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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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Does a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together? Write the n
olganol [36]

Answer : Yes, a precipitate form when a solution of calcium chloride and a solution of mercury(I) nitrate are mixed together.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

Explanation :

In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

The given balanced ionic equation will be,

CaCl_2(aq)+2HgNO_3(aq)\rightarrow Ca(NO_3)_2(aq)+Hg_2Cl_2(s)

The ionic equation in separated aqueous solution will be,

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In this equation, Ca^{2+}\text{ and }NO_3^- are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

2Hg^{+}(aq)+2Cl^{-}(aq)\rightarrow Hg_2Cl_2(s)

7 0
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What happens to liquid when it releases enough energy
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8 0
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Read 2 more answers
ammonia gas reacts with oxygen gas, o2(g), to produce nitrogen dioxide and water. when 28.5 g of ammonia gas reacts with 83.4 g
jolli1 [7]

The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.

A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)

NH₃ + O₂ → NO₂ + H₂O

The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O

Calculate the number of moles from the reactant

  • Ammonia gas
    Molar mass N = 14 gr/mol
    Molar mass H = 1 gr/mol
    Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
    mass = 28.5 grams
    n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol
  • Oxygen gas
    Molar mass O = 16 gr/mol
    Molar mass O₂ = 16 × 2 = 32 gr/mol
    mass = 83.4 grams
    n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol
  • n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
    n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
    0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.

According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.

  • Theoretically the number moles of NO₂
    n O₂ : n NO₂ = 7 : 4
    2.61 : n NO₂ = 7 : 4
    n NO₂ = 4 x 2.61 : 7 = 1.49 mol
  • The actual number of moles NO₂
    Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
    n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol

The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.

P = (1.35 ÷ 1.49) × 100%

P = 0.9052 × 100%

P = 90.52%

Learn more about stoichiometry here: brainly.com/question/13691565

#SPJ4

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