Integration and Relative Numbers of Protons

Introduction

Counting the signals in a ¹H NMR spectrum tells you how many proton environments a molecule has, but not how many hydrogens sit in each one. That extra information comes from integration : the area under each signal is proportional to the number of protons that produce it. A tall, narrow peak and a short, broad multiplet can represent the same number of hydrogens, so it is the area — not the height — that matters.

Core explanation

Why area measures number. Every ¹H nucleus in the sample absorbs radio-frequency energy with the same efficiency under standard acquisition conditions. A signal produced by three equivalent protons therefore absorbs three times as much energy in total as a signal from one proton. The total intensity is spread across the whole signal, including every line of a split multiplet, so the area under the signal, not its height, is proportional to the number of protons.

The integration trace. Older spectra show a stepped line drawn above the peaks. As the trace passes over each signal it rises by an amount proportional to that signal's area. You measure each vertical step with a ruler and compare them. Modern software instead prints a number under or above each signal; these numbers are relative, not absolute.

Relative, not absolute. An integral of "2.47" means nothing on its own. Integrals only tell you ratios. To turn them into numbers of hydrogens, divide every integral by the smallest one, then multiply by a small integer if necessary to clear decimals.

Example: ethanol. Ethanol, CH₃CH₂OH, gives three signals. If the integrals are 1.52, 1.01 and 0.50, dividing by 0.50 gives 3.0 : 2.0 : 1.0. The signals therefore come from the CH₃, CH₂ and OH groups respectively.

Using the molecular formula. The ratio from integration is only the simplest ratio. If a compound C₄H₈O₂ gives signals in the ratio 3 : 2 : 3, the sum is 8, which matches the eight hydrogens in the formula, so the actual numbers are 3, 2 and 3. If a compound with ten hydrogens gave a ratio of 2 : 3, the true numbers would be 4 and 6 — the ratio must be scaled to the total.

Typical integration patterns:

Group Protons Integral contribution --- --- --- CH₃ 3 3 CH₂ 2 2 Two equivalent CH₃ groups 6 6 tert-Butyl, C(CH₃)₃ 9 9 Monosubstituted benzene ring 5 5

Accuracy. Integrals are usually reliable to about ±5–10%. Values such as 2.9 or 3.1 should be read as 3. Broad signals from O–H or N–H protons often integrate less reliably, and residual solvent or water peaks must be excluded from the count.

Formulae

Number of protons in signal = (integral of signal ÷ sum of all integrals) × total number of hydrogens in the molecular formula.

Step-by-step reasoning

To turn integrals into numbers of protons:

1. Read the integral (or step height) for every signal from the compound. 2. Divide each by the smallest value. 3. Round to whole numbers, multiplying through if you get halves or thirds. 4. Add the ratio numbers and compare the total with the hydrogen count in the molecular formula. 5. Scale the ratio up if necessary so that it sums to that total.

Visual explanation

In the simulation, switch on the integration trace for ethyl ethanoate. The trace climbs in three steps: a step of 2 units over the quartet near δ 4.1, then 3 units over the singlet near δ 2.0 and 3 units over the triplet near δ 1.3. The steps are equal for the two CH₃ signals even though one is a single tall line and the other is split into three shorter lines.

Real-world analogy

Imagine weighing bags of sugar that have been poured into piles of different shapes. A tall, narrow pile and a wide, flat pile can contain the same amount of sugar. Judging by height alone would mislead you; measuring the total amount — the area — gives the true comparison.

Real-world example

Pharmaceutical analysts use quantitative NMR (qNMR) to measure the purity of drug substances. A precisely weighed internal standard with a known number of protons is added, and comparing its integral with a signal from the drug gives the drug's amount directly, without needing a calibration curve.

Why?

Why do we divide by the smallest integral? The spectrometer's absolute intensity scale depends on sample concentration, instrument gain and settings, so raw values vary from run to run. Dividing by one signal cancels these factors and leaves only the ratio of protons, which depends on the structure alone.

Common misconception

"The tallest peak has the most protons." Height depends on how the signal is split and how broad it is. A singlet from three protons can be taller than a quartet from two protons and yet both areas reflect the true counts. Always compare areas or integral values.

Worked example

Question: A compound C₅H₁₀O gives two ¹H signals with integrals 0.84 and 1.26. Find the number of protons in each environment.

Reasoning: Divide by 0.84: 1.00 : 1.50. Multiply by 2: 2 : 3. The sum is 5, but the molecule has 10 hydrogens, so double again: 4 : 6.

Answer: One environment contains 4 protons and the other 6, consistent with pentan-3-one, (CH₃CH₂)₂CO.

Quick check

1. Three signals have integrals 0.9, 0.6 and 1.8. What is the simplest whole-number proton ratio? Answer: Dividing by 0.6 gives 1.5 : 1 : 3, so the whole-number ratio is 3 : 2 : 6.

Exam focus

Show your working when converting integrals: divide by the smallest, round, then check against the molecular formula. Examiners penalise answers that quote a simplest ratio without scaling it to the total number of hydrogens when the formula is given.

Advanced insight

Accurate integration requires the nuclei to relax fully between pulses. If the delay between scans is too short, protons with slow relaxation, such as isolated aromatic or quaternary-adjacent protons, are under-represented. Quantitative NMR work therefore uses long relaxation delays, often five times the longest relaxation time. In routine ¹³C spectra this condition is not met, which is why carbon peaks are not normally integrated.

Summary

The area under a ¹H NMR signal is proportional to the number of protons producing it. Integration traces or integral values give only relative numbers, so divide by the smallest value, round to whole numbers and scale to the total hydrogen count from the molecular formula. Area, not height, is what counts.

Practice questions

1. Why is peak area rather than peak height used to compare numbers of protons? Answer: Splitting and broadening spread the intensity over different widths, so only the total area is proportional to the number of protons. 2. A compound C₃H₆O₂ gives integrals of 1.0 and 1.0 for its two signals. How many protons are in each environment? Answer: The ratio is 1 : 1 with six hydrogens in total, so each signal represents three protons. 3. The integration steps for a compound measure 12 mm, 8 mm and 12 mm. Give the proton ratio. Answer: Dividing by 8 gives 1.5 : 1 : 1.5, which doubles to 3 : 2 : 3. 4. A compound with 12 hydrogens gives integrals in the ratio 1 : 3. How many protons are in each environment? Answer: The ratio sums to 4, so multiply by 3 to get 3 and 9 protons.