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3 years ago

Which of the following statements are True about the experimental process used in the Diels Alder reaction?

Chemistry

1 answer:

vovikov84 [41]3 years ago

8 0

Here we have to get the correct statements among the given, applicable for Diels-Alder reaction.

The true statements in case Diels-Alder reaction are-

1. An excess of Maleic anhydride is used.

2. The I.R. of the products are indistinguishable.

The Diels-Alder reaction is the most is the most important cyclo-addition reaction in organic chemistry. These are addition reactions in which ring systems are formed without eliminating any compounds.

There remains one diene and one dienophile. The reaction is reversible in nature and requires elevated temperature to obtain its transition state. The reaction rate become faster in certain condition like using of polar solvents.

Among the given statements the following statements are true-

1. An excess of maleic anhydride (the most effective di-enophile) is used to process the reaction in forward direction.

2. The products obtain in this reaction are stereoisomers thus are indistinguishable by infrared spectroscopy (IR).

The statements which are not true for the Diels-Alder reaction:

3. The re-crystallization of the products by any polar solvent like methanol is not feasible as it will cause the retro reaction due to stability of the transition state in polar solvent.

4. Cleaning of glassware are compulsory for any reaction it is not specifically true for Diels-Alder reaction.

5. The reaction occurs at elevated temperature thus flame is required.

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Answer:

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Explanation:

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3 years ago

The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3

9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

$\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )$

Where,

$k_1\ is\ the\ rate\ constant\ at\ T_1$

$k_2\ is\ the\ rate\ constant\ at\ T_2$

$E_a$ is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol

$k_2=2.3\times 10^8$

$T_2=280\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (280 + 273.15) K = 553.15 K

$T_2=553.15\ K$

$k_1=4.8\times 10^8$

$T_1=376\ ^0C$

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (376 + 273.15) K = 649.15 K

$T_1=649.15\ K$

So,

$\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)$