Noesis
Level 2 — RecallSpectroscopy & Analysis (Intro)

Spectroscopy & Analysis (Intro)

40 marksprintable — key stays hidden on paper

Level: 2 (Recall & Standard Problems) Time: 30 minutes Total marks: 40

Use ...... notation where needed. Constants: h=6.626×10−34 J sh = 6.626\times10^{-34}\ \text{J s}, c=3.00×108 m s−1c = 3.00\times10^8\ \text{m s}^{-1}, NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\ \text{mol}^{-1}.

Q1. Arrange the following regions of the electromagnetic spectrum in order of increasing energy: IR, X-ray, microwave, UV, visible. (3 marks)

Q2. State the Beer–Lambert law, defining every symbol. (3 marks)

Q3. A solution of a dye has an absorbance of 0.600.60 in a 1.0 cm1.0\ \text{cm} cell. The molar absorptivity is ε=1.5×104 L mol−1cm−1\varepsilon = 1.5\times10^4\ \text{L mol}^{-1}\text{cm}^{-1}. Calculate the molar concentration of the dye. (3 marks)

Q4. Explain briefly how extending conjugation in a molecule affects its λmax⁡\lambda_{\max} in UV–Vis spectroscopy, and why. (4 marks)

Q5. Give the approximate characteristic IR absorption range (in cm−1\text{cm}^{-1}) for each of the following bonds: (a) O–H (alcohol), (b) C=O (carbonyl), (c) C≡N (nitrile), (d) N–H (amine). (4 marks)

Q6. In 1^1H NMR: (a) State the n+1n+1 rule and explain what it predicts. (2 marks) (b) An ethyl group (–CH2_2CH3_3) appears as two signals. Predict the multiplicity of each signal and give the integration ratio. (3 marks)

Q7. Calculate the energy (in J) of one photon of UV radiation with wavelength 250 nm250\ \text{nm}. (3 marks)

Q8. The mass spectrum of a compound shows a molecular ion peak at m/z=46m/z = 46 and a strong fragment at m/z=31m/z = 31. (a) State what the molecular ion represents. (1 mark) (b) The compound is ethanol (CH3_3CH2_2OH). Identify the fragment lost to give m/z=31m/z = 31 and name the ion at m/z=31m/z = 31. (3 marks)

Q9. For 13^{13}C NMR / DEPT: (a) Why is 13^{13}C NMR generally less sensitive than 1^1H NMR? (2 marks) (b) In a standard DEPT-135 experiment, state how CH3_3, CH2_2, CH, and quaternary carbons appear. (4 marks)

Q10. State the principle of separation for (a) TLC and (b) GC, naming the mobile phase in each. (2 marks + 2 marks = 4 marks)

End of paper

Answer keyMark scheme & solutions

Q1. (3 marks) Increasing energy: microwave < IR < visible < UV < X-ray. Energy increases as wavelength decreases (E=hc/λE = hc/\lambda). Marks: 2 for fully correct order; 1 if only one pair transposed. +1 justification (energy ∝ 1/λ).

Q2. (3 marks) A=εclA = \varepsilon c l where AA = absorbance (no units), ε\varepsilon = molar absorptivity (L mol−1cm−1\text{L mol}^{-1}\text{cm}^{-1}), cc = concentration (mol L−1\text{mol L}^{-1}), ll = path length (cm). Marks: 1 for equation, 2 for defining all symbols (½ each, rounded).

Q3. (3 marks) c=Aεl=0.601.5×104×1.0=4.0×10−5 mol L−1c = \dfrac{A}{\varepsilon l} = \dfrac{0.60}{1.5\times10^4 \times 1.0} = 4.0\times10^{-5}\ \text{mol L}^{-1}. Marks: 1 rearrange, 1 substitution, 1 answer with units. Why: Beer–Lambert is linear in cc; solve for cc.

Q4. (4 marks)

  • Extending conjugation increases λmax⁡\lambda_{\max} (shifts to longer wavelength / lower energy — a red/bathochromic shift). (2)
  • Reason: more conjugation raises the HOMO and lowers the LUMO, decreasing the HOMO–LUMO energy gap ΔE\Delta E. Since ΔE=hc/λ\Delta E = hc/\lambda, a smaller gap means longer wavelength absorbed. (2)

Q5. (4 marks, 1 each) (a) O–H (alcohol): ~32003200–3550 cm−13550\ \text{cm}^{-1} (broad). (b) C=O: ~16801680–1750 cm−11750\ \text{cm}^{-1} (strong). (c) C≡N: ~22002200–2260 cm−12260\ \text{cm}^{-1}. (d) N–H: ~33003300–3500 cm−13500\ \text{cm}^{-1}. Marks: 1 each; accept values within reasonable range.

Q6. (5 marks) (a) The n+1n+1 rule: a proton with nn equivalent neighbouring (non-equivalent to itself) protons is split into n+1n+1 peaks (a multiplet). (2) (b) –CH2_2– has 3 neighbours on CH3_3 → 3+1=43+1 = 4 → quartet; –CH3_3 has 2 neighbours on CH2_2 → 2+1=32+1 = 3 → triplet. Integration ratio CH2_2 : CH3_3 = 2:32:3. (1 quartet + 1 triplet + 1 ratio)

Q7. (3 marks) E=hcλ=(6.626×10−34)(3.00×108)250×10−9E = \dfrac{hc}{\lambda} = \dfrac{(6.626\times10^{-34})(3.00\times10^8)}{250\times10^{-9}} =1.9878×10−252.50×10−7=7.95×10−19 J= \dfrac{1.9878\times10^{-25}}{2.50\times10^{-7}} = 7.95\times10^{-19}\ \text{J}. Marks: 1 formula, 1 substitution (λ in m), 1 answer.

Q8. (4 marks) (a) The molecular ion is the radical cation formed by loss of one electron from the whole molecule; m/zm/z = relative molecular mass (M+∙M^{+\bullet}). (1) (b) Loss = 46−31=1546 - 31 = 15, corresponds to loss of ·CH3_3 (methyl radical, mass 15). (2) The ion at m/z=31m/z = 31 is CH2_2OH+^+ (protonated formaldehyde / hydroxymethyl cation). (1)

Q9. (6 marks) (a) 13^{13}C has low natural abundance (~1.1%) and a smaller gyromagnetic ratio than 1^1H, giving a much weaker signal → lower sensitivity. (2) (b) DEPT-135: CH3_3 positive (up); CH2_2 negative (down); CH positive (up); quaternary carbons (no attached H) absent (not observed). (1 each)

Q10. (4 marks) (a) TLC: separation by differential adsorption/partition between a solid stationary phase (silica/alumina on plate) and a liquid mobile phase (solvent); components move at different rates. (2) (b) GC: separation by differential partition/volatility between a stationary phase (liquid coating on column) and a gaseous mobile phase (inert carrier gas, e.g. N2_2/He); components have different retention times. (2)

[
  {"claim":"Q3 concentration = 4.0e-5 mol/L", "code":"A=0.60; eps=1.5e4; l=1.0; c=A/(eps*l); result = abs(c - 4.0e-5) < 1e-7"},
  {"claim":"Q7 photon energy approx 7.95e-19 J", "code":"h=6.626e-34; cl=3.00e8; lam=250e-9; E=h*cl/lam; result = abs(E - 7.95e-19) < 1e-21"},
  {"claim":"Q6 ethyl multiplicities: CH2 quartet (4), CH3 triplet (3)", "code":"ch2 = 3+1; ch3 = 2+1; result = (ch2==4) and (ch3==3)"},
  {"claim":"Q8 lost fragment mass = 15 (methyl)", "code":"M=46; frag=31; loss=M-frag; result = loss==15"}
]