Questions: NMR Spectroscopy: Chemical Shifts and Spin Coupling

Question 1

The ¹H NMR spectrum of a compound shows a signal at δ 9.5 ppm. Which functional group most likely accounts for this chemical shift?

• A. Aromatic ring proton (δ 6–8 ppm range)
• B. Alkyl CH₃ group adjacent to a carbonyl
• C. Aldehyde C–H proton
• D. Vinylic proton on an isolated double bond

Answer: (C)

Aldehyde protons (CHO) resonate at δ 9–10 ppm — one of the most downfield positions in common ¹H NMR spectra. The extreme deshielding occurs because the carbonyl oxygen withdraws electron density through both induction and anisotropy, leaving the aldehyde proton in a very electron-poor environment. Aromatic protons appear at δ 6–8 ppm (option A). Alkyl protons adjacent to a carbonyl appear around δ 2–3 ppm. Vinylic protons appear around δ 4.5–6 ppm.

Question 2

In the ¹H NMR spectrum of ethanol (CH₃CH₂OH), the CH₂ group is adjacent to both CH₃ (3 protons) and OH (1 exchangeable proton). Assuming the OH proton does not couple under typical conditions, what splitting pattern does the CH₂ signal show?

• A. A doublet, because the CH₂ sees two equivalent neighbors
• B. A quartet, because the CH₂ sees three CH₃ protons and splits into n+1 = 4 lines
• C. A quintet, because the CH₂ sees four total neighboring protons (3 from CH₃ + 1 from OH)
• D. A singlet, because both neighboring groups cancel each other's splitting

Answer: (B)

Under typical NMR conditions, the OH proton exchanges rapidly with solvent or trace water, so it does not couple to adjacent protons and appears as a broad singlet or is averaged out. The CH₂ therefore only couples to the three equivalent CH₃ protons, giving n+1 = 3+1 = 4 lines — a quartet. Option C would be correct if the OH proton were coupled, but fast exchange typically eliminates this coupling in routine ¹H NMR of alcohols in common solvents.

Question 3

The coupling constant J measured from a doublet signal in proton A is identical to the J measured from the doublet in proton B when A and B are mutually coupled.

• T. True
• F. False

Answer: True

This is a fundamental property of J-coupling: the coupling constant between two nuclei is the same regardless of which partner you measure it from. If proton A shows a doublet with J = 7 Hz, proton B (to which A is coupled) will also show its splitting with J = 7 Hz. This symmetry arises because J reflects the interaction through shared bonding electrons, which is a property of the A–B bond, not of either nucleus alone. This fact allows you to identify which peaks are coupled to each other by matching J values across signals.

Question 4

Increasing the external magnetic field strength (moving from a 300 MHz to a 600 MHz spectrometer) will increase the coupling constant J between two neighboring protons.

• T. True
• F. False

Answer: False

Coupling constants J (in Hz) are independent of the external magnetic field strength. J-coupling is transmitted through bonding electrons and reflects the intrinsic spin-spin interaction between nuclei — a property of molecular structure, not of the spectrometer. Chemical shifts in Hz do scale with field strength (which is why moving to higher-field spectrometers improves resolution of overlapping signals), but J stays constant. This independence of J from field strength is one way to distinguish coupling from other line-broadening effects.

Question 5

How does spin-spin coupling reveal information about molecular connectivity, and why does this complement (rather than duplicate) the structural information provided by chemical shifts?

Model answer: Chemical shift tells you the electronic environment of each proton (shielding, nearby electron-withdrawing or -donating groups) but cannot tell you which protons are bonded to adjacent carbons. Spin-spin coupling provides this connectivity: the splitting pattern of a signal reveals how many non-equivalent protons are attached to neighboring atoms (via the n+1 rule), and the shared J value identifies which pairs of signals are coupled. Together, shift (identity of functional environment) and splitting (identity of neighboring groups) allow reconstruction of the full molecular skeleton.

Chemical shift alone would identify the types of environments present but could not tell you how they are connected. For example, two compounds could have identical sets of chemical shifts but different connectivity. Coupling resolves this because the splitting pattern — and specifically which signals share the same J value — reveals the graph of connections between proton-bearing carbons. NMR structure determination therefore reads connectivity from coupling first, then assigns functional group context from chemical shifts.