Why IR Absorption Appears as a Downward Trough on Some Spectra

At \(1715\ \mathrm{cm}^{-1}\), an organic compound absorbs infrared radiation. Yet, on some IR spectra, the graph drops sharply downwards at this exact position. Why does stronger absorption make the graph go down rather than up?

The answer is in what the vertical axis measures. Many infrared spectra plot percentage transmittance (%T), which tells you how much infrared radiation passes through the sample. When molecules absorb radiation, less reaches the detector, so the percentage transmittance falls. The absorption therefore appears as a downward trough.

If the same spectrum is plotted using absorbance instead, the very same absorption appears as an upward peak. The chemistry has not changed. Only the way the measurement is displayed has changed.

01Why does absorption produce a downward trough?

Imagine shining infrared radiation through a sample of an organic compound.

Some of the radiation passes straight through. However, at particular wavenumbers, molecules absorb infrared energy through their vibrational motions. Less radiation then reaches the detector.

Before looking at the graph, make a prediction: if a sample absorbs strongly at one particular wavenumber, will the detector record more or less transmitted radiation?

It records less. That reduction is the experimental evidence of absorption.

What does the spectrometer actually measure?

To understand the shape of the graph, follow the radiation through a basic transmission infrared experiment.

1. An infrared source produces radiation across a range of wavenumbers.
2. The radiation travels towards the sample.
3. Molecules in the sample absorb radiation at particular wavenumbers corresponding to their infrared-active vibrations.
4. The remaining radiation passes through the sample.
5. A detector measures the transmitted radiation, and the instrument processes the measurements to produce a spectrum.

There is one important detail: the instrument needs to know how much radiation was available before the sample was introduced.

It therefore takes a background or reference measurement. This records the incoming radiation and accounts for contributions from the instrument, sample holder, and surrounding environment.

The sample measurement can then be compared with this reference.

The experiment makes a clear prediction:

• Where the sample does not absorb significantly, most of the reference radiation passes through.
• Where the sample absorbs strongly, substantially less radiation reaches the detector.

The resulting reduction in transmitted radiation is what produces the trough.

Why is a reference measurement necessary?

Suppose the detector records a low signal at \(1500\ \mathrm{cm}^{-1}\).

Does that automatically mean the sample absorbed strongly at this wavenumber?

Not necessarily. The infrared source, detector, and optical components do not necessarily produce or measure the same intensity at every wavenumber.

A low raw detector reading could therefore reflect the instrument rather than the sample.

Comparing the sample measurement with a background measurement helps remove these contributions. The resulting transmittance spectrum is much more representative of the sample’s absorption.

This is an important distinction between the raw detector signal and the processed spectrum.

Modern Fourier-transform infrared (FTIR) instruments use additional optical components and signal processing, but the same basic comparison between background and sample measurements underlies the transmittance calculation.

For a more technical explanation, see Thermo Fisher’s FT-IR glossary.

02Transmittance versus absorbance: the mathematical explanation

The downward trough becomes much easier to understand once you know the relationship between transmittance and absorbance.

1. Percentage transmittance

Transmittance measures the fraction of incident infrared radiation that passes through the sample.

It is represented by:

\[
T=\frac{I}{I_0}
\]

Where:

• \(T\) = transmittance, a dimensionless quantity.
• \(I\) = transmitted infrared intensity.
• \(I_0\) = reference infrared intensity.

Percentage transmittance is:

\[
\%T=\frac{I}{I_0}\times100
\]

For example, suppose the reference intensity is 800 units and the transmitted intensity is 200 units.

\[
\%T=\frac{200}{800}\times100
\]

\[
\%T=25\%
\]

Only 25% of the reference intensity has passed through the sample.

The spectrum therefore records a transmittance value of 25% at that wavenumber.

If the surrounding baseline is 100%, this appears as a downward trough.

2. Absorbance

Absorbance measures how strongly the sample attenuates the incident radiation.

It is related to transmittance through:

\[
A=-\log_{10}(T)
\]

Where:

• \(A\) = absorbance, a dimensionless quantity.
• \(T\) = transmittance expressed as a decimal, not a percentage.

The negative sign is important.

When absorption occurs, transmittance is less than 1. Its logarithm is therefore negative. Multiplying by negative one gives a positive absorbance.

Using the previous example:

\[
T=0.25
\]

\[
A=-\log_{10}(0.25)
\]

\[
A=0.602
\]

The same measurement can therefore be represented in two ways:

• Percentage transmittance: 25%, producing a downward trough.
• Absorbance: 0.602, producing an upward peak.

Neither graph is wrong. They simply display different mathematical representations of the same measurement.

Comparing the two representations

The following values are illustrative.

Notice the pattern.

As percentage transmittance decreases, absorbance increases.

This is why the same absorption produces opposite-looking features on the two graphs.

One common mistake is to assume that 25% transmittance means an absorbance of 75%.

It does not.

The 75% represents the fraction of reference radiation that was not transmitted. Absorbance is logarithmic, so its value is 0.602, not 75%.

The distinction matters whenever you compare the strength of absorption bands.

03How do the spectrum axes affect the appearance?

The direction of an absorption band depends on the vertical axis, not the horizontal axis.

The horizontal axis of a typical IR spectrum represents wavenumber, measured in \(\mathrm{cm}^{-1}\).

It commonly decreases from approximately \(4000\ \mathrm{cm}^{-1}\) on the left to \(400\ \mathrm{cm}^{-1}\) on the right.

The vertical axis can represent either percentage transmittance or absorbance.

A useful rule is to check the vertical axis before interpreting any absorption.

Do not assume that every upward peak represents emission or that every downward feature represents weaker absorption.

An upward peak on an absorbance spectrum represents absorption. An upward peak on a graph of emitted radiation intensity represents emission. The axis label tells you which physical quantity is being measured.

The horizontal axis is a separate issue. A trough at \(1715\ \mathrm{cm}^{-1}\) still represents absorption at \(1715\ \mathrm{cm}^{-1}\), regardless of whether it points upwards or downwards.

If you need to revise the meaning and direction of the wavenumber axis, see How to Read Wavenumber and Convert It into Wavelength.

04What do the depth and width of a downward trough tell you?

Recognising a downward trough is only the first step. You also need to understand what its shape tells you.

Three features are particularly useful.

Position: where the absorption occurs

The horizontal position identifies the wavenumber at which the sample absorbs radiation.

For example, a strong absorption near \(1715\ \mathrm{cm}^{-1}\) is consistent with a carbonyl stretching vibration.

The vertical direction does not change this position. A peak and a trough at the same wavenumber can represent exactly the same absorption.

Depth: how much radiation is transmitted

For a percentage-transmittance spectrum, a deeper trough means a lower percentage of radiation passes through at that wavenumber.

Consider two illustrative measurements:

• Sample A has a minimum transmittance of 60%.
• Sample B has a minimum transmittance of 15%.

Sample B allows less radiation through and therefore has greater absorbance at that wavenumber, assuming comparable baselines and measurement conditions.

However, do not automatically conclude that Sample B contains four times as many absorbing bonds.

Absorption strength can also depend on concentration, sample thickness, and the absorptivity of the particular vibration.

Width: how much of the wavenumber range is affected

A broad trough extends across a larger range of wavenumbers than a narrow trough.

This can reflect a range of molecular environments or overlapping vibrational absorptions.

For example, hydrogen bonding can produce a broad O-H absorption because O-H bonds experience different local environments and therefore absorb across a range of wavenumbers.

A relatively narrow absorption can occur when a vibration produces a more concentrated band.

However, a broad trough is not automatically stronger than a narrow one. Width and depth describe different aspects of the spectrum.

A useful distinction is:

• Position helps identify the type of vibration.
• Depth indicates the strength of absorption at a particular wavenumber.
• Width describes the range of wavenumbers over which the absorption occurs.

You need to consider these features together rather than treating one measurement as proof of a particular structure.

For more practice combining different absorption features, see How to Choose Between Organic Structures Using Two IR Features.

05Worked examples

06Questions and solutions

The following questions use illustrative data rather than experimental measurements. They progress from reading a basic transmittance spectrum to interpreting more complicated measurements.

Question 1

An infrared spectrometer records a reference intensity of 500 units at \(1720\ \mathrm{cm}^{-1}\). When an organic sample is introduced, the transmitted intensity falls to 100 units.

Calculate:

a) The percentage transmittance.

b) The absorbance.

c) Whether the feature appears as an upward peak or downward trough on a percentage-transmittance spectrum.

Solution 1

The percentage transmittance is 20%, the absorbance is 0.699, and the feature appears as a downward trough.

Part a: Calculate percentage transmittance.

Use:

\[
\%T=\frac{I}{I_0}\times100
\]

Substitute:

\[
\%T=\frac{100}{500}\times100
\]

\[
\boxed{\%T=20\%}
\]

Part b: Calculate absorbance.

First convert percentage transmittance into a decimal:

\[
T=0.20
\]

Then:

\[
A=-\log_{10}(0.20)
\]

\[
\boxed{A=0.699}
\]

Absorbance is dimensionless.

Part c: Determine the appearance of the feature.

The vertical axis represents percentage transmittance.

Because only 20% of the reference radiation passes through the sample, the graph falls from its surrounding baseline towards 20%.

This produces a downward trough, indicating strong absorption at \(1720\ \mathrm{cm}^{-1}\).

The same absorption would appear as an upward peak if absorbance were plotted instead.

Question 2

An illustrative infrared spectrum has:

• A horizontal axis running from \(4000\ \mathrm{cm}^{-1}\) on the left to \(500\ \mathrm{cm}^{-1}\) on the right.
• An absorbance vertical axis ranging from 0 at the bottom to 1.2 at the top.
• An upward absorption feature centred at \(1715\ \mathrm{cm}^{-1}\).

A student makes two claims:

1. The upward feature must represent emission because it points towards the top of the graph.
2. The absorption occurs at a high wavenumber because the feature is on the left-hand side of the spectrum.

Evaluate both claims.

Solution 2

Both claims are incorrect. The upward feature represents absorption, not emission, and \(1715\ \mathrm{cm}^{-1}\) is a lower wavenumber than \(2000\ \mathrm{cm}^{-1}\).

Claim 1: The upward feature represents emission.

This is incorrect because the vertical axis measures absorbance.

A higher absorbance value means that more infrared radiation has been absorbed by the sample relative to the reference.

Therefore, an upward peak on this graph represents stronger absorption.

An emission spectrum would instead measure emitted radiation intensity. The direction of a peak alone cannot distinguish these two processes.

Claim 2: The absorption occurs at a high wavenumber.

This is also incorrect.

The horizontal axis decreases from left to right. The feature at \(1715\ \mathrm{cm}^{-1}\) is positioned between \(2000\) and \(1500\ \mathrm{cm}^{-1}\).

It is therefore at a lower wavenumber than \(2000\ \mathrm{cm}^{-1}\).

The important lesson is that the horizontal and vertical axes describe different quantities. The horizontal position identifies the wavenumber, while the vertical position describes absorbance.

Question 3

Two samples of the same organic compound are measured using different infrared sample-cell path lengths.

The measurements are:

Assume that:

• Both samples contain the same absorbing compound.
• The measurements have been correctly baseline-corrected.
• The molar absorptivity is identical.
• Both samples are within the linear Beer-Lambert range.

A student concludes that Sample B must have a higher concentration because it has lower transmittance.

Determine whether this conclusion is correct. Support your answer with calculations.

Solution 3

The student’s conclusion is incorrect. Both samples have the same concentration, despite having different transmittance values.

The key is that absorbance depends on both concentration and path length.

The Beer-Lambert relationship is:

\[
A=\varepsilon \ell c
\]

Where:

• \(A\) = absorbance.
• \(\varepsilon\) = molar absorptivity.
• \(\ell\) = path length in cm.
• \(c\) = concentration.

Step 1: Calculate the absorbance of Sample A.

Convert 60% transmittance into a decimal:

\[
T_A=0.60
\]

Then:

\[
A_A=-\log_{10}(0.60)
\]

\[
A_A=0.222
\]

Step 2: Calculate the absorbance of Sample B.

Convert 36% transmittance into a decimal:

\[
T_B=0.36
\]

Then:

\[
A_B=-\log_{10}(0.36)
\]

\[
A_B=0.444
\]

Sample B has twice the absorbance of Sample A.

Step 3: Account for the different path lengths.

Rearrange the Beer-Lambert relationship:

\[
\frac{A}{\ell}=\varepsilon c
\]

For Sample A:

\[
\frac{A_A}{\ell_A}
=
\frac{0.222}{1.0}
=0.222\ \mathrm{cm}^{-1}
\]

For Sample B:

\[
\frac{A_B}{\ell_B}
=
\frac{0.444}{2.0}
=0.222\ \mathrm{cm}^{-1}
\]

The absorbance per unit path length is identical.

Since both samples have the same molar absorptivity, their concentrations must also be equal.

Why the tempting answer fails:

Sample B has a deeper transmittance trough because the infrared radiation travels through twice as much sample material.

A longer path length produces greater absorption even when concentration remains unchanged.

Comparing trough depths without accounting for path length would therefore produce an incorrect conclusion.

Question 4

An illustrative percentage-transmittance spectrum contains two absorption features:

• Band A is broad, extending from \(3200\) to \(3600\ \mathrm{cm}^{-1}\), with a minimum transmittance of 60%.
• Band B is narrow, extending from \(1680\) to \(1740\ \mathrm{cm}^{-1}\), with a minimum transmittance of 20%.

Both bands have a surrounding baseline of 100%.

A student argues that Band A must represent greater absorption because it covers a wider range of wavenumbers.

Evaluate this claim.

In your answer:

a) Calculate the peak absorbance of each band.

b) Identify which band has the greater peak absorbance.

c) Explain whether the width of a band determines the total absorption or the number of bonds responsible for it.

Solution 4

The student’s claim is not justified. Band B has the greater peak absorbance, even though Band A is broader. Band width alone does not establish total absorption or the number of absorbing bonds.

Part a: Calculate the peak absorbance of each band.

For Band A:

\[
T_A=0.60
\]

\[
A_A=-\log_{10}(0.60)
\]

\[
\boxed{A_A=0.222}
\]

For Band B:

\[
T_B=0.20
\]

\[
A_B=-\log_{10}(0.20)
\]

\[
\boxed{A_B=0.699}
\]

Part b: Compare peak absorbance.

Band B has the greater peak absorbance:

\[
0.699>0.222
\]

It absorbs more strongly at its central wavenumber under the stated measurement conditions.

Its narrow shape does not make the absorption weaker.

Part c: Consider band width and total absorption.

Band width describes the range of wavenumbers over which an absorption occurs.

A broad band can arise when molecular vibrations occur across a range of environments. Hydrogen bonding is one example of a factor that can broaden an O-H absorption.

However, a broader band does not automatically have a greater peak absorbance.

Nor can we determine the total absorption across a band from its width and minimum transmittance alone. We would need the complete band shape and its integrated absorbance.

Finally, peak height and width do not directly reveal the number of bonds present. Different vibrations can have different molar absorptivities.

The correct approach is to distinguish band position, peak strength, and band width rather than treating them as interchangeable measurements.

Question 5

Two samples of the same organic compound are measured using different sample cells.

The resulting percentage-transmittance spectra have the following features at the same diagnostic wavenumber:

The instrument has left a smooth multiplicative baseline offset in Sample B. Assume this offset is independent of wavenumber across the absorption band and is not caused by additional sample absorption.

Both spectra have been measured at the same temperature, and the compound has the same molar absorptivity in both cells.

A student argues:

“Sample B must have a higher concentration because its trough reaches 20%, compared with 50% for Sample A.”

Determine whether the samples have the same concentration or different concentrations.

Your answer must:

a) Correct for the different local baselines.

b) Calculate the absorbance contributed by the absorption band in each sample.

c) Account for the different path lengths.

d) Explain why comparing the two minimum transmittance values directly gives a misleading answer.

Solution 5

The two samples have the same concentration.

Although Sample B has a lower minimum transmittance, its baseline is also lower, and its path length is twice as large. Once both effects are accounted for, the absorbance per unit path length is identical.

Part a: Correct for the different baselines.

The absorption band must be compared with its own local baseline, rather than assuming both spectra start at 100%.

For Sample A:

\[
T_{A,\mathrm{relative}}=\frac{50}{100}
\]

\[
T_{A,\mathrm{relative}}=0.50
\]

For Sample B:

\[
T_{B,\mathrm{relative}}=\frac{20}{80}
\]

\[
T_{B,\mathrm{relative}}=0.25
\]

The corrected relative transmittance values are therefore 0.50 and 0.25.

Part b: Calculate the absorbance contributed by each band.

Use:

\[
A_{\mathrm{band}}
=
-\log_{10}(T_{\mathrm{relative}})
\]

For Sample A:

\[
A_A=-\log_{10}(0.50)
\]

\[
\boxed{A_A=0.301}
\]

For Sample B:

\[
A_B=-\log_{10}(0.25)
\]

\[
\boxed{A_B=0.602}
\]

Sample B has twice the band absorbance of Sample A.

Part c: Account for path length.

Use the Beer-Lambert relationship:

\[
A=\varepsilon \ell c
\]

Rearrange:

\[
\frac{A}{\ell}=\varepsilon c
\]

For Sample A:

\[
\frac{A_A}{\ell_A}
=
\frac{0.301}{1.0}
\]

\[
=0.301\ \mathrm{cm}^{-1}
\]

For Sample B:

\[
\frac{A_B}{\ell_B}
=
\frac{0.602}{2.0}
\]

\[
=0.301\ \mathrm{cm}^{-1}
\]

The absorbance per unit path length is identical.

Since the molar absorptivity is also identical, the concentrations must be equal.

Part d: Explain the misleading comparison.

The student’s mistake is comparing the raw minimum transmittance values without considering the baseline and path length.

Sample B appears to have a deeper trough because its displayed transmittance falls to 20%, compared with 50% for Sample A.

However, Sample B starts from a different local baseline. Its relative transmittance must first be corrected to isolate the absorption band.

Even after this correction, Sample B has twice the band absorbance. That is explained entirely by its twice-as-long sample path.

The final comparison must therefore use absorbance per unit path length, not the raw trough depth.

This question illustrates why a deeper downward trough does not automatically mean a more concentrated sample. You must understand how the spectrum was measured before drawing a chemical conclusion.

07Where to go next

Once you understand why an IR absorption appears as a downward trough, the next task is identifying which molecular vibration produces it.

Start by locating the absorption using the horizontal wavenumber axis. Then consider its position, strength, and shape, and check whether other bands support the same interpretation.

For example, a strong downward trough in the carbonyl region can indicate a C=O stretching vibration, but you need additional evidence to distinguish a ketone, aldehyde, ester, or carboxylic acid.

That is the next step in interpreting an IR spectrum: moving from recognising an absorption to using a combination of absorptions to identify an organic structure.