# Analysis of Organic Substances

> Chemical tests for functional groups, then mass spectrometry, infrared spectroscopy, carbon-13 NMR and proton NMR, and how to combine them to identify an unknown compound.

From HSC Chemistry on WhizzNotes, free study notes for NSW students: https://whizznotes.com/chemistry/analysing-organic-substances/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The detective and the molecule
A forensic chemist receives a tiny drop of clear liquid from a crime scene. It could be one of a hundred different compounds. She cannot taste it or smell it safely, and there is far too little for many test-tube reactions.

Instead, she puts a speck into three machines. One weighs the molecule and its broken pieces. One shows which bonds it has. One maps where every hydrogen and carbon sits. Like clues in a detective story, each result rules some suspects out. Put together, they point to just one molecule.

**In short:** Organic compounds are identified with chemical tests and instruments. Mass spectrometry gives the molar mass, infrared spectroscopy shows functional groups like O−H and C=O, and NMR spectroscopy shows the different carbon and hydrogen environments, which together reveal the structure.

## Chemical tests for functional groups

| Group | Test | Positive result |
|---|---|---|
| C=C (alkene) | Bromine water | Orange-brown to colourless, quickly |
| −OH (1° or 2° alcohol) | Acidified $\mathrm{K_2Cr_2O_7}$, warm | Orange to green |
| −OH (3° alcohol) | Acidified $\mathrm{K_2Cr_2O_7}$, warm | No change |
| −COOH (carboxylic acid) | $\mathrm{Na_2CO_3}$ or $\mathrm{NaHCO_3}$ | Fizzes ($\mathrm{CO_2}$) |
| −COOH | Blue litmus | Turns red |
| −NH₂ (amine) | Red litmus | Turns blue |
| Aldehyde vs ketone | Acidified dichromate | Aldehyde: orange to green. Ketone: no change |

Chemical tests are cheap and quick, but they use up the sample and only tell you about one group at a time.

## Mass spectrometry (MS)

A mass spectrometer measures the **mass** of a molecule and its fragments.

1. **Ionisation**: a beam of high-energy electrons knocks an electron off each molecule, making a positive ion, $\mathrm{M^+}$.
2. **Acceleration**: the ions are sped up by an electric field.
3. **Deflection**: a magnetic field bends their paths. Lighter ions bend more.
4. **Detection**: ions are counted at each mass-to-charge ratio, $m/z$.

The sample is **destroyed** in the process.

**Reading a mass spectrum:**

- The **molecular ion peak** ($\mathrm{M^+}$) is the peak with the highest $m/z$ (ignore tiny peaks at M+1 from carbon-13). It gives the **molar mass**.
- The **base peak** is the tallest peak, set to 100%. It is the most stable fragment.
- Other peaks are **fragments**. The difference from M tells you what broke off.

| $m/z$ (or mass lost) | Fragment |
|---|---|
| 15 | $\mathrm{CH_3^+}$ (loss of 15 = lost $\mathrm{CH_3}$) |
| 29 | $\mathrm{C_2H_5^+}$ or $\mathrm{CHO^+}$ |
| 31 | $\mathrm{CH_2OH^+}$ |
| 43 | $\mathrm{C_3H_7^+}$ or $\mathrm{CH_3CO^+}$ |
| 45 | $\mathrm{COOH^+}$ |
| Loss of 17 | lost OH |
| Loss of 18 | lost $\mathrm{H_2O}$ |

**Chlorine and bromine** show two molecular ion peaks, 2 apart, because of their isotopes: Cl gives M and M+2 in a 3 : 1 ratio; Br gives M and M+2 in a 1 : 1 ratio.

## Infrared (IR) spectroscopy

Bonds **vibrate** (stretch and bend). Each type of bond absorbs infrared radiation at its own wavenumbers (in cm⁻¹). An IR spectrum shows **troughs** (low transmittance) where the sample absorbs.

| Bond | Wavenumber (cm⁻¹) | Look |
|---|---|---|
| O−H (alcohol) | 3230 to 3550 | Broad |
| O−H (carboxylic acid) | 2500 to 3000 | Very broad |
| N−H (amine, amide) | 3300 to 3500 | Medium, sharper |
| C−H | 2850 to 3300 | Sharp |
| C=O (aldehyde, ketone, acid, ester, amide) | 1680 to 1750 | Strong and sharp |
| C=C | 1620 to 1680 | |
| C−O | 1000 to 1300 | |

The region below 1500 cm⁻¹ is the **fingerprint region**: complex, but unique to each molecule. It can confirm an identity by matching a known spectrum.

IR tells you **which functional groups** are present, but not how many or where.

## Carbon-13 NMR

Some nuclei, including $^{13}\mathrm{C}$ and $^1\mathrm{H}$, behave like tiny magnets. In a strong magnetic field they absorb radio waves. The exact frequency depends on the atom's **chemical environment** (what it is bonded to). This is **nuclear magnetic resonance**.

In a $^{13}\mathrm{C}$ NMR spectrum:

- The **number of peaks** = the number of **different carbon environments**.
- The **chemical shift** $\delta$ (in ppm) shows what each carbon is near.
- Peak heights are **not** used for counting.

| Carbon type | Shift (ppm) |
|---|---|
| C−C (alkane) | 5 to 40 |
| C−Cl or C−Br | 10 to 70 |
| C−O (alcohol, ether, ester) | 50 to 90 |
| C=C | 90 to 150 |
| C=O (acid, ester, amide) | 160 to 185 |
| C=O (aldehyde, ketone) | 190 to 220 |

Carbons in symmetrical positions are **equivalent**. Propanone, $\mathrm{CH_3COCH_3}$, has only **2** environments (both $\mathrm{CH_3}$ are the same).

## Proton (¹H) NMR

A ¹H NMR spectrum gives **three** pieces of information:

1. **Number of peak sets** = number of different **hydrogen environments**.
2. **Area under each set** (or the given ratio) = the **number of H** in that environment.
3. **Splitting**: a set of peaks is split by the H on **neighbouring carbons**. The **n + 1 rule**: $n$ neighbouring H atoms split a signal into $n + 1$ peaks.

| Neighbouring H | Peaks | Name |
|---|---|---|
| 0 | 1 | Singlet |
| 1 | 2 | Doublet |
| 2 | 3 | Triplet |
| 3 | 4 | Quartet |
| 6 | 7 | Septet |

H on O−H or N−H is usually a **singlet** and does not split its neighbours.

| Hydrogen type | Shift (ppm) |
|---|---|
| R−CH₃, R−CH₂−R | 0.9 to 1.7 |
| $\mathrm{CH_3}$−C=O | 2.0 to 2.7 |
| −CH−O (next to O) | 3.3 to 4.5 |
| R−OH | 1 to 6 (variable) |
| −CHO (aldehyde) | 9 to 10 |
| −COOH | 9 to 13 |

### Worked example: Ethanol, CH₃CH₂OH
- 3 environments: $\mathrm{CH_3}$, $\mathrm{CH_2}$, OH. Areas 3 : 2 : 1.
- $\mathrm{CH_3}$ has 2 neighbouring H (on $\mathrm{CH_2}$): a **triplet** near 1.2 ppm.
- $\mathrm{CH_2}$ has 3 neighbouring H (on $\mathrm{CH_3}$): a **quartet** near 3.7 ppm (next to O).
- OH: a **singlet**.

A triplet (area 3) with a quartet (area 2) is the classic sign of an **ethyl group**, $\mathrm{CH_3CH_2-}$.

## Putting it together

| Technique | Tells you |
|---|---|
| Chemical tests | Presence of a functional group |
| Mass spectrometry | Molar mass and fragments |
| Infrared | Functional groups (bond types) |
| ¹³C NMR | Number and type of carbon environments |
| ¹H NMR | Number, type and neighbours of hydrogen environments |

No single technique is enough to be sure. Use them together:

1. Find the molar mass from MS (and the formula, if percentage composition is given).
2. Find the functional groups from IR and chemical tests.
3. Count environments with ¹³C and ¹H NMR.
4. Use splitting to join the pieces.
5. Check your final structure against every piece of data.

**Mind map**

```text
Organic analysis
  Chemical tests
    Bromine water: C=C
    Dichromate: 1°, 2° alcohol, aldehyde
    Carbonate: acid
  Mass spectrometry
    Molecular ion: molar mass
    Base peak: tallest
    Fragments
    Cl 3:1, Br 1:1
  Infrared
    O−H broad
    C=O 1680 to 1750
    Fingerprint region
  13C NMR
    Peaks = C environments
    Shift shows type
  1H NMR
    Sets = H environments
    Area = number of H
    n + 1 splitting
  Combine
    Check all data
```

### Year 11: Empirical and molecular formulas
If you are given the percentage composition, find the **empirical formula** first:

1. Assume 100 g, so each percentage becomes grams.
2. Divide by each atomic mass to get moles.
3. Divide by the smallest number of moles.

**Example.** 54.5% C, 9.1% H, 36.4% O. Moles: C 4.54, H 9.03, O 2.28. Ratio 2 : 4 : 1, so the empirical formula is $\mathrm{C_2H_4O}$ (44 g/mol).

If MS shows the molecular ion at 88, the molecular formula is $\mathrm{C_4H_8O_2}$ ($88 \div 44 = 2$).

### Year 12: Solving a full problem
An unknown has formula $\mathrm{C_4H_8O_2}$.

- **IR**: strong peak at 1740 cm⁻¹ (C=O), **no** broad O−H. So it is an **ester**, not an acid.
- **¹³C NMR**: 4 peaks, one at 171 ppm (ester C=O) and one at 60 ppm (C−O).
- **¹H NMR**: a singlet (area 3) at 2.0 ppm, a quartet (area 2) at 4.1 ppm, a triplet (area 3) at 1.3 ppm.

The singlet of 3 H at 2.0 ppm is a $\mathrm{CH_3}$ next to C=O with no H neighbours. The quartet and triplet are an **ethyl group**, and the quartet at 4.1 ppm means the $\mathrm{CH_2}$ is next to O. The compound is **ethyl ethanoate**, $\mathrm{CH_3COOCH_2CH_3}$.

Its isomer methyl propanoate, $\mathrm{CH_3CH_2COOCH_3}$, would show the singlet near 3.7 ppm instead (the $\mathrm{CH_3}$ on O) and the quartet near 2.3 ppm.

**Common fragments.** The mass lost tells you the piece: 15 is $\mathrm{CH_3}$, 17 is OH, 29 is $\mathrm{C_2H_5}$ or CHO, 45 is COOH. Butanoic acid (88) giving a peak at 73 has lost a $\mathrm{CH_3}$. Propanone gives a large peak at 43 ($\mathrm{CH_3CO^+}$); propanal gives a peak at 29 ($\mathrm{CHO^+}$ or $\mathrm{C_2H_5^+}$) and 57 (lost one H).

**More halogen atoms, more peaks.** Each Cl can be 35 or 37, so 4 chlorines give **five** molecular ion peaks: M, M+2, M+4, M+6 and M+8.

**Shapes to recognise in ¹H NMR.**
- A doublet (6 H) with a septet (1 H) is an **isopropyl** group, $\mathrm{(CH_3)_2CH-}$.
- A singlet at 11 to 12 ppm is a carboxylic acid O−H.
- A peak at 9 to 10 ppm is an aldehyde H, split by H on the next carbon.
- A singlet near 8.0 ppm (1 H) in an ester is the H of a **methanoate**, $\mathrm{HCOO-}$.

**Telling isomers apart.** Count environments. 1,1,1-trichloropropane has 2 H environments; 1,2,3-trichloropropane also has 2 but with different splitting and areas (4 : 1). ¹H NMR separates them. In ¹³C NMR, a ketone C=O is near 205 to 220 ppm, while acids and esters are near 165 to 185 ppm.

## Quick summary

- Chemical tests: bromine water (C=C), dichromate (1° and 2° alcohols, aldehydes), carbonate (acids).
- MS: molecular ion gives molar mass; fragments show pieces. Sample is destroyed.
- IR: broad O−H at 3230 to 3550 (alcohol) or 2500 to 3000 (acid); C=O at 1680 to 1750 cm⁻¹.
- ¹³C NMR: number of peaks = number of carbon environments.
- ¹H NMR: number of sets, peak areas and n + 1 splitting.
- Combine all the data and check the structure fits every piece.

## Quiz

Try these questions from the chapter quiz.

1. Which test shows a C=C double bond?
   - A. Bromine water turns colourless
   - B. Sodium carbonate fizzes
   - C. Litmus turns blue
   - D. Dichromate stays orange
   - Answer: A. Bromine water turns colourless. Bromine adds across the double bond.

2. In mass spectrometry, the molecular ion peak gives:
   - A. The number of carbon environments
   - B. The molar mass
   - C. The functional groups
   - D. The boiling point
   - Answer: B. The molar mass. $\mathrm{M^+}$ has the mass of the whole molecule.

3. The base peak in a mass spectrum is:
   - A. The peak with the highest $m/z$
   - B. The tallest peak
   - C. The first peak
   - D. Always at $m/z$ 15
   - Answer: B. The tallest peak. It is the most abundant ion, set to 100%.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/chemistry/analysing-organic-substances/#quiz