Sulfonation of naphthalene, C10H8, results in two products. One product is kinetically favored and predominates in the beginning of the reaction. Because the reaction is reversible, eventually the kinetically slower but thermodynamically favored product predominates. Draw the structure of these two products. (The naphthalene ring is already drawn for you. Do not change the double bond configuration in the given structures.)

Answers

Answer 1
Answer:

Explanation:

The sulfonation of the naphthalene yield 2 products under different conditions:

When the reaction is carried at 80 °C, 1-naphthalenesulfonic acid is the major product because it is kinetically favoured product as arenium ion formed in the transition state corresponding to 1-naphthalenesulfonic acid is more stable due to better resonance stabilization.

When the reaction is carried at 160 °C, 2-naphthalenesulfonic acid is the major product as it is more stable than 1-naphthalenesulfonic acid because of  steric interaction of the sulfonic acid group in 1-position and the hydrogen in 8-position.

The products are shown in image below.


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Answers

D volcanic corruption

For the reaction ? NO + ? O2 → ? NO2 , what is the maximum amount of NO2 which could be formed from 16.42 mol of NO and 14.47 mol of O2? Answer in units of g. 003 1.0 points For the reaction ? C6H6 + ? O2 → ? CO2 + ? H2O 37.3 grams of C6H6 are allowed to react with 126.1 grams of O2. How much CO2 will be produced by this reaction? Answer in units of gram

Answers

Answer:

1a. The balanced equation is given below:

2NO + O2 → 2NO2

The coefficients are 2, 1, 2

1b. 755.32g of NO2

2a. The balanced equation is given below:

2C6H6 + 15O2 → 12CO2 + 6H2O

The coefficients are 2, 15, 12, 6

2b. 126.25g of CO2

Explanation:

1a. Step 1:

Equation for the reaction. This is given below:

NO + O2 → NO2

1a. Step 2:

Balancing the equation. This is illustrated below:

NO + O2 → NO2

There are 2 atoms of O on the right side and 3 atoms on the left side. It can be balance by putting 2 in front of NO and 2 in front of NO2 as shown below:

2NO + O2 → 2NO2

The equation is balanced.

The coefficients are 2, 1, 2

1b. Step 1:

Determination of the limiting reactant. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO required 1 mole of O2.

Therefore, 16.42 moles of NO will require = 16.42/2 = 8.21 moles of O2.

From the calculations made above, there are leftover for O2 as 8.21 moles out of 14.47 moles reacted. Therefore, NO is the limiting reactant and O2 is the excess reactant.

1b. Step 2:

Determination of the maximum amount of NO2 produced. This is illustrated below:

2NO + O2 → 2NO2

From the balanced equation above, 2 moles of NO produced 2 moles of NO2.

Therefore, 16.42 moles of NO will also produce 16.42 moles of NO2.

1b. Step 3:

Conversion of 16.42 moles of NO2 to grams. This is illustrated below:

Molar Mass of NO2 = 14 + (2x16) = 14 + 32 = 46g/mol

Mole of NO2 = 16.42 moles

Mass of NO2 =?

Mass = number of mole x molar Mass

Mass of NO2 = 16.42 x 46

Mass of NO2 = 755.32g

Therefore, the maximum amount of NO2 produced is 755.32g

2a. Step 1:

The equation for the reaction.

C6H6 + O2 → CO2 + H2O

2a. Step 2:

Balancing the equation:

C6H6 + O2 → CO2 + H2O

There are 6 atoms of C on the left side and 1 atom on the right side. It can be balance by 6 in front of CO2 as shown below:

C6H6 + O2 → 6CO2 + H2O

There are 6 atoms of H on the left side and 2 atoms on the right. It can be balance by putting 3 in front of H2O as shown below:

C6H6 + O2 → 6CO2 + 3H2O

There are a total of 15 atoms of O on the right side and 2 atoms on the left. It can be balance by putting 15/2 in front of O2 as shown below:

C6H6 + 15/2O2 → 6CO2 + 3H2O

Multiply through by 2 to clear the fraction.

2C6H6 + 15O2 → 12CO2 + 6H2O

Now, the equation is balanced.

The coefficients are 2, 15, 12, 6

2b. Step 1:

Determination of the mass of C6H6 and O2 that reacted from the balanced equation. This is illustrated below:

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of C6H6 = (12x6) + (6x1) = 72 + 6 = 78g/mol

Mass of C6H6 from the balanced equation = 2 x 78 = 156g

Molar Mass of O2 = 16x2 = 32g/mol

Mass of O2 from the balanced equation = 15 x 32 = 480g

2b. Step 2:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

156g of C6H6 required 480g of O2.

Therefore, 37.3g of C6H6 will require = (37.3x480)/156 = 114.77g of O2.

From the calculations made above, there are leftover for O2 as 114.77g out of 126.1g reacted. Therefore, O2 is the excess reactant and C6H6 is the limiting reactant.

2b. Step 3:

Determination of mass of CO2 produced from the balanced equation. This is illustrated belowb

2C6H6 + 15O2 → 12CO2 + 6H2O

Molar Mass of CO2 = 12 + (2x16) = 12 + 32 = 44g/mol

Mass of CO2 from the balanced equation = 12 x 44 = 528g

2b. Step 4:

Determination of the mass of CO2 produced by reacting 37.3g of C6H6 and 126.1g O2. This is illustrated below:

From the balanced equation above,

156g of C6H6 produced 528g of CO2.

Therefore, 37.3g of C6H6 will produce = (37.3x528)/156 = 126.25g of CO2

What mass of Nz will be needed to produce 31.5 grams of N2O5?4N2 + 502 --> 2N2O5
a) 158.3 grams
b) 38.64 grams
c) 4.96 grams
d) 16.34 grams

Answers

Answer: d) 16.34 grams

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}  

\text{Moles of} N_2O_5=(31.5g)/(108.01g/mol)=0.292moles

The balanced chemical reaction given is:

4N_2+5O_2\rightarrow 2N_2O_5  

According to stoichiometry :

As 2 moles of N_2O_5 are produced by= 4 moles of N_2

Thus 0.292 moles of N_2O_5 are produced by=  =(4)/(2)* 0.292=0.584moles of N_2

Mass of N_2=moles* {\text {Molar mass}}=0.584moles* 28g/mol=16.34g

Thus 16.34 g of N_2 will be needed

Learning Task 1. Read each item carefully. Choose the letter of your answer.1. Which type of soil is characterized as having the finest particles holding greater amount of water?
A. Loam
B. Clay
C. Sand
2. Which type of soil is best for planting?
A. Loam
B. Clay
C. Sand
3. How does each soil types differ?
A. Texture
B. Color
C. Both A & B
4. Which type of soil do you usually expect if the community is along the seashore?
A. Loam
B. Clay
C. Sand
5. Why is soil important to living things?
A. Forms part of the earth where animals live
B. Provides the necessary nutrients needed by plants
C. Serves as a place where people live
D. All of the above​

Answers

Answer:

1. B

2. A

3. C

4. C

5. D

Explanation:

Soil is regarded as the solid unconsolidated material of the earth crust. Soil is of three different types namely: Sandy soil, clay soil and loamy soil. These three different soil types possess different properties that distinguish them. Some of them are:

- CLAY soil is characterized as having the finest particles and can hold greater amount of water i.e. have a high water holding capacity.

- LOAMY SOIL is the best soil type for planting agricultural crops because it has the highest concentration of nutrients that suited for plant growth.

- loamy, Sandy and clay differ in how we feel when touched i.e. texture, and colour.

- SANDY soils are the kind of soils that are found in Sea shores and beaches.

- Soil is important to living things as it forms part of the earth where animals live, provides the necessary nutrients needed by plants, serves as a place where people live.

Metal vapor deposition is a process used to deposit a very thin layer of metal on a target substrate. A researcher puts a glass slide in the metal vapor deposition chamber and coats the slide with copper. After deposition, the glass slide had increased in mass by 2.26 milligrams. Approximately how many copper atoms were deposited on the glass slide

Answers

Answer:

2.14x10¹⁹ atoms of Cu were deposited

Explanation:

The increased in mass of the glass slide is due the deposition of copper.

That means the mass of copper deposited is 2.26mg = 2.26x10⁻³g Cu

To know the copper atoms we need to convert this mass to moles of Cu using molar mass of copper (63.546g/mol), and these moles are converted to atoms using Avogadro's number (6.022x10²³ atoms = 1 mole)

Moles Cu:

2.26x10⁻³g Cu * (1 mol / 63.546g) = 3.556x10⁻⁵ moles Cu

Atoms Cu:

3.556x10⁻⁵ moles Cu * (6.022x10²³ atoms / 1 mole) =

2.14x10¹⁹ atoms of Cu were deposited

A 14 wt% solution of cacl2 (110.98 g/mol) has a density of 1.126 g/ml. what is the mass (in milligrams) of a 20.0-ml solution of 14.0 wt% cacl2?

Answers

Density = mass / volume

Thus, Mass of the solution can be expressed as:

Mass of solution = Density of solution × volume of solution

Given- Density of solution = 1.126 g/ml

volume of solution = 20.0 ml

∴ Mass of 14 wt% solution of CaCl2 = (1.126 g/ml) × (20 ml)

                                                           = 22.52 g

                                                           = 22520 mg