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iVinArrow [24]
3 years ago
13

Draw the structure of the product of each step in the following three-step synthesis. Show the formal charges, if applicable. As

a start, the benzene ring is drawn for you in each product. Although the first step produces a mixture of isomers, the para isomer is isolated as the sole product of interest and used in the second step. Give only the major product for the second and third steps.

Chemistry
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

Second step: 4-bromo-1-methyl-2-nitrobenzene.

Third step: 1.5-dibromo-2-methyl-3-nitrobenzene.

Explanation:

To solve this exercise I will use the concepts of electrophilic substitution. In these reactions, a functional group is displaced by an electrophile. In the attached image are the two main products.

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Formula: H2O Formula Weight: 18.02 CAS No.: 7732-18-5 Density: 1.000 g/mL at 3.98 °C(lit.)

Explanation:

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4 years ago
A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change
MrRa [10]

Answer:

-54 kJ/mol

Explanation:

Given that:

A student mixed 50 ml of 1.0 M HCl and 50 ml of 1.0 M NaOH in a coffee cup calorimeter and calculated the molar enthalpy change of the acid-base neutralization reaction to be –54 kJ/mol

i.e

50 ml of 1.0 M HCl +  50 ml of 1.0 M NaOH -----> -54 kJ/mol

If he repeat the same experiment with :

100 ml of 1.0 M HCl + 100 ml of 1.0 M NaOH. ------> ????

From The experiment; the molar enthalpy of change of the acid-base neutralization reaction will be -54 kJ/mol

This is because : The second reaction requires 50 ml in order to neutralize the reaction, then the remaining 50 ml will be excess, Hence, there is no change in the enthalpy of the reaction.

Similarly; we can assume that :

In the first reaction;  P moles of  is used to liberate Q kJ heat ; then  the change in molar enthalpy will be Q/P (kJ/mol).

SO; when he used 100 ml ;

then the amount of moles used is double, likewise the heat liberated will be doubled ;

So;

2P moles is used to liberate 2Q kJ heat ;

2P/2Q = Q/P ( kJ/mol) = -54 kJ/mol

8 0
3 years ago
A city planner needs a new method for cleaning a town’s water supply. He hires a scientist to research the issue. Because the sc
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3 years ago
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In nature, one common strategy to make thermodynamically unfavorable reactions proceed is to couple them chemically to reactions
maks197457 [2]

Answer:

31.3 kJ/mol is the free energy, ΔG, for the overall reaction, A⇌D.

Explanation:

A⇌B, \Delta G_1 = 11.9 kJ/mol  ...[1]

B⇌C, \Delta G_2= -26.7 kJ/mol  ...[2]

C⇌D, \Delta G_3= 7.30 kJ/mol  ...[3]

A⇌D, \Delta G_4 = ?...[4]

[4] = [1] - [2] - [3]  (Using Hess's law)

\Delta G_4=\Delta G_1-(\Delta G_2+\Delta G_3)

=11.9 kJ/mol - (-26.7 kJ/mol+7.30 kJ/mol)

=\Delta G_4=31.3 kJ/mol

31.3 kJ/mol is the free energy, ΔG, for the overall reaction, A⇌D.

7 0
4 years ago
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