Organic Chemistry: Techniques & Spectra
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Organic Chemistry: Techniques & Spectra
Chemists purify and identify unknown organic compounds using physical lab techniques (reflux, distillation, drying) to make and clean up a compound, then "fingerprint" it using mass spectrometry (which tells you the mass and pieces of the molecule) and infrared spectroscopy (which tells you which bonds/functional groups are present).
📋 Summary — What This Chapter Covers
- Heating under reflux lets a reaction go as far as possible without losing any reactants, products, or solvent by evaporation.
- Distillation separates compounds by boiling point, letting you collect a specific product (e.g. an aldehyde) as it forms and boils off.
- Boiling point determination is used to check the purity and identity of a liquid — impure samples boil over a range and at a higher-than-expected temperature.
- Solvent extraction (separating funnel) separates an organic product from an aqueous layer based on the fact the two liquids don't mix.
- Drying agents (anhydrous salts) remove leftover traces of water from an organic liquid product.
- Mass spectrometry bombards molecules with electrons to form a molecular ion, which fragments into smaller charged pieces — the m/z values of the peaks reveal the molecular mass and structural clues.
- Isotopes show up as multiple peaks (e.g. in halogenoalkanes) because different isotopes have different masses but the same charge.
- Infrared (IR) spectroscopy identifies functional groups by which frequencies of IR radiation are absorbed by particular bonds, shown as absorption peaks at characteristic wavenumbers.
🧪 Part 1: Organic Chemistry Techniques
Heating Under Reflux
Here's the problem reflux solves: lots of organic reactions are slow at room temperature. The obvious fix is to heat the mixture up — heat speeds up reactions. But if you just heat an open flask, your volatile reactants, solvent, and product will simply evaporate and escape before the reaction finishes. You'd be boiling away the very things you need.
Reflux solves this by trapping the vapour and sending it back down. A vertical condenser is attached to the top of the flask. As the mixture boils, vapour rises into the condenser, where cold water flowing through the outer "water jacket" cools it back into a liquid — and that liquid drips straight back down into the flask. Nothing escapes; the reaction mixture just keeps circulating and reacting at a raised, controlled temperature.
Reflux = keep everything IN the flask (goal: complete the reaction).
Distillation = get a specific compound OUT of the flask (goal: separate/collect a product).
Typical setup: reaction mixture + anti-bumping granules in a pear-shaped or round-bottomed flask → heated in a heating mantle or water bath → vertical condenser clamped on top (water enters at the bottom, exits at the top — this "counter-current" flow keeps the condenser fully cooled) → joints lightly greased to seal and ease disassembly.
Classic reflux examples: oxidising a primary alcohol all the way to a carboxylic acid using acidified potassium dichromate(VI) (needs excess oxidant and full reaction time), or making an ester from an alcohol + acid with an acid catalyst.
Explain why anti-bumping granules are used, and why a reaction mixture is heated under reflux rather than in an open flask, when preparing ethanoic acid from ethanol.
Distillation
Distillation is used when a reaction doesn't go to completion, or produces more than one chemical alongside your target — you need to physically separate the mixture. The separation principle is simple: different compounds have different boiling points, so the one with the lowest boiling point evaporates and gets collected first.
The go-to example is oxidising a primary alcohol → aldehyde (not all the way to a carboxylic acid). Here, distillation is essential, not just convenient: the aldehyde has a lower boiling point than the alcohol it came from (because the aldehyde has lost the alcohol's O–H group, and with it, hydrogen bonding). As soon as aldehyde forms, it evaporates out of the hot reaction mixture and is distilled off — which conveniently also stops it being oxidised further to a carboxylic acid, since it doesn't hang around in the flask.
Notice the apparatus is slightly different from reflux: the condenser is angled downward (not vertical) and connects via a side-arm "still head", so the condensed liquid drips out into a separate collecting vessel rather than falling back into the flask. A thermometer bulb sits right where vapour enters the condenser, so you can monitor the temperature of what's currently distilling over.
Ethanol boils at 78 °C and ethanal boils at 21 °C. Explain, in terms of intermolecular forces, why ethanal can be separated from the reaction mixture by distillation while the reaction is still ongoing.
Boiling Point Determination
A compound's boiling point is like a fingerprint — it's characteristic of that specific substance, so measuring it (via distillation) lets you check both the identity and purity of a liquid.
A pure liquid boils sharply at one exact temperature that matches the literature/database value.
An impure liquid boils over a range of temperatures, and that range is typically higher than the literature value.
Why does an impurity raise the boiling point and spread it into a range? Impurity particles disrupt the way solvent molecules pack and interact, meaning slightly more energy (and a spread of energies as the mixture composition changes as it boils) is needed to get all the molecules into the vapour phase — this is the same underlying idea as boiling point elevation you may have met with solutions.
Solvent Extraction & Drying
Using a separating funnel
Many organic preparations end up with your organic product mixed together with water. Since organic solvents and water are usually immiscible (they don't mix — think oil and water), they form two distinct layers, and a separating funnel lets you cleanly split them apart.
Method, step by step: transfer the mixture into the funnel, add a stopper, invert and shake to mix the layers thoroughly (opening the stopcock periodically to release pressure — repeated roughly 15–20 times), then let the funnel stand so the two layers fully separate. Open the stopcock to drain the bottom (aqueous) layer away, then collect the organic layer separately.
Sometimes the reaction mixture also needs neutralising before purification (e.g. if acid catalyst or acidic reagents were used) — this is done by adding sodium carbonate solution to the vessel or funnel, which also helps wash out other impurities. When neutralising, open the stopcock slowly, since carbon dioxide gas is released and you don't want to lose product in the fizzing.
Using drying agents
Even after solvent extraction, tiny traces of water can remain dissolved in the organic layer. Drying agents mop this up. They are anhydrous inorganic salts that readily absorb (hydrate with) water.
| Drying Agent | Typical Use |
|---|---|
| Anhydrous calcium chloride | Commonly used to dry hydrocarbons |
| Anhydrous calcium sulfate / magnesium sulfate | More general-purpose drying agents |
Method: add a spatula of drying agent to the organic liquid and swirl. If it clumps together, there's still water present — keep adding more drying agent until some grains remain freely dispersed as a fine powder (this shows all the water has been absorbed). The dry liquid should now look clear, and can be decanted or filtered away from the solid drying agent. If the liquid has a low boiling point, keep it stoppered/lidded during this process to reduce evaporation loss.
A student adds anhydrous magnesium sulfate to a cloudy organic liquid. After swirling, the solid clumps together. What does this tell the student, and what should they do next?
⚛️ Part 2: Interpreting Mass Spectra
How a Mass Spectrometer Works
Mass spectrometry identifies unknown compounds by essentially "weighing" molecules and their fragments. Here's the logical chain of what happens:
- A tiny sample of the compound is vaporised and bombarded with a beam of high-energy electrons.
- This knocks an electron out of a molecule, leaving behind a positively charged species with one unpaired electron — the molecular ion, written M+•.
- The molecular ion is often unstable and breaks apart (fragments) into smaller ions, neutral molecules, and radicals. Crucially, only the positively charged fragments are detected — neutral fragments and radicals just aren't picked up by the instrument.
- The positive ions are accelerated by an electric field, then deflected by a magnetic field based on their mass-to-charge ratio (m/z).
- A detector records how many ions arrive at each m/z value, producing the spectrum — a set of peaks at different m/z positions.
m/z = (mass of the ion) ÷ (charge of the ion)
Most ions carry a charge of just +1, so for these, m/z simply equals the ion's mass (e.g. an ion of mass 12 with charge 1+ gives m/z = 12). But watch out for higher charges: an ion of mass 16 with charge 2+ gives m/z = 16 ÷ 2 = 8.
Deflection rule: smaller and/or more highly charged ions have a smaller m/z, and these are deflected the most by the magnetic field (they're detected "first" in the sense of needing the least field strength to redirect them into the detector, being most strongly attracted toward the magnet's negative pole).
The base peak is simply the tallest peak in the spectrum — it corresponds to the most abundant (most commonly formed) ion, which isn't necessarily the molecular ion.
In a sample of iron, the ions ⁵⁴Fe²⁺ and ⁵⁶Fe³⁺ are detected. Calculate their m/z ratios and determine which ion is deflected more.
Isotopes in Mass Spectra
Isotopes are atoms of the same element (same number of protons/electrons) but with a different number of neutrons, giving them different masses. Mass spectrometry is actually one of the best tools for finding the relative abundance of isotopes in nature — the proportion of each isotope present in a natural sample.
In a mass spectrum of an element with isotopes, you'll see separate peaks for each isotope, and the height of each peak is proportional to how abundant that isotope is naturally. For example, boron shows two peaks: one at m/e = 10 with height 19.9% (boron-10) and a taller one at m/e = 11 with height 80.1% (boron-11) — meaning about 80% of natural boron atoms are the heavier isotope.
Deducing the Molecular Formula (M⁺ and M+1 peaks)
The single most important peak in a spectrum for identifying the compound's mass is the one with the highest m/z value — this is the molecular ion peak (M+), and it directly gives you the compound's relative molecular mass, because it represents the whole molecule having simply lost one electron (no fragmentation yet).
Just next to it (one unit higher) you'll often spot a small extra peak called the [M+1] peak. This isn't a different compound — it's caused by the natural presence of the carbon-13 isotope. Since only about 1% of naturally occurring carbon is ¹³C, this peak is always small. But its size is meaningful: the more carbon atoms a molecule has, the bigger the [M+1] peak (more chances for one of those carbons to be ¹³C). So an M+1 peak for hexane (6 carbons) would be noticeably taller than for ethane (2 carbons).
A mass spectrum shows a molecular ion peak at m/z = 70 (with a small extra peak at m/z = 71). Does this correspond to but-1-ene or pent-1-ene?
Fragmentation Patterns
Two different compounds can share the same molecular mass (and so the same M⁺ peak position) — so on its own, the molecular ion peak can't fully identify a structure. This is where the smaller fragment peaks become useful: they act like a structural "barcode" unique to each compound, because fragmentation tends to break at particular, chemically sensible points, and different structures break apart differently.
Fragments generally arise in two ways: forming a characteristic fragment ion directly (e.g. m/z = 29 is very characteristic of C₂H₅⁺), or through the loss of a small, stable molecule from the molecular ion (common losses: 18 for H₂O, 28 for CO, 44 for CO₂).
Alkanes — breaking C–C bonds
Simple alkanes fragment by breaking C–C bonds, generating a family of predictable ions:
| Fragment | m/e |
|---|---|
| CH₃⁺ | 15 |
| C₂H₅⁺ | 29 |
| C₃H₇⁺ | 43 |
| C₄H₉⁺ | 57 |
| C₅H₁₁⁺ | 71 |
| C₆H₁₃⁺ | 85 |
Notice the pattern: each fragment differs from the next by 14 (one CH₂ unit) — very handy for quickly checking or predicting peak positions.
Halogenoalkanes — double molecular ion peaks
Halogens have their own naturally-occurring isotopes (e.g. bromine exists as ⁷⁹Br and ⁸¹Br in roughly equal abundance), so a halogenoalkane's mass spectrum shows two molecular ion peaks close together — e.g. at m/e = 108 and 110 for a bromoalkane — roughly equal in height because the two bromine isotopes are similarly abundant in nature.
Alcohols — losing water or CH₂OH
Alcohols have a signature fragmentation behaviour: they readily lose a whole water molecule, producing a peak 18 mass units below the molecular ion. They also commonly show a peak at m/e = 31, corresponding to the CH₂OH⁺ fragment.
Take propan-1-ol (C₃H₇OH, M = 60) as the model example — it fragments in four distinct, explainable ways:
| m/e | What it is | How it forms |
|---|---|---|
| 60 | Molecular ion | CH₃CH₂CH₂OH loses one electron |
| 59 | C₃H₇O⁺ | Loss of H• |
| 42 | C₃H₆⁺ | Loss of a whole H₂O molecule |
| 31 | CH₂OH⁺ | Loss of •C₂H₅ |
| 29 | C₂H₅⁺ | Loss of •CH₂OH |
Which of the following statements about the mass spectrum of CH₃Br is correct?
A. One peak for the molecular ion at m/e = 44
B. One peak for the molecular ion at m/e = 95
C. Last two peaks in ratio 3:1 at m/e = 94 and 96
D. Last two peaks equal in size at m/e = 94 and 96
Which alcohol is not likely to have a fragment ion at m/e = 43?
A. (CH₃)₂CHCH₂OH B. CH₃CH(OH)CH₂CH₂CH₃ C. CH₃CH₂CH₂CH₂OH D. CH₃CH₂CH(OH)CH₃
🌈 Part 3: Using Infrared Spectrometry
How IR Spectroscopy Works
Where mass spectrometry tells you about the mass and pieces of a molecule, infrared spectroscopy tells you about the types of bonds present — meaning it's the go-to technique for identifying functional groups.
The underlying idea: bonds between atoms aren't rigid sticks — they constantly vibrate (stretching, bending, twisting), a bit like a spring. Each type of bond has its own natural "resonance frequency" it likes to vibrate at. When you shine infrared radiation of exactly that matching frequency onto the molecule, the bond absorbs that energy and vibrates more vigorously. A spectrophotometer shines a whole range of IR frequencies through the sample and measures how much of each frequency makes it through (the ones that get absorbed show up as dips/peaks in the spectrum).
Rather than wavelength, IR spectra are plotted using wavenumber — the reciprocal of the wavelength, measured in cm⁻¹. This is just a convention, but you must recognise it and read spectra correctly (note: wavenumber decreases left-to-right along the x-axis of a typical IR spectrum!).
Each type of absorption peak has a characteristic shape, described by:
- Width — broad or sharp
- Intensity — strong or weak
For instance, O–H bonds in alcohols and carboxylic acids give broad peaks (because hydrogen bonding causes a spread of slightly different vibrational energies), while the C=O bond in carbonyl groups gives a strong, sharp peak. Learning to recognise these "shapes" by eye, not just the position, is a key exam skill.
Absorption Ranges to Memorise
| Bond | Found in | Wavenumber range (cm⁻¹) |
|---|---|---|
| C–O | Hydroxy, ester | 1040 – 1300 |
| C=C | Aromatic compound, alkene | 1500 – 1680 |
| C=O | Amide carbonyl, carboxyl, ester | 1640 – 1690 / 1670 – 1740 / 1710 – 1750 |
| C≡N | Nitrile | 2200 – 2250 |
| C–H | Alkane | 2850 – 2950 |
| N–H | Amine, amide | 3300 – 3500 |
| O–H | Carboxyl, hydroxyl | 2500 – 3000 / 3200 – 3600 |
Reading a Spectrum in Practice
When comparing an IR spectrum against known compounds, look for the most distinctive peak(s) rather than trying to match every wiggle. A strong, sharp peak around 1710 cm⁻¹ (C=O) is a dead giveaway for a ketone/aldehyde/carboxylic acid, while a strong, broad peak spanning roughly 3200–3600 cm⁻¹ (O–H) points to an alcohol or carboxylic acid.
Two IR spectra are provided — one for propanone (CH₃COCH₃) and one for propan-2-ol ((CH₃)₂CHOH). Spectrum A shows a strong, sharp absorption around 1710 cm⁻¹ and no broad peak near 3200–3600 cm⁻¹. Spectrum B shows a strong, broad absorption around 3200–3500 cm⁻¹. Identify each spectrum.
An unknown liquid compound shows a strong, broad absorption at 2500–3000 cm⁻¹ AND a strong, sharp absorption at 1710–1750 cm⁻¹ in its IR spectrum, but no peak in the 3200–3600 cm⁻¹ range. What functional group is likely present, and why can you rule out a simple alcohol?
🧠 What to Memorise
✅ Concepts Checklist
🎯 Exam Tips & Common Mistakes
- 🎯 Exam Tips & Common Mistakes
- Solvent Extraction & Drying
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