Questions: ¹H NMR Spectroscopy: Chemical Shift and Coupling Patterns

Question 1

A compound shows a signal at δ 9.8 ppm that integrates for 1H and appears as a doublet. Which interpretation is most consistent with these data?

• A. An aldehyde proton with one adjacent CH group
• B. An aromatic proton with one adjacent ring proton
• C. An alcohol OH proton coupled to a neighboring CH
• D. A methyl group next to a carbonyl

Answer: (A)

δ ~9.8 ppm is the hallmark of an aldehyde proton (strongly deshielded by the adjacent C=O). A doublet (n+1 = 2, so n = 1 neighbor) is consistent with a CHO proton next to a single CH group — the classic pattern for a branched aldehyde like isobutyraldehyde. Aromatic protons appear around δ 6.5–8, not δ 9.8. Alcohol OH protons typically appear around δ 1–5 and often exchange rapidly, giving broad singlets rather than clean doublets.

Question 2

In ethanol (CH₃CH₂OH), you observe a quartet and a triplet in the ¹H NMR. A student claims the quartet must have 4 times more area (integration) than the triplet. Is the student correct?

• A. No — the quartet arises from the CH₂ group (2H) and the triplet from the CH₃ group (3H), so the triplet integrates for more
• B. Yes — quartets by definition represent more protons than triplets
• C. No — the quartet arises from the CH₃ group (3H) and the triplet from the CH₂ group (2H), so the quartet integrates for more
• D. Integration cannot be determined without knowing the molecular weight

Answer: (C)

Splitting pattern and integration are independent pieces of information. The CH₃ group has 2 CH₂ neighbors → 2+1 = triplet, and represents 3H. The CH₂ group has 3 CH₃ neighbors → 3+1 = quartet, and represents 2H. So the triplet (3H) integrates for MORE than the quartet (2H). The common misconception confuses multiplicity (from coupling) with integration (from proton count) — a bigger multiplet does not mean more protons.

Question 3

Integration in ¹H NMR reports the absolute number of hydrogen atoms at each chemical shift.

• T. True
• F. False

Answer: False

Integration gives *relative* ratios, not absolute counts. A signal integrating for '3 units' relative to another at '2 units' tells you the 3:2 ratio — consistent with CH₃ vs. CH₂, or C₂H₆ vs. C₄H₈ at corresponding positions in a larger symmetric molecule. To assign absolute proton counts, you must combine integration ratios with the molecular formula (from mass spectrometry) or another independent constraint.

Question 4

The coupling constant (J value) measured in a doublet from proton A must equal the coupling constant measured in the corresponding multiplet of proton B, when A and B are coupled to each other.

• T. True
• F. False

Answer: True

Coupling is mutual: the same J value appears in both coupled partners' signals. If proton A is split into a doublet by B with J = 7 Hz, then proton B's signal will also show a splitting of 7 Hz from A. This matching of J values is a key tool for confirming which protons are neighbors — it provides independent verification beyond chemical shift and integration.

Question 5

Why does a proton next to a carbonyl group (like the alpha-H in acetaldehyde, CH₃CHO) appear at a higher δ value than a simple alkyl CH proton?

Model answer: The carbonyl group is strongly electron-withdrawing. It pulls electron density away from the adjacent proton, reducing the local shielding at that nucleus. With less electron density to shield it from the external magnetic field, the proton resonates at a higher δ (is 'deshielded' and appears downfield).

Chemical shift directly reflects electron density around the proton. Electron-withdrawing groups (carbonyls, halogens, aromatic rings) deshield nearby protons by reducing electron density, shifting signals downfield (higher δ). Electron-donating groups shield protons, shifting signals upfield (lower δ). Understanding this lets you read chemical shift as a direct map of electronic environment and functional group neighborhood.