C₃H₆O₂ accommodates CH₃CH₂COOH.

Answer: Propanoic acid, CH₃CH₂COOH.

Quick check

1. In which δ range would you expect to find the ring protons of methylbenzene, and what causes the shift? Answer: About δ 6.5–8.0, caused by deshielding from the aromatic ring current.

Exam focus

You will normally be given a shift table, so the skill tested is using it: match each δ value to possible environments, then use the molecular formula, integration and splitting to decide. Know that O–H and N–H signals vary and are often broad, and that the table ranges can overlap.

Advanced insight

Modern prediction software combines large databases of measured shifts with increment rules and quantum-chemical calculations to estimate spectra of proposed structures. Comparing predicted and measured spectra is now a routine check in structure determination, but chemists still rely on the underlying reasoning to spot errors in predictions.

Summary

The ¹H chemical shift table summarises typical δ ranges: alkyl protons 0.9–1.7; protons near C=O, C=C, rings or nitrogen 2–3; protons on carbon bonded to O or halogen 3–4.3; alkene 4.5–6.5; aromatic 6.5–8; aldehyde 9–10; carboxylic acid 10–12. O–H and N–H shifts vary. The order reflects shielding, electronegativity and anisotropy, and assignments must be confirmed with integration, splitting and other evidence.

Practice questions

1. Give the typical δ ranges for aromatic protons and aldehyde protons. Answer: Aromatic protons δ 6.5–8.0; aldehyde protons δ 9.0–10.0. 2. Why is the range given for alcohol O–H protons so wide? Answer: The shift depends on concentration, solvent and temperature because of hydrogen bonding and proton exchange. 3. A compound with no oxygen, nitrogen or halogen shows a signal at δ 5.2. What type of proton is most likely responsible? Answer: An alkene proton on a C=C double bond. 4. Methyl ethanoate shows two singlets at δ 2.0 and δ 3.7. Assign them using the table. Answer: δ 2.0 is the CH₃ bonded to the carbonyl carbon; δ 3.7 is the OCH₃ bonded to the ester oxygen.