Noesis
Level 4 — ApplicationSpectroscopy & Analysis (Intro)

Spectroscopy & Analysis (Intro)

⏱ 60 minutes60 marksprintable — key stays hidden on paper

Time limit: 60 minutes Total marks: 60 Instructions: Show all reasoning. Use ...... for mathematical expressions. Constants: c=3.00×108 m s−1c = 3.00 \times 10^8\ \text{m s}^{-1}, h=6.63×10−34 J sh = 6.63 \times 10^{-34}\ \text{J s}, NA=6.02×1023 mol−1N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}.

Question 1 — Beer–Lambert & conjugation (13 marks)

A dye has molar absorptivity ε=1.8×104 L mol−1cm−1\varepsilon = 1.8 \times 10^4\ \text{L mol}^{-1}\text{cm}^{-1} at its λmax=505 nm\lambda_{max} = 505\ \text{nm}.

(a) A solution in a 1.00 cm1.00\ \text{cm} cell gives an absorbance A=0.72A = 0.72. Calculate the molar concentration. (3)

(b) The solution is diluted by a factor of 4 and measured in the same cell. State the new absorbance and the new percentage transmittance (%T)(\%T). (4)

(c) Calculate the energy (in J\text{J}) of one photon at 505 nm505\ \text{nm}, and hence the energy in kJ mol−1\text{kJ mol}^{-1}. (4)

(d) A structurally related dye has one fewer C=C in its conjugated chain. Predict, with reasoning, whether its λmax\lambda_{max} is greater or smaller than 505 nm505\ \text{nm}. (2)

Question 2 — Combined structure elucidation (16 marks)

An unknown compound X has molecular formula C4H8O2\text{C}_4\text{H}_8\text{O}_2.

Data:

  • MS: molecular ion at m/z=88m/z = 88; strong fragment at m/z=43m/z = 43; fragment at m/z=45m/z = 45.
  • IR: strong sharp band at 1740 cm−11740\ \text{cm}^{-1}; no broad band near 30003000–3500 cm−13500\ \text{cm}^{-1}.
  • ¹H NMR: singlet δ 2.0\delta\ 2.0 (3H); quartet δ 4.1\delta\ 4.1 (2H); triplet δ 1.3\delta\ 1.3 (3H).

(a) Calculate the degree of unsaturation of X. (2)

(b) Identify the functional group indicated by the IR band at 1740 cm−11740\ \text{cm}^{-1} and explain why the absence of a broad 30003000–3500 cm−13500\ \text{cm}^{-1} band rules out a carboxylic acid. (3)

(c) Deduce the full structure of X, using the NMR multiplicities and integrations to justify each fragment. (6)

(d) Assign the MS fragments at m/z=43m/z = 43 and m/z=45m/z = 45 to specific ions. (3)

(e) Predict the number of signals in the ¹³C NMR spectrum of X. (2)

Question 3 — NMR prediction (12 marks)

Consider 1-chloropropane, CH3CH2CH2Cl\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}.

(a) State the number of distinct ¹H environments and predict the multiplicity of each signal using the n+1n+1 rule. (6)

(b) Give the expected integration ratio of the three signals. (2)

(c) The isomer 2-chloropropane, (CH3)2CHCl(\text{CH}_3)_2\text{CHCl}, gives only two ¹H signals. Predict their multiplicities and integration ratio, and explain how ¹H NMR alone distinguishes the two isomers. (4)

Question 4 — Chromatography & spectrum matching (11 marks)

A student separates a two-component mixture by TLC. On a plate of length used for a 10.0 cm10.0\ \text{cm} solvent run, component P travels 7.5 cm7.5\ \text{cm} and component Q travels 3.0 cm3.0\ \text{cm}.

(a) Calculate the RfR_f value of each component. (2)

(b) State which component is more polar (silica stationary phase, non-polar mobile phase) and justify. (3)

(c) The student wants to purify grams of the mixture. State which technique (column chromatography or GC) is appropriate and why, and state one physical property required of the sample for GC to be usable. (3)

(d) Explain in terms of interaction with the two phases why increasing mobile-phase polarity would change the RfR_f values. (3)

Question 5 — Electromagnetic spectrum reasoning (8 marks)

(a) Arrange the following in order of increasing photon energy: microwave, UV, IR, radiofrequency (NMR), X-ray. (2)

(b) Match each region to the molecular process it probes: (i) bond vibrations, (ii) nuclear spin transitions, (iii) electronic transitions in conjugated systems. (3)

(c) An IR transition occurs at wavenumber ν~=1700 cm−1\tilde{\nu} = 1700\ \text{cm}^{-1}. Calculate the corresponding frequency in Hz. (3)

Answer keyMark scheme & solutions

Question 1 (13 marks)

(a) Beer–Lambert: A=εclA = \varepsilon c l. c=Aεl=0.721.8×104×1.00=4.0×10−5 mol L−1c = \frac{A}{\varepsilon l} = \frac{0.72}{1.8\times10^4 \times 1.00} = 4.0\times10^{-5}\ \text{mol L}^{-1}

  • Correct rearrangement (1); substitution (1); answer 4.0×10−54.0\times10^{-5} M (1).

(b) Absorbance is proportional to concentration, so dilution by 4 → A=0.72/4=0.18A = 0.72/4 = 0.18. %T\%T: A=−log⁡10T⇒T=10−A=10−0.18=0.661A = -\log_{10}T \Rightarrow T = 10^{-A} = 10^{-0.18} = 0.661, so %T=66.1%\%T = 66.1\%.

  • New A=0.18A = 0.18 (2); conversion formula (1); %T≈66%\%T \approx 66\% (1).

(c) λ=505 nm=5.05×10−7 m\lambda = 505\ \text{nm} = 5.05\times10^{-7}\ \text{m}. E=hcλ=6.63×10−34×3.00×1085.05×10−7=3.94×10−19 JE = \frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3.00\times10^8}{5.05\times10^{-7}} = 3.94\times10^{-19}\ \text{J} Per mole: Em=3.94×10−19×6.02×1023=2.37×105 J mol−1=237 kJ mol−1E_m = 3.94\times10^{-19}\times6.02\times10^{23} = 2.37\times10^5\ \text{J mol}^{-1} = 237\ \text{kJ mol}^{-1}.

  • Formula/λ conversion (1); photon energy 3.94×10−193.94\times10^{-19} J (2); per mole ≈237\approx 237 kJ/mol (1).

(d) Fewer C=C → shorter conjugation → larger HOMO–LUMO gap → higher energy absorption → shorter (smaller) λmax\lambda_{max} (< 505 nm).

  • Correct direction (1); conjugation/energy-gap reasoning (1).

Question 2 (16 marks)

(a) DoU =2(4)+2−82=10−82=1= \dfrac{2(4)+2-8}{2} = \dfrac{10-8}{2} = 1.

  • Formula (1); answer 1 (1).

(b) 1740 cm−11740\ \text{cm}^{-1} = C=O stretch (ester/carbonyl). A carboxylic acid would show a broad O–H stretch (25002500–3300 cm−13300\ \text{cm}^{-1}); its absence rules out –COOH, so the compound is an ester.

  • C=O identification (1); O–H absence reasoning (1); conclusion ester (1).

(c) Structure is ethyl acetate, CH3COOCH2CH3\text{CH}_3\text{COOCH}_2\text{CH}_3.

  • Singlet 3H at δ\delta2.0 = CH3\text{CH}_3 next to C=O (no neighbouring H) (2);
  • quartet 2H at δ\delta4.1 = OCH2\text{OCH}_2 (deshielded by O, coupled to 3H → 3+1=43+1=4) (2);
  • triplet 3H at δ\delta1.3 = terminal CH3\text{CH}_3 (coupled to 2H → 2+1=32+1=3) (1); correct full structure (1).

(d) m/z=43m/z=43 = acylium CH3CO+\text{CH}_3\text{CO}^+ (loss of OC2H5\text{OC}_2\text{H}_5, mass 45) (1.5); m/z=45m/z=45 = C2H5O+/OCH2CH3+\text{C}_2\text{H}_5\text{O}^+ / \text{OCH}_2\text{CH}_3^+ (or COOH\text{COOH}-type) — accept ethoxy cation (1.5).

(e) Four distinct carbons: CH3\text{CH}_3(CO), C=O, OCH2\text{OCH}_2, CH3\text{CH}_3. → 4 signals.

  • Count (2) (1 if minor error).

Question 3 (12 marks)

(a) Three environments:

  • CH3\text{CH}_3 (adjacent to CH₂, 2H): 2+1=32+1 = 3 → triplet (2)
  • middle CH2\text{CH}_2 (adjacent to CH₃ (3H) and CH₂Cl (2H); non-equivalent neighbours total 5): 5+1=65+1 = 6 → sextet (2)
  • CH2Cl\text{CH}_2\text{Cl} (adjacent to CH₂, 2H): 2+1=32+1 = 3 → triplet (2)

(b) Integration ratio CH3:CH2:CH2Cl=3:2:2\text{CH}_3 : \text{CH}_2 : \text{CH}_2\text{Cl} = 3:2:2. (2)

(c) 2-chloropropane: two 6H equivalent CH3\text{CH}_3 groups (adjacent to 1 CH → 1+1=21+1=2 = doublet) and CH (adjacent to 6H → 6+1=76+1=7 = septet); ratio 6:16:1.

  • multiplicities (1); ratio 6:1 (1); distinguishing statement: 1-chloropropane shows 3 signals (3:2:2) vs 2-chloropropane 2 signals (6:1), so number of signals and integration differ (2).

Question 4 (11 marks)

(a) Rf(P)=7.5/10.0=0.75R_f(P) = 7.5/10.0 = 0.75; Rf(Q)=3.0/10.0=0.30R_f(Q) = 3.0/10.0 = 0.30.

  • each (1).

(b) Q is more polar. On silica (polar stationary phase) with a non-polar mobile phase, polar compounds adsorb more strongly and travel less → lower RfR_f.

  • identify Q (1); polar stationary phase interaction (1); links low RfR_f to strong adsorption (1).

(c) Column chromatography — preparative, handles gram quantities and non-volatile samples; GC requires the sample to be volatile (and thermally stable), and is analytical/small-scale.

  • column chosen (1); scale/preparative reason (1); volatility requirement for GC (1).

(d) Increasing mobile-phase polarity strengthens interaction between solute and mobile phase (and competes with adsorption sites), so polar solutes spend more time in the mobile phase → they migrate further → RfR_f increases.

  • effect on partition/adsorption (1); more time in mobile phase (1); RfR_f increases (1).

Question 5 (8 marks)

(a) Increasing energy: radiofrequency (NMR) < microwave < IR < UV < X-ray. (2) (1 for one misplacement).

(b) (i) bond vibrations → IR; (ii) nuclear spin transitions → NMR (radiofrequency); (iii) electronic transitions → UV-Vis. (3) (1 each).

(c) ν~=1700 cm−1=1.70×105 m−1\tilde{\nu} = 1700\ \text{cm}^{-1} = 1.70\times10^5\ \text{m}^{-1}. ν=cν~=3.00×108×1.70×105=5.10×1013 Hz\nu = c\tilde{\nu} = 3.00\times10^8 \times 1.70\times10^5 = 5.10\times10^{13}\ \text{Hz}

  • unit conversion cm⁻¹→m⁻¹ (1); formula ν=cν~\nu=c\tilde\nu (1); answer 5.10×10135.10\times10^{13} Hz (1).
[
  {"claim":"Q1a concentration = 4.0e-5 mol/L","code":"A=0.72; eps=1.8e4; l=1.00; c=A/(eps*l); result = abs(c-4.0e-5) < 1e-7"},
  {"claim":"Q1b diluted A=0.18 gives %T approx 66.1%","code":"A=0.18; T=10**(-A); result = abs(T*100-66.07) < 0.5"},
  {"claim":"Q1c photon energy at 505 nm approx 3.94e-19 J and 237 kJ/mol","code":"h=6.63e-34; cc=3.00e8; lam=5.05e-7; E=h*cc/lam; Em=E*6.02e23/1000; result = abs(E-3.94e-19)<2e-21 and abs(Em-237)<3"},
  {"claim":"Q2a degree of unsaturation of C4H8O2 = 1","code":"C=4; H=8; DoU=(2*C+2-H)/2; result = DoU==1"},
  {"claim":"Q5c frequency for 1700 cm-1 = 5.10e13 Hz","code":"cc=3.00e8; wn=1700*100; nu=cc*wn; result = abs(nu-5.10e13) < 1e11"}
]