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4 years ago

Write the mechanism of the reaction of methyl benzoate to form methyl 2-nitrobenzoate.

Chemistry

2 answers:

Galina-37 [17]4 years ago

8 0

The question is partially incorrect, because nitration of <span> methyl benzoate results in generation of methyl 3-nitrobenzoate, and not methyl 2-benzoate.

This a because of the present of ester group, which deactivated benzene ring at ortho and para position. Due to this, the electrophile (NO2+) attackes on meta position.

The detailed mechanism is attached below.</span>

mylen [45]4 years ago

3 0

Answer:

See the attached picture.

Explanation:

Hello,

Nitration of alkyl benzenes is carried out adding both nitric and sulfuric acid to the organic reactant in order to promote the inclusion of the nitro group into the ring. Nonetheless, due to the steric hindrance such inclusion is not at the second carbon but at the third one. The addition of nitric acid promotes the attaching of the nitro group and the sulfuric acid shifts the equilibrium towards the nitration is it ionizes the nitric acid to nitronium ion NO₂⁺. The mechanism is shown on the attached picture.

Best regards.

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Nitrogen dioxide is a red-brown gas responsible for the brown color of smog. A container of nitrogen dioxide that is at low pres

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Answer:

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

Explanation:

Volume of the container at low pressure and at room temperature = $V_1=3.4 L$

Number of moles in the container = $n_1$

After more addition of nitrogen gas at the same pressure and temperature.

Volume of the container after addition = $V_2=5.11 L$

Number of moles in the container after addition= $n_2=2.28\times 10^{-2} mol$

Applying Avogadro's law:

$\frac{Volume}{Moles}=constant$ (at constant pressure and temperature)

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$

$n_1=\frac{V_1\times n_2}{V_2}=\frac{3.4 L\times 2.28\times 10^{-2} mol}{5.11 L}$

$n_1=1.51\times 10^{-2} mol$

Initially $1.51\times 10^{-2} moles$ of nitrogen dioxide were in the container .

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3 years ago

Calculate the density of a solid in g/cm3 if it weighs 38.3 kg and has a volume of 0.00463 m3.

RideAnS [48]

Answer: D=8.27 g/cm³

Explanation:

Density is mass/volume. Mass is in grams and volume is in liters. In this case, the problem wants our volume to be in cm³. All we need to do is to make some conversions to convert kg/m³ to g/cm³.

$D=\frac{38.3kg}{0.00463m^3} *\frac{(1m)^3}{(100cm)^3} *\frac{1000g}{1kg}$

With this equation, the m³ and kg cancel out, and we are left with g/cm³.

D=8.27 g/cm³

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3 years ago

The observed brightness of a star depends on which factors?

dimulka [17.4K]

Answer:

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Explanation:

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3 years ago

Which postulate of Dalton's atomic theory is the result of the law of conservation of mass

creativ13 [48]

The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is: Atoms are indivisible particles, which can neither be created nor destroyed in a chemical reaction.

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3 years ago

How many grams of NH3 can be produced from 2.51 mil of N2 and excess H2 ?

salantis [7]

Answer:

85.34g of NH3

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

N2 + 3H2 —> 2NH3

Step 2:

Determination of the number of moles of NH3 produced by the reaction of 2.51 moles of N2. This is illustrated below:

From the balanced equation above,

1 mole of N2 reacted to produce 2 moles of NH3.

Therefore, 2.51 moles of N2 will react to produce = (2.51 x 2)/1 = 5.02 moles of NH3.

Therefore, 5.02 moles of NH3 is produced from the reaction.

Step 3:

Conversion of 5.02 moles of NH3 to grams. This is illustrated below:

Molar mass of NH3 = 14 + (3x1) = 17g/mol

Number of mole of NH3 = 5.02 moles

Mass of NH3 =..?

Mass = mole x molar Mass

Mass of NH3 = 5.02 x 17

Mass of NH3 = 85.34g

Therefore, 85.34g of NH3 is produced.

3 0

3 years ago