Mass spectrometry, IR and NMR
Spectroscopy identifies molecules from the way they respond to energy. A mass spectrometer gives the relative molecular mass and fragments of a molecule; infrared spectra show which bonds (and so which functional groups) are present; NMR spectra show how many different environments the carbon and hydrogen atoms are in, and how they are arranged.
Key facts
- In a mass spectrum, Ar = Σ(m/z × abundance) ÷ Σ(abundance). For a compound, the molecular ion peak M⁺ (furthest right, apart from a small M+1 peak) gives Mr; other peaks are fragments such as CH₃⁺ (15), C₂H₅⁺ (29) and CH₃CO⁺ (43).
- One chlorine atom gives M and M+2 peaks in a 3 : 1 ratio (³⁵Cl and ³⁷Cl); one bromine atom gives them in a 1 : 1 ratio. High-resolution mass spectrometry gives precise masses that tell apart formulas with the same whole-number mass.
- Infrared absorptions (cm⁻¹): O–H in alcohols 3230–3550 (broad), O–H in acids 2500–3000 (very broad), C=O 1680–1750, C–H 2850–3100, N–H 3300–3500. Below about 1500 cm⁻¹ is the fingerprint region.
- In ¹³C NMR, each different carbon environment gives one peak; carbons that are the same by symmetry give one peak.
- In ¹H NMR, the number of peaks is the number of hydrogen environments, and the area of each peak is proportional to the number of hydrogens in it.
- Splitting follows the n + 1 rule: a peak is split into n + 1 lines by n hydrogens on the neighbouring carbon atoms (singlet, doublet, triplet, quartet). O–H hydrogens do not usually split, and disappear when D₂O is added.
- TMS, Si(CH₃)₄, is the reference at δ = 0; samples are dissolved in solvents such as CDCl₃ that have no ¹H atoms.
Worked examples
Example 1
The mass spectrum of chlorine atoms shows peaks at m/z 35 (abundance 75) and m/z 37 (abundance 25). Calculate the relative atomic mass of chlorine.
- Multiply each mass by its abundance and add. 35 × 75 + 37 × 25 = 3550
- Divide by the total abundance. 3550 ÷ 100 = 35.5
Answer: 35.5
Example 2
A compound C₃H₆O₂ has this infrared spectrum. What type of compound is it?
- A very broad absorption from about 2500 to 3300 cm⁻¹: O–H in a carboxylic acid.
- A strong, sharp absorption near 1710 cm⁻¹: C=O.
- Both together mean –COOH: a carboxylic acid, CH₃CH₂COOH.
Answer: a carboxylic acid (⟦propanoic acid⟧)
Example 3
Predict the ¹H NMR spectrum of ethanol, CH₃CH₂OH.
- There are three hydrogen environments: CH₃, CH₂ and OH. three peaks, areas 3 : 2 : 1
- CH₃ has 2 hydrogens on the neighbouring carbon. 2 + 1 = 3: a triplet
- CH₂ has 3 hydrogens on the neighbouring carbon (the OH does not split it). 3 + 1 = 4: a quartet
- The OH hydrogen is a singlet; it disappears if D₂O is added.
Answer: Three peaks, areas 3 : 2 : 1: a triplet (CH₃), a quartet (CH₂) and a singlet (OH)
Example 4
How many peaks are in the ¹³C NMR spectrum of ⟦butanone⟧, CH₃COCH₂CH₃?
- The four carbons are all in different environments: CH₃ next to C=O, the C=O carbon, CH₂, and the CH₃ at the end.
- There is no symmetry to make any two the same. 4 peaks
Answer: 4