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satela [25.4K]
3 years ago
5

Write a mechanism for the esterification of propanoic acid with 18O-labeled ethanol. Show clearly the fate of the 18O label. (b)

Acid-catalyzed hydrolysis of an unlabeled ester with 18O-labeled water (H218O) leads to incorporation of some 18O into both oxygens of the carboxylic acid product. Explain by a mechanism. (Hint: You must use the fact that all steps in the mechanism are reversible.)
Chemistry
1 answer:
tatuchka [14]3 years ago
5 0

Answer:

See explanation and images attached

Explanation:

a) In the mechanism for the acid catalysed esterification of propanoic acid using ethanol, we can see that the first step is the protonation of the acid followed by nucleophillic attack of the alcohol. Loss of water and consequent deprotonation regenerates the acid catalyst. We can see the fate of the 18O labelled ethanol in the mechanism shown.

b)  In the second mechanism, an unnamed ester is hydrolysed using an acid catalyst. The attack of the acid and subsequent nucleophillic attack of water labelled with 18O leads to the incorporation of this 18O into the product acid as shown in the mechanism attached to this answer.

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No

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3 0
3 years ago
The reaction (CH3)3CBr + OH- (CH3)3COH + Br- in a certain solvent is first order with respect to (CH3)3CBr and zero order with r
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Answer and Explanation:

The rate constant (K) is related to activation energy (Ea), frequency factor (A) and temperature (T) in Kelvin by the equation

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In K = In A - (Ea/RT)

In K = (-Ea/R)(1/T) + In A

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y = In K, slope, m = (-Ea/R), x = (1/T) and intercept, c = In A

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Ea = -11000 × 8.314 = 91454 J/mol = 91.454 KJ/mol

b) c = In A = 33.5

A = e^33.5 = (3.54 × (10^14))/s

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A = (3.54 × (10^14))/s, Ea = 91454 J/mol, T = 25°C = 298.15 K, R = 8.314 J/K.mol

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QED!

5 0
3 years ago
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Igoryamba

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