Chemical shift tells us where a proton environment appears in a 1H NMR spectrum. Splitting can tell us about neighbouring protons. But there is still another important question: how many protons belong to each environment?
This information comes from integration. Instead of treating every signal as equally important, we measure its total area. The relative areas then act like a molecular head count. Once we can turn those areas into proton numbers, an NMR spectrum becomes far more useful for working out molecular structure.
Try these five questions before reading the chapter. They focus on the central idea behind integration: converting areas on a 1H NMR spectrum into relative numbers of protons.
- 5 diagnostic questions
- NMR spectra and integration diagrams where useful
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What does peak area mean in 1H NMR?
An NMR spectrum is usually drawn with chemical shift, δ, along the horizontal axis and signal intensity along the vertical axis. It is tempting to look at the tallest point of a signal and use its height to decide how many protons are present. That is not the correct method.
Instead, we consider the total area associated with the complete signal. Mathematically, this can be thought of as integrating signal intensity across the chemical-shift range occupied by that signal.
Here, I(δ) represents the signal intensity at a particular chemical shift. You are not normally expected to perform this integration mathematically in an A-Level examination. The NMR instrument or processing software provides the integral values for you.
For signals in the same 1H NMR spectrum, integrated area is proportional to the number of protons contributing to the signal.
Why peak height is not the same as peak area
Peak height describes only how high the signal reaches at one particular point. Peak area takes the whole shape of the signal into account. Two signals can therefore reach similar heights yet have very different total areas.
Real NMR signals can also be split into several lines. A proton environment might appear as a doublet, triplet, quartet or more complex multiplet, but all of those lines together still form one signal for that proton environment. Its total integrated area is what is compared with the areas of the other signals.
Peak height: the vertical intensity at one point. Do not use this by itself for proton counting.
Integrated peak area: the total contribution over the complete signal or multiplet. This is the quantity used to obtain proton ratios.
Why peak area is related to the number of protons
Suppose one signal is produced by a set of three equivalent protons while another is produced by one proton. Under the assumptions used for normal examination spectra, the first signal will have approximately three times the integrated area of the second.
This gives us a molecular counting system. The spectrum does not need to tell us immediately that a signal represents exactly 3 H. It may instead report relative integrals such as 15 and 5, 3.0 and 1.0, or 6.0 and 2.0. All three pairs describe the same 3 : 1 proton ratio.
Equivalent protons contribute together
Protons that are chemically equivalent experience the same chemical environment and contribute to the same NMR signal. For example, the three hydrogen atoms within an ordinary methyl group, CH₃, are often chemically equivalent.
If that CH₃ group gives one signal, its integral represents all three protons together. The integral is therefore not counting the number of signals; it is counting the relative number of protons responsible for each signal.
A spectrum containing signals with proton counts of 3 H, 2 H and 1 H would have relative integrated areas of 3 : 2 : 1. The instrument might display these as 6.0 : 4.0 : 2.0 or 1.5 : 1.0 : 0.5 instead. The absolute numbers are unimportant; their ratio is what matters.
Reading an integration trace
Some NMR spectra show integration graphically as an integration trace, sometimes called an integral curve. Instead of giving only numerical values, the trace rises as it passes across each signal.
The important measurement is the change in vertical height of the integration trace across a signal. A larger vertical step represents a larger integrated area and therefore more protons.
Measure the rise of the integral trace across each complete signal, not its absolute vertical position on the page.
Why the height change matters
Imagine that the integration trace begins at an arbitrary height of 20 units. After the first signal it rises to 30 units, and after the second signal it rises to 50 units. The two relevant step heights are therefore:
Second step = 50 − 30 = 20
The proton ratio is therefore 10 : 20 = 1 : 2. The absolute values 20, 30 and 50 are not themselves proton counts.
How to calculate the proton ratio from peak areas
Most examination questions make the integration calculation quite mechanical once you know what to look for. The goal is to convert the measured values into the smallest chemically sensible whole-number ratio.
Step 1: Write down the integral for every signal
Use the numerical integral printed on the spectrum, or calculate the change in height of each step of the integration trace.
Step 2: Find a common scale
If the values have an obvious common factor, divide by it. For example:
= 3 : 2 : 1
A useful first attempt is often to divide all the values by the smallest integral. But this does not always give whole numbers immediately.
Step 3: Convert sensible fractions to whole numbers
Suppose the integrals are:
Multiplying both values by 2 gives:
This is perfectly legitimate because multiplying every member of a ratio by the same number does not change the relative proportion.
1.5 : 1: chemically represents the same relative amounts as 3 : 2.
2.5 : 1: can be multiplied by 2 to give 5 : 2.
1.33 : 1: may, allowing for rounding, indicate approximately 4 : 3.
Step 4: Check the molecular formula
The simplest integral ratio does not necessarily give the actual number of protons immediately. It gives the relative proportion. If a molecular formula is supplied, use the total number of hydrogens as a consistency check and, when appropriate, to scale the ratio.
Suppose a compound has molecular formula C₄H₁₀O and two relevant integrals of 15 and 10. The ratio is 15 : 10 = 3 : 2. These values account for only 5 ratio units, whereas the formula contains 10 H. Multiplying the entire ratio by 2 gives 6 H : 4 H.
A complete worked example
Consider a 1H NMR spectrum containing three signals with relative integration values:
Start by dividing every value by the smallest value, 1.6:
3.2 ÷ 1.6 = 2
1.6 ÷ 1.6 = 1
The proton ratio is therefore:
If the molecular formula contains six hydrogens represented by these signals, no further scaling is needed because:
The three signals can therefore be assigned integrals of 3 H, 2 H and 1 H.
Never round each integral independently before looking at the complete ratio. Compare all of the values first and look for the simplest chemically sensible whole-number relationship.
Integration applies to the whole multiplet
One of the easiest mistakes is to confuse splitting with integration. If a CH₃ group appears as a triplet, the three lines of that triplet do not represent three separate proton environments.
All three lines belong to the same signal. The area of the entire triplet represents the protons in that environment.
| Feature of the spectrum | Main information obtained | Typical question answered |
|---|---|---|
| Chemical shift, δ | Chemical environment | What type of environment are these protons in? |
| Integrated area | Relative number of protons | How many protons contribute to this signal? |
| Splitting pattern | Information about neighbouring non-equivalent protons | What proton environments are nearby? |
Signals that need special care
Examination spectra are normally designed so that their major signals can be interpreted cleanly. Real spectra, however, may contain other peaks, and not every proton signal necessarily integrates perfectly.
TMS
Tetramethylsilane, Si(CH₃)₄, is used as the reference for the chemical shift scale and is assigned δ = 0 ppm. Its signal does not belong to the compound being analysed and must not be included in the analyte proton ratio.
Solvent and impurity signals
Residual solvent, water or other impurities can produce additional peaks. A peak should not be included merely because it appears on the spectrum. In an examination question, use the information supplied to decide which signals belong to the compound.
Exchangeable O–H and N–H protons
O–H and N–H protons can undergo proton exchange. Their signals may be broad and their measured integrals may be less reliable than ordinary C–H proton signals. For this reason, do not automatically force an imperfect O–H or N–H integral into an otherwise convincing integer ratio unless the question clearly expects you to do so.
C–H proton integrations: normally form the main basis of straightforward proton-counting questions.
O–H and N–H integrations: can sometimes be used, but proton exchange can make them less quantitatively reliable.
Using integration in structural determination
Integration becomes most powerful when combined with the other pieces of NMR information. A signal with an integral of 3 H immediately suggests a set of three equivalent protons; a 2 H signal suggests two equivalent protons; and a 1 H signal suggests a single proton in that environment.
But an integral of 3 H does not, by itself, prove that the signal comes from a particular CH₃ group. The assignment must also fit its chemical shift, splitting pattern and the overall molecular formula.
Structural determination is strongest when chemical shift, integration, splitting and molecular formula all support the same structure.
Symmetry can produce larger integrals
Sometimes several protons in apparently separate parts of a displayed structure are chemically equivalent because the molecule is symmetrical. They then contribute to the same signal.
For example, an integration of 6 H often points toward two equivalent CH₃ groups, although the rest of the spectrum must still be checked before making that assignment.
Treat an integral as a clue rather than a complete structure. A 6 H signal tells you that six equivalent protons contribute to that environment; the molecular structure explains why those six protons are equivalent.
Common Mistakes
| Common mistake | Why it is wrong | Better approach |
|---|---|---|
| Using the tallest peak height to count protons | Height depends on the shape and width of the signal and is not equivalent to integrated area. | Use the numerical integral or the change in height of the integration trace across the complete signal. |
| Integrating each line of a triplet or quartet separately | The individual lines are produced by splitting of one signal; they do not represent separate proton environments. | Integrate the entire multiplet as one signal. |
| Stopping at a ratio such as 1.5 : 1 | Actual numbers of protons must ultimately be whole numbers. | Multiply the entire ratio by the same factor; 1.5 : 1 becomes 3 : 2. |
| Including TMS in the proton count | TMS is an external reference substance and is not part of the compound being analysed. | Exclude the TMS signal at δ = 0 ppm. |
| Assuming every extra peak must belong to the compound | Solvent, water and impurities can also give NMR signals. | Use the supplied chemical information and the complete spectrum before deciding which signals belong to the analyte. |
| Forcing every O–H or N–H integral into an exact ratio | Exchangeable protons may produce broad or imperfectly integrated signals. | Treat exchangeable proton integrations cautiously and use the context of the question. |
What you should be able to do
Identify the correct quantity
Recognise that integrated signal area, rather than peak height, is used for relative proton counting.
Read an integration trace
Determine the change in vertical height across each signal and compare the resulting steps.
Calculate a proton ratio
Convert numerical integrals into the simplest chemically sensible whole-number ratio.
Use the molecular formula
Scale a relative integral ratio to the actual number of protons when sufficient molecular-formula information is available.
Integrate complete signals
Treat a doublet, triplet, quartet or multiplet as one signal when determining its total integrated area.
Recognise misleading peaks
Exclude reference, identified solvent and impurity signals and treat exchangeable O–H or N–H integrations with appropriate caution.
These ten questions test whether you can convert NMR integrals into proton ratios and combine integration with the other information found in a proton NMR spectrum.
- 10 mastery questions
- Numerical integration and spectrum interpretation
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Frequently asked questions
Questions students often ask about NMR integration
Once you understand that integration measures relative proton numbers, most questions reduce to careful ratio reasoning. A few details are still worth clearing up.
Is peak area always exactly equal to the number of protons?
Not necessarily as a raw numerical value. An instrument might report integrals of 6.0 and 4.0 for signals representing 3 H and 2 H. What matters is that the relative areas are proportional to the relative proton numbers. In real spectra, small experimental deviations from perfect integer ratios can also occur.
Why do we not simply use the height of an NMR peak?
Peak height measures only the maximum intensity at one point, whereas proton counting requires the total signal contribution. Differences in signal width and shape mean that two signals can have similar heights but different areas. Integration therefore uses the complete area of each signal.
Do I integrate every individual peak in a triplet separately?
No. A triplet is one signal that has been split into three lines. Add the contribution of the whole triplet together. Its total integrated area represents the protons in that chemical environment.
What should I do if dividing by the smallest integral gives a decimal?
Look for a simple fractional relationship and multiply the whole ratio by the same number. For example, 1.5 : 1 becomes 3 : 2 on multiplying both terms by 2. Similarly, approximately 1.33 : 1 may indicate 4 : 3 after allowing for experimental rounding.
Why do I need the molecular formula if the integrals already give a ratio?
Integration gives relative proton numbers. A ratio of 3 : 2 could describe 3 H : 2 H, 6 H : 4 H or another common multiple. The molecular formula can tell you which scale is consistent with the actual number of hydrogens in the molecule.
Should the TMS peak be included in the integration?
No. Tetramethylsilane is a reference substance used to define δ = 0 ppm. Its protons are not part of the compound being analysed, so its signal must not be counted when determining the analyte's proton ratio.
Can an integral of 6 H tell me that there are two CH₃ groups?
It can be a strong clue, especially if symmetry makes two methyl groups chemically equivalent. However, integration alone cannot prove that assignment. The chemical shift, splitting pattern, molecular formula and the rest of the spectrum must also agree.
Why might an O–H signal not integrate perfectly to 1 H?
O–H protons can undergo exchange, which may broaden their signals and make their measured integrations less quantitatively reliable. N–H signals can behave similarly. In a structural question, use the full set of evidence rather than forcing an exchangeable proton integral into an exact ratio.
Is integration enough to determine a molecular structure?
Usually not by itself. Integration tells you the relative number of protons in each environment. Chemical shift gives information about those environments, while splitting provides information about neighbouring protons. Structural determination comes from combining all of these clues with the molecular formula and any other spectroscopic information supplied.
Integration turns a 1H NMR spectrum from a collection of peaks into a molecular head count. Once you can convert peak areas into sensible proton ratios, you can combine those numbers with chemical shift and splitting to begin assembling the actual structure behind the spectrum.