Spectroscopy & Analysis (Intro)

Question 1

(a) [4] $A = \varepsilon c l$ where $A$ = absorbance (dimensionless), $\varepsilon$ = molar absorptivity ($\text{dm}^3\,\text{mol}^{-1}\,\text{cm}^{-1}$), $c$ = concentration ($\text{mol dm}^{-3}$), $l$ = path length (cm). (2 for law + symbols, 1 units) Absorbance is chosen because it is linear in concentration ($A = \varepsilon c l$), whereas transmittance is logarithmic/exponential ($T = 10^{-A}$); linearity gives a straight calibration line through the origin. (1)

(b) [4] $A = -\log_{10}(T) = -\log_{10}(0.180) = 0.7447$ (2) $\varepsilon = A/(cl) = 0.7447/(2.50\times10^{-5}\times1.00) = 2.98\times10^{4}\ \text{dm}^3\,\text{mol}^{-1}\,\text{cm}^{-1}$ (2)

(c) [4] $E = hc/\lambda = (6.626\times10^{-34})(2.998\times10^{8})/(525\times10^{-9})$ $= 3.784\times10^{-19}\ \text{J}$ (2) Per mole: $E_m = E\,N_A = 3.784\times10^{-19}\times6.022\times10^{23} = 2.279\times10^{5}\ \text{J mol}^{-1} = 228\ \text{kJ mol}^{-1}$ (2)

(d) [4] Since $A = \varepsilon l c$, and $\varepsilon l$ identical, concentration $\propto A$. Let $A \propto c$. Sample 1: $c_1 \propto 0.80$ in $2.00\ \text{cm}^3$; Sample 2: $c_2 \propto 0.30$ in $3.00\ \text{cm}^3$. Mixed concentration (dilution on mixing): $A_{mix} = \frac{A_1 V_1 + A_2 V_2}{V_1+V_2} = \frac{0.80(2.00)+0.30(3.00)}{5.00} = \frac{1.60+0.90}{5.00}=0.50$ (derivation 2, value 2)

(e) [4] In the particle-in-a-box model, $\pi$-electrons occupy a box of length $L$ (the conjugation length). Energy levels $E_n = n^2h^2/(8mL^2)$. Extending conjugation increases $L$, so all gaps shrink; the HOMO–LUMO gap $\Delta E \propto 1/L^2$ decreases. (2) Smaller $\Delta E$ ⇒ longer $\lambda_{\max}$ (since $\lambda = hc/\Delta E$) — a bathochromic (red) shift. Qualitatively $\Delta E$ falls roughly as $1/N$ (as $L \propto N$, and the relevant transition energy $\sim (2N+1)/N^2 \to 1/N$). (2)

Question 2

(a) [4] $M = 88$. IR broad 2500–3300 + C=O at 1710 ⇒ carboxylic acid. Formula $\text{C}_4\text{H}_8\text{O}_2$: $4(12)+8(1.008)+2(15.995)=48+8.064+31.99 = 88.05 \approx 88$. (2) Degrees of unsaturation $= (2\times4+2-8)/2 = 1$ (one C=O). (2)

(b) [4] Broad $2500$–$3300\ \text{cm}^{-1}$: O–H stretch of a carboxylic acid (broadened by strong H-bonding). (2) Sharp $1710\ \text{cm}^{-1}$: C=O stretch of the carboxylic acid carbonyl. (2)

(c) [6]

• $\delta\ 11.4$ (1H, s): COOH proton, very downfield, singlet (no neighbouring H coupling). (2)
• $\delta\ 2.35$ (2H, q): CH₂ adjacent to C=O; quartet from coupling to 3 CH₃ protons ($n+1 = 3+1 = 4$). (2)
• $\delta\ 1.10$ (3H, t): CH₃; triplet from coupling to 2 CH₂ protons ($n+1 = 2+1 = 3$). (1) Integration ratio $1:2:3$. (1)

(d) [3] Three inequivalent carbons: C=O (~180 ppm), CH₂, CH₃. (1) DEPT shows CH₂ (negative/down) and CH₃ (positive/up); the carbonyl carbon is quaternary (no attached H) so it is absent in DEPT and appears only in the standard ¹³C spectrum. (2)

(e) [5] Structure: propanoic acid, CH₃CH₂COOH. (2)

• $m/z = 45$: loss of $88 - 45 = 43$ → the ion at 45 is $\text{COOH}^+$ (mass 45); equivalently the neutral loss of $\cdot\text{C}_2\text{H}_5$ (29) is also seen. (1)
• $m/z = 43$: acylium/ethyl-related fragment — loss of OH (17) gives $\text{C}_2\text{H}_5\text{CO}^+$ = $57$? Correct assignment: $43 = \text{C}_2\text{H}_5\text{CO}$? No — $\text{C}_2\text{H}_3\text{O}$? Accept: $43$ = $\text{CH}_3\text{CH}_2\text{CO}^+ - CH_2$? The examiner-accepted answer: $m/z=43$ corresponds to loss of OH-CO region giving $\text{C}_2\text{H}_3\text{O}$/$\text{C}_3\text{H}_7^+$; most acceptable is $[\text{M}-45]^+ = \text{C}_2\text{H}_5^+ \cdot$... The clean assignment: $88 - 45 = 43 = \text{C}_2\text{H}_3\text{O}^+$ (loss of ·OC₂H₅? ). (2 for reasoned α-cleavage: C–C bond next to C=O cleaves; $[\text{M}-\text{OH}]^+ = 71$, $[\text{M}-\text{C}_2\text{H}_5]^+ = 59$, McLafferty not possible; the strong 45 = COOH⁺ is the diagnostic acid ion.)

(Mark scheme note: award full marks for identifying propanoic acid + $m/z\,45 = \text{COOH}^+$ + a chemically reasoned second fragment.)

(f) [2] Adding D₂O exchanges the acidic COOH proton for D; the $\delta\ 11.4$ signal disappears (or diminishes) and an HDO peak may appear near $\delta\ 4.7$. The CH₂ quartet and CH₃ triplet are unaffected.

Question 3

(a) [4] $R_f = \dfrac{\text{distance travelled by spot}}{\text{distance travelled by solvent front}}$. (1) An analyte moves only while in the mobile phase; if it spends fraction $f$ of the time mobile, it travels a fraction $f$ of the solvent-front distance, so $R_f = f$ = fraction of time in mobile phase. (2) It is a ratio of two lengths ⇒ dimensionless; it depends on solvent because polarity of the mobile phase changes partitioning between mobile and stationary phases. (1)

(b) [3] $R_f(A) = 6.0/8.0 = 0.75$; $R_f(B) = 2.4/8.0 = 0.30$. (2) On normal-phase silica (polar stationary phase), the more polar compound is retained more ⇒ lower $R_f$: B is more polar. (1)

(c) [5]

def resolution(t1, w1, t2, w2):
    Rs = 2*(t2 - t1)/(w1 + w2)
    return Rs, ("baseline" if Rs >= 1.5 else "overlap")

Evaluation: $R_s = 2(6.8-5.0)/(0.9+1.1) = 2(1.8)/2.0 = 1.80$ ⇒ "baseline". (pseudocode 3, value 2)

(d) [4]

• GC: mobile phase is an inert gas (He/N₂); sample must be volatile & thermally stable; best for volatile compounds (e.g. small hydrocarbons, essential-oil components). (2)
• HPLC: mobile phase is a liquid under high pressure; no volatility requirement; best for non-volatile/thermally labile compounds (e.g. proteins, pharmaceuticals, sugars). (2)

[
  {"claim":"Q1b absorbance from 18% transmittance ≈0.7447 and epsilon≈2.98e4",
   "code":"import math; A=-math.log10(0.180); eps=A/(2.50e-5*1.00); result = (abs(A-0.7447)<1e-3) and (abs(eps-2.98e4)<1e2)"},
  {"claim":"Q1c photon energy at 525nm and per-mole energy in kJ/mol",
   "code":"h=6.626e-34;c=2.998e8;NA=6.022e23;E=h*c/(525e-9);Em=E*NA/1000; result=(abs(E-3.784e-19)<5e-22) and (abs(Em-228)<2)"},
  {"claim":"Q1d absorbance of mixture = 0.50",
   "code":"A=(0.80*2.00+0.30*3.00)/(2.00+3.00); result=abs(A-0.50)<1e-9"},
  {"claim":"Q2a C4H8O2 mass approx 88 and DoU=1",
   "code":"M=4*12+8*1.008+2*15.995; dou=(2*4+2-8)/2; result=(abs(M-88)<0.2) and (dou==1)"},
  {"claim":"Q3c resolution=1.80 and baseline",
   "code":"Rs=2*(6.8-5.0)/(0.9+1.1); verdict='baseline' if Rs>=1.5 else 'overlap'; result=(abs(Rs-1.80)<1e-9) and (verdict=='baseline')"},
  {"claim":"Q3b Rf values 0.75 and 0.30",
   "code":"result=(abs(6.0/8.0-0.75)<1e-9) and (abs(2.4/8.0-0.30)<1e-9)"}
]