40 Advanced Multiple-Choice Questions on Nuclear Magnetic Resonance Spectroscopy
Instructions:
Choose the best answer for each question. Explanations are provided below each answer.
1. Which of the following nuclei is NOT typically a focus of routine NMR experiments in organic chemistry due to its lack of nuclear spin? A) 1H B) 13C C) 12C D) 31P
Answer: C) 12C Explanation: Nuclei with even mass and even atomic numbers (like 12C and 16O) have a nuclear spin quantum number I=0, meaning they do not possess nuclear spin and are therefore NMR-inactive.
2. A nucleus with a spin quantum number I=1 (e.g., 2H) would exhibit how many allowed spin states in an applied magnetic field? A) 1 B) 2 C) 3 D) 4
Answer: C) 3 Explanation: The number of allowed spin states is given by 2I+1. For I=1, it is 2(1)+1=3 spin states (typically +1, 0, -1).
3. In an NMR experiment, the energy difference (ΔE) between nuclear spin states is directly proportional to which of the following? A) The Planck’s constant (h) B) The magnetogyric ratio (γ) and the applied magnetic field strength (B0) C) The Larmor frequency only D) The temperature of the sample
Answer: B) The magnetogyric ratio (γ) and the applied magnetic field strength (B0) Explanation: The fundamental equation relating energy difference to field strength is ΔE=γ(2πh)B0. The Larmor frequency is derived from this relationship.
4. What is the primary reason why modern NMR instruments operate at increasingly higher magnetic field strengths (e.g., 600 MHz vs. 300 MHz)? A) To decrease sample heating. B) To reduce the cost of the instrument. C) To increase the energy separation between spin states, enhancing sensitivity and resolution. D) To simplify the acquisition of 2D NMR spectra.
Answer: C) To increase the energy separation between spin states, enhancing sensitivity and resolution. Explanation: Higher fields increase ΔE, leading to a greater population difference between spin states (Boltzmann distribution), which significantly improves signal-to-noise ratio (sensitivity). Higher fields also increase chemical shift dispersion (in Hz), leading to better resolution of closely spaced signals and often simpler first-order spectra.
5. What is the physical phenomenon that leads to the “wobbling” motion of a spinning nucleus in an external magnetic field? A) Resonance B) Relaxation C) Precession D) Saturation
Answer: C) Precession Explanation: Precession is the characteristic conical motion of the magnetic moment of a spinning nucleus around the axis of the applied magnetic field.
6. If a proton experiences increased diamagnetic shielding, how will its resonance frequency (in Hz) and chemical shift (δ) change relative to a less shielded proton on the same spectrometer? A) Higher resonance frequency, higher δ. B) Lower resonance frequency, higher δ. C) Higher resonance frequency, lower δ. D) Lower resonance frequency, lower δ.
Answer: D) Lower resonance frequency, lower δ. Explanation: Diamagnetic shielding opposes the applied field, meaning the nucleus experiences a weaker net field. This results in a lower precessional frequency and thus a lower resonance frequency. Since TMS is highly shielded and set to 0 ppm, more shielded protons appear upfield (lower δ values).
7. Why is Tetramethylsilane (TMS) an ideal internal standard for both 1H and 13C NMR spectroscopy? A) It is highly reactive, allowing for easy removal. B) Its protons and carbons are generally deshielded. C) It is volatile, chemically inert, and its protons/carbons are highly shielded, resonating at one extreme of the typical chemical shift ranges. D) It has multiple non-equivalent protons, providing a good internal reference.
Answer: C) It is volatile, chemically inert, and its protons/carbons are highly shielded, resonating at one extreme of the typical chemical shift ranges. Explanation: TMS meets several criteria: it’s inert, doesn’t interfere with the sample, is easily removable, and its protons/carbons are typically more shielded than most organic compounds, defining the 0 ppm reference point.
8. Which effect causes protons on an aromatic ring to be significantly deshielded (e.g., 6.5-8 ppm)? A) Inductive effect of the ring carbons. B) Presence of a strong σ-bond framework. C) Magnetic anisotropy due to the ring current effect. D) Spin-spin coupling with neighboring carbons.
Answer: C) Magnetic anisotropy due to the ring current effect. Explanation: The delocalized π-electrons in an aromatic ring generate an induced magnetic field (ring current) that aligns with the applied field in the region of the aromatic protons, causing strong deshielding.
9. Why do protons involved in hydrogen bonding often exhibit broad NMR signals and concentration-dependent chemical shifts? A) They are rapidly exchanging with the solvent. B) Their environments are static. C) They undergo rapid exchange processes, leading to averaged signals and variable shielding. D) They are not subject to anisotropic effects.
Answer: C) They undergo rapid exchange processes, leading to averaged signals and variable shielding. Explanation: Hydrogen-bonded protons can exchange quickly with other protons (e.g., with solvent or other H-bonding molecules), causing the NMR spectrometer to “see” an average environment. This exchange rate often influences linewidth and chemical shift.
10. What type of magnet is used in modern high-field NMR spectrometers to achieve very strong and stable magnetic fields? A) Electromagnets B) Permanent magnets C) Superconducting magnets D) Rare-earth magnets
Answer: C) Superconducting magnets Explanation: Superconducting magnets, cooled by liquid helium and nitrogen, are essential for generating the high, stable magnetic fields required for modern high-resolution NMR.
11. What is the primary purpose of “shimming” in an NMR spectrometer? A) To adjust the sample spinning rate. B) To finely tune the homogeneity of the magnetic field. C) To change the RF pulse duration. D) To convert the FID into a spectrum.
Answer: B) To finely tune the homogeneity of the magnetic field. Explanation: Shimming uses small gradient coils to correct for tiny imperfections in the magnetic field, ensuring a uniform field across the sample and leading to sharper, better-resolved peaks.
12. What is the main advantage of using a cryoprobe in an NMR spectrometer? A) It allows for high-temperature NMR experiments. B) It significantly reduces thermal noise in the receiver, increasing sensitivity. C) It eliminates the need for deuterated solvents. D) It provides a stronger main magnetic field.
Answer: B) It significantly reduces thermal noise in the receiver, increasing sensitivity. Explanation: By cooling the receiver coils to very low temperatures, a cryoprobe minimizes thermal noise, allowing for the detection of weaker signals and boosting overall sensitivity.
13. In a Pulsed Fourier Transform (FT) NMR experiment, what does a “90° pulse” primarily achieve? A) It inverts the net magnetization vector by 180 degrees. B) It brings the net magnetization vector into the transverse plane. C) It saturates all the nuclear spins. D) It eliminates all spin-spin coupling.
Answer: B) It brings the net magnetization vector into the transverse plane. Explanation: A 90° pulse, specifically calibrated, rotates the equilibrium net magnetization vector (aligned along B0) into the xy (transverse) plane, where it can precess and induce a signal in the detector coil.
14. What is the Free-Induction Decay (FID) signal in FT-NMR primarily a representation of? A) The final, processed frequency-domain spectrum. B) The time-domain signal generated by relaxing nuclei after an RF pulse. C) The raw data from a Continuous-Wave (CW) experiment. D) The strength of the applied magnetic field.
Answer: B) The time-domain signal generated by relaxing nuclei after an RF pulse. Explanation: The FID is the raw data collected in FT-NMR, a complex decaying waveform in the time domain, representing the combined precessing signals of all excited nuclei.
15. What is the mathematical operation that transforms the time-domain FID signal into a frequency-domain NMR spectrum? A) Integration B) Differentiation C) Fourier Transform D) Deconvolution
Answer: C) Fourier Transform Explanation: The Fourier Transform is a powerful mathematical tool that deconvolutes the complex FID signal into its individual frequency components, yielding the conventional NMR spectrum.
16. Which of the following is true about the coupling constant (J) in spin-spin coupling? A) It is dependent on the applied magnetic field strength. B) It is measured in parts per million (ppm). C) It provides information about the number of equivalent protons. D) It is a measure of the splitting between peaks within a multiplet and is independent of field strength.
Answer: D) It is a measure of the splitting between peaks within a multiplet and is independent of field strength. Explanation: J values are fixed for a given coupling interaction, regardless of the spectrometer frequency. They are crucial for determining connectivity and sometimes stereochemistry.
17. A proton signal exhibiting a “triplet” multiplicity indicates coupling to how many equivalent neighboring protons? A) 1 B) 2 C) 3 D) 4
Answer: B) 2 Explanation: According to the n+1 rule, a triplet (3 peaks) arises from coupling to n=2 equivalent neighboring protons.
18. What characteristic indicates a second-order NMR spectrum? A) All peaks are perfectly sharp singlets. B) Peak intensities precisely follow Pascal’s Triangle. C) Peak intensities deviate from Pascal’s Triangle, often showing a “roofing effect,” and new peaks may appear. D) The coupling constant (J) is zero.
Answer: C) Peak intensities deviate from Pascal’s Triangle, often showing a “roofing effect,” and new peaks may appear. Explanation: Second-order effects occur when the chemical shift difference (Δν) is not significantly larger than the coupling constant (J), leading to distorted multiplets and sometimes more complex peak patterns.
19. What information does the integral (area under the peak) of an NMR signal directly provide? A) The chemical shift of the proton. B) The multiplicity of the signal. C) The number of equivalent protons contributing to that signal. D) The coupling constant (J).
Answer: C) The number of equivalent protons contributing to that signal. Explanation: The area under an NMR peak is directly proportional to the number of nuclei responsible for that resonance, providing quantitative information about the relative abundance of different proton types.
20. What is the primary mechanism of Spin-Lattice (T1) relaxation? A) Loss of phase coherence among precessing nuclei. B) Transfer of energy from the excited spin system to the surrounding molecular lattice. C) Absorption of RF energy by the nuclei. D) Random fluctuations in the applied magnetic field.
Answer: B) Transfer of energy from the excited spin system to the surrounding molecular lattice. Explanation: T1 relaxation involves the non-radiative dissipation of energy from the excited nuclear spins to the thermal motions of the surrounding molecules (the “lattice”). This process governs the return of the net magnetization to its equilibrium alignment along B0.
21. A very broad NMR peak is often indicative of a very short value of which relaxation time? A) T1 B) T2 C) Both T1 and T2 D) Neither T1 nor T2
Answer: B) T2 Explanation: Spin-spin (T2) relaxation dictates the linewidth of an NMR signal. Rapid dephasing of nuclear spins due to a short T2 leads to broad peaks.
22. What is the main challenge in 13C NMR spectroscopy compared to 1H NMR? A) The chemical shift range is too narrow. B) The nucleus has a very large magnetic moment. C) The low natural abundance of the 13C isotope and its small magnetic moment, leading to low sensitivity. D) 13C nuclei do not undergo spin-spin coupling.
Answer: C) The low natural abundance of the 13C isotope and its small magnetic moment, leading to low sensitivity. Explanation: Only 1.1% of natural carbon is 13C, and its magnetogyric ratio is much smaller than that of 1H, making 13C NMR intrinsically much less sensitive and requiring signal averaging.
23. What is the primary effect of “broad-band proton decoupling” in 13C NMR? A) It causes all 13C signals to appear as multiplets. B) It removes all C-H spin-spin coupling, leading to singlet 13C signals and enhances signal intensity via NOE. C) It broadens all 13C signals. D) It differentiates between CH, CH$_2$, and CH$_3$ carbons.
Answer: B) It removes all C-H spin-spin coupling, leading to singlet 13C signals and enhances signal intensity via NOE. Explanation: By continuously irradiating the proton frequencies, proton decoupling collapses all C-H coupling to singlets, simplifying the spectrum, and simultaneously enhances the carbon signals through the Nuclear Overhauser Effect (NOE).
24. Which advanced NMR technique would you use to differentiate between CH, CH$_2$, and CH$_3$ carbons in a 13C NMR spectrum? A) COSY B) HMBC C) DEPT D) NOESY
Answer: C) DEPT Explanation: Distortionless Enhancement by Polarization Transfer (DEPT) is a specific pulse sequence designed to determine the number of protons attached to each carbon (CH, CH$_2$, CH$_3$, and quaternary carbons).
25. In a DEPT-135 experiment, which carbon signals appear as negative (inverted) peaks? A) CH$_3$ carbons B) CH$_2$ carbons C) CH carbons D) Quaternary carbons
Answer: B) CH$_2$ carbons Explanation: DEPT-135 shows CH and CH$_3$ as positive, CH$_2$ as negative, and quaternary carbons are absent.
26. Which 2D NMR experiment is specifically used to correlate protons that are spin-spin coupled through two or three bonds? A) HSQC B) HMBC C) NOESY D) DOSY
Answer: B) HMBC (Heteronuclear Multiple Bond Correlation) Explanation: HMBC provides correlation peaks between carbons and protons separated by 2 or 3 bonds (and sometimes 4), which is crucial for identifying quaternary carbons and establishing long-range connectivity.
27. To determine the spatial proximity of protons in a molecule, regardless of bond connectivity, which 2D NMR technique would be most appropriate? A) COSY B) HSQC C) NOESY D) DEPT
Answer: C) NOESY (Nuclear Overhauser Effect SpectroscopY) Explanation: NOESY relies on the Nuclear Overhauser Effect, which is a through-space interaction, thus revealing protons that are close in 3D space, even if they are not directly bonded or coupled. It’s excellent for stereochemical and conformational analysis.
28. What type of information does a COSY spectrum primarily provide? A) Correlation between directly bonded carbons and protons. B) Correlation between spatially close protons. C) Correlation between spin-spin coupled protons. D) The diffusion coefficients of molecules.
Answer: C) Correlation between spin-spin coupled protons. Explanation: COSY (COrrelation SpectroscopY) reveals off-diagonal cross peaks between protons that are coupled to each other, allowing the tracing of spin systems.
29. What is the primary purpose of using deuterated solvents in NMR spectroscopy? A) To make the sample more soluble. B) To shift the chemical shifts of the analyte. C) To avoid overwhelming solvent proton signals in the spectrum and provide a lock signal. D) To increase the relaxation times of the nuclei.
Answer: C) To avoid overwhelming solvent proton signals in the spectrum and provide a lock signal. Explanation: Deuterated solvents (e.g., CDCl$_3$) have deuterium (2H) instead of hydrogen (1H). Deuterium is NMR-active but resonates at a very different frequency than protons and is typically used for the spectrometer’s field/frequency lock, preventing the solvent’s huge proton signal from obscuring the analyte.
30. What term describes the condition when the chemical shift difference (Δν) between coupled nuclei is small relative to their coupling constant (J), leading to complex multiplets with distorted intensities? A) First-order spectrum B) High-resolution spectrum C) Saturated spectrum D) Second-order spectrum
Answer: D) Second-order spectrum Explanation: A second-order spectrum arises when Δν/J is small, causing peak intensities to deviate from Pascal’s triangle and sometimes introducing additional peaks, requiring more complex analysis.
31. Protons alpha (α) to a carbonyl group (C=O) typically experience deshielding due to a combination of which two effects? A) Hydrogen bonding and inductive effects. B) Anisotropic effects and steric hindrance. C) Inductive effects and magnetic anisotropy of the C=O bond. D) Spin-spin coupling and relaxation.
Answer: C) Inductive effects and magnetic anisotropy of the C=O bond. Explanation: The electronegative oxygen inductively withdraws electron density, and the carbonyl π-system creates an anisotropic magnetic field that deshields nearby protons, especially those in the deshielding cone.
32. What is the approximate chemical shift range for carbons in a typical 13C NMR spectrum (relative to TMS)? A) 0-15 ppm B) 0-50 ppm C) 0-220 ppm D) 200-500 ppm
Answer: C) 0-220 ppm Explanation: 13C NMR has a much wider chemical shift range than 1H NMR, extending up to approximately 220 ppm, which helps in resolving distinct carbon environments.
33. If a 13C NMR experiment is performed with broad-band proton decoupling, what will be the typical multiplicity of each carbon signal? A) Depends on the number of attached protons. B) All signals will be singlets. C) All signals will be triplets. D) All signals will be quartets.
Answer: B) All signals will be singlets. Explanation: Broad-band decoupling irradiates all proton frequencies, collapsing all C-H coupling and making every carbon signal appear as a singlet, simplifying the spectrum for direct carbon counting.
34. Which relaxation process primarily governs the return of the net magnetization vector along the B0 (longitudinal) axis after an RF pulse? A) Spin-Spin (T2) relaxation B) Spin-Lattice (T1) relaxation C) Cross-relaxation D) Chemical exchange
Answer: B) Spin-Lattice (T1) relaxation Explanation: T1 relaxation describes the recovery of the magnetization along the longitudinal axis (z-axis, aligned with B0), as the system returns to thermal equilibrium.
35. What phenomenon is responsible for the enhancement of 13C signal intensity observed during broad-band proton decoupling? A) Chemical shift anisotropy B) Spin-spin coupling C) Nuclear Overhauser Effect (NOE) D) Ring current effect
Answer: C) Nuclear Overhauser Effect (NOE) Explanation: When protons are decoupled, their spin populations are perturbed, leading to a transfer of polarization to nearby carbons, which significantly enhances the carbon signal intensity via the NOE.
36. If a sample contains both water (H$_2O)andasmallamountofanorganicanalytedissolvedinD_2O,whichtechniquewouldbeusefultoobservetheanalyte′ssignalswithoutbeingoverwhelmedbytheresidualH_2$O signal? A) DEPT B) COSY C) Solvent suppression techniques D) HMBC
Answer: C) Solvent suppression techniques Explanation: Specific pulse sequences and data processing methods can selectively suppress the large signal from residual protons in deuterated solvents (like H$_2OinD_2$O) to reveal much weaker analyte signals.
37. What does Dynamic NMR (DNMR) primarily investigate? A) Molecular weight determination. B) Rates of chemical reactions. C) Conformational changes and other dynamic processes in molecules using variable temperature experiments. D) The crystal structure of solids.
Answer: C) Conformational changes and other dynamic processes in molecules using variable temperature experiments. Explanation: DNMR involves acquiring spectra at different temperatures to study processes where molecules interconvert between different forms, leading to changes in signal broadening and coalescence.
38. Which 2D NMR experiment correlates carbons with their directly attached protons via the one-bond coupling constant (1JCH)? A) COSY B) HSQC C) HMBC D) NOESY
Answer: B) HSQC (Heteronuclear Single Quantum Coherence) Explanation: HSQC provides cross peaks connecting each carbon to its directly bonded proton(s), which is highly valuable for assigning both carbon and proton signals simultaneously.
39. Protons of an aldehyde group (R-CHO) are typically found in which chemical shift range in 1H NMR? A) 2-3 ppm B) 4-5 ppm C) 6-8 ppm D) 9-10 ppm
Answer: D) 9-10 ppm Explanation: Aldehyde protons are strongly deshielded due to the combined inductive and anisotropic effects of the carbonyl group, placing them in a characteristic far downfield region.
40. What is a key difference between T1 and T2 relaxation processes? A) T1 involves loss of phase coherence, while T2 involves energy transfer to the lattice. B) T1 influences signal linewidth, while T2 governs signal repetition rate. C) T1 involves energy transfer to the lattice and governs recovery of longitudinal magnetization, while T2 involves loss of phase coherence and influences linewidth. D) T1 is always shorter than T2.
Answer: C) T1 involves energy transfer to the lattice and governs recovery of longitudinal magnetization, while T2 involves loss of phase coherence and influences linewidth. Explanation: This concisely summarizes the distinct mechanisms and effects of the two primary relaxation processes: T1 (spin-lattice) is about energy dissipation and longitudinal recovery, while T2 (spin-spin) is about dephasing and transverse magnetization decay, directly impacting linewidth.