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Len [333]
3 years ago
15

Cumene (isopropylbenzene) is a relatively inexpensive commercially available starting material. Show how you could prepare m-iso

propylnitrobenzene from cumene.

Chemistry
1 answer:
mafiozo [28]3 years ago
6 0

Answer:

Nitration is a general chemical process for the introduction of a nitro group into a chemical compound through a chemical reaction.

Aromatic nitration occurs with aromatic compounds thanks to an aromatic electrophilic substitution mechanism that includes the attack of an electron-rich benzene ring by the nitronium ion.

Explanation:

<h3> Reaction mechanism </h3>

Cumene reacts with concentrated and hot nitric acid giving

m- isopropylnitrobenzene. The reaction has two drawbacks: it is slow and also the concentrated and hot nitric acid can oxidize any organic compound by an explosive reaction. A safer procedure is to use a mixture of nitric acid and sulfuric acid. Sulfuric acid acts as a catalyst, allowing the reaction to take place more quickly and at lower temperatures.

Sulfuric acid reacts with nitric acid generating nitronium ion (NO2 +), which is the electrophile of the aromatic electrophilic substitution reaction.

The nitronium ion reacts with cumene forming the sigma complex that loses a proton that is trapped by the bisulfate ion to give rise to m-isopropylnitrobenzene.

Nitration selectivity can be found by examining ring substituents and the effect they have on the reaction rate of this aromatic electrophilic substitution. Deactivation groups such as other nitro groups have a withdrawal effect of the electron pair that deactivates the reaction (creating difficulty in the formation of polynitrated products) and directs the electrophilic nitronium ion to attack the aromatic meta position.

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What volume of a 6.0m hcl solution is required to make 250.0 milliters of 1.5 m hcl
Georgia [21]

Answer : The volume for 6.0m HCl solution required = 62.5 ml

Solution : Given,

Initial concentration of HCl solution = 6.0m

Final concentration of HCl solution = 1.5m

Final volume of HCl solution = 250 ml

Initial volume of HCl solution = ?

Formula used for dilution is,

M_1V_1=M_2V_2

where,

M_1 = initial concentration

M_2 = final concentration

V_1 = initial volume

V_2 = final volume

Now put all the given values in above formula, we get the initial volume of HCl solution.

6.0m\times V_1=1.5m\times 250ml

V_1 = 62.5 ml

Therefore, the volume for 6.0m HCl solution required = 62.5 ml


8 0
3 years ago
When magnesium is ignited in air, the magnesium reacts with oxygen and nitrogen.
vampirchik [111]
51: The only coefficient is in the first one i think and it would be a 3
52: Neon (Ne)
53: Mg has a low nuclear charge so all it's electrons would be lost in the electron shell, making the ion smaller than the atom.
8 0
3 years ago
Read 2 more answers
15. If the moon is 384,400km away. Does it take more than a second for light to travel from the Earth to the moon?​
marishachu [46]

Answer:

Yes

Explanation:

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3 0
1 year ago
The following equation shows the correct molecules formed in the reaction, but the products are incorrect. CH4 +202=H20+ CO2 How
Hunter-Best [27]
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5 0
3 years ago
Two samples of the same compound are compared. what does the data represent? sample 1: 24.22 g carbon and 32.00 g oxygen sample
goblinko [34]

According to law of definite proportion:

In a compound, elements are always arranged in fixed ratio by mass.

Here, sample 1 has 23.22 g Carbon and 32.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{24.22 g}{12 g/mol}\approx 2 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{32 g}{16 g/mol}=2 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=2:2=1:1

Therefore, formula of compound will be CO.

Sample 2:

It has 36.22 g Carbon and 48.00 g Oxygen.

Converting mass into number of moles:

Molar mass of carbon is 12 g/mol and that of oxygen is 16 g/mol thus,

n_{C}=\frac{m_{C}}{M_{C}}=\frac{36.22 g}{12 g/mol}\approx 3 mol

Similarly, number of moles of oxygen will be:

n_{O}=\frac{m_{O}}{M_{O}}=\frac{48 g}{16 g/mol}=3 mol

The ratio of number of moles of carbon and oxygen will be:

C:O=n_{C}:n_{O}=3:3=1:1

The formula of compound will be CO.

Therefore, it is proved that carbon and oxygen are present in fixed ratios in both the samples.


4 0
3 years ago
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