Arrange the following chloroarenes in increasing order of their reactivity in nucleophilic substitution to form their correspond
ing phenols
1 answer:
The question is incomplete. Complete question is attached below
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Correct Answer: Option C i.e. I ~ III < IV < V < II
Reason:
During a nucleophilic subsitution reaction of chloroarenes, Cl- group is replaced by an nucleophile like OH-.
Order of reactivity, during such reactions depends on the electron density on carbon atom that is attached to Cl. Lower the electron density, greater will be the reactivity.
Among the provided chloroarenes, electron density on C atom will be minimum in case of compound II, because of presence of electron withdrawing group (-NO2) at ortho and para position. Due to this, there will be large number of resonating structures. This signifies greater electron de-localization, and hence largest reactivity for nucleophilic substitution reaction.
Followed by this, compound V will show greater reactivity, due to presence of -NO2 group at para and one of the ortho position. Compound IV will have less number of resonating structures as compared to compound II and V, hence it will display poor reactivity towards nucleophilic substitution reaction.
Finally, compound 1 and III will minimum reactivity towards nucleophilic substitution reaction, because -NO2 group present at meta position (compound III) will not participate in resonance.
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Correct Answer: Option C i.e. I ~ III < IV < V < II
Reason:
During a nucleophilic subsitution reaction of chloroarenes, Cl- group is replaced by an nucleophile like OH-.
Order of reactivity, during such reactions depends on the electron density on carbon atom that is attached to Cl. Lower the electron density, greater will be the reactivity.
Among the provided chloroarenes, electron density on C atom will be minimum in case of compound II, because of presence of electron withdrawing group (-NO2) at ortho and para position. Due to this, there will be large number of resonating structures. This signifies greater electron de-localization, and hence largest reactivity for nucleophilic substitution reaction.
Followed by this, compound V will show greater reactivity, due to presence of -NO2 group at para and one of the ortho position. Compound IV will have less number of resonating structures as compared to compound II and V, hence it will display poor reactivity towards nucleophilic substitution reaction.
Finally, compound 1 and III will minimum reactivity towards nucleophilic substitution reaction, because -NO2 group present at meta position (compound III) will not participate in resonance.
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<u>Answer:</u> The correct answer is Option b.
<u>Explanation:</u>
Reducing agents are defined as the agents which help the other substance to get reduced and itself gets oxidized. They undergo oxidation reaction.
For determination of reducing agents, we will look at the oxidation potentials of the substance. Oxidation potentials can be determined by reversing the standard reduction potentials.
For the given options:
- <u>Option a:</u>
This ion cannot be further oxidized because +1 is the most stable oxidation state of silver.
- <u>Option b:</u>
This metal can easily get oxidized to ion and the standard oxidation potential for this is 0.13 V
- <u>Option c:</u>
This metal can easily get oxidized to ion and the standard oxidation potential for this is 0.0 V
- <u>Option d:</u>
This metal can easily get oxidized to ion and the standard oxidation potential for this is -0.80 V
- <u>Option e:</u>
This ion cannot be further oxidized because +2 is the most stable oxidation state of magnesium.
By looking at the standard oxidation potential of the substances, the substance having highest positive potential will always get oxidized and will undergo oxidation reaction. Thus, considered as strong reducing agent.
From the above values, the correct answer is Option b.
<u>Answer:</u> The concentration of in the cell is
<u>Explanation:</u>
We are given:
<u>Oxidation half reaction:</u>
<u>Reduction half reaction:</u>
The substance having highest positive potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.
Here, tin will undergo reduction reaction and will get reduced.
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
where,
= electrode potential of the cell = 0.660 V
= standard electrode potential of the cell = +0.624 V
n = number of electrons exchanged = 2
= ?
Putting values in above equation, we get:
Hence, the concentration of ions is
Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y = = 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>