Ch 1 · Infrared spectroscopy · AS & A Level Chemistry 9701
Topic 1 of 2 in Analytical techniques — AS & A Level Chemistry 9701 — 48 lessons.
Introduction
Infrared spectroscopy identifies the bonds present in a molecule from the frequencies of radiation it absorbs, giving a rapid, non-destructive way of confirming a functional group before further chemical tests.
Key ideas — How an infrared spectrum is read
An infrared spectrum plots percentage transmittance against wavenumber, measured in mathrmcm^-1, typically over the range 400–4000\ mathrmcm^-1; a downward peak (trough in transmittance) at a given wavenumber shows absorption by a bond vibrating at that frequency. The region below about 1500\ mathrmcm^-1, the fingerprint region, is unique to each compound and is not usually interpreted bond-by-bond, but the region above 1500\ mathrmcm^-1 contains recognisable peaks for specific bonds.
🧑🏫 Teacher's explanation
Imagine a molecule is like a collection of balls connected by springs. These springs are the chemical bonds. If you shake the molecule at just the right frequency, the bonds will vibrate more intensely. In infrared spectroscopy, we shine infrared light through a sample. If a bond is capable of vibrating at a specific frequency, it will 'soak up' (absorb) that energy, causing a dip in the light that reaches the detector.
📖 Definitions
🔍 Worked example — Easy
A sample shows a broad, strong absorption peak at 3200 cm^-1. Identify the functional group likely present.
🔍 Worked example — Standard
An unknown compound shows a sharp, strong peak at 1715 cm^-1. What functional group is present?
🔍 Worked example — Exam-style
A spectrum shows a peak at 2950 cm^-1 and another at 1720 cm^-1. Identify the functional groups.
⚠️ Common errors
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🎯 Exam question
Key ideas — Characteristic absorption regions
From the Cambridge data booklet, the key regions are: mathrmO-H (carboxylic acid), broad, 2500–3300\ mathrmcm^-1; mathrmO-H (alcohol), broad, 3230–3550\ mathrmcm^-1; mathrmN-H, 3300–3500\ mathrmcm^-1; mathrmC-H, 2850–3300\ mathrmcm^-1; and mathrmC=O, sharp, 1650–1750\ mathrmcm^-1.
🧑🏫 Teacher's explanation
Think of a chemical bond like a spring connecting two masses. If you pluck a guitar string, it vibrates at a specific frequency depending on how tight the string is and how heavy it is. In a molecule, the 'strings' are the chemical bonds and the 'masses' are the atoms. Infrared (IR) radiation provides the energy needed to make these bonds vibrate.
📖 Definitions
🔍 Worked example — Easy
A student is given an IR spectrum with a broad, strong absorption peak located at 3400 cm^-1. Identify the most likely functional group present.
🔍 Worked example — Standard
An unknown organic compound shows a sharp, intense peak at 1715 cm^-1. Identify the functional group.
🔍 Worked example — Exam-style
A spectrum shows two distinct broad peaks: one centered at 3000 cm^-1 (very wide) and another sharp peak at 1710 cm^-1. Identify the functional group.
⚠️ Common errors
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🎯 Exam question
Key ideas — Distinguishing related functional groups
The precise position and shape of a peak distinguishes related compounds: an alcohol's mathrmO-H peak is broad and centred around 3300\ mathrmcm^-1, clearly separated from the broader, lower carboxylic acid mathrmO-H peak, which extends down toward 2500\ mathrmcm^-1 and often overlaps the C-H region. A spectrum showing both a mathrmC=O peak near 1700\ mathrmcm^-1 and a broad mathrmO-H peak below 3000\ mathrmcm^-1 indicates a carboxylic acid rather than a simple alcohol, aldehyde or ketone, since only the acid combines both features.
🧑🏫 Teacher's explanation
Think of infrared spectroscopy like a musical instrument. Every chemical bond is like a specific string on a guitar. When you pluck the string (hit it with infrared radiation), it vibrates at a specific frequency. If the frequency of the light matches the natural vibration frequency of the bond, the molecule absorbs that energy, creating a 'dip' or a peak in the spectrum.
📖 Definitions
🔍 Worked example — Easy
Identify the functional group present if a spectrum shows a strong, broad absorption peak at 3300 cm^-1.
🔍 Worked example — Standard
A molecule has a sharp peak at 1715 cm^-1 and a broad peak centered at 3000 cm^-1 that overlaps the C-H region. Identify the functional group.
🔍 Worked example — Exam-style
Distinguish between Propan-1-ol and Propanoic acid using IR data. Propan-1-ol has a peak at 3300 cm^-1 (broad) and no peak at 1700 cm^-1. Propanoic acid has a peak at 3000 cm^-1 (very broad) and a peak at 1710 cm^-1.
⚠️ Common errors
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🎯 Exam question
Key ideas — Why different bonds absorb at different frequencies
Covalent bonds continually vibrate by stretching and bending, and each bond absorbs infrared radiation only at the specific frequency matching its own natural vibrational frequency, exciting it to a higher vibrational energy level. This frequency depends on the masses of the two bonded atoms and the strength of the bond: lighter atoms and stronger bonds vibrate at higher frequencies, which is why the light hydrogen atom in an mathrmO-H or mathrmN-H bond absorbs at a much higher wavenumber than the heavier-atom bonds found in the fingerprint region.
🧑🏫 Teacher's explanation
Imagine a heavy, thick rubber band stretched between two heavy bowling balls, versus a thin, tight rubber band stretched between two ping-pong balls. If you pluck the bowling ball setup, it vibrates slowly and deeply. If you pluck the ping-pong ball setup, it vibrates very fast and quickly.
📖 Definitions
🔍 Worked example — Easy
A molecule contains a C=O bond. If the bond strength increases, does the absorption frequency increase or decrease? Explain using the relationship between bond strength and frequency.
🔍 Worked example — Standard
Compare the absorption frequency of an O-H bond to a C-O bond. Which one absorbs at a higher wavenumber?
🔍 Worked example — Exam-style
A sample shows a broad, strong absorption peak at 3200--3600 cm^-1. Identify the functional group and explain why this peak is found in this specific region rather than the fingerprint region.
⚠️ Common errors
🧠 Quick check 1
🧠 Quick check 2
🎯 Exam question
(a) Identify the two functional groups present. [2] (b) Explain why the peak at 3300 cm^-1 occurs at a much higher wavenumber than the peak at 1715 cm^-1. [2] [4 marks]
Worked example
Question. An infrared spectrum of compound Z, molecular formula C_3H_6O_2, shows a strong sharp absorption at 1715\ mathrmcm^-1 and a very broad absorption between 2500 and 3300\ mathrmcm^-1. (a) Identify the bonds responsible for each absorption. [2] (b) Suggest the functional group present in Z, and hence a possible structure. [2] (c) Explain why the absorption at 2500–3300\ mathrmcm^-1 is so much broader than a typical mathrmC=O absorption. [1]
🔍 Worked example — Easy
An IR spectrum shows a single, sharp, strong absorption peak at 1740 cm^-1. Identify the functional group likely present.
🔍 Worked example — Standard
A molecule has the formula C_2H_4O. Its IR spectrum shows a peak at 1720 cm^-1 but no absorption above 3000 cm^-1. Identify the functional group.
🔍 Worked example — Exam-style
An IR spectrum of compound Z, molecular formula C_3H_6O_2, shows a strong sharp absorption at 1715 cm^-1 and a very broad absorption between 2500 and 3300 cm^-1. (a) Identify the bonds responsible for each absorption. (b) Suggest the functional group and a possible structure. (c) Explain why the 2500--3300 cm^-1 absorption is so broad.
⚠️ Common errors
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