Combining IR and Mass Spectrometry

Introduction

Infrared spectroscopy tells you what kind of compound you have — an alcohol, a ketone, a carboxylic acid — but says little about its size. Mass spectrometry tells you how heavy the molecule is and how it breaks, but a single mass can fit many different compounds. Put the two together and the list of possible structures often shrinks to one or two. This page shows how to combine them efficiently.

Core explanation

What each technique contributes.

Technique Main information Main limitation --- --- --- IR Functional groups present or absent Little information about chain length or Mᵣ Mass spectrometry Mᵣ, isotope clues, fragments Isomers can share Mᵣ and some fragments

Using IR first. Scan the diagnostic region above 1500 cm⁻¹. Key questions are: is there a strong C=O stretch near 1680–1750 cm⁻¹? Is there a broad O–H band (3200–3550 cm⁻¹ for alcohols, very broad 2500–3300 cm⁻¹ for carboxylic acids)? Are there N–H stretches near 3300–3500 cm⁻¹, or the two aldehyde C–H bands near 2720 and 2820 cm⁻¹? The absence of a band is just as useful as its presence: no C=O means no ketone, aldehyde, acid or ester.

Using the mass spectrum next. The molecular ion gives Mᵣ. Knowing the functional group, you can subtract its mass and work out what remains. For example, a carboxylic acid contains COOH (mass 45), so an acid with Mᵣ 74 has 74 − 45 = 29 left, which is C₂H₅: propanoic acid.

Using fragments to decide between isomers. Once IR has fixed the functional group, fragments often settle the remaining choice. Ketones cleave next to the C=O to give acylium ions (RCO⁺); primary alcohols give CH₂=OH⁺ at m/z 31; carboxylic acids often show a loss of OH (M − 17) or COOH (M − 45).

Consistency check. The final structure must fit all the data: the correct Mᵣ, every strong IR band accounted for, no expected band missing, and the main fragments explained.

Step-by-step reasoning

1. From IR, list the functional groups present and those clearly absent. 2. From the mass spectrum, read Mᵣ. 3. Subtract the mass of the functional group to find the size of the remaining carbon framework. 4. Draw every isomer that fits. 5. Use fragment peaks (and any remaining IR detail) to pick one.

Visual explanation

Imagine two overlapping circles. One contains every compound with a C=O and a broad acid O–H; the other contains every compound with Mᵣ 60. Only the compounds in the overlap — here just ethanoic acid — satisfy both sets of evidence. Adding more techniques adds more circles and shrinks the overlap.

Real-world analogy

Identifying a person from "wears a red coat" alone is hopeless in a busy station, and so is "about 1.8 m tall". Combine the two and very few people match. IR and mass spectrometry work like two independent descriptions of the same suspect.

Real-world example

In quality control of solvents, a laboratory may confirm that a drum contains ethyl ethanoate by checking the ester C=O stretch near 1740 cm⁻¹ and C–O bands in the IR spectrum, and then confirming Mᵣ 88 and the CH₃CO⁺ fragment at m/z 43 by GC–MS. Contamination by ethanol would add an O–H band and extra peaks.

Why?

Why are IR and mass spectrometry so complementary? IR measures bond vibrations, which depend mainly on local groups of atoms, while mass spectrometry measures the mass of the whole molecule and its pieces. One is local and one is global, so their information overlaps very little.

Common misconception

"A strong band near 1700 cm⁻¹ plus a molecular mass is always enough." Ketones and aldehydes of the same Mᵣ, such as butanone and butanal (both 72), both show C=O. You must look for the aldehyde C–H bands near 2720 and 2820 cm⁻¹ or use fragment ions to separate them.

Worked example

Question: A compound containing C, H and O has M⁺ at m/z 60. Its IR spectrum shows a very broad band from 2500 to 3300 cm⁻¹ and a strong band at 1710 cm⁻¹. Identify it.

Reasoning: The very broad O–H band together with a C=O stretch indicates a carboxylic acid, COOH (mass 45). 60 − 45 = 15, which is CH₃. Propan-1-ol (also Mᵣ 60) is ruled out because it has no C=O; methyl methanoate (Mᵣ 60) is ruled out because it has no O–H.

Answer: Ethanoic acid, CH₃COOH.

Quick check

1. An IR spectrum shows no absorption between 1650 and 1800 cm⁻¹. Which compound classes can be ruled out? Answer: All carbonyl compounds, including aldehydes, ketones, carboxylic acids, esters and amides, because none has a C=O stretch.

Exam focus

State which evidence comes from which technique and quote wavenumber ranges and m/z values. When several isomers fit, explicitly rule out the others with a reason — marks are often given for the elimination as well as for the final answer.

Advanced insight

In modern laboratories the two techniques are often coupled to a separation step. GC–MS separates a mixture and records a mass spectrum for each component, while GC–IR records an IR spectrum as each compound leaves the column. Matching retention time, IR spectrum and mass spectrum against authentic standards gives very high confidence in identification.

Summary

IR identifies functional groups; mass spectrometry gives Mᵣ and fragment information. Use IR to fix the functional group, subtract its mass from Mᵣ to find the rest of the molecule, then use fragments to choose between isomers. The final structure must be consistent with every piece of evidence.

Practice questions

1. A compound shows a broad IR band at 3350 cm⁻¹, no C=O band, M⁺ at 60 and a base peak at m/z 31. Identify it. Answer: Propan-1-ol, CH₃CH₂CH₂OH: the broad O–H shows an alcohol, and CH₂=OH⁺ at 31 indicates a primary alcohol. 2. A carbonyl compound with Mᵣ 72 shows IR bands at 2720 and 2820 cm⁻¹. Name it. Answer: Butanal, CH₃CH₂CH₂CHO — the two C–H bands are characteristic of an aldehyde. 3. A compound shows a C=O band at 1740 cm⁻¹, no O–H band and M⁺ at 60. Identify it. Answer: Methyl methanoate, HCOOCH₃, an ester of Mᵣ 60. 4. Explain why IR alone cannot distinguish propanoic acid from butanoic acid. Answer: Both contain the same COOH group and give very similar diagnostic bands; they differ mainly in Mᵣ (74 and 88), which mass spectrometry reveals.