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Archy [21]
3 years ago
15

As noted in Assignment 1.4.1., alkylation of benzene with 1-chlorobutane in the presence of AlCl3 gives both butylbenzene and (1

-methylpropyl)benzene as products. Propose a route to butylbenzene from benzene that does not also give the (1-methylpropyl)benzene side-product.
Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0

Answer:

Use the Bromotriflouride catalyst, BF₃

Explanation:

The BF₃ is most likely to yield less desired side products. The effect lies in the reaction mechanism.

BF₃ is a Lewis acid. Its role is to promote the ionization of the HF. This is achieved through the electrophilic mechanism. The reaction mechanism is as follows:

2 - methylpropene + H-F-BF₃ → H-F + H₃C + benzene

butylbenzene + F-BF₃ → tert-butylbenzene + H-F + BF₃ (regenerated catalyst)

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Covalent bonds are in the air you breathe and the water you drink. How are covalent bonds formed?
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Covalent bond is the term that is being used to describe the bonds in the compounds that are created due to the sharing of one or more electrons. One of the best example of the simplest covalent bond is the bond that is being created when two isolated hydrogen atoms come together to form an H2 molecule. An isolated hydrogen atom has one proton and electron being combined by the force of attraction from the opposite-charged articles. When a pair of isolated hydrogen atom combines, two forces of attraction are created coming from each of the isolated hydrogen atom.
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3 years ago
All but one of the following is an important component of soil.
Sholpan [36]

Answer: B is the answer i believe.

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3 years ago
At 100 degrees Celsius, the ion product for pure water is Kw= 51.3*10^-14. What is the pOH of water at this temperature? A) 7.00
Levart [38]

7.86 is the pOH of water at this temperature of 100 degrees celsius.

Option E is the right answer.

Explanation:

Data given:

Kw = 51.3 x 10^{-14}

pOH = ?

we know that pure water is neutral and will have pH pf 7.

The equation for relation between Kw and H+ and OH- ion is given by:

Kw = [H+] [OH-}

here the concentration of H+ ion and OH- ion is equal

so, [H+]= [OH-]

Putting the values in the equation of Kw

pKw = -log[Kw]

pKw = -log [51.3 x 10^{-14}]

pKw = 12.28

since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14

Now, pOH is calculated by using the equation:

14 = pOH + pH

14- 6.14 = pOH

pOH = 7.86

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3 years ago
How did atomic theory develop and change?
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There ya have it!
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3 years ago
According to this balanced equation, how many grams of water (H.0) form in
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Answer:

your simpal answer is 177.32

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