Organic Chemistry
Instrumentation & Spectroscopy (HL)
5th Year · 6th Year (Leaving Cert)
- ✓Explain the principles and applications of mass spectrometry in determining relative molecular mass and identifying fragments.
- ✓Describe the principles of infrared (IR) spectroscopy and use IR spectra to identify characteristic functional groups.
- ✓Outline the principles of nuclear magnetic resonance (NMR) spectroscopy and interpret simple ¹H NMR spectra to deduce structural information.
- ✓Correlate data from different spectroscopic techniques to determine the structure of organic compounds.
Key concepts
Mass spectrometry is an analytical technique used to determine the relative molecular mass of a compound and to identify structural fragments. A sample is first vaporised and then ionised, typically by electron impact, to form positively charged ions (molecular ions, M⁺). These ions are then accelerated through an electric field and deflected by a magnetic field. The extent of deflection depends on the mass-to-charge ratio (m/z) of the ion; lighter ions and more highly charged ions are deflected more. A detector records the m/z values and their relative abundances, producing a mass spectrum. The molecular ion peak (M⁺) corresponds to the relative molecular mass of the compound. Fragmentation of the molecular ion can occur, leading to peaks at lower m/z values, which provide valuable information about the compound's structure. The most intense peak in the spectrum is called the base peak.
Infrared spectroscopy is a technique used to identify the functional groups present in a molecule. Molecules absorb infrared radiation at specific frequencies, causing their bonds to vibrate (stretch or bend) at characteristic energy levels. Different functional groups have unique absorption frequencies, which are typically measured in wavenumbers (cm⁻¹). An IR spectrum plots the percentage transmittance of IR radiation against wavenumber. A strong absorption at a particular wavenumber indicates the presence of a specific functional group. For example, a strong absorption around 1700 cm⁻¹ indicates a carbonyl (C=O) group, while a broad absorption between 3200-3600 cm⁻¹ suggests an alcohol O-H group, and a very broad absorption between 2500-3300 cm⁻¹ indicates a carboxylic acid O-H group. The region below 1500 cm⁻¹ is known as the 'fingerprint region' and is unique to each compound, but it is often too complex for detailed analysis at Leaving Cert level.
Nuclear Magnetic Resonance (NMR) spectroscopy provides detailed information about the carbon-hydrogen framework of a molecule, specifically focusing on the environment of hydrogen nuclei (protons, ¹H NMR). Certain atomic nuclei, such as ¹H and ¹³C, possess nuclear spin. When placed in a strong external magnetic field, these nuclei can align either with or against the field. Applying radiofrequency radiation causes these nuclei to 'flip' their spin state, absorbing energy at specific frequencies. The exact frequency of absorption, known as the chemical shift (δ), is measured in parts per million (ppm) relative to a standard reference (tetramethylsilane, TMS, at δ = 0 ppm). The chemical shift depends on the electronic environment of the nucleus, providing clues about the functional groups nearby. A ¹H NMR spectrum provides four key pieces of information: 1. Number of signals: Indicates the number of different types of chemically equivalent protons in the molecule. 2. Chemical shift (δ): Provides information about the electronic environment of the protons (e.g., protons near electronegative atoms are 'deshielded' and appear at higher δ values). 3. Integration (area under peak): The relative area under each signal is proportional to the number of protons giving rise to that signal. 4. Splitting pattern (multiplicity): Caused by spin-spin coupling with neighbouring non-equivalent protons. The 'n+1 rule' states that a signal for a proton with 'n' equivalent neighbouring protons will be split into 'n+1' peaks (e.g., a singlet, doublet, triplet, quartet).
Key facts to remember
- 1Mass spectrometry determines the relative molecular mass (from the M⁺ peak) and provides structural information through fragmentation patterns.
- 2Infrared (IR) spectroscopy identifies functional groups by detecting characteristic bond vibrations (stretches and bends) at specific wavenumbers.
- 3A strong absorption around 1700 cm⁻¹ in an IR spectrum indicates the presence of a carbonyl (C=O) group.
- 4Broad O-H absorptions in IR spectra are diagnostic: 3200-3600 cm⁻¹ for alcohols, and 2500-3300 cm⁻¹ for carboxylic acids.
- 5¹H NMR spectroscopy provides information on the number of different proton environments, their relative numbers (integration), and the number of neighbouring protons (splitting pattern).
- 6The chemical shift (δ, in ppm) in NMR indicates the electronic environment of the protons.
- 7The 'n+1 rule' in ¹H NMR predicts that a signal for a proton with 'n' equivalent neighbouring protons will be split into 'n+1' peaks.
- 8Tetramethylsilane (TMS) is used as the reference standard (δ = 0 ppm) in NMR spectroscopy.
Worked examples
Example 1
The mass spectrum of an organic compound shows a molecular ion peak (M⁺) at m/z 46. Significant fragment peaks are observed at m/z 45, m/z 31, and m/z 29. Suggest a possible structure for the compound and explain the origin of the fragment peaks.
Answer
The compound is likely ethanol (CH₃CH₂OH).
The molecular ion peak (M⁺) is crucial for determining the overall molecular mass, while fragment peaks provide clues about the compound's structure.
Example 2
An organic compound has the molecular formula C₃H₆O₂. Its IR spectrum shows a strong absorption at approximately 1710 cm⁻¹ and a very broad absorption between 2500-3300 cm⁻¹. Suggest the functional groups present and a possible structure for the compound.
Answer
The compound contains a carboxylic acid functional group. A possible structure is propanoic acid (CH₃CH₂COOH).
Pay close attention to the shape and exact wavenumber range of O-H absorptions in IR spectra to differentiate between alcohols and carboxylic acids.
Example 3
A compound with molecular formula C₂H₄Br₂ shows only one singlet in its ¹H NMR spectrum. Deduce its structure.
Answer
The structure of the compound is 1,2-dibromoethane (BrCH₂CH₂Br).
Symmetry is a key factor in determining the number of signals in an NMR spectrum. Equivalent protons do not split each other.
Common mistakes
- ✗Confusing the molecular ion peak (M⁺) with the base peak in a mass spectrum.
- ✗Misinterpreting the 'fingerprint region' (below 1500 cm⁻¹) in IR spectra for specific functional groups, which is generally too complex for Leaving Cert analysis.
- ✗Not distinguishing between the broad O-H absorption of an alcohol and the very broad, lower wavenumber O-H absorption of a carboxylic acid in IR.
- ✗Incorrectly applying the n+1 rule in ¹H NMR, especially by counting equivalent protons as neighbours or failing to recognise symmetry.
- ✗Failing to combine information from all available spectroscopic techniques to deduce a complete structure, instead relying on just one technique.
Exam tips
- ★Always identify the molecular ion peak (M⁺) first in a mass spectrum to establish the relative molecular mass of the compound.
- ★For IR spectroscopy, focus on the diagnostic region (above 1500 cm⁻¹) for characteristic functional group absorptions. Refer to the data booklet for typical wavenumber ranges.
- ★When interpreting ¹H NMR spectra, systematically analyse the number of signals, their chemical shifts, integration values, and splitting patterns. Use the n+1 rule carefully.
- ★Remember that equivalent protons do not split each other in NMR. Look for molecular symmetry to identify equivalent protons.
- ★Practise correlating data from MS, IR, and NMR to propose a single, consistent structure for unknown organic compounds.
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