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timama [110]
3 years ago
5

Using the following reaction:

Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0

Answer:

Mn is the oxidizing agent.

N is the reducing agent.

Explanation:

Hello!

In this case, according to the undergoing chemical reaction, it is seen that the manganese in KMnO4 has an oxidation state of 7+, in MnSO4 of 2+ and nitrogen in KNO2 is 3+ and in KNO3 is 5+; thus we have the following half-reactions:

Mn^{7+}+5e^-\rightarrow Mn^{2+}\\\\N^{3+}\rightarrow N^{5+}+2e^-

Thus, since manganese is undergoing a decrease in the oxidation state, we infer it is the oxidizing agent whereas nitrogen, undergoing an increase in the oxidation state is the reducing agent.

Best regards!

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Read 2 more answers
The product of the nitration reaction will have a nitro group at which position with respect to the methyl group? Group of answe
zimovet [89]

Answer:

Mostly Para

Explanation:

First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).

Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.

Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.

Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.

4 0
3 years ago
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