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Pepsi [2]
3 years ago
14

Provide the structure of the major product which results from 1,4-addition of br2 to the diene shown below.

Chemistry
1 answer:
Vlad1618 [11]3 years ago
7 0
The image provided shows the diene that is used for this question. We are told that the major product of the reaction is the 1,4-addition product. The result is the addition of a bromine atom to the first carbon and the fourth carbon of the diene. However, Br₂ can also add to just one alkene of the diene in a 1,2-addition to get the other product shown in the image.

As the first bromine atom adds to one of the alkenes, it adds to the first carbon which leads to the formation of a carbocation. The carbocation can be a stable tertiary center at the 2 carbon of the diene, or the less stable secondary center of the 4 carbon. To addition to the 4-carbon has a higher activation barrier which, but the product has a lower energy than the 1,2-product. Therefore, the 1,4-addition is the thermodynamic product and will form at higher temperatures. The 1,2-product is the kinetic product that will form at lower temperatures.

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Answer : The partial pressure of N_2,H_2\text{ and }NH_3 at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar

Solution :  Given,

Initial pressure of N_2 = 1.42 bar

Initial pressure of H_2 = 2.87 bar

K_p = 0.036

The given equilibrium reaction is,

                              N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)

Initially                   1.42      2.87             0

At equilibrium    (1.42-x)  (2.87-3x)     2x

The expression of K_p will be,

K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}

Now put all the values of partial pressure, we get

0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}

By solving the term x, we get

x=0.287\text{ and }3.889

From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.

Thus, the partial pressure of NH_3 at equilibrium = 2x = 2 × 0.287 = 0.574 bar

The partial pressure of N_2 at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar

The partial pressure of H_2 at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar

The total pressure at equilibrium = Partial pressure of N_2 + Partial pressure of H_2 + Partial pressure of NH_3

The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar

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Explanation:

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