Living nature is homochiral: almost all natural amino acids have L-configuration, sugars – D-configu — Analytical Chemistry Chemistry Question
Asymmetric autocatalysis – amplification of chiral asymmetry
Living nature is homochiral: almost all natural amino acids have L-configuration, sugars – D-configuration. One of the possible explanations of this phenomenon is based on the concept of asymmetric autocatalysis. In some reactions chiral products can serve as catalysts of their own formation: the larger is the content of one of the enantiomers the faster is its synthesis.
The simplest equation for autocatalysis is: A + P → 2 P, where P is product.
Reaction can be performed under various conditions: either in a closed system when reagents are mixed only once, or in an open system where reagent A is being continuously added to the mixture so that its concentration is maintained constant.
Write the kinetic equations and draw the kinetic curves for product P in the closed and open systems. Assume that the initial concentration of P is non-zero but small.
Model Answer
a) The closed system. The kinetic equation:
d[P]/dt = k[A][P]
Taking into account the mass balance [A] + [P] = [A]0 + [P]0, we get:
At early stages the rate of P formation increases, but after some accumulation of the product reaction becomes more slow and finally its rate approaches zero.
b) The open system. The kinetic equation:
d[P]/dt = k[A][P]
d[P]/dt = k([A]0 + [P]0 - [P])[P]
Both the rate of reaction and concentration [P] increase with time:
[P] = [P]0 exp(k[A]0t)
The first reaction of asymmetric autocatalysis was discovered in the early 1990-s. Addition of diisopropylzinc to pyrimidine-5-carbaldehyde in toluene leads to the mixture of enantiomers X1 and X2, which after hydrolysis is transformed to enantiomeric alcohols Y1 and Y2:
Draw the structure of enantiomeric pairs X and Y, and show the configuration of the stereocenter.
Model Answer
Diisopropylzinc is added across the C=O bond. Subsequent hydrolysis leads to a mixture of enantiomeric secondary alcohols:
It turned out that the presence of small amounts of any product (Y1 or Y2) selectively accelerates the formation of that specific product which leads to enantiomeric enrichment of the reaction mixture. Suppose that the yield of each product is proportional to the square of its molar fraction in the mixture of alcohols prior to synthesis.
To 1 mmol of mixture Y1 and Y2, containing 55 % of Y1, 1 mmol of aldehyde and 1 mmol of diisopropylzinc are added several times. Assuming that total reaction yield is 100 %, calculate how many times we should add the reagents to enrich the mixture of alcohols up to: a) 70 %, b) 90 %, c) 99 % of Y1.
Note. You need to write a small iteration program.
Model Answer
After the (n – 1)th addition the system will contain n mmol of mixture of alcohols. Let the fraction of (S)-isomer be an, and that of (R)-isomer – bn. Let us add one more mmol of reagents. The yield of each alcohol is proportional to its fraction, hence additionaly an^2 / (an^2 + bn^2) mmol of (S- and bn^2 / (an^2 + bn^2) mmol of (R)-isomer are formed. The new fraction of (S)-isomer is:
a_{n+1} = (n * an + an^2 / (an^2 + bn^2)) / (n+1)
Now we need to solve the inequalities a_{n+1} > 0.7; 0.9; 0.99 with the initial condition a1 = 0.55. It is easily done numerically. The iteration program can be written in any language...
Applying recurrence formula, we obtain: a9 > 0.7, a40 > 0.9, a437 > 0.99.
Answer. a) n = 8; b) n = 39; c) n = 436.