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vredina [299]
3 years ago
8

From the lists of available reagents select the one(s) you would use to in a preparation of acetophenone (phenyl methyl ketone)

from methyl benzoate. (In no case are more than three steps necessary.) Reagents Available a. CH COCI, AICI, d. CH CHO g. Mg 1. SOCI b. (CH3),Culi e. Hg(OC), H,Othen NaBH4 h. NaOH, H2O C. CH, MgBr, then H30* . KMnO4 H20 i. Dess-Martin periodinane List reagents by letter in the order that they are used; example:
Reagents used: ______
Chemistry
1 answer:
VARVARA [1.3K]3 years ago
3 0

Answer:

Step 1) hydrolysis using NaOH/H2O to form benzylalcohol

Step2) oxidation to Carboxylic acid using KMnO4 followed by decarboxylation to form benzene

3) friedel craft acylation using CH3COCl/AlCl3

Explanation:

The above 3 steps will yield acetophenone from methylbenzoate

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When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
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Answer:

2

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

Explanation:

Half-life

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

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

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

So

\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}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

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