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,

\ce{(2S,3S)-2-amino-3-methylpentanoic acid}

, epimerizes at the

\alpha

-carbon to D-allo-isoleucine,

\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

K_{ep} = 1.38

at 374 K. Taking the standard Gibbs energy of L-isoleucine as 0 kJ mol

^{-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

k_1(374\ \text{K}) = 9.02\times10^{-5}

^{-1}

and

k_1(421\ \text{K}) = 1.18\times10^{-2}

^{-1}

. Define the diastereomeric excess

de = \frac{[L]-[D]}{[L]+[D]}\times100\%

. Boiling L-isoleucine for 1943 h at 374 K: what is

de

(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 = [\ce{L}] - [\ce{L}]_{eq}

decays as

x = x(0)e^{-(k_1+k_2)t}

. For 1.00 mol dm

^{-3}

L-isoleucine boiled 1943 h at 374 K, find

[\ce{L}]_{eq}

and

de

Step 6 of 6: Reversible kinetics: equilibrium and

de

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k_2 = \frac{k_1}{K_{ep}} = 6.54\times10^{-5}\ \text{h}^{-1};\; [\ce{L}]_{eq} = \frac{k_2}{k_1+k_2}[\ce{L}]_0 = 0.420\ \text{mol dm}^{-3};\; de = \frac{k_2 - k_1 + 2k_1 e^{-(k_1+k_2)t}}{k_1+k_2}\times100\% = 69.8\%

Analysis

The reverse constant follows from $K_{ep} = k_1/k_2$ : $k_2 = 6.54\times10^{-5}$ h $^{-1}$ , so $[\ce{L}]_{eq} = 0.420$ mol dm $^{-3}$ . With $x(0) = [\ce{D}]_{eq} = 0.580$ and $e^{-(k_1+k_2)t} = e^{-0.302} = 0.739$ , $[\ce{L}] = 0.420 + 0.580\times0.739 = 0.849$ , giving $de = (0.849 - 0.151)/1.00 = 69.8\%$ .

Common pitfall. The equilibrium mixture is not 50:50 — the D-allo form is slightly more stable, so

de

approaches

(K_{ep}-1)/(K_{ep}+1) \times 100\% \approx 16\%

, not zero, at long times.