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SVEN [57.7K]
3 years ago
8

Compound a is an alkene that was treated with ozone (followed by dms) to yield only 4-heptanone. Draw the major product that is

expected when compound a is treated with mcpba followed by aqueous acid (h3o+).

Chemistry
1 answer:
olga nikolaevna [1]3 years ago
5 0

Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive  cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.

Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.

The reaction is as shown in the image.

The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid H_3O^{+} is shown in the image.

m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.

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Answer:

It would be an element my dear.

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Pure gold is an element. That is, it is not a combination of other materials. However, gold used in jewelry is usually a “mixture”, called an alloy mixed with other metals to harden it, to change the color as in copper makes the pink or rose gold that is currently enjoying great polarity.

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3 years ago
A reaction vessel is charged with hydrogen iodide, which partially decomposes to molecular hydrogen and iodine: 2HI (g) H2(g) +
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Answer:

The value of the equilibrium constant: K_{p} = 0.25

Explanation:

Given reaction: 2HI (g) ⇌ H₂(g) + I₂(g)

Number of moles of- HI: n₁ = 2 mole; H₂: n₂ = 1 mole; I₂: n₃ = 1 mole

Total number of moles: n = n₁ + n₂ + n₃ = 2 + 1 + 1 = 4 moles

The equilibrium constant for the given reaction is given as:

K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}

Given: Temperature: T = 425 °C = 425 + 273 = 698 K

The partial pressure: pHI = 0.708 atm,

and, pH₂ = pI₂

 

∵ <em>partial pressure of a given gas</em>: pₐ = Χₐ . P

Here, P is the total pressure

Χₐ is the <em>mole fraction</em> of given gas and is given by the equation

\chi_{a} = \frac{number \, of \,moles \,of \,given \,gas (n_{a})}{total \,number \,of \,moles (n)}

Mole fraction for HI: \chi_{1} = \frac {n_{1}}{n} = \frac {2}{4} = 0.5

Mole fraction for H₂: \chi_{2} = \frac {n_{2}}{n} = \frac {1}{4} = 0.25

Mole fraction for I₂: \chi_{3} = \frac {n_{3}}{n} = \frac {1}{4} = 0.25

Thus, Χ₂ = Χ₃ = 0.25

The partial pressure of HI is given by;

pHI = Χ₁ P

0.708 atm = 0.5 × P

⇒ P = 1.416 atm

 

As the partial pressures: pH₂ = pI₂

∴ pH₂ = pI₂ = Χ₂ P = Χ₃ P = 0.25 × 1.416 atm = 0.354 atm

Therefore, the value of Kp can be calculated as:

K_{p} = \frac{pH_{2}\; pI_{2}}{(pHI)^{2}}

K_{p} = \frac{0.354 atm\times 0.354 atm}{(0.708 atm)^{2}} = 0.25

<u>Therefore, the value of the equilibrium constant: </u>K_{p} = 0.25

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