# AHL: Spectroscopic identification of organic compounds

> IB Chemistry HL · IB Chemistry HL (2025 Syllabus)
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u6-ahl-spectroscopic-identification-of-organic/

This module covers the three core spectroscopic techniques used to identify unknown organic compounds: mass spectrometry, infrared (IR) spectroscopy, and proton nuclear magnetic resonance (¹H NMR) spectroscopy. You will learn to combine data from all three to deduce full molecular structures for exam questions.

**Prerequisites:** Organic functional group nomenclature and properties; Covalent bond polarity and vibrational modes; Basic nuclear spin properties

## Learning objectives

- Interpret mass spectra to find molecular mass and molecular formula of organic compounds
- Correlate IR absorption peaks to specific organic functional groups
- Analyze ¹H NMR spectra using chemical shift, integration, and splitting patterns
- Combine data from multiple spectroscopic techniques to deduce full organic structures

## Mass Spectrometry for Organic Compounds

**Mass Spectrometry** — A technique that ionizes organic molecules, separates ions by mass-to-charge ratio ($m/z$), and produces a spectrum of ion abundance vs $m/z$. The highest significant $m/z$ peak (the molecular ion peak) gives the molecular mass of the intact compound.

*Example:* A molecular ion peak at $m/z = 72$ corresponds to a molar mass of 72 g mol⁻¹ for the compound.

Molecules fragment during ionization, producing smaller fragment ions that give clues about the carbon skeleton and structure of the original molecule. The tallest peak in the spectrum is called the base peak, which corresponds to the most stable (most abundant) fragment.

**Worked example:** A straight-chain alkane has a molecular ion peak at $m/z = 58$. Determine its molecular formula.

1. Write the general formula for an alkane and its molar mass:
2. $$C_nH_{2n+2}, \quad M = 12n + (2n+2) = 14n + 2$$
3. Set the molar mass equal to 58 and solve for $n$:
4. $$14n + 2 = 58 \implies 14n = 56 \implies n = 4$$
5. The molecular formula is therefore $C_4H_{10}$ (butane).

> **Exam tip:** Never confuse the base peak (tallest peak) with the molecular ion peak. Only the highest significant $m/z$ peak gives the molecular mass of the full molecule.

## Infrared (IR) Spectroscopy

**IR Spectroscopy** — A technique that measures absorption of infrared radiation by covalent bonds. Each bond type has a characteristic vibrational frequency (stretching/bending) that corresponds to a specific absorption wavenumber, allowing identification of functional groups.

*Example:* A strong absorption peak near 1700 cm⁻¹ almost always indicates the presence of a carbonyl (C=O) group.

The region below 1500 cm⁻¹ is called the fingerprint region. This region is unique to every compound, so it can be used to confirm the identity of a known compound by matching to a reference spectrum.

**Worked example:** An organic compound has a broad absorption peak between 3200-3600 cm⁻¹ and no absorption peak near 1700 cm⁻¹. Identify the most likely functional group.

1. Broad absorption in the 3200-3600 cm⁻¹ range is characteristic of O-H bond stretching, which points to a hydroxyl group.
2. A peak near 1700 cm⁻¹ is required for a carbonyl group; the absence of this peak rules out carboxylic acids (which have both O-H and C=O).
3. The only functional group matching this data is an alcohol hydroxyl (-OH) group.

## Proton (¹H) NMR Spectroscopy

**¹H NMR Spectroscopy** — A technique that detects hydrogen nuclei (protons) in a magnetic field. Three key properties of NMR peaks reveal the structure: chemical shift, integration, and splitting.

1. **Chemical shift (δ, ppm):** Indicates the electron environment of the proton; adjacent electronegative groups shift peaks to higher δ values.
2. **Integration:** The area under a peak, proportional to the number of equivalent protons producing that peak.
3. **Spin-spin splitting:** Follows the $n+1$ rule: $n$ equivalent adjacent protons produce $n+1$ peaks in the splitting pattern.

> **mnemonic**
>
> n+1 = neighbors + one: just add 1 to the number of equivalent adjacent protons to get the number of split peaks.

**Worked example:** Predict the number of peaks and splitting pattern for ¹H NMR of propane ($CH_3CH_2CH_3$).

1. Identify equivalent protons: the two end methyl groups are identical (6 protons total), and the middle methylene group is unique (2 protons total). This gives **2 distinct peaks**.
2. For the end methyl (6H) protons: adjacent carbon has 2 equivalent protons, so $n=2$, splitting = $2+1 = 3$ → triplet.
3. For the middle methylene (2H) protons: adjacent carbons have 6 equivalent total protons, so $n=6$, splitting = $6+1 = 7$ → septet.
4. Final NMR: 2 peaks: 6H triplet and 2H septet.

> **Exam tip:** The peak at δ = 0 ppm is always the TMS reference standard, not a peak from your unknown compound. Ignore it when counting peaks.

## Combining Spectroscopic Data

No single spectroscopic technique gives enough information to fully determine an unknown structure. Exam questions will always require you to combine data from all three techniques following a standard workflow.

**Worked example:** A compound has a molecular ion peak at 60 $m/z$, an IR peak at ~1710 cm⁻¹, and ¹H NMR with two peaks: a 3H singlet and a 1H singlet. Deduce the structure.

1. Calculate molecular formula from 60 $m/z$: $C_2H_4O_2$ = (2×12) + (4×1) + (2×16) = 60, which fits.
2. IR at 1710 cm⁻¹ confirms a C=O carbonyl group, which matches the formula with two oxygen atoms.
3. NMR has two peaks (two sets of equivalent protons), both singlets meaning no adjacent protons ($n=0$, $0+1=1$ peak).
4. The only matching structure is ethanoic acid: $CH_3COOH$. The $CH_3$ protons are equivalent with no adjacent protons on the carbonyl carbon, giving a 3H singlet. The O-H proton gives a 1H singlet, matching all data.

## Common pitfalls

- **Wrong:** Confusing the base peak with the molecular ion peak in mass spectrometry
  - Why it fails: The base peak is the tallest peak from the most abundant fragment, not the intact molecule
  - Correct: Use the highest significant $m/z$ peak as the molecular ion to get the full molecular mass
- **Wrong:** Assuming a 3200-3600 cm⁻¹ O-H peak always means an alcohol
  - Why it fails: Carboxylic acids also have an O-H peak in this range, but additionally have a C=O peak near 1700 cm⁻¹
  - Correct: Always check for the presence of a 1700 cm⁻¹ C=O peak to distinguish alcohol from carboxylic acid O-H
- **Wrong:** Counting the TMS peak at δ=0 ppm as a proton peak from the unknown compound
  - Why it fails: TMS is added as a reference standard and is not part of the unknown compound
  - Correct: Ignore the δ=0 ppm peak when analyzing the unknown compound's NMR spectrum
- **Wrong:** Applying the n+1 rule to non-equivalent adjacent protons
  - Why it fails: The n+1 rule only works for equivalent adjacent protons; non-equivalent protons produce more complex splitting
  - Correct: Only use n+1 when all adjacent protons are in identical chemical environments
- **Wrong:** Counting equivalent protons on the same carbon for splitting calculations
  - Why it fails: Equivalent protons do not split each other, only non-equivalent adjacent protons cause splitting
  - Correct: Only count non-equivalent adjacent protons when applying the n+1 rule

## Cheatsheet

| Technique | Key Information | Exam Reminder |
| --- | --- | --- |
| Mass Spectrometry | Molecular mass from molecular ion peak, fragment clues for carbon skeleton | Highest m/z = molecular mass, base peak = most abundant fragment |
| IR Spectroscopy | Functional group identification from bond absorption | Broad 3200-3600 cm⁻¹ = O-H, ~1700 cm⁻¹ = C=O, use Data Booklet |
| ¹H NMR | Chemical shift = environment, Integration = number of protons, n+1 = adjacent protons | δ=0 = TMS reference (ignore it), equivalent protons do not split each other |

## What's next

Spectroscopic identification of organic compounds is one of the most heavily tested topics in IB Chemistry HL, appearing in nearly every exam session as a full structured question. Mastery of this skill is not only required for your IB exam, but also forms the foundation for organic structure determination in university chemistry and related fields like medicine, pharmacology, and materials science. The workflow you learned here for combining multiple data types is a core scientific skill that transfers to many other areas of chemistry. Next, you can extend your knowledge to carbon-13 NMR spectroscopy, which gives additional information about the carbon skeleton of organic molecules, and apply your structure deduction skills to organic reaction mechanism problems, where identifying reaction products is a common exam requirement.

- [Practical and investigative skills](https://www.owlsprep.com/study/ib-chemistry-hl-u7-overview/)
- [Experimental Design](https://www.owlsprep.com/study/ib-chemistry-hl-u7-experimental-design/)
- [Data collection and processing](https://www.owlsprep.com/study/ib-chemistry-hl-u7-data-collection-and-processing/)

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