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Varvara68 [4.7K]
3 years ago
13

Why is it unnecessary to include an aluminum trihalide in electrophilic aromatic bromination reaction of acetanilide wih molecul

ar bromine?
Chemistry
1 answer:
Helen [10]3 years ago
4 0
Many electrophilic aromatic halogenations require the presence of an aluminum trihalide as a catalyst. We generally acetylated the amino group as protection. Now, this acetanilide can be brominated at Ortho or para position. An atom that is attached to an aromatic system usually hydrogen is replaced by an electrophile is an organic reaction which is called Electrophilic aromatic substitution. There are what you called important electrophilic aromatic substitutions they are aromatic nitration, aromatic sulfonation, aromatic halogenation and acylation and alkylating Friedel-Crafts reaction. Aromatic bromination is an electrophilic aromatic substitution (EAS) reaction, which will require benzene to act as a nucleophile to acquire an electrophile. Therefore, any directing groups that activate the ring will make it react more quickly with respect to aromatic bromination. Acetanilide is a moderately-activated ring <span>having a decent EWG.</span>
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The vapor pressure of diethyl ether (ether) is 463.57 mm Hg at 25 °C. A nonvolatile, nonelectrolyte that dissolves in diethyl et
Alexxx [7]

<u>Answer:</u> The vapor pressure of solution is 459.17 mmHg

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For testosterone:</u>

Given mass of testosterone = 7.752 g

Molar mass of testosterone = 288.4 g/mol

Putting values in equation 1, we get:

\text{Moles of testosterone}=\frac{7.752g}{288.4g/mol}=0.027mol

  • <u>For diethyl ether:</u>

Given mass of diethyl ether = 208.0 g

Molar mass of diethyl ether = 74.12 g/mol

Putting values in equation 1, we get:

\text{Moles of diethyl ether}=\frac{208.0g}{74.12g/mol}=2.81mol

Mole fraction of a substance is calculated by using the equation:

\chi_A=\frac{n_A}{n_A+n_B}

\chi_{\text{testosterone}}=\frac{n_{\text{testosterone}}}{n_{\text{testosterone}}+n_{\text{diethyl ether}}}

\chi_{\text{testosterone}}=\frac{0.027}{0.027+2.81}\\\\\chi_{\text{testosterone}}=0.0095

The formula for relative lowering of vapor pressure will be:

\frac{p^o-p_s}{p^o}=i\times \chi_{\text{solute}}

where,

p^o = vapor pressure of solvent (diethyl ether) = 463.57 mmHg

p^s = vapor pressure of the solution = ?

i = Van't Hoff factor = 1 (for non electrolytes)

\chi_{\text{solute}} = mole fraction of solute (testosterone) = 0.0095

Putting values in above equation, we get:

\frac{463.57-p^s}{463.57}=1\times 0.0095\\\\p^s=459.17mmHg

Hence, the vapor pressure of solution is 459.17 mmHg

7 0
3 years ago
Prform the following operation and express the answer in scientific notation 6.0 x 10^3-2.310^3
Marrrta [24]

The value 6.0 x 10^3- 2.3 × 10^3 in scientific notation is 3.7 × 10^3.

<h3> What is scientific notation?</h3>

Scientific notation is a way to write very large or very small numbers so that they are easier to read and work with.

You express a number as the product of a number greater than or equal to 1 but less than 10 and an integral power of 10 .

<h3>Why it is used? </h3>

There are two reasons to use scientific notation.

  • The first is to reveal honest uncertainty in experimental measurements.
  • The second is to express very large or very small numbers so they are easier to read.

Given,

= 6.0 x 10^3- 2.3 × 10^3

= (6.0 - 2.3) × 10^3

= 3.7 × 10^3

Thus, we find that the value 6.0 x 10^3- 2.3 × 10^3 in scientific notation is 3.7 × 10^3.

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3 0
2 years ago
Chlorofluorocarbons (CFCs) are no longer used as refrigerants because they destroy the ozone layer. Trichlorofluoromethane (CCl₃
Daniel [21]

Answer:

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

Explanation:

Entropy :It is defined as amount of energy which is unable to do work or the measurement of randomness or disorderedness in a system.

S=\frac{Q}{T(Kelvins)}

Molar heat of molar vaporization of Trichlorofluoromethane = 24.8 kJ/mol

Temperature at which Trichlorofluoromethan boils , T= 296.95 K

The molar entropy of the evaporation of Trichlorofluoromethan :

=\frac{24.8 kJ/mol}{296.95 K}=0.083516 kJ/mol K = 83.516 J/molK

The molar entropy of the evaporation of Trichlorofluoromethan is 83.516 J/molK.

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How many places are there for electrons in the third shell of an atom?
Dafna1 [17]
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Which of the following statements about C3 carbon fixation is true? a. C3 carbon fixation is an adaptation for plants exposed to
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Answer: The answer is B

Explanation:

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