Which carboxylic acid is used to prepare the ester shown?

Chemistry

1 answer:

valentinak56 [21]3 years ago

3 0

Answer:

Explanation:

The general equation for the reaction of a carboxylic acid with an alkanol to form an ester is shown below;

RCOOH + ROH ------> RCOOR + H2O

Hence; the reactant carboxylic acid can only be the compound (CH3)2-CH-CH2-COOH in accordance with the general reaction equation shown above.

Hence the reaction is;

(CH3)2-CH-CH2-COOH + CH3-CH2OH -------> CH3CH2 OCO-CH2-CH-(CH3)2

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Answer:

$0.4167\ \text{M}^{-1}\text{min}^{-1}$

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${t_{1/2}}A=2\ \text{min}$

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We have the relation

$t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}$

$\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}$

Comparing the exponents we get

$1=n-1\\\Rightarrow n=2$

The order of the reaction is 2.

$t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}$