O–H Absorptions in Alcohols and Carboxylic Acids

Introduction

Of all the bands in an infrared spectrum, the O–H stretch is one of the easiest to spot. In an alcohol it is a broad, rounded trough centred near 3300 cm⁻¹; in a carboxylic acid it becomes an enormous, ragged band spreading from about 3300 cm⁻¹ down to 2500 cm⁻¹. Its breadth is not an accident — it is a direct signature of hydrogen bonding between molecules.

Core explanation

Where O–H absorbs. The oxygen–hydrogen bond is strong and involves the very light hydrogen atom, so it stretches at high wavenumber. An isolated, non-hydrogen-bonded (free) O–H group absorbs sharply at about 3600–3650 cm⁻¹. Such free O–H bands are seen only in the gas phase or in very dilute solutions in non-polar solvents.

Hydrogen bonding lowers and broadens the band. In a pure liquid or solid alcohol, the hydrogen of each O–H group is attracted to a lone pair on the oxygen of a neighbouring molecule. This hydrogen bond pulls electron density along the O–H axis and slightly weakens the O–H bond, lowering its force constant. A weaker bond vibrates at a lower wavenumber, so the band moves down to around 3200–3550 cm⁻¹.

At any instant the sample contains a huge variety of hydrogen-bonded arrangements: short and long hydrogen bonds, chains, rings and molecules with no partner at all. Each arrangement weakens the O–H bond by a slightly different amount, so each absorbs at a slightly different wavenumber. The spectrum records the sum of them all — a broad band rather than a sharp line.

Alcohols. A typical alcohol, such as ethanol, shows a strong, broad band centred near 3300–3400 cm⁻¹. It is usually accompanied by a strong C–O stretch between 1000 and 1300 cm⁻¹, and there is no C=O band.

Carboxylic acids. Carboxylic acids form especially strong hydrogen bonds. Two molecules pair up into a cyclic dimer , each O–H hydrogen-bonded to the C=O oxygen of its partner. The O–H bond is weakened more than in alcohols, so the band is centred lower and is very broad, typically 2500–3300 cm⁻¹. It is so wide that the sharper C–H stretches near 2900–3000 cm⁻¹ often appear as small spikes sitting on top of it. Acids also show a strong C=O band at about 1700–1725 cm⁻¹.

Phenols behave much like alcohols, with a broad O–H band near 3200–3550 cm⁻¹, together with aromatic C=C bands near 1500–1600 cm⁻¹.

Step-by-step reasoning

To classify a broad band above 2500 cm⁻¹:

1. Note where the band is centred and how wide it is. 2. A broad band centred about 3300 cm⁻¹ suggests an alcohol or phenol O–H. 3. A very broad band spanning roughly 2500–3300 cm⁻¹ suggests a carboxylic acid O–H. 4. Check for a strong C=O band near 1700 cm⁻¹; its presence supports an acid. 5. Check the 1000–1300 cm⁻¹ region for C–O absorptions.

Visual explanation

Picture an alcohol spectrum as a wide, smooth valley on the left, reaching down near 3300 cm⁻¹. Now picture an acid spectrum: the valley is wider and deeper, stretching right across to 2500 cm⁻¹, with a sharp, deep C=O notch near 1710 cm⁻¹ further to the right.

Real-world analogy

A single tuning fork gives one clear note. A choir of slightly out-of-tune singers all trying to sing the same note produces a broad smear of sound. Free O–H groups are the tuning fork; hydrogen-bonded O–H groups, each in a slightly different environment, are the choir.

Real-world example

When a wine is exposed to air, bacteria can oxidise ethanol to ethanoic acid, turning it to vinegar. An IR spectrum of the dried residue shows the change clearly: the broad alcohol band near 3300 cm⁻¹ gives way to the very broad acid band reaching 2500 cm⁻¹, and a strong C=O band appears near 1710 cm⁻¹.

Why?

Why does a stronger hydrogen bond give a lower O–H wavenumber? The more strongly the hydrogen is attracted to the neighbouring oxygen, the more its bond to its own oxygen is stretched and weakened. A smaller force constant means a lower vibrational frequency, so the absorption moves to lower wavenumber.

Common misconception

"A broad band near 3300 cm⁻¹ always means an alcohol." Water in a wet sample, or an N–H group, can also absorb here. A broad O–H band should be supported by other evidence, such as a C–O band, and the sample must be dry.

Worked example

Question: Compound A shows a broad band at 3350 cm⁻¹ and a strong band at 1050 cm⁻¹. Compound B shows a very broad band from 2500 to 3300 cm⁻¹ and a strong band at 1710 cm⁻¹. Both have formula C₃H₆O₂ or C₃H₈O. Identify the class of each.

Reasoning: A has an alcohol O–H and a C–O band but no C=O, so it is an alcohol (propan-1-ol or propan-2-ol, C₃H₈O). B has the acid O–H pattern plus C=O at 1710 cm⁻¹.

Answer: A is an alcohol; B is a carboxylic acid, propanoic acid.

Quick check

1. Why is the O–H band of a carboxylic acid broader and lower in wavenumber than that of an alcohol? Answer: Acids form strong hydrogen-bonded dimers, which weaken the O–H bond more and in more varied ways.

Exam focus

Always describe the O–H band with both a range and its shape: "broad absorption at 2500–3300 cm⁻¹ due to O–H in a carboxylic acid". Examiners reward linking two bands, such as O–H plus C=O for an acid, rather than relying on one band alone.

Advanced insight

Diluting an alcohol in tetrachloromethane progressively breaks its hydrogen bonds. As the concentration falls, a sharp free O–H band near 3630 cm⁻¹ grows while the broad bonded band shrinks. Intramolecular hydrogen bonds, as in 2-nitrophenol, do not change on dilution, so this experiment distinguishes the two types.

Summary

Free O–H groups absorb sharply near 3600–3650 cm⁻¹, but hydrogen bonding weakens the O–H bond and spreads the absorption into a broad band. Alcohols and phenols give a broad band centred near 3300 cm⁻¹, often with a C–O band. Carboxylic acids form strongly hydrogen-bonded dimers and give a very broad band from about 2500 to 3300 cm⁻¹, accompanied by a strong C=O band.

Practice questions

1. State the approximate range of the O–H absorption in a hydrogen-bonded alcohol. Answer: About 3200–3550 cm⁻¹, as a strong, broad band. 2. Explain why the O–H band of an alcohol is broad in the pure liquid. Answer: Molecules are hydrogen-bonded in many different arrangements, each weakening the O–H bond by a different amount, so absorption occurs over a spread of wavenumbers. 3. How would the spectrum of butanoic acid differ from that of butan-1-ol above 1500 cm⁻¹? Answer: Butanoic acid shows a very broad O–H band from about 2500 to 3300 cm⁻¹ and a strong C=O band near 1710 cm⁻¹; butan-1-ol shows a narrower O–H band near 3300 cm⁻¹ and no C=O band. 4. Predict what happens to the O–H band of ethanol when it is highly diluted in a non-polar solvent. Answer: Hydrogen bonds are broken, so a sharp free O–H band appears near 3600–3650 cm⁻¹ and the broad band weakens.