Difficulty: Intermediate | Prerequisites: General Chemistry, introductory organic chemistry (functional groups, bonding, Lewis structures). Familiarity with basic atomic structure (isotopes, atomic mass) is assumed.
In organic chemistry, identifying an unknown compound is one of the core practical skills. You are given spectral data (mass spectrum, infrared spectrum, ¹H and ¹³C NMR spectra) plus elemental analysis, and your job is to piece together the molecular formula, the functional groups, and ultimately the full structure. This is the detective work of organic chemistry: each technique gives you a different kind of clue, and none of them alone is enough. If you have been away from the course for a few weeks, this topic pulls together material from molecular formula determination, functional group chemistry, and aromaticity, so brush up on those first.
You identify an unknown organic compound by combining four spectral techniques. Mass spectrometry gives you the molecular weight and molecular formula. IR spectroscopy tells you which functional groups are present (or absent). ¹H NMR reveals how many distinct hydrogen environments exist, how many protons are in each, and what neighbours they have. ¹³C NMR tells you how many unique carbon environments there are and whether each carbon carries hydrogens. Put them together with the degree of unsaturation and you can pin down a single structure.
Molecular ion (parent ion, M⁺)
The peak in a mass spectrum corresponding to the intact molecule after it has lost one electron. Its m/e value equals the molecular weight. Think of it as the "whole molecule" peak before anything breaks apart.
Base peak
The tallest (most intense) peak in the mass spectrum, set to 100% relative abundance. It represents the most stable fragment ion. In simple terms, it is the fragment the molecule most readily breaks into.
Fragment ion
Any ion formed when the molecular ion breaks apart. Fragment masses tell you which pieces the molecule contains. Think of these as the jigsaw pieces you get when the molecule shatters.
Isotope peaks (M+1, M+2)
Smaller peaks just above the molecular ion caused by heavier isotopes of the constituent atoms (¹³C, ³⁷Cl, ⁸¹Br). These are especially diagnostic for halogens: bromine gives a near-equal M and M+2 pair, chlorine gives an M+2 about one-third the height of M.
Degree of unsaturation (hydrogen deficiency index, HDI)
A count of how many rings and/or pi bonds a molecule contains, calculated from its molecular formula. Each ring or double bond adds one unit; a triple bond adds two. The formula is (2C + 2 + N − H − X) / 2 for a molecule with C carbons, N nitrogens, H hydrogens, and X halogens.
Chemical shift (δ, ppm)
The position of an NMR signal on the horizontal axis, measured in parts per million relative to TMS (tetramethylsilane). It reflects the electronic environment around a nucleus. In simple terms, the more electron-poor a proton or carbon is, the further downfield (higher ppm) it appears.
Splitting pattern (multiplicity)
The shape of an NMR signal caused by neighbouring non-equivalent protons. A proton next to n equivalent neighbours appears as an n+1 line pattern: singlet (0 neighbours), doublet (1), triplet (2), quartet (3), and so on.
Integration (¹H NMR)
The area under a ¹H NMR peak, proportional to the number of protons producing that signal. Often shown as a number printed above or beside the peak.
DEPT / off-resonance decoupling (¹³C NMR)
Techniques that reveal how many hydrogens are attached to each carbon. In off-resonance decoupled ¹³C spectra: singlet = C (no H), doublet = CH, triplet = CH₂, quartet = CH₃.
Wavenumber (cm⁻¹)
The x-axis unit of an IR spectrum, proportional to the frequency (and energy) of the absorbed light. Higher wavenumber means higher energy. Functional groups absorb at characteristic wavenumber ranges.
Locating the molecular ion (M⁺)
Look for the highest m/e cluster in the spectrum. This is typically the molecular ion region.
The molecular ion gives you the molecular weight directly.
If two different halogens are present, the molecular ion region will show a characteristic cluster of peaks separated by 2 m/e units, reflecting different isotope combinations.
Reading isotope patterns for halogens
One bromine atom: M and M+2 peaks in roughly 1:1 ratio (⁷⁹Br and ⁸¹Br are nearly equally abundant).
One chlorine atom: M and M+2 peaks in roughly 3:1 ratio (³⁵Cl is about three times as abundant as ³⁷Cl).
Both Br and Cl present: the molecular ion region shows a more complex cluster (M, M+2, M+4), because each halogen contributes its own isotope pair.
Identifying the base peak and fragment ions
The base peak is the tallest peak in the spectrum; it is assigned 100% relative intensity.
Fragment ions result from bonds breaking in the molecular ion. Common losses include loss of a halogen (loss of 35 for Cl, loss of 79 for Br), loss of CH₂X groups, and loss of small stable molecules.
The difference between the molecular ion and a fragment tells you the mass of the piece that was lost.
Combining mass spec with elemental analysis
Elemental analysis gives percentage composition (e.g., C 40.92%, H 2.94%).
Divide each percentage by the element's atomic mass to get mole ratios. This gives the empirical formula.
Compare the empirical formula mass with the molecular ion mass. If they match, the molecular formula equals the empirical formula. If the molecular ion is a whole-number multiple, scale up accordingly.
Key diagnostic regions
3200 to 3600 cm⁻¹: O-H stretch (broad) or N-H stretch (sharper). Absence of a broad peak here rules out alcohols, carboxylic acids, and amines.
2850 to 3000 cm⁻¹: C-H stretches (sp³). Nearly always present in organic molecules.
~3030 cm⁻¹: C-H stretch (sp², aromatic or vinyl). Look for this alongside aromatic C=C peaks.
~1700 cm⁻¹: C=O stretch (strong, sharp). One of the most distinctive peaks in IR. Its absence rules out ketones, aldehydes, esters, carboxylic acids, and amides.
1450 to 1600 cm⁻¹: aromatic C=C stretches (medium intensity, often two or three peaks in this region).
1600 to 2000 cm⁻¹: overtone/combination region for substituted aromatics. The pattern of weak peaks here can indicate substitution pattern (mono, ortho, meta, para).
600 to 800 cm⁻¹: C-Cl stretch (strong).
500 to 600 cm⁻¹: C-Br stretch (strong). This region is sometimes cut off on printed spectra.
Strategy: confirm presence and absence
IR is most powerful for confirming what functional groups are not present. No broad O-H, no sharp C=O at 1700, no N-H peaks: you have eliminated several functional group families immediately. Then look for positive evidence of the groups that remain consistent with your molecular formula.
Three things to extract from every ¹H NMR spectrum
Number of signals: tells you how many chemically distinct proton environments the molecule has.
Integration (the number printed above or beside each peak): tells you the ratio of protons in each environment.
Splitting pattern: tells you how many non-equivalent neighbouring protons each group has (n+1 rule).
Chemical shift ranges to know
0.8 to 1.5 ppm: alkyl CH₃, CH₂ (no electronegative neighbours).
1.5 to 2.5 ppm: allylic or benzylic protons, CH next to C=O.
2.5 to 4.0 ppm: protons on carbon bonded to electronegative atoms (CH₂-X where X = halogen, O, N).
4.5 to 6.5 ppm: vinyl protons (C=C-H).
6.5 to 8.5 ppm: aromatic protons.
9.0 to 10.0 ppm: aldehyde protons.
10 to 12 ppm: carboxylic acid protons.
Recognising a para-disubstituted benzene ring
A para-disubstituted ring has two pairs of equivalent protons. At lower field strength, the four aromatic protons may appear as a single peak. At higher resolution (e.g. 300 MHz), that apparent singlet resolves into two closely spaced doublets, each integrating for 2H. This pattern is a strong indicator of 1,4-disubstitution.
Singlets from CH₂ between two groups with no H
A CH₂ group flanked on both sides by atoms that carry no hydrogens (e.g., a carbon bearing a halogen and an aromatic ring carbon) appears as a singlet, because there are no adjacent protons to cause splitting. Its chemical shift (typically 3.5 to 4.5 ppm when next to a halogen) confirms the electronic environment.
What each signal type tells you (off-resonance or DEPT)
Singlet: a carbon with no hydrogens attached (quaternary carbon, or C bonded to halogen/oxygen with no H).
Doublet: a CH group (one hydrogen on that carbon).
Triplet: a CH₂ group (two hydrogens).
Quartet: a CH₃ group (three hydrogens).
Chemical shift ranges for ¹³C
0 to 50 ppm: sp³ carbons (alkyl). Carbons bonded to halogens tend toward 20 to 50 ppm depending on the halogen.
100 to 160 ppm: aromatic and vinyl carbons (sp²). Carbons directly bonded to electronegative substituents on the ring shift downfield within this range.
170 to 220 ppm: carbonyl carbons (C=O). Absence of signals in this region rules out ketones, aldehydes, esters, acids.
Counting signals to determine symmetry
The number of ¹³C signals equals the number of chemically distinct carbon environments. A molecule with five ¹³C signals and seven carbons in its formula must have symmetry: two pairs of equivalent carbons. For a para-disubstituted benzene ring (six ring carbons), the four CH carbons appear as two signals (each representing two equivalent carbons) and the two substituted ring carbons each give one signal, for a total of four ring-carbon signals.
The formula
\text{HDI} = \frac{2C + 2 + N - H - X}{2}where C = number of carbons, N = number of nitrogens, H = number of hydrogens, X = number of halogens. Oxygen and sulfur do not appear in the formula (they do not change the hydrogen count).
What the number means
HDI = 0: no rings, no double bonds. A fully saturated, acyclic molecule.
HDI = 1: one ring or one double bond.
HDI = 4: the classic value for a benzene ring (three double bonds + one ring = 4 HDI). Any time you calculate HDI = 4, consider whether an aromatic ring is present.
HDI = 5 or more: likely an aromatic ring plus additional unsaturation.
Worked example with C₇H₆BrCl
HDI = (2 × 7 + 2 - 6 - 2) / 2 = (14 + 2 - 6 - 2) / 2 = 8 / 2 = 4. Halogens count the same as hydrogens in this formula (each subtracts 1). An HDI of 4 is consistent with a benzene ring.
Students often think the base peak is the molecular ion. It is not. The base peak is simply the most intense peak in the spectrum, which is usually a fragment, not the intact molecule.
Students confuse the n+1 rule with the number of hydrogens on the carbon itself. The splitting tells you about neighbouring protons, not the protons on that carbon. A CH₂ group next to no other H-bearing carbons gives a singlet, not a triplet.
In ¹³C NMR, students often assume every carbon gives a separate signal. Equivalent carbons (related by symmetry) give a single signal. A para-disubstituted ring has only four distinct carbon signals, not six.
Students sometimes include oxygen or sulfur in the HDI formula. These atoms do not appear in the calculation: they do not change the hydrogen requirement of a molecule.
⚠️ You will almost certainly be asked to determine a molecular formula from elemental analysis + molecular ion mass. Practise converting percentages to mole ratios quickly.
⚠️ Recognising halogen isotope patterns in the mass spectrum is a frequently tested skill. Know the approximate ratios for Cl (3:1) and Br (1:1) at M vs M+2.
⚠️ The HDI calculation appears on nearly every spectroscopy exam. Make sure you know how halogens enter the formula (they subtract, like H).
⚠️ Being able to distinguish ortho, meta, and para substitution patterns from ¹H NMR splitting is a common exam question. Para gives two closely spaced doublets; ortho and meta give more complex patterns.
⚠️ IR "absence" evidence is tested just as often as "presence" evidence. Saying "no carbonyl peak at 1700 cm⁻¹, therefore no ketone/aldehyde/ester/acid" is valid and expected.
True or false: The base peak in a mass spectrum always corresponds to the molecular ion. False. The base peak is the most intense fragment; the molecular ion may be much smaller.
Fill in the blank: A compound with molecular formula C₆H₅Cl has an HDI of ___. 4 (consistent with a benzene ring).
True or false: A ¹H NMR singlet means the proton has no neighbouring protons on adjacent carbons. True.
Fill in the blank: In an IR spectrum, a strong absorption near 1700 cm⁻¹ indicates the presence of a ___ bond. C=O (carbonyl).
True or false: In ¹³C NMR, a triplet in the off-resonance decoupled spectrum means the carbon has three hydrogens attached. False. A triplet means CH₂ (two hydrogens). Three hydrogens give a quartet.
Q: A compound has elemental analysis C 40.92%, H 2.94%, with the remainder being halogens. The molecular ion appears at m/e = 204 with significant M+2 and M+4 peaks. What is the molecular formula?
A: C₇H₆BrCl. The percentages for C and H account for about 43.86% of the mass, leaving 56.14% for halogens. The isotope pattern (M, M+2, M+4) indicates both Br and Cl. Working from the molecular ion mass of 204: 7 × 12 + 6 × 1 + 79 + 35 = 204.
Q: You calculate HDI = 4 for a compound. What structural feature should you consider first, and why?
A: A benzene ring. A benzene ring contributes exactly 4 units of HDI (three C=C double bonds + one ring). HDI = 4 is a strong hint that the molecule is aromatic.
Q: An IR spectrum shows no absorption near 1700 cm⁻¹ and strong peaks in the 600 to 800 cm⁻¹ region. What can you conclude?
A: There is no carbonyl group (no ketone, aldehyde, ester, or carboxylic acid). The strong peaks in the 600 to 800 cm⁻¹ region suggest a C-Cl bond.
Q: A ¹H NMR spectrum shows two signals: a singlet integrating for 2H at 4.3 ppm and what appears to be a singlet integrating for 4H at 7.15 ppm that resolves into two closely spaced doublets at higher field strength. What does this tell you?
A: The 4.3 ppm singlet (2H) is a CH₂ group with no neighbouring protons, in an environment deshielded by an electronegative atom (consistent with CH₂ bonded to a halogen). The 7.15 ppm signal (4H) resolving into two doublets indicates a para-disubstituted aromatic ring, where two pairs of equivalent aromatic protons couple to each other.
Q: In a ¹³C NMR spectrum, a signal appears as a triplet at 29.3 ppm. What type of carbon is this?
A: A CH₂ carbon (triplet = two attached hydrogens) in an sp³ environment. The chemical shift of 29.3 ppm is consistent with a carbon bearing a halogen substituent such as bromine.
This material connects directly to aromatic chemistry (Chapter topics on benzene, electrophilic aromatic substitution, and substituent effects on the ring). Understanding how electron-donating and electron-withdrawing groups shift NMR signals ties back to resonance and induction concepts from earlier in the course. The halogen chemistry here also links to nucleophilic substitution (SN1, SN2) and elimination reactions, since the C-halogen bonds you identify by spectroscopy are the same bonds that react in those mechanisms.
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