Difficulty: Intermediate | Prerequisites: Spectroscopy Interpretation Methods study notes (Doc 1). You should be comfortable with the HDI formula, basic mass spec / IR / NMR interpretation, and aromatic chemistry before working through this example.
This is a full worked example of identifying an unknown compound from spectral data, the kind of problem you will see on an Organic Chemistry II exam. Compound 76 is a low-melting solid containing two different halogens, and you are given its elemental analysis, mass spectrum, IR spectrum, ¹H NMR, and ¹³C NMR data. The exercise walks through every step of the reasoning, from calculating the molecular formula to assigning the final structure. If you can follow the logic here and reproduce it on a similar problem, you are in good shape for exam-style spectral unknowns.
Compound 76 is 4-chlorobenzyl bromide (1-bromomethyl-4-chlorobenzene), molecular formula C₇H₆BrCl. It is a para-disubstituted benzene ring with a chlorine on one ring carbon and a CH₂Br group on the carbon directly opposite. The identification comes from combining elemental analysis with a molecular ion at m/e 204, an HDI of 4 (aromatic ring), IR confirmation of C-Cl and aromatic bonds, and NMR data showing a CH₂ singlet at 4.3 ppm alongside two closely spaced doublets at 7.15 ppm (the para substitution fingerprint).
Para-disubstituted benzene
A benzene ring where two substituents sit on opposite carbons (positions 1 and 4). This arrangement creates a plane of symmetry through the ring, making two pairs of equivalent ring hydrogens and two pairs of equivalent ring carbons.
Benzyl group
A CH₂ group bonded directly to a benzene ring. In "benzyl bromide," the bromine is on the CH₂, not on the ring itself.
Alkoxide displacement
A reaction where an alkoxide ion (RO⁻) displaces a leaving group. The problem states that alkoxide can displace only one of the two halogens, yielding an ether. This tells you one halogen is on a reactive benzylic carbon (the CH₂Br) and the other is on the aromatic ring, where it resists nucleophilic substitution.
Given data
Elemental analysis: C 40.92%, H 2.94%. The remainder (56.14%) is halogens.
The compound contains two different halogens.
Molecular ion region: m/e = 204 to 208.
Working out the formula
Moles of C: 40.92 / 12.01 = 3.41. Moles of H: 2.94 / 1.008 = 2.92.
Ratio C:H is approximately 3.41 : 2.92, which simplifies to roughly 7 : 6.
This gives a partial formula of C₇H₆, with mass 7(12) + 6(1) = 90.
Remaining mass from molecular ion: 204 - 90 = 114.
Two different halogens summing to 114: Br (79) + Cl (35) = 114. This is the only combination of two different common halogens that works.
Molecular formula: C₇H₆BrCl (mass = 204).
Confirmation from isotope peaks
The cluster at m/e 204 to 208 is consistent with both Br and Cl isotopes contributing. The M+2 peaks arise from ⁸¹Br and ³⁷Cl; M+4 arises from both heavy isotopes being present simultaneously. This pattern would not appear with, for example, two chlorines or two bromines.
For C₇H₆BrCl, halogens count the same as hydrogen in the HDI formula:
\text{HDI} = \frac{2(7) + 2 - 6 - 2}{2} = \frac{8}{2} = 4HDI = 4 is the signature value for a benzene ring (three double bonds + one ring). At this point, an aromatic ring is the leading hypothesis.
Molecular ion region (m/e 204 to 208)
The molecular ion (parent ion) appears at m/e = 204, representing the intact C₇H₆⁷⁹Br³⁵Cl molecule. The peaks at 206 and 208 are isotope peaks from combinations of ⁸¹Br and ³⁷Cl.
Base peak at m/e = 125
This is the tallest peak in the spectrum. Loss of 79 from the molecular ion (204 - 79 = 125) corresponds to loss of ⁷⁹Br. The resulting fragment, C₇H₆Cl⁺ (mass 125), is the chlorotoluene-type cation, which is stabilised by the aromatic ring.
Fragment at m/e = 127
This corresponds to loss of ⁷⁷ from a heavier isotope molecular ion (206 - 79 = 127, or equivalently the ³⁷Cl-containing fragment at 125 + 2 = 127). It confirms the chlorine isotope pattern in the fragment.
Reading the fragments together
The dominant fragmentation is loss of Br (not Cl), which makes chemical sense: the C-Br bond in a benzylic CH₂Br position is weaker and breaks more readily than the C-Cl bond on the aromatic ring.
What is absent (and why that matters)
No broad O-H stretch around 3200 to 3600 cm⁻¹: rules out alcohols and carboxylic acids.
No strong, sharp C=O peak near 1700 cm⁻¹: rules out ketones, aldehydes, esters, amides, and carboxylic acids.
These absences are consistent with a halogenated aromatic hydrocarbon, which has no oxygen-containing functional groups.
What is present
C-H stretches near 2850 to 3000 cm⁻¹ (sp³ C-H from the CH₂ group) and near 3030 cm⁻¹ (aromatic sp² C-H).
Medium peaks at 1450 to 1600 cm⁻¹: aromatic C=C ring stretches, confirming the benzene ring hypothesis from the HDI calculation.
Overtone/combination peaks in the 1600 to 2000 cm⁻¹ region, consistent with a disubstituted aromatic.
Strong peak in the 600 to 800 cm⁻¹ region: C-Cl stretch.
The C-Br stretch (expected around 500 to 600 cm⁻¹) falls at the edge or beyond the printed spectrum range and is not clearly visible.
¹H NMR
Two signals are observed:
Singlet at ~4.3 ppm, integration 2H: a CH₂ group. It appears as a singlet because neither neighbouring carbon carries any hydrogens (one side is the aromatic ring carbon, the other side is the bromine). The chemical shift of 4.3 ppm is consistent with CH₂ bonded to a halogen (expected range roughly 2.5 to 4.5 ppm for CH₂-X), shifted slightly downfield by the adjacent aromatic ring.
Apparent singlet at ~7.15 ppm, integration 4H, resolving into two closely spaced doublets at 300 MHz: four aromatic protons in a para-disubstituted ring. The two doublets arise because the ring has two pairs of equivalent protons (H₂/H₆ and H₃/H₅), and each pair couples to the adjacent pair.
¹³C NMR (off-resonance decoupled)
Five signals, corresponding to five distinct carbon environments in a seven-carbon molecule (two pairs of equivalent ring carbons account for the discrepancy):
Singlet at 146.0 ppm: a quaternary aromatic carbon (no H). The high chemical shift suggests it is bonded to the electronegative chlorine.
Doublet at 129.8 ppm: aromatic CH, one of the two pairs of equivalent ring carbons. The doublet means one hydrogen is attached.
Singlet at 120.1 ppm: a quaternary aromatic carbon (no H). This is the ring carbon bearing the CH₂Br substituent.
Doublet at 114.3 ppm: the other pair of equivalent aromatic CH carbons.
Triplet at 29.3 ppm: the CH₂ carbon (two hydrogens attached). This is the benzylic carbon bonded to bromine. Its chemical shift in the sp³ range confirms it is not part of the ring.
Putting it together
The ¹H NMR confirms a para-disubstituted aromatic ring (two closely spaced doublets at 7.15 ppm, 4H total) plus a CH₂ group (singlet, 2H). The ¹³C data confirms five unique carbon environments, two of which are quaternary ring carbons (one bearing Cl, the other bearing CH₂Br), two are equivalent-pair CH ring carbons, and one is the sp³ CH₂.
The compound is 4-chlorobenzyl bromide (also called 1-bromomethyl-4-chlorobenzene or p-chlorobenzyl bromide). It is a benzene ring with a chlorine atom at position 1 (or 4) and a -CH₂Br group at position 4 (or 1), giving a para arrangement.
This structure is consistent with all the data:
Molecular formula C₇H₆BrCl, MW 204.
HDI = 4 (benzene ring).
Mass spectrum: molecular ion at 204; loss of Br gives the base peak at 125.
IR: aromatic C=C, sp³ and sp² C-H, C-Cl stretch, no carbonyl.
¹H NMR: CH₂ singlet (2H) at 4.3 ppm; para-disubstituted aromatic pattern (4H) at 7.15 ppm.
¹³C NMR: five environments, two quaternary ring carbons, two equivalent-pair CH ring carbons, one CH₂.
Reactivity: alkoxide displaces the benzylic Br (good SN2 substrate) but not the aryl Cl.
Students sometimes assume both halogens must be on the ring. In this compound, only the chlorine is on the ring. The bromine is on the benzylic CH₂ carbon. The chemical reactivity clue (alkoxide displaces only one halogen) is designed to catch this mistake.
Students occasionally confuse "closely spaced doublets" with a singlet. At lower resolution they look like one peak, but the splitting is real and carries structural information (para substitution).
When the C-Br stretch is not visible on the IR, students sometimes conclude bromine is absent. It is more likely that the spectrum simply does not extend low enough in wavenumber. Always cross-reference with other data before ruling out a functional group.
Students sometimes assign the ¹³C triplet at 29.3 ppm to an aromatic carbon. Aromatic carbons appear at 100 to 160 ppm. A signal at 29.3 ppm is firmly in the sp³ region.
⚠️ This is a representative exam-level spectral unknown. Being able to walk through each technique in order and cross-check conclusions is the core skill being tested.
⚠️ The elemental analysis to molecular formula conversion is almost always the first part of these problems. Practise it until the arithmetic is fast.
⚠️ The "alkoxide displaces only one halogen" clue is a reactivity-based hint, not a spectral one. Exams often mix reactivity and spectral data; do not ignore non-spectral clues.
⚠️ Knowing that a benzylic halide is reactive toward SN2 while an aryl halide is not connects this problem back to substitution and elimination chemistry.
True or false: In Compound 76, both halogens are bonded directly to the aromatic ring. False. Only chlorine is on the ring; bromine is on the benzylic CH₂.
Fill in the blank: The base peak at m/e = 125 corresponds to loss of ___ from the molecular ion. Bromine (⁷⁹Br, mass 79).
True or false: The two closely spaced doublets at 7.15 ppm indicate ortho substitution. False. They indicate para (1,4-) substitution.
Fill in the blank: The ¹³C singlet at 146.0 ppm represents a ring carbon bonded to ___. Chlorine.
True or false: The CH₂ appears as a singlet in ¹H NMR because it has no neighbouring protons on adjacent carbons. True.
Q: Why does the mass spectrum show a cluster of peaks at m/e 204 to 208 rather than a single molecular ion peak?
A: Because both bromine and chlorine have naturally occurring heavier isotopes (⁸¹Br and ³⁷Cl). Different combinations of the light and heavy isotopes produce molecular ions at 204, 206, and 208.
Q: The base peak is at 125, not at 169. What does this tell you about which bond breaks more easily?
A: Loss of Br (204 - 79 = 125) is favoured over loss of Cl (204 - 35 = 169). The benzylic C-Br bond breaks more readily than the aryl C-Cl bond, producing the more stable benzylic/chlorotropylium-type cation.
Q: If this compound were ortho-disubstituted instead of para, how would the ¹H NMR aromatic region differ?
A: In an ortho-disubstituted ring, the four aromatic protons are no longer in two equivalent pairs. You would see a more complex multiplet rather than two neat doublets.
Q: The problem states that alkoxide can displace only one of the two halogens. Which one, and why?
A: The bromine on the CH₂ (benzylic position). Benzylic halides undergo SN2 readily. Aryl halides (the Cl on the ring) do not undergo standard nucleophilic substitution because the C-Cl bond is strengthened by resonance with the ring.
Q: If you saw a ¹³C signal at 170 ppm, what functional group would you suspect, and would it be consistent with Compound 76?
A: A signal around 170 ppm suggests a carbonyl carbon (C=O). This would not be consistent with Compound 76, which has no oxygen in its molecular formula.
The reactivity clue (alkoxide displaces only one halogen) ties this problem directly to nucleophilic substitution: benzylic halides are excellent SN2 substrates, while aryl halides resist standard substitution (you would need a strong nucleophile and special conditions, or a transition-metal catalyst, to substitute on an aromatic ring). The IR and NMR interpretation here also builds on concepts from electrophilic aromatic substitution, where substituent effects on the ring influence chemical shifts and substitution patterns.
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