Chemistry Labs

Problem 2

Repatriation of remains in the Middle Ages. Racemization at ambient temperature is slow and can be used for dating and for studying the thermal history of biological objects. L-isoleucine, (2 S, 3 S)-2-amino-3-methylpentanoic acid\ce{(2S,3S)-2-amino-3-methylpentanoic acid}, epimerizes at the α\alpha-carbon to D-allo-isoleucine, (2 R, 3 S)-2-amino-3-methylpentanoic acid\ce{(2R,3S)-2-amino-3-methylpentanoic acid}. (a) The four stereoisomers are L-isoleucine (2S,3S), D-isoleucine (2R,3R), L-allo-isoleucine (2S,3R), D-allo-isoleucine (2R,3S). Which statement is true about D-allo-isoleucine vs L-isoleucine: identical/opposite/different specific rotations; same or different melting points? (b) The equilibrium constant of epimerization is Kep=1.38K_{ep} = 1.38 at 374 K. Taking the standard Gibbs energy of L-isoleucine as 0 kJ mol−1^{-1}, give the Gibbs energies of all four stereoisomers at 374 K. (c) What is the maximum number of stereoisomers of the tripeptide Ile–Ile–Ile? (d) Neglecting the reverse reaction, epimerization is first-order with k1(374 K)=9.02×10−5k_1(374\ \text{K}) = 9.02\times10^{-5} h−1^{-1} and k1(421 K)=1.18×10−2k_1(421\ \text{K}) = 1.18\times10^{-2} h−1^{-1}. Define the diastereomeric excess de=[L]−[D][L]+[D]×100%de = \frac{[L]-[D]}{[L]+[D]}\times100\%. Boiling L-isoleucine for 1943 h at 374 K: what is dede (i) before and (ii) after boiling? (e) How long does conversion of 10% of L-isoleucine to D-allo-isoleucine take at 298 K? (f) Accounting for the reverse reaction, x=[L]−[L]eqx = [\ce{L}] - [\ce{L}]_{eq} decays as x=x(0)e−(k1+k2)tx = x(0)e^{-(k_1+k_2)t}. For 1.00 mol dm−3^{-3} L-isoleucine boiled 1943 h at 374 K, find [L]eq[\ce{L}]_{eq} and dede.
Step 5 of 6: Extrapolate to 298 K and compute the time
Ea=Rln⁡k421k374/(1374−1421)=136 kJ mol−1;  k(298)=1.25×10−9 h−1;  t=ln⁡1.111k=8.4×107 h≈9600 yearsE_a = R\ln\frac{k_{421}}{k_{374}}\bigg/\left(\tfrac{1}{374}-\tfrac{1}{421}\right) = 136\ \text{kJ mol}^{-1};\; k(298) = 1.25\times10^{-9}\ \text{h}^{-1};\; t = \frac{\ln 1.111}{k} = 8.4\times10^{7}\ \text{h} \approx 9600\ \text{years}
Analysis

The Arrhenius plot from the two rate constants gives Ea=136E_a = 136 kJ mol−1^{-1} and ln⁡A=34.4\ln A = 34.4; hence k(298 K)=e34.4−136000/(8.314×298)=1.25×10−9k(298\ \text{K}) = e^{34.4 - 136000/(8.314\times298)} = 1.25\times10^{-9} h−1^{-1}. For 10% conversion, t=ln⁡(1/0.90)/k=8.4×107t = \ln(1/0.90)/k = 8.4\times10^7 h ≈\approx 9610 years.