1. Infrared (IR) Spectroscopy
Covalent bonds vibrate naturally by stretching and bending. Each bond absorbs infrared radiation matching its natural vibrational frequency. Absorption peaks appear as dips on the transmittance scale:
| Bond Type | Functional Group | Wavenumber Range (cm⁻¹) | Peak Appearance & Diagnostics |
|---|---|---|---|
| O-H | Alcohols | 3230 - 3550 | Broad, smooth trough due to hydrogen bonding. |
| O-H | Carboxylic acids | 2500 - 3000 | Very broad, jagged trough overlapping C-H stretching. |
| C=O | Aldehydes, Ketones, Carboxylic acids, Esters | 1680 - 1750 | Sharp, intense, deep spike (unmistakable carbonyl peak). |
| C=C | Alkenes | 1620 - 1680 | Moderate sharp absorption. |
| C-H | Alkanes, Alkenes, Arenes | 2850 - 3300 | Sharp, multi-pointed absorption. |
2. The Fingerprint Region and Impurities
The complex region of an infrared spectrum below 1500 cm⁻¹, caused by complex bending vibrations of the whole molecule.
While the functional group region (1500-4000 cm⁻¹) identifies specific bonds, the fingerprint region is unique to an individual compound. Unknown molecules are positively confirmed by computer comparison of their fingerprint spectrum against a library of known reference spectra.
Using IR to Identify Impurities
A pure compound gives only the peaks expected for its bonds. Extra peaks show an impurity.
- The peak at 1730 cm⁻¹ is C=O (1680 to 1750 cm⁻¹), so the aldehyde has formed.
- An aldehyde has no O-H bond. The broad O-H peak at 3230 to 3550 cm⁻¹ shows unreacted alcohol in the sample.
3. Infrared Radiation and Global Warming
- The Earth's surface absorbs energy from the Sun and emits infrared radiation.
- CO₂, CH₄ and water vapour have bonds (C=O, C-H and O-H) that absorb infrared radiation.
- The absorbed energy is re-emitted in all directions. Some goes back towards the Earth, which warms the atmosphere.
- More of these gases in the atmosphere means more infrared is absorbed, which contributes to global warming.
4. Mass Spectrometry in Organic Analysis
In electron impact mass spectrometry, an intact organic molecule loses one electron to form a radical cation known as the molecular ion (\(M^{\bullet+}\)):
\[ \text{M} + e^- \rightarrow M^{\bullet+} + 2e^- \]
The \(m/z\) ratio of the molecular ion peak gives the relative molecular mass (\(M_r\)) of the intact molecule.
High-Resolution Mass Spectrometry
Standard low-resolution mass spectrometry measures \(m/z\) to the nearest integer. High-resolution spectrometers measure masses to four or five decimal places. This differentiates compounds that have identical integer \(M_r\) values (using 5 d.p. masses):
- Propanal (C₃H₆O): Accurate mass = 3(12.00000) + 6(1.00782) + 15.99491 = 58.04183.
- Butane (C₄H₁₀): Accurate mass = 4(12.00000) + 10(1.00782) = 58.07820.
| Formula | Working | Precise Mr |
|---|---|---|
| C₃H₆O | 36.0000 + 6.0468 + 15.9949 | 58.0417 |
| C₄H₁₀ | 48.0000 + 10.0780 | 58.0780 |
| C₂H₂O₂ | 24.0000 + 2.0156 + 31.9898 | 58.0054 |
Only C₃H₆O matches 58.0417, so the compound is C₃H₆O.
5. The [M+1] Peak and Carbon Counting (Beyond the OxfordAQA specification)
Beyond the OxfordAQA specification: 9620 does not require the [M+1] peak. This section is background only.
Directly to the right of the molecular ion peak \(M\), a tiny peak is observed at \([M+1]\). This peak arises from the natural 1.1% abundance of the carbon-13 isotope (¹³C) in the molecule.
The number of carbon atoms (\(n\)) in the molecule can be deduced by comparing the peak heights:
\[ n = \frac{\text{Height of }[M+1]\text{ peak}}{\text{Height of }M\text{ peak}} \times \frac{100}{1.1} \]
6. Chemical Identification Tests Summary
| Functional Group | Test Reagent | Positive Observation |
|---|---|---|
| Alkene (C=C) | Bromine water (\(\text{Br}_2(aq)\)) | Orange to colourless. |
| Halogenoalkane (R − X) | Warm with \(\text{NaOH}(aq)\), acidify with HNO₃, add \(\text{AgNO}_3(aq)\) | AgCl white ppt; AgBr cream ppt; AgI yellow ppt. |
| Primary or secondary alcohol (−OH) | Acidified potassium dichromate(VI), warm | Orange to green. (Tertiary alcohol: no change, stays orange.) |
| Aldehyde (−CHO) | Tollens' reagent, warm or Fehling's solution, warm |
Silver mirror forms on the glass. Blue solution gives a brick-red precipitate. |
| Carboxylic acid (−COOH) | Sodium hydrogencarbonate (\(\text{NaHCO}_3\)) or sodium carbonate (\(\text{Na}_2\text{CO}_3\)) | Effervescence; the gas (CO₂) turns limewater cloudy. |
Required practical 5: tests for alcohols, aldehydes, alkenes and carboxylic acids
See how each test is carried out, with the exact observations to write, on the RP5 page.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
7. Practice Questions
Show Answer & Mark Scheme
Deduction:
- Absorption at 1715 cm⁻¹ indicates the presence of a carbonyl group (C=O). (1 mark)
- Absence of absorption between 3200-3600 cm⁻¹ confirms compound X is not an alcohol. (1 mark)
- Negative Tollens' test confirms X is a ketone, not an aldehyde. (1 mark)
- With formula C₃H₆O, the only possible ketone is propanone (CH₃COCH₃). (1 mark)
Show Answer & Mark Scheme
Answer:
- C₂H₄O = 24.0000 + 4.0312 + 15.9949 = 44.0261. (1 mark)
- C₃H₈ = 36.0000 + 8.0624 = 44.0624 and CO₂ = 12.0000 + 31.9898 = 43.9898, so neither matches. (1 mark)
Show Answer & Mark Scheme
Answer:
- 1715 cm⁻¹ is C=O, so propanone has formed. (1 mark)
- The broad peak at 3400 cm⁻¹ is O-H (alcohol). Propanone has no O-H, so the product contains unreacted propan-2-ol as an impurity. (1 mark)
Show Answer & Mark Scheme
Answer:
- The C-H bonds in methane absorb infrared radiation emitted by the Earth's surface. (1 mark)
- The energy is re-emitted in all directions. (1 mark)
- Some is re-emitted back towards the Earth, which warms the atmosphere. (1 mark)
Show Answer & Mark Scheme
Answer:
- Add sodium hydrogencarbonate (or sodium carbonate). (1 mark)
- Propanoic acid: effervescence, and the gas turns limewater cloudy. Propan-1-ol: no effervescence. (1 mark)