Welcome to Spectroscopic Techniques
Imagine you are a forensic scientist or a pharmaceutical chemist handed a vial containing a mystery clear liquid or white powder. How do you find out exactly what molecules are inside without tasting or touching it? You use spectroscopy!
In this chapter of A2 2: Organic Chemistry, we explore two complementary analytical tools:
• Infrared (IR) Spectroscopy: Identifies the types of chemical bonds and functional groups present in a molecule.
• Mass Spectrometry (MS): Measures the molecular mass of a compound and reveals its structural framework by breaking it into pieces.
Together, these techniques allow chemists to deduce unknown structures, track chemical reactions (such as the synthesis of aspirin), and test products for purity.
Part 1: Infrared (IR) Spectroscopy
How Does IR Spectroscopy Work?
Covalent bonds are not rigid sticks; they behave like tiny springs that vibrate constantly. When you shine infrared radiation onto a sample:
• The bonds absorb specific frequencies of IR radiation that match their natural vibrational frequencies (stretching and bending).
• Different bonds (such as \(\text{C–H}\), \(\text{C=O}\), or \(\text{O–H}\)) vibrate at different frequencies because of their bond strength and the masses of the bonded atoms.
• The absorption is measured in units called wavenumber, expressed in \(\text{cm}^{-1}\).
Everyday Analogy: Think of guitar strings. A thick, loose string vibrates at a different pitch (frequency) than a thin, tight string. Similarly, a double bond (\(\text{C=O}\)) vibrates at a higher frequency than a single bond (\(\text{C–O}\)).
Diagnostic Absorption Bands to Know
When looking at an IR spectrum, you will see a graph plotting percentage transmittance against wavenumber (\(\text{cm}^{-1}\)). Deep dips (absorptions) tell you which functional groups are present:
• \(\text{C–H}\) stretch (\(\sim 2850–3000\text{ cm}^{-1}\)): A sharp series of dips present in almost all organic compounds due to alkyl chains.
• \(\text{O–H}\) (alcohol) stretch (\(\sim 3200–3600\text{ cm}^{-1}\)): A smooth, broad band (curved like a smooth bowl) caused by extensive hydrogen bonding in alcohols such as ethanol and propanol.
• \(\text{O–H}\) (carboxylic acid) stretch (\(\sim 2500–3300\text{ cm}^{-1}\)): A very broad, ragged band that extends widely and visibly overlaps the \(\text{C–H}\) region.
• \(\text{C=O}\) (carbonyl) stretch (\(\sim 1680–1750\text{ cm}^{-1}\)): A very strong, sharp, prominent "dagger-like" dip. Essential for identifying carbonyls, carboxylic acids, and esters (such as aspirin).
• \(\text{C=C}\) (alkene) stretch (\(\sim 1620–1680\text{ cm}^{-1}\)): A weak-to-medium dip that confirms unsaturation in alkenes.
• \(\text{C–O}\) stretch (\(\sim 1000–1300\text{ cm}^{-1}\)): A strong absorption found in alcohols, carboxylic acids, and esters.
Spot the Difference: Alcohol \(\text{O–H}\) vs Carboxylic Acid \(\text{O–H}\)
Don't worry if this seems tricky at first—this is one of the most common exam traps! Here is how to tell them apart easily:
• Alcohol \(\text{O–H}\): Smooth, rounded trough between \(3200–3600\text{ cm}^{-1}\). It is clearly separated from the \(\text{C–H}\) dips at \(2850–3000\text{ cm}^{-1}\).
• Carboxylic Acid \(\text{O–H}\): Broad, uneven trough extending all the way from \(2500–3300\text{ cm}^{-1}\). It "swallows" the \(\text{C–H}\) peaks, giving a messy, distorted baseline.
The Fingerprint Region
The region of the spectrum below \(1500\text{ cm}^{-1}\) is called the fingerprint region.
• It contains a complex series of overlapping absorptions caused by bending vibrations unique to the entire molecular skeleton.
• Examiner Tip: You do not need to identify every single peak below \(1500\text{ cm}^{-1}\). Instead, this pattern is compared directly against a computer database of known reference spectra to confirm the exact identity of a molecule.
Key Takeaway: IR Spectroscopy
IR spectroscopy detects bond vibrations. A peak at \(\sim 1680–1750\text{ cm}^{-1}\) confirms a \(\text{C=O}\) bond, a broad peak at \(3200–3600\text{ cm}^{-1}\) confirms an alcohol \(\text{O–H}\), and a very broad band at \(2500–3300\text{ cm}^{-1}\) confirms a carboxylic acid \(\text{O–H}\). The region below \(1500\text{ cm}^{-1}\) is the unique fingerprint region.
Part 2: Mass Spectrometry (MS)
How Does Mass Spectrometry Work?
While IR spectroscopy looks at how bonds vibrate, mass spectrometry works by "weighing" molecules and their broken fragments.
1. Vaporisation & Ionisation: The sample is vaporised and bombarded with high-energy electrons, knocking out an electron to form a positively charged molecular ion: \(\text{M} \rightarrow \text{M}^{\bullet+} + \text{e}^-\).
2. Acceleration & Deflection: The positive ions are accelerated by an electric field and deflected by a magnetic field according to their mass-to-charge ratio (\(m/z\)).
3. Detection: A detector records the abundance of each ion arriving with a specific \(m/z\) value.
Note: Since the charge (\(z\)) is almost always \(+1\), the \(m/z\) value directly equals the mass of the ion.
Key Peaks on a Mass Spectrum
1. The Molecular Ion Peak (\(\text{M}^+\) or \(\text{M}^{\bullet+}\)):
• This is the peak representing the intact, unfragmented molecule with one electron removed.
• It is found at the highest nominal \(m/z\) value on the spectrum (ignoring tiny isotopic peaks).
• Significance: The \(m/z\) of the molecular ion peak gives the relative molecular mass (\(M_r\)) of the compound!
2. The Base Peak:
• This is the tallest peak in the spectrum and is assigned an arbitrary relative abundance of \(100\%\).
• It represents the most stable and most abundant positive fragment ion formed during fragmentation.
Understanding Fragmentation Patterns
Inside the mass spectrometer, excess energy causes molecular ions to break apart into smaller pieces:
\(\text{M}^{\bullet+} \rightarrow \text{Positive Fragment Ion } (\text{X}^+) + \text{Neutral Free Radical } (\text{Y}^{\bullet})\)
Only the positive ions are accelerated, detected, and appear as peaks on the spectrum. Uncharged neutral radicals are pumped away undetected.
By calculating the mass difference (\(\Delta m/z\)) between the molecular ion peak and other fragment peaks, you can determine what pieces were lost:
• Loss of \(\Delta m/z = 15\): Loss of a methyl group (\(-\text{CH}_3\)), leaving a fragment peak.
• Loss of \(\Delta m/z = 17\): Loss of a hydroxyl radical (\(-\text{OH}\)).
• Loss of \(\Delta m/z = 18\): Loss of a water molecule (\(-\text{H}_2\text{O}\)), common in alcohols.
• Loss of \(\Delta m/z = 29\): Loss of an ethyl group (\(-\text{C}_2\text{H}_5\)).
CRITICAL EXAM RULE: Always Include the Positive Charge!
When an exam question asks: "Identify the species responsible for the peak at \(m/z = 15\)":
• Correct answer: \(\text{CH}_3^+\)
• Incorrect answer (0 marks): \(\text{CH}_3\)
Remember: Neutral radicals do not hit the detector. Every fragment species you write must carry a positive charge (\(+\))!
Key Takeaway: Mass Spectrometry
The molecular ion peak (\(\text{M}^+\)) at the high \(m/z\) end gives the \(M_r\) of the molecule. The base peak is the tallest peak (\(100\%\) abundance). Fragment peaks tell you the structural pieces, and all detected species must be written with a positive charge (e.g., \(\text{C}_2\text{H}_5^+\)).
Part 3: Applications in Organic Synthesis & Purity
Spectroscopic methods are invaluable in the laboratory for monitoring reactions and assessing the purity of synthesized products, such as in the preparation of aspirin (acetylsalicylic acid) from 2-hydroxybenzoic acid (salicylic acid).
1. Monitoring Reaction Progress
When synthesizing aspirin, 2-hydroxybenzoic acid reacts with ethanoic anhydride. Spectroscopic changes confirm the reaction has occurred:
• Phenol/Alcohol \(\text{O–H}\) disappearance: The phenolic \(\text{O–H}\) group in 2-hydroxybenzoic acid is esterified, so its characteristic absorption diminishes.
• Ester \(\text{C=O}\) appearance: A distinct ester carbonyl (\(\text{C=O}\)) absorption band appears in the region \(\sim 1680–1750\text{ cm}^{-1}\) alongside the existing carboxylic acid \(\text{C=O}\) band.
2. Assessing Product Purity
• IR Spectroscopy: Pure aspirin will show only the expected absorption bands (e.g., carboxylic acid \(\text{O–H}\), \(\text{C=O}\), and aromatic stretches). The presence of extra peaks (such as an unreacted phenolic \(\text{O–H}\) stretch) indicates an impure product.
• Mass Spectrometry: A pure product will exhibit a clear molecular ion peak corresponding to aspirin's \(M_r\) without rogue peaks from unreacted starting materials or side products.
Part 4: Common Pitfalls & Examiner Tips
Keep these four frequent mistakes in mind to protect your marks:
1. Confusing the Base Peak with the Molecular Ion Peak:
The base peak is the tallest peak (\(100\%\) abundance). The molecular ion peak is the heaviest primary intact ion found at the right-hand end (highest \(m/z\)). They can occasionally be the same peak, but do not assume they always are!
2. Confusing the Two Types of \(\text{O–H}\) Stretch:
Always state whether an \(\text{O–H}\) belongs to an alcohol (\(3200–3600\text{ cm}^{-1}\)) or a carboxylic acid (\(2500–3300\text{ cm}^{-1}\)). Quoting the wrong range will cost marks.
3. Forgetting the Positive Charge on Fragment Ions:
Never write neutral formulae for mass spec fragments. Write \(\text{CH}_3^+\), \(\text{C}_2\text{H}_5^+\), etc.
4. Over-analysing the Fingerprint Region:
Do not waste time trying to assign every peak below \(1500\text{ cm}^{-1}\). Simply state that this is the fingerprint region and that it is compared against reference database spectra to confirm compound identity.
Quick Knowledge Check
Q1: What physical property is measured on the horizontal axis of an IR spectrum?
Answer: Wavenumber (in \(\text{cm}^{-1}\)).
Q2: An unknown alcohol has a molecular ion peak at \(m/z = 46\) and a major fragment peak at \(m/z = 31\). What neutral species was lost to give the \(m/z = 31\) peak, and what is the formula of the ion at \(m/z = 31\)?
Answer: The mass lost is \(\Delta m/z = 46 - 31 = 15\), which corresponds to the loss of a \(-\text{CH}_3\) radical. The ion responsible for the peak at \(m/z = 31\) is \(\text{CH}_2\text{OH}^+\) (or \(\text{CH}_3\text{O}^+\)).
Q3: How does the IR spectrum of a carboxylic acid differ from that of an ester?
Answer: Both have a strong \(\text{C=O}\) peak at \(\sim 1680–1750\text{ cm}^{-1}\) and \(\text{C–O}\) stretch at \(\sim 1000–1300\text{ cm}^{-1}\), but the carboxylic acid also contains a very broad \(\text{O–H}\) absorption band across \(2500–3300\text{ cm}^{-1}\), which is absent in the ester.