AHL: Spectroscopic identification of organic compounds

IB Chemistry HL· Unit 6: Organic Chemistry, AHL Topic 11· 35 min read

1. Mass Spectrometry for Organic Compounds★★☆☆☆⏱ 8 min

📘 Definition

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 $C_{4} H_{10}$ (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.

2. Infrared (IR) Spectroscopy★★★☆☆⏱ 10 min

📘 Definition

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.

3. Proton (¹H) NMR Spectroscopy★★★★★⏱ 12 min

📘 Definition

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.

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

📐 Worked Example

Predict the number of peaks and splitting pattern for ¹H NMR of propane ( $C H_{3} C H_{2} C H_{3}$ ).

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.

4. Combining Spectroscopic Data★★★★☆⏱ 5 min

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 $m / z$ : $C_{2} H_{4} O_{2}$ = (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: $C H_{3} C O O H$ . The $C H_{3}$ 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.

5. Common Pitfalls

Wrong move:

Confusing the base peak with the molecular ion peak in mass spectrometry

Why:

The base peak is the tallest peak from the most abundant fragment, not the intact molecule

Correct move:

Use the highest significant $m / z$ peak as the molecular ion to get the full molecular mass

Wrong move:

Assuming a 3200-3600 cm⁻¹ O-H peak always means an alcohol

Why:

Carboxylic acids also have an O-H peak in this range, but additionally have a C=O peak near 1700 cm⁻¹

Correct move:

Always check for the presence of a 1700 cm⁻¹ C=O peak to distinguish alcohol from carboxylic acid O-H

Wrong move:

Counting the TMS peak at δ=0 ppm as a proton peak from the unknown compound

Why:

TMS is added as a reference standard and is not part of the unknown compound

Correct move:

Ignore the δ=0 ppm peak when analyzing the unknown compound's NMR spectrum

Wrong move:

Applying the n+1 rule to non-equivalent adjacent protons

Why:

The n+1 rule only works for equivalent adjacent protons; non-equivalent protons produce more complex splitting

Correct move:

Only use n+1 when all adjacent protons are in identical chemical environments

Wrong move:

Counting equivalent protons on the same carbon for splitting calculations

Why:

Equivalent protons do not split each other, only non-equivalent adjacent protons cause splitting

Correct move:

Only count non-equivalent adjacent protons when applying the n+1 rule

6. Quick Reference 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

7. Frequently Asked

Do I need to memorize all absorption and chemical shift values?

No, most values are provided in the IB Data Booklet. However, common peaks like O-H (3200-3600 cm⁻¹) and C=O (~1700 cm⁻¹) are worth memorizing for speed.

What is the standard workflow for deducing an unknown structure?

Get molecular mass from mass spec to find molecular formula. 2. Use IR to identify functional groups. 3. Use NMR to map proton arrangement and carbon skeleton. 4. Confirm all data matches your proposed structure.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 2
Structure deduction from combined spectra
2024 · Paper 3 A
NMR splitting pattern interpretation
2023 · Paper 2
Functional group identification from IR

Going deeper

syllabus resourceIB Chemistry Data BookletUse for IR and NMR data tables during practice

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.