40 Advanced Multiple-Choice Questions on Combined Structure Problems
Instructions:
Choose the best answer for each question. Explanations are provided below each answer.
1. A compound with a molecular formula of C8H10O is analyzed by HRMS, yielding an exact mass of 122.0732. What is its Index of Hydrogen Deficiency (IHD)? A) 1
B) 2
C) 3
D) 4
Answer: D) 4
Explanation: For C$_cH_hN_nO_oX_x$, IHD = c+1−2h−x+n. For C$8H{10}$O, IHD = 8+1−210−0+0=9−5=4. An IHD of 4 strongly suggests the presence of an aromatic ring (3 double bonds + 1 ring).
2. Which of the following statements about HRMS is most accurate for molecular formula determination? A) It primarily identifies functional groups.
B) It measures nominal mass, which is less precise than elemental analysis.
C) It provides a precise mass measurement (e.g., to 4 decimal places) allowing unambiguous determination of the molecular formula.
D) It only detects compounds with nitrogen atoms.
Answer: C) It provides a precise mass measurement (e.g., to 4 decimal places) allowing unambiguous determination of the molecular formula.
Explanation: HRMS’s high mass accuracy allows differentiation of isobaric compounds (compounds with the same nominal mass but different exact masses due to different elemental compositions), leading to unambiguous formula assignment.
3. A mass spectrum shows an M:M+2 peak ratio of approximately 1:1. This is characteristic of the presence of which element in the molecule? A) Chlorine
B) Sulfur
C) Bromine
D) Nitrogen
Answer: C) Bromine
Explanation: Naturally occurring bromine consists of two isotopes, $^{79}$Br and $^{81}Br,inroughlya1:1ratio,leadingtoadistinctiveM:M+2peakratioof1:1inthemassspectrum.Chlorinehasa3:1M:M+2ratio(^{35}$Cl: $^{37}$Cl).
4. In the mass spectrum of an alcohol, which two fragmentation pathways are commonly observed? A) McLafferty rearrangement and aromatic ring current.
B) α-cleavage and dehydration (loss of H$_2$O).
C) Retro-Diels-Alder and tropylium ion formation.
D) Loss of methyl and ethyl radicals.
Answer: B) α-cleavage and dehydration (loss of H$_2$O).
Explanation: α-cleavage adjacent to the hydroxyl group forms a resonance-stabilized oxonium ion, while dehydration results in a M-18 peak due to the loss of a water molecule.
5. An IR spectrum shows a very broad absorption band stretching from 2500-3300 cm$^{-1}$ that overlaps with C-H stretches, along with a strong band at 1710 cm$^{-1}$. This combination is most indicative of which functional group? A) Alcohol
B) Amide
C) Carboxylic acid
D) Aldehyde
Answer: C) Carboxylic acid
Explanation: The very broad O-H stretch (due to strong hydrogen bonding dimerization) combined with a C=O stretch in the 1700-1725 cm$^{-1}$ range is characteristic of a carboxylic acid. Alcohols have a broader, but usually less intensely overlapping O-H band.
6. Which of the following functional groups would exhibit an IR absorption above 3000 cm$^{-1}$ for C-H stretching? A) Only sp$^3$ C-H in alkanes.
B) sp C-H in terminal alkynes and sp$^2$ C-H in alkenes/aromatics.
C) Only aromatic C-H.
D) Only aldehyde C-H.
Answer: B) sp C-H in terminal alkynes and sp$^2$ C-H in alkenes/aromatics.
Explanation: These are the C-H stretches for unsaturated carbons. sp$^3$ C-H stretches are found below 3000 cm$^{-1}.AldehydeC−Hhasadistinctivepairofpeaksnear2720and2820cm^{-1}$.
7. How does conjugation of a C=O group with a C=C bond typically affect its IR stretching frequency? A) It increases the C=O frequency.
B) It has no effect on the C=O frequency.
C) It lowers the C=O frequency by approximately 20-40 cm$^{-1}$.
D) It causes the C=O absorption to disappear.
Answer: C) It lowers the C=O frequency by approximately 20-40 cm$^{-1}$.
Explanation: Conjugation allows for resonance delocalization of electron density, which reduces the double bond character of the C=O bond, leading to a weaker bond and thus a lower stretching frequency.
8. Which of the following UV-Vis transitions is typically strong and occurs at lower wavelengths? A) n →π∗ transitions in ketones.
B) π→π∗ transitions in conjugated dienes.
C) σ→σ∗ transitions in alkanes.
D) n →σ∗ transitions in alcohols.
Answer: B) π→π∗ transitions in conjugated dienes.
Explanation: π→π∗ transitions are typically high-intensity absorptions with large molar absorptivities (ϵ), characteristic of systems with π bonds (e.g., alkenes, aromatics, carbonyls). Conjugation further extends the chromophore and increases λmax.
9. In 1H NMR, what is the primary reason for the significant deshielding of aldehyde protons (R-CHO) to 9-10 ppm? A) Strong inductive effect of the alkyl group.
B) Spin-spin coupling with neighboring carbons.
C) Combined inductive and anisotropic effects of the carbonyl group.
D) Hydrogen bonding with the solvent.
Answer: C) Combined inductive and anisotropic effects of the carbonyl group.
Explanation: The electronegative oxygen exerts an inductive effect, and the carbonyl’s π-system generates an anisotropic magnetic field that strongly deshields the proton directly attached to the carbonyl carbon.
10. A 1H NMR spectrum shows a signal integrating to 2 protons, appearing as a triplet. This suggests that these two protons are adjacent to a group containing how many equivalent protons? A) 1
B) 2
C) 3
D) 4
Answer: C) 3
Explanation: A triplet (n+1 rule = 3) indicates coupling to n=2 equivalent neighboring protons. The question states the signal is integrating to 2 protons, meaning the two protons themselves are equivalent, and they are coupled to two equivalent neighbors to form a triplet.
11. A proton on an alkene shows a large coupling constant of 15 Hz to another proton. This is highly indicative of which geometric relationship? A) Geminal coupling
B) cis-vicinal coupling
C) trans-vicinal coupling
D) Long-range coupling
Answer: C) trans-vicinal coupling
Explanation: trans-vicinal coupling constants (3JHH) in alkenes are typically large (12-18 Hz) due to the nearly 180° dihedral angle, whereas cis-vicinal couplings are smaller (6-12 Hz).
12. When is a 1H NMR spectrum considered “second-order” (or strongly coupled)? A) When all signals are singlets.
B) When the chemical shift difference (Δν) between coupled nuclei is much larger than their coupling constant (J).
C) When the chemical shift difference (Δν) between coupled nuclei is comparable to their coupling constant (J).
D) When only a very high field spectrometer is used.
Answer: C) When the chemical shift difference (Δν) between coupled nuclei is comparable to their coupling constant (J).
Explanation: Second-order effects (distorted multiplets, “roofing effect”) occur when the ratio Δν/J is small, typically less than ≈7.
13. A compound has a formula C$6H{12}$. Its $^{13}$C NMR spectrum (broadband decoupled) shows only three signals. What does this immediately suggest about the molecule? A) It contains a carbonyl group.
B) It is highly symmetrical.
C) It has no sp$^2$ hybridized carbons.
D) It is a branched alkane.
Answer: B) It is highly symmetrical.
Explanation: If there are 6 carbons but only 3 unique signals, it means there are equivalent sets of carbons due to molecular symmetry (e.g., cyclohexane, 2,3-dimethylbutane, or certain C$6H{12}$ alkenes).
14. In a DEPT-135 $^{13}$C NMR spectrum, which type of carbon signal would appear as an inverted (negative) peak? A) CH$_3$
B) CH$_2$
C) CH
D) Quaternary carbon (C)
Answer: B) CH$_2$
Explanation: In DEPT-135, CH and CH$_3$ carbons appear as positive peaks, CH$_2$ carbons appear as negative peaks, and quaternary carbons are completely suppressed (absent).
15. Which 2D NMR experiment is most useful for connecting a proton signal to its directly bonded carbon signal? A) COSY
B) HMBC
C) HSQC
D) NOESY
Answer: C) HSQC (Heteronuclear Single Quantum Coherence)
Explanation: HSQC (or HMQC) provides correlations between protons and the carbons to which they are directly attached (one-bond coupling, 1JCH).
16. You suspect a quaternary carbon in your molecule. Which 2D NMR experiment would be most crucial for establishing its connectivity to neighboring protons? A) COSY
B) DEPT-90
C) HMBC
D) HSQC
Answer: C) HMBC (Heteronuclear Multiple Bond Correlation)
Explanation: Quaternary carbons have no directly attached protons, so they don’t show up in 1H NMR, DEPT-90, or HSQC. HMBC detects long-range (2-4 bond) couplings between carbons and protons, making it ideal for identifying and assigning quaternary carbons by showing correlations to nearby protons.
17. What information does a cross peak in a NOESY spectrum provide? A) That two protons are directly bonded.
B) That two protons are spin-spin coupled through bonds.
C) That two protons are spatially close in three-dimensional space.
D) That a proton is bonded to a specific carbon.
Answer: C) That two protons are spatially close in three-dimensional space.
Explanation: NOESY relies on the Nuclear Overhauser Effect, which is a through-space interaction. Cross peaks indicate that the two protons are within close proximity (typically ≈5 Å), providing conformational and stereochemical information.
18. You are trying to distinguish between cis– and trans-1,2-dimethylcyclohexane isomers. Which NMR technique would be most informative? A) IR spectroscopy to identify C-H bonds.
B) Mass spectrometry to compare fragmentation.
C) Analysis of vicinal coupling constants (3JHH) in 1H NMR and/or NOESY correlations.
D) Broadband decoupled $^{13}$C NMR.
Answer: C) Analysis of vicinal coupling constants (3JHH) in 1H NMR and/or NOESY correlations.
Explanation: The dihedral angles between vicinal protons will differ significantly between cis and trans isomers, leading to different J values (Karplus relationship). NOESY would show through-space correlations between methyl protons and specific axial/equatorial protons that differ for cis vs trans.
19. Why are D$_2$O shakes performed in 1H NMR spectroscopy? A) To increase the solubility of the compound.
B) To shift the chemical shifts of aromatic protons.
C) To identify and confirm the presence of exchangeable protons (e.g., OH, NH, COOH).
D) To remove all spin-spin coupling.
Answer: C) To identify and confirm the presence of exchangeable protons (e.g., OH, NH, COOH).
Explanation: Acidic protons on heteroatoms (like OH, NH, COOH) rapidly exchange with deuterium from D$_2$O, causing their corresponding 1H NMR signals to disappear or significantly diminish.
20. A prominent fragment ion at m/z 91 in a mass spectrum is highly suggestive of which structural feature? A) A primary alcohol
B) An ethyl ester
C) A benzylic group (e.g., alkylbenzene)
D) A cyclic ketone
Answer: C) A benzylic group (e.g., alkylbenzene)
Explanation: The tropylium ion (C$_7H_7^+$) at m/z 91 is a highly characteristic and stable fragment formed from the loss of a radical from an alkylbenzene via benzylic cleavage and subsequent rearrangement.
21. In an IR spectrum, a strong absorption at 1740 cm$^{-1}$ combined with two C-H stretching bands near 2720 cm$^{-1}$ and 2820 cm$^{-1}$ points to the presence of a(n): A) Ketone
B) Ester
C) Aldehyde
D) Carboxylic acid
Answer: C) Aldehyde
Explanation: The 1740 cm$^{-1}$ is consistent with a C=O stretch. The two characteristic C-H stretches in the 2700-2800 cm$^{-1}$ region are unique to the aldehyde C-H bond.
22. How does the presence of an electron-withdrawing group (EWG) typically affect the chemical shift of nearby protons in 1H NMR? A) It causes shielding and an upfield shift (lower δ).
B) It causes deshielding and a downfield shift (higher δ).
C) It has no effect on chemical shift.
D) It only affects the coupling constant.
Answer: B) It causes deshielding and a downfield shift (higher δ).
Explanation: EWGs pull electron density away from neighboring protons, reducing their shielding and causing them to resonate at higher chemical shift values (downfield).
23. Which of the following is NOT a common reason for signal broadening in NMR spectroscopy? A) Rapid chemical exchange.
B) Presence of paramagnetic impurities.
C) High homogeneity of the magnetic field.
D) Short T2 relaxation times.
Answer: C) High homogeneity of the magnetic field.
Explanation: High magnetic field homogeneity (good shimming) leads to sharper, not broader, signals. Rapid exchange, paramagnetic species, and short T2 times all contribute to peak broadening.
24. For a molecule with formula C$5H{10}$O, if the $^{13}$C NMR shows 5 distinct signals and the DEPT-135 shows 2 positive peaks and 2 negative peaks, how many quaternary carbons are present? A) 0
B) 1
C) 2
D) 5
Answer: B) 1
Explanation:
- Total carbons = 5.
- DEPT-135 positive peaks: CH and CH$_3$. (2 signals)
- DEPT-135 negative peaks: CH$_2$. (2 signals)
- Total identified by DEPT-135 = 2 (CH/CH$_3$) + 2 (CH$_2$) = 4 carbons.
- Since the total number of carbon signals is 5, the remaining 5 – 4 = 1 carbon must be a quaternary carbon, as quaternary carbons are absent in DEPT-135.
25. A compound has an IHD of 5. Which combination of structural features is consistent with this IHD? A) One double bond and one ring.
B) One triple bond and one double bond.
C) An aromatic ring and one additional double bond.
D) Two rings and one double bond.
Answer: C) An aromatic ring and one additional double bond.
Explanation: An aromatic ring accounts for 4 IHD (3 π bonds + 1 ring). Adding one additional double bond (1 π bond) brings the total IHD to 4 + 1 = 5.
26. Why is a highly concentrated sample generally preferred for advanced 2D NMR experiments like HMBC? A) To minimize solvent interference.
B) To increase sample spinning stability.
C) To improve signal-to-noise ratio due to the inherent low sensitivity of 2D experiments and long-range couplings.
D) To avoid second-order coupling effects.
Answer: C) To improve signal-to-noise ratio due to the inherent low sensitivity of 2D experiments and long-range couplings.
Explanation: 2D NMR experiments, especially those detecting long-range correlations (like HMBC), are much less sensitive than 1D NMR. Higher sample concentration increases the number of nuclei available for resonance, improving the signal intensity relative to noise.
27. In 1H NMR, a “W-coupling” (4JHH) is often observed in rigid cyclic systems. What is the typical magnitude of this coupling? A) > 10 Hz
B) 6-8 Hz
C) < 1 Hz
D) It only occurs in aromatic rings.
Answer: C) < 1 Hz
Explanation: “W-coupling” is a small, long-range coupling (4JHH) that occurs when four bonds are in a planar “W” (zigzag) conformation, typically seen in rigid systems, and usually has a magnitude of less than 1 Hz.
28. Which of the following statements is true regarding the relationship between T1 and T2 relaxation times? A) T1 is always shorter than T2.
B) T2 is always shorter than or equal to T1.
C) T1 and T2 are independent of each other.
D) T1 influences linewidth, and T2 influences signal repetition.
Answer: B) T2 is always shorter than or equal to T1.
Explanation: T2 (spin-spin relaxation, loss of phase coherence) includes all mechanisms contributing to T1 (spin-lattice relaxation, energy transfer) plus additional dephasing mechanisms. Therefore, T2 can never be longer than T1.
29. What is the primary use of DOSY (Diffusion-Ordered SpectroscopY) in structure elucidation? A) To determine relative stereochemistry.
B) To measure diffusion coefficients, allowing separation of signals from different components in a mixture based on size.
C) To identify quaternary carbons.
D) To enhance the signal of insensitive nuclei like $^{13}$C.
Answer: B) To measure diffusion coefficients, allowing separation of signals from different components in a mixture based on size.
Explanation: DOSY spreads out the NMR spectrum into a second dimension based on the diffusion rates of molecules. Larger molecules diffuse slower, separating their signals from smaller molecules, making it useful for analyzing mixtures.
30. You have an unknown compound with a molecular formula C$_7H_8O.TheIRshowsabroadO−Hstretcharound3300cm^{-1}$ and C-H stretches above 3000 cm$^{-1}$. The 1H NMR shows signals between 6.5-7.5 ppm integrating to 5H and a singlet at 4.5 ppm integrating to 1H. What type of compound is it most likely to be? A) Aldehyde
B) Ketone
C) Phenol
D) Carboxylic acid
Answer: C) Phenol
Explanation:
- IHD = 7+1−8/2=8−4=4, consistent with an aromatic ring.
- IR: O-H stretch and sp$^2$ C-H (above 3000 cm$^{-1}$) confirm aromatic ring and -OH.
- 1H NMR: 5H in the aromatic region (6.5-7.5 ppm) indicates a monosubstituted benzene ring. The 1H singlet at 4.5 ppm (exchangeable with D$_2$O, not directly coupled to ring protons) is characteristic of a phenolic OH proton.
31. When analyzing the McLafferty rearrangement in MS, which of the following is a key requirement? A) Aromatic ring
B) Presence of a β-hydrogen.
C) Presence of a γ-hydrogen.
D) A bromine atom.
Answer: C) Presence of a γ-hydrogen.
Explanation: The McLafferty rearrangement requires a hydrogen atom on the carbon gamma (γ) to the carbonyl oxygen, allowing for a six-membered ring transition state.
32. What is the typical effect of hydrogen bonding on the chemical shift of an amide N-H proton? A) It causes an upfield shift.
B) It causes a downfield shift and often broadens the signal.
C) It has no effect on the chemical shift.
D) It leads to splitting into a quartet.
Answer: B) It causes a downfield shift and often broadens the signal.
Explanation: Hydrogen bonding deshields the N-H proton, shifting its signal downfield. Rapid proton exchange mediated by hydrogen bonding also often leads to broadening of the signal.
33. If an IR spectrum shows a strong C=O absorption at 1715 cm$^{-1}$ and no O-H stretch, this suggests a(n): A) Carboxylic acid
B) Ester
C) Ketone
D) Amide
Answer: C) Ketone
Explanation:
- Carboxylic acids have O-H.
- Esters typically absorb at higher wavenumbers (1735-1750 cm$^{-1}$).
- Amides absorb at lower wavenumbers (1630-1690 cm$^{-1}$) due to resonance.
- A ketone’s C=O typically absorbs around 1710-1720 cm$^{-1}$ with no O-H.
34. A compound has a molecular formula C$_3H_6$O. Its 1H NMR shows a singlet at 2.1 ppm (3H) and a singlet at 2.4 ppm (3H). Its $^{13}$C NMR shows three signals: ≈20 ppm, ≈30 ppm, ≈200 ppm. What is the structure? A) Propanal
B) Acetone
C) Cyclopropanol
D) Propene oxide
Answer: B) Acetone
Explanation:
- IHD = 3+1−6/2=4−3=1. (One π bond or ring).
- ^{13}$C NMR: One carbon at $\approx$200 ppm (carbonyl). Two carbons at aliphatic shifts ($\approx20 ppm, ≈30 ppm).
- Given the molecular formula C$_3H_6Owithacarbonyl,Acetone(CH_3COCH_3$) is the most plausible structure. Acetone has two equivalent methyl groups and a carbonyl carbon. In a perfectly symmetrical environment, the 1H NMR would show one 6H singlet. The reporting of two distinct 3H singlets at 2.1 ppm and 2.4 ppm might indicate a minor experimental artifact or a slight deviation from perfect equivalence, but Acetone is the only option that fits the overall carbon count and functional group.
35. Which factor is LEAST likely to affect the chemical shift of a proton in 1H NMR? A) Hydrogen bonding.
B) Inductive effects from distant substituents.
C) Magnetic anisotropy of adjacent π-systems.
D) The spin state of an uncoupled, non-neighboring proton on a different molecule.
Answer: D) The spin state of an uncoupled, non-neighboring proton on a different molecule.
Explanation: Chemical shifts are determined by local electronic environment, inductive effects, and anisotropic effects. While intermolecular interactions (like H-bonding) can affect chemical shifts, the spin state of a distant, uncoupled proton (especially on a different molecule not involved in strong interactions) has no direct bearing on the chemical shift of the observed proton.
36. A compound (C$_8H_8O)hasanIHDof5.ItsIRshowsC−Hstretchesabove3000cm^{-1}$ and a strong absorption at 1680 cm$^{-1}$. Its 1H NMR shows a multiplet around 7.5 ppm (5H) and a singlet at 9.9 ppm (1H). What is the structure? A) Acetophenone
B) Benzyl alcohol
C) Benzaldehyde
D) Phenylacetylene
Answer: C) Benzaldehyde
Explanation:
- IHD = 8+1−8/2=5. Consistent with a benzene ring (4 IHD) and one additional π bond (e.g., C=O).
- IR: C-H > 3000 cm$^{-1}$ indicates aromatic/vinylic. Strong 1680 cm$^{-1}$ is a conjugated C=O.
- 1H NMR: 5H in aromatic region (7.5 ppm) suggests monosubstituted benzene. 1H singlet at 9.9 ppm is characteristic of an aldehyde proton.
- Combined, this perfectly fits Benzaldehyde (C$_6H_5$CHO).
37. Which technique is best suited for identifying if a molecule contains a hydroxyl group (-OH) through an exchange experiment? A) Mass spectrometry
B) IR spectroscopy
C) 1H NMR with D$_2$O exchange
D) UV-Vis spectroscopy
Answer: C) 1H NMR with D$_2$O exchange
Explanation: The D$_2$O shake test in 1H NMR causes the disappearance of exchangeable protons (like -OH, -NH, -COOH), definitively confirming their presence. IR can show an O-H stretch, but cannot confirm exchangeability.
38. You are trying to determine the structure of a complex natural product. Your initial 1D NMR spectra are highly overlapped. Which 2D NMR technique would be the most logical starting point to assign protons to their directly bonded carbons? A) HMBC
B) COSY
C) HSQC
D) NOESY
Answer: C) HSQC
Explanation: HSQC (or HMQC) directly correlates 1H and $^{13}$C chemical shifts for directly bonded C-H pairs. This is an excellent starting point for assigning signals in complex spectra, as it resolves overlaps by spreading data into two dimensions and directly links proton and carbon assignments.
39. A compound has the molecular formula C$_4H_8O_2$. The IR shows a strong absorption at 1735 cm$^{-1}$ and no O-H stretch. The 1H NMR shows a triplet (3H), a quartet (2H), and a singlet (3H). What is the structure? A) Butanoic acid
B) Ethyl acetate
C) Methyl propanoate
D) Diethyl ether
Answer: B) Ethyl acetate
Explanation:
- IHD = 4+1−8/2=5−4=1. (One π bond).
- IR: 1735 cm$^{-1}$ (strong C=O), no O-H. This indicates an ester or possibly a saturated ketone/aldehyde with other features. Given two oxygens and IHD 1, an ester is highly likely.
- 1H NMR:
- Triplet (3H) and Quartet (2H): This is characteristic of an ethyl group (-CH$_2CH_3$).
- Singlet (3H): A methyl group not coupled to any neighbors.
- Combining these fragments with C=O and two oxygens: The only structure that fits is CH$_3COOCH_2CH_3$ (Ethyl acetate). The 3H singlet is the acetyl methyl, and the triplet/quartet is the ethyl group.
40. In determining the stereochemistry of an alkene, why are vicinal coupling constants (3JHH) particularly valuable? A) They indicate the number of hydrogens on an adjacent carbon.
B) They are highly dependent on the dihedral angle between the coupled protons.
C) They are always very large for cis alkenes.
D) They are the same for cis and trans alkenes.
Answer: B) They are highly dependent on the dihedral angle between the coupled protons.
Explanation: The Karplus relationship dictates that vicinal coupling constants are a function of the dihedral angle. For alkenes, the cis (0° angle) and trans (180° angle) isomers have characteristically different dihedral angles, resulting in distinct J values (e.g., cis 6-12 Hz, trans 12-18 Hz), allowing for unambiguous stereochemical assignment.