Difficulty: Intermediate | Prerequisites: Basic ¹H NMR concepts (chemical shift, integration, splitting patterns), molecular formula and degrees of unsaturation.
NMR spectroscopy is the primary tool for determining what an organic molecule actually looks like. This section of the exam tests whether you can work backwards from a ¹H NMR spectrum and a molecular formula to deduce a structure. You should be comfortable calculating degrees of unsaturation, interpreting chemical shift regions, reading integration values, and recognising common patterns (aromatic protons, benzylic protons, methyl groups).
TL;DR
Given a molecular formula and a ¹H NMR spectrum, calculate degrees of unsaturation first, then use chemical shifts and integration to assign groups of protons to structural fragments. Piece those fragments together into a structure consistent with all the data. The exam pairs this with a reaction (HBr addition), so you need to identify both the starting material and the product.
Chemical shift (δ, in ppm)
The position of a signal on the NMR spectrum's horizontal axis. It reflects the electronic environment of the proton. Shielded protons (electron-rich) appear upfield (lower ppm); deshielded protons (electron-poor, near electronegative atoms or π systems) appear downfield (higher ppm).
In simple terms, chemical shift tells you what neighbourhood a proton lives in.
Integration
The area under an NMR signal, proportional to the number of equivalent protons producing that signal. A signal integrating for 5H next to a signal integrating for 2H tells you there are 5 protons in one environment and 2 in another.
Splitting pattern (multiplicity)
The shape of a signal caused by neighbouring non-equivalent protons. Follows the n+1 rule: n equivalent neighbours produce n+1 peaks (singlet, doublet, triplet, quartet, etc.).
Degrees of unsaturation (DoU)
Also called the index of hydrogen deficiency (IHD). Calculated from the molecular formula: DoU = (2C + 2 + N – H – X) / 2 for CₓHₙNₘOₕXₖ. Each degree represents one ring or one π bond. A benzene ring accounts for 4 degrees (3 double bonds + 1 ring).
Aromatic protons
Protons directly on a benzene ring appear in the 6.5–8.5 ppm region. A monosubstituted benzene ring gives 5 aromatic H; a disubstituted ring gives 4, and so on.
Benzylic protons
Protons on a carbon directly attached to a benzene ring. They appear in roughly the 2.0–4.5 ppm region, depending on what else is attached to that carbon.
Step 1: Calculate degrees of unsaturation
For CₓHₙ: DoU = (2x + 2 – y) / 2. For a molecule with the formula C₁₀H₁₂:
DoU = (2(10) + 2 – 12) / 2 = (22 – 12) / 2 = 5.
Five degrees of unsaturation strongly suggests a benzene ring (4 DoU) plus one additional ring or double bond.
Step 2: Identify the aromatic region
Look for signals between 6.5 and 8.5 ppm. If the integration is 5H, you have a monosubstituted benzene ring (C₆H₅–). That accounts for C₆H₅ of your formula and 4 of your 5 DoU.
Step 3: Work through remaining signals
Subtract the aromatic fragment from the molecular formula. For C₁₀H₁₂ minus C₆H₅ = C₄H₇ remaining, with 1 DoU left. That remaining degree of unsaturation could be a ring or a double bond in the C₄H₇ fragment.
Use chemical shifts to assign each signal:
0–1.0 ppm: CH₃ groups (methyl), typically in alkyl environments
1.0–2.0 ppm: CH₂ or CH groups in alkyl chains
2.0–3.0 ppm: benzylic CH₂ or CH₃ (protons on a carbon next to an aromatic ring)
4.5–6.5 ppm: vinyl protons (C=C–H) or benzylic CH next to electronegative groups
Step 4: Piece together and check
Assemble the fragments. Verify that the total number of C and H atoms matches the molecular formula, that the DoU is accounted for, and that the structure is consistent with every signal (chemical shift, integration and splitting).
Step 5: If a reaction is given, apply it
The exam asks you to identify both the starting material (C₁₀H₁₂) and its product after reaction with HBr. Once you know the starting material, apply HBr addition (Markovnikov) to get the product and check it against the product's NMR spectrum.
Given information:
Starting material molecular formula: C₁₀H₁₂
Its NMR has a single resonance between 4.5 and 6.5 ppm (a vinyl proton region)
Product A (C₁₀H₁₃Br) has a ¹H NMR spectrum with signals at roughly: ~7 ppm (5H), ~3 ppm (a CH–Br type signal), ~2 ppm (2H, CH₂), and ~1 ppm (3H, CH₃)
Solving for the starting material (3 pts):
DoU for C₁₀H₁₂ = (22 – 12)/2 = 5. A monosubstituted benzene accounts for 4 DoU and 5 aromatic H.
Remaining: C₄H₇, with 1 DoU. This fragment has one double bond.
The NMR shows a single resonance between 4.5 and 6.5 ppm, meaning there is one type of vinyl proton. Combined with a phenyl group and C₄H₇ with one double bond, the starting material is likely a phenyl-substituted butene.
The answer from the exam key: the starting material is an allylbenzene-type compound (C₆H₅–CH₂–CH=CH–CH₃ or similar, any valid alkene with C₁₀H₁₂). Full credit is awarded for any correct structure matching the molecular formula and NMR data.
Solving for Product A (9 pts):
HBr adds across the double bond (Markovnikov). The product has the formula C₁₀H₁₃Br.
From the product NMR:
~7 ppm, 5H: monosubstituted benzene ring (unchanged)
~3 ppm: a CH bearing Br (deshielded by the bromine)
~2 ppm, 2H: a CH₂ group (benzylic or adjacent to the CHBr)
~1 ppm, 3H: a CH₃ group
The product structure: PhCH₂–C(Br)(CH₃)– with the remainder completing the molecular formula. Partial credit is given for correct fragments (2 pts per correctly assigned signal/fragment, 2 pts for correct stereochemistry notation if applicable).
Key grading notes from the exam key:
2 points per correct NMR signal assignment
1 point for each correctly identified structural fragment
Stereochemistry is not necessarily required for full credit on this particular problem
Students forget to calculate degrees of unsaturation before looking at the spectrum. DoU is your first filter: it immediately tells you whether to expect a benzene ring, a triple bond, or just double bonds and rings.
Students miscount aromatic protons. A monosubstituted benzene always has 5 aromatic H. If the integration near 7 ppm says 5H, that is a monosubstituted ring, full stop.
Students assume a signal between 4.5 and 6.5 ppm must be aromatic. Vinyl protons (on a C=C) also appear in this region. Check the chemical shift carefully: aromatic protons are typically above 6.5 ppm.
Students neglect to check that their proposed structure matches all the data. After drawing a structure, go back through every signal and verify the integration, chemical shift and splitting all agree.
⚠️ The NMR section is worth 12 points (3 for the starting material, 9 for the product). The product is worth three times as much, so spend your time there.
⚠️ Partial credit is generous: 2 points per statement about the structure that is supported by the NMR data. Even if you cannot get the full structure, identifying the aromatic ring, a CH₃ group, or a CHBr unit earns marks.
⚠️ The exam provides chemical shift and IR reference tables on the last page. Use them.
Fill in the blank: The degrees of unsaturation for C₆H₆ is ____.
4 (benzene)
True or false: A signal at 7.2 ppm integrating for 5H indicates a disubstituted benzene ring.
False. 5H is a monosubstituted ring.
Fill in the blank: Protons on a carbon bearing a bromine atom appear at roughly – ppm.
3–4 ppm (deshielded by the electronegative bromine)
Q: A compound with the formula C₈H₁₀ has 4 degrees of unsaturation. Its ¹H NMR shows a 5H signal at 7.3 ppm and a 5H signal at 2.3 and 1.2 ppm. What is a likely structure?
A: 4 DoU with 5 aromatic H = monosubstituted benzene (C₆H₅–). Remaining: C₂H₅, with 0 DoU left. C₂H₅ = an ethyl group. Structure: ethylbenzene (PhCH₂CH₃). The 2.3 ppm signal (2H) is the benzylic CH₂, and the 1.2 ppm signal (3H) is the CH₃.
Q: How do you distinguish a vinyl proton (~5–6 ppm) from an aromatic proton (~7–8 ppm) when they are in nearby regions?
A: Check the exact chemical shift. Aromatic protons are generally above 6.5 ppm, often in the 7.0–7.5 ppm range for simple aromatics. Vinyl protons typically appear between 4.5 and 6.5 ppm. Integration also helps: if 5H are at 7.2 ppm, that is an aromatic ring; a 1H signal at 5.5 ppm is more likely a vinyl proton.
Q: After determining that a C₁₀H₁₂ starting material is a phenyl-substituted alkene, how do you predict the product of HBr addition?
A: Identify the double bond in the structure. Apply Markovnikov's rule: H adds to the less substituted end of the double bond, Br to the more substituted (or benzylic) carbon. Draw the product, add up the atoms (should be C₁₀H₁₃Br), and verify against the product NMR.
NMR is the thread that ties organic chemistry together. Every reaction you learn can be verified by NMR: did the starting material's alkene signal disappear? Did a new CHBr signal appear at 3–4 ppm? Thinking in terms of "what would the NMR look like" is an excellent way to check your synthesis and mechanism answers.
Degrees of unsaturation connect directly to the nomenclature section: every double bond and ring you learned to name corresponds to one degree of unsaturation.
Structural elucidation connects forward to Organic Chemistry II, where you will combine ¹H NMR with ¹³C NMR, IR spectroscopy and mass spectrometry for more complex unknowns.
Related Terms / Search Tags
NMR spectroscopy, 1H NMR, proton NMR, chemical shift, ppm, integration, splitting pattern, multiplicity, degrees of unsaturation, index of hydrogen deficiency, DoU, IHD, aromatic protons, vinyl protons, benzylic protons, structural elucidation, NMR problem solving, C10H12, monosubstituted benzene, HBr addition NMR, CHEM 2301, Pomerantz, Exam 4