Level 1 — RecognitionSpectroscopy & Analysis (Intro)

Spectroscopy & Analysis (Intro)

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time limit: 20 minutes Total marks: 30

Section A — Multiple Choice (1 mark each)

Q1. Which region of the electromagnetic spectrum has the highest photon energy? (a) Microwave (b) Infrared (c) Ultraviolet (d) Radio (NMR)

Q2. The Beer–Lambert law is correctly written as: (a) $A = \varepsilon c l$ (b) $A = \varepsilon / (cl)$ (c) $A = cl/\varepsilon$ (d) $A = \varepsilon + c + l$

Q3. Extending conjugation in an organic molecule generally causes $\lambda_{max}$ in the UV-Vis spectrum to: (a) decrease (b) increase (c) stay constant (d) disappear

Q4. A strong, broad IR absorption at ~3300 cm⁻¹ is most consistent with: (a) C=O stretch (b) O–H stretch (c) C≡N stretch (d) C–Cl stretch

Q5. A sharp, strong IR band at ~1715 cm⁻¹ is characteristic of a: (a) C–H bend (b) O–H stretch (c) C=O stretch (d) N–H stretch

Q6. In ¹H NMR, a proton with three equivalent neighbouring protons appears as a: (a) singlet (b) doublet (c) triplet (d) quartet

Q7. In ¹H NMR the integration (relative area under a peak) tells you the: (a) number of neighbouring protons (b) chemical shift (c) relative number of protons giving that signal (d) coupling constant

Q8. A ¹³C NMR DEPT-135 experiment is primarily used to distinguish: (a) solvents (b) CH₃/CH₂/CH carbons (c) molecular mass (d) aromatic vs aliphatic only

Q9. In mass spectrometry, the molecular ion (M⁺) peak corresponds to: (a) the smallest fragment (b) the whole molecule minus an electron (c) the base peak always (d) the solvent

Q10. Which technique separates volatile compounds using a carrier gas? (a) TLC (b) HPLC (c) GC (d) Column chromatography using gravity

Q11. In TLC, the $R_f$ value is defined as: (a) distance moved by solvent ÷ distance moved by spot (b) distance moved by spot ÷ distance moved by solvent (c) distance moved by spot × distance moved by solvent (d) mass of spot ÷ mass of plate

Section B — Matching (1 mark each; total 5)

Q12. Match the functional group to its approximate IR stretching frequency.

Group		Frequency (cm⁻¹)
(i) O–H (alcohol)		(P) ~1650
(ii) C≡N		(Q) ~3300 (broad)
(iii) C=O		(R) ~2250
(iv) C=C		(S) ~1715
(v) N–H		(T) ~3400 (sharp, 1–2 bands)

Section C — True / False WITH Justification (2 marks each: 1 statement, 1 justification)

Q13. "NMR spectroscopy uses radio-frequency radiation, which is lower in energy than the UV radiation used in UV-Vis spectroscopy."

Q14. "In the n+1 rule, a CH₂ group adjacent to a CH₃ group will split the CH₃ signal into a triplet."

Q15. "The fingerprint region of an IR spectrum (below ~1500 cm⁻¹) is useless because no information can be obtained from it."

Q16. "For a molecule of relative molecular mass 72, an M⁺ peak at m/z = 72 and a fragment peak at m/z = 57 corresponds to loss of a CH₃ group."

Q17. "In HPLC the stationary phase is a gas and the mobile phase is a solid."

Answer keyMark scheme & solutions

Section A (1 mark each)

Q1. (c) Ultraviolet — energy increases with frequency; UV > IR > microwave > radio. [1]

Q2. (a) $A = \varepsilon c l$ — absorbance = molar absorptivity × concentration × path length. [1]

Q3. (b) increase — greater conjugation lowers the HOMO–LUMO gap, so absorption shifts to longer wavelength (bathochromic shift). [1]

Q4. (b) O–H stretch — broad band ~3200–3550 cm⁻¹ due to hydrogen bonding. [1]

Q5. (c) C=O stretch — carbonyls absorb strongly near 1700–1750 cm⁻¹. [1]

Q6. (d) quartet — n+1 = 3+1 = 4. [1]

Q7. (c) relative number of protons giving that signal. [1]

Q8. (b) CH₃/CH₂/CH carbons — DEPT distinguishes number of attached H (CH, CH₂ up/down, quaternary absent). [1]

Q9. (b) the whole molecule minus an electron — gives relative molecular mass. [1]

Q10. (c) GC — gas chromatography uses an inert carrier gas as mobile phase. [1]

Q11. (b) distance moved by spot ÷ distance moved by solvent. [1]

Section B — Matching (1 mark each; 5 total)

Q12.

(i) O–H → (Q) ~3300 broad
(ii) C≡N → (R) ~2250
(iii) C=O → (S) ~1715
(iv) C=C → (P) ~1650
(v) N–H → (T) ~3400 sharp [5 × 1 = 5]

Section C — True/False + Justification (2 marks each)

Q13. TRUE. [1] Justification: NMR uses radio-frequency radiation (long wavelength, low frequency); by $E = h\nu$ , lower frequency = lower energy than UV. [1]

Q14. FALSE. [1] Justification: A CH₃ group (3 protons) splits the adjacent CH₂ signal, not the other way in this claim's logic — but specifically, the CH₃ is split by the CH₂ (2 protons), giving n+1 = 3, a triplet? No: CH₃ split by 2 CH₂ protons → 2+1 = triplet is correct only if CH₂ neighbours the CH₃. The statement says "CH₂ ... will split the CH₃ signal into a triplet." CH₂ has 2 protons → n+1 = 3 → triplet. ✔ The multiplicity is right, so mark based on stated reasoning: TRUE — CH₃ has 2 neighbouring protons (from CH₂), 2+1 = 3 = triplet. [1] (Accept TRUE with correct n+1 reasoning.)

(Examiner note: award full marks for TRUE + correct 2+1=3 justification.)

Q15. FALSE. [1] Justification: The fingerprint region is highly complex and unique to each compound; it is used to confirm identity by comparison with reference spectra, so it does provide information. [1]

Q16. TRUE. [1] Justification: 72 − 57 = 15, which is the mass of a CH₃ group; loss of 15 = loss of methyl. [1]

Q17. FALSE. [1] Justification: In HPLC the stationary phase is a solid (packed column) and the mobile phase is a liquid pumped under high pressure — the statement reverses them. [1]

[
  {"claim":"Q6 quartet: n+1 for 3 neighbours = 4","code":"n=3; result = (n+1 == 4)"},
  {"claim":"Q14 triplet: CH3 split by 2 CH2 protons gives 2+1=3","code":"n=2; result = (n+1 == 3)"},
  {"claim":"Q16 mass loss 72-57 = 15 (CH3)","code":"result = (72 - 57 == 15)"},
  {"claim":"Beer-Lambert A = eps*c*l with eps=100,c=0.01,l=1 gives A=1","code":"eps,c,l=100,0.01,1; result = (eps*c*l == 1)"}
]