Optically active p-chlorobenhydryl p-nitrobenzoate: specifically labeled with ^1

Question

Optically active p-chlorobenhydryl p-nitrobenzoate: specifically labeled with ^18O at the carbonyl oxygen, was prepared and its solvolysis is aqueous acetone was studied. Three processes can be followed to produce three rate constants: k_t for the rate of formation of product, k_o for the rate of loss of optical activity; k_eq and for the rate of equilibration of the isotopic label between the ether and carbonyl positions. The process measured by involves both racemization of starting ester and formation of racemic solvolysis product. The rate of racemization of starting ester can be calculated as k_rac = k - k_t. Why are all of the rates faster in 80% acetone than in 90% acetone? Explain why k_t

Explanation / Answer

Ans:

a) Larger the activation energy, lesser the value of rate constant.

b) Keq is constant. change in volume, concentartion & Pressure does not change the keq. Only change in temperature impact the Keq.

c) At 90% Acetone

= 3.5 / (1.4 + 3.5) = 3.5 / 4.9 = 0.71 (About 71% ion pairs return to esters, remaining 19% is reversible reaction)

Reversible reaction = 3.5 / 0.186 = 18.8 equ 19%.

At 80% Acetone

= 1.8 / (1.2+1.8) = 1.8/3 = 0.60 (About 60% ion pairs return to esters, remaining 22% is reversible reaction)

Reversible reaction = 0.8/(2 x 1.8) = 0.22 (22%)

d) Kintetic changes is equivalent to applied temperature during solvolysis im presense of Sodium azide.

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