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3 years ago

Which reactant should be used to convert propene to 1,2-dichloropropane?

Chemistry

2 answers:

raketka [301]3 years ago

3 0

Propene +cl2 -> 1,2 dichloropropane

babunello [35]3 years ago

3 0

Answer:

Reactant is shown below

Explanation:

To convert propene to 1,2-dichloropropane, $Cl_{2}$ should be used in a polar aprotic slovent like $CH_{2}Cl_{2}$ .
Here, $Cl_{2}$ adds onto double bond and forms a chloronium cation followed by nucleophilic attack by a chloride ion.
Chloronium ion intermediate has transient carbocationic character, Therefore chloride ion attack at more substituted position.
Reaction mechanism has been shown below.

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The bond enthalpy of the oxygen-oxygen bond in O2 is 498 kJ/mol. Based on the enthalpy of the reaction represented above, what i

kirill115 [55]

The given question is incomplete. The complete question is as follows.

The enthalpy of formation of ozone is 142.7 kJ / mol. The bond energy of $O_{2}$ is 498 kJ / mol. What is the average O=O bond energy of the bent ozone molecule O=O=O?

Explanation:

The given reaction is as follows.

$3O_{2}(g) \rightarrows 2O_{3}$

The value of $\Delta H$ for 2 moles of ozone is $2 \times 142.7 kJ/mol$ = 285.4 kJ/mol.

So, in this reaction three O=O bonds are broken down and four O-O bonds of ozone are formed.

Expression for the enthalpy of reaction is as follows.

$\Delta H = B.E_{reactants} - B.E_{products}$

285.4 kJ/mol = $(3 \times 498) - (2 \times O_{3})$

285.4 kJ/mol = $1494 - 2 \times O_{3})$

-1208.6 kJ/mol = $-2O_{3}$

$O_{3}$ = 604.3 kJ/mol

Now, the average bond enthalpy of O-O bond in $O_{3}$ is as follows.

= $\frac{604.3}{3}$ kJ/mol

= 201.43 kJ/mol

Thus, we can conclude that for the given reaction average bond enthalpy of an oxygen-oxygen bond in $O_{3}$ is 201.43 kJ/mol.

7 0

4 years ago

What is the function of hemoglobin in the body?

aleksley [76]

Main function of haemoglobin in the body is to transport oxygen to every cell/organ of the body

Hope this helps!!

7 0

4 years ago

Read 2 more answers

Steam reforming of methane ( CH4 ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas, which is the sta

Hatshy [7]

Answer: The pressure equilibrium constant for the reaction is 1473.8

Explanation:

We are given:

Initial partial pressure of methane gas = 2.4 atm

Initial partial pressure of water vapor = 3.9 atm

Equilibrium partial pressure of hydrogen gas = 6.5 atm

The chemical equation for the reaction of methane gas and water vapor follows:

$CH_4+H_2O\rightleftharpoons CO+3H_2$

Initial: 2.4 3.9

At eqllm: 2.4-x 3.9-x x 3x

Evaluating the value of 'x':

$\Rightarrow 3x=6.5\\\\x=2.167$

So, equilibrium partial pressure of methane gas = (2.4 - x) = [2.4 - 2.167] = 0.233 atm

Equilibrium partial pressure of water vapor = (3.9 - x) = [3.9 - 2.167] = 1.733 atm

Equilibrium partial pressure of carbon monoxide gas = x = 2.167 atm

The expression of $K_p$ for above equation follows:

$K_p=\frac{p_{CO}\times (p_{H_2})^3}{p_{CH_4}\times p_{H_2O}}$

Putting values in above equation, we get:

$K_p=\frac{2.167\times (6.5)^3}{0.233\times 1.733}\\\\K_p=1473.8$

Hence, the pressure equilibrium constant for the reaction is 1473.8

5 0

4 years ago

Am i adopted? and why

vredina [299]

Answer:

because you are a living organism. you also have feeling.

6 0

3 years ago

Read 2 more answers

A Group VIIA atom combines with a Group IA atom. The bond formed between them will most likely be:

ad-work [718]

Answer: The correct answer is option B.

Explanation:

Ionic bond is formed when there is complete transfer of electrons from one element to another.

Covalent bond is formed by the sharing of electrons between two elements.

Metallic bond is formed when two metals combine together.

Electronic Configuration of Group VII-A : $ns^2np^5$

This group require 1 electron to attain stable electronic configuration.

Electronic configuration of Group I-A : $ns^1$

This group will loose 1 electron to attain stable electronic configuration.

So, there will be complete transfer of electron from Group VII-A to Group I-A and hence, will form ionic bond between them.

Hence, the correct option is B.

8 0

3 years ago