Toluene with potassium permanganate in an alkaline environment. Organic Oxidation Reactions

Toluene is a colorless liquid with a specific odor. Toluene is lighter than water and does not dissolve in it, but it is easily soluble in organic solvents - alcohol, ether, acetone. Toluene is a good solvent for many organic substances. It burns with a smoky flame due to the high carbon content of its molecule.

The physical properties of toluene are presented in the table.

Table. Physical properties of toluene.

Chemical properties of toluene

I. The oxidation reaction.

1. Burning (smoking flame):

2C 6 H 5 CH 3 + 16O 2 t → 14CO 2 + 8H 2 O + Q

2. Toluene is oxidized by potassium permanganate (potassium permanganate is discolored):

A) in an acidic environment to benzoic acid

When potassium permanganate and other strong oxidizing agents act on toluene, the side chains are oxidized. No matter how complex the substituent chain, it breaks down, with the exception of the a-carbon atom, which is oxidized to the carboxyl group. Toluene gives benzoic acid:

B) in neutral and slightly alkaline to salts of benzoic acid

C 6 H 5 -CH 3 + 2KMnO 4 → C 6 H 5 COOC + KOH + 2MnO 2 + H 2 O

II. ADDITION REACTIONS

1. Halogenation

FROM 6 N 5 CH 3 + Vg 2  FROM 6 N 5 CH 2 Vg + NVg

C 6 H 5 CH 3 + Cl 2 h ν → C 6 H 5 CH 2 Cl + HCl

2. Hydrogenation

C 6 H 5 CH 3 + 3H 2 t , Pt or Ni → C 6 H 11 CH 3 (methylcyclohexane)

III. SUBSTITUTION REACTIONS- ionic mechanism (lighter than alkanes)

1. Halogenation -

In chemical properties, alkyl radicals are similar to alkanes. The hydrogen atoms in them are replaced by halogen by the free-radical mechanism. Therefore, in the absence of a catalyst during heating or UV irradiation, a radical substitution reaction occurs in 4 side chains. The influence of the benzene ring on alkyl substituents leads to the fact that the hydrogen atom is always replaced at the carbon atom directly bonded to the benzene ring (a-carbon atom).

C 6 H 5 -CH 3 + Cl 2 h ν → C 6 H 5 -CH 2 -Cl + HCl

in the presence of a catalyst

C 6 H 5 -CH 3 + Cl 2 AlCl 3 → (mixture of orth, a pair of derivatives) + HCl

2. Nitration (with nitric acid)

C 6 H 5 -CH 3 + 3HO-NO 2 t , H 2 SO 4 → CH 3 -C 6 H 2 (NO 2) 3 + 3H 2 O

2,4,6-trinitrotoluene (tol, trotyl)

The use of toluene.

Toluene C 6 H 5 –CH 3 is a solvent used in the manufacture of dyes, drugs, and explosives (TNT (tol), or 2,4,6-trinitrotoluene TNT).

2.2. Being in nature

Toluene was first obtained by distillation of pine resin in 1835. Peltier P., later it was isolated from toluan balsam (resin from the bark of the tree Myraxylo, growing in Central America). This substance was named after the city of Tolu (Colombia).

2.3. Anthropogenic sources of toluene entry into the biosphere.

The main sources are coal distillation and a number of petrochemical processes, in particular catalytic reforming, crude oil distillation and alkylation of lower aromatic hydrocarbons. Polycyclic hydrocarbons are present in urban smoke.

The source of air pollution may be the metallurgical industry, motor vehicles.

The background level of toluene in the atmosphere is 0.75 μg / m 3 (0,00075 mg / m 3).

The main sources of toluene release into the environment are the chemical production of explosives, epoxies, varnishes and paints, etc.

This material can be difficult to master with self-study, due to the large amount of information, many nuances, all kinds of NO and IF. Read carefully!

What exactly will it be about?

In addition to complete oxidation (combustion), some classes of organic compounds are characterized by incomplete oxidation reactions, while they turn into other classes.

There are specific oxidizing agents for each class: CuO (for alcohols), Cu (OH) 2 and OH (for aldehydes) and others.

But there are two classical oxidizing agents, which, so to speak, are universal for many classes.

This is potassium permanganate - KMnO 4. And the potassium dichromate (dichromate) is K 2 Cr 2 O 7. These substances are strong oxidizing agents due to manganese in the oxidation state of +7, and chromium in the oxidation state of +6, respectively.

Reactions with these oxidizing agents are quite common, but nowhere is there a complete guide on how to choose the products of such reactions.

In practice, a lot of factors affect the course of the reaction (temperature, environment, concentration of reagents, etc.). Often a mixture of products is obtained. Therefore, to predict the product that is formed is almost impossible.

But for the USE this is not good: you can’t write "maybe either this, or like this, or otherwise, or a mixture of products." They need specifics.

The compilers of the tasks invested a certain logic, a certain principle by which a certain product should be written. Unfortunately, they did not share with anyone.

This question in most manuals is rather slippery aside: two or three reactions are given as an example.

I present in this article, what can be called the results of a study-analysis of the tasks of the exam. The logic and principles of the preparation of oxidation reactions with permanganate and dichromate are solved quite accurately (in accordance with the Unified State Examination standards). First things first.

Determination of oxidation state.

The first, when dealing with redox reactions, there is always an oxidizing agent and a reducing agent.

The oxidizing agent is manganese in permanganate or chromium in dichromate, and the reducing agent is atoms in the organic (namely, carbon atoms).

It is not enough to identify products, the reaction should be equalized. For equalization, the electronic balance method is traditionally used. To apply this method, it is necessary to determine the oxidation state of reducing agents and oxidizing agents before and after the reaction.

Inorganic substances, we are able to oxidize from the 9th grade:

But in organics, probably in the 9th grade was not determined. Therefore, before learning to write OVR in organic chemistry, you need to learn how to determine the degree of oxidation of carbon in organic substances. This is done in a slightly different way, other than in inorganic chemistry.

Carbon has a maximum oxidation state of +4, a minimum of -4. And he can exhibit any degree of oxidation of this gap: -4, -3, -2, -1, 0, +1, +2, +3, +4.

First you need to remember what is the degree of oxidation.

The oxidation state is a conditional charge arising on an atom, under the assumption that electron pairs are shifted completely toward a more electronegative atom.

Therefore, the oxidation state is determined by the number of displaced electron pairs: if it shifts to a given atom, then it acquires an excess minus (-) charge, if from an atom, then it acquires an excess plus (+) charge. In principle, this is the whole theory that you need to know to determine the degree of oxidation of a carbon atom.

To determine the degree of oxidation of a specific carbon atom in a compound, we need to consider EVERY bond and look in which direction the electron pair will shift and what excess charge (+ or -) will arise from this on the carbon atom.

Let's analyze specific examples:

Carbon three bonds with hydrogen. Carbon and hydrogen - who is more electronegative? Carbon, therefore, along these three bonds, the electron pair will shift towards carbon. Carbon takes away from each hydrogen one negative charge: it turns out -3

Fourth bond with chlorine. Carbon and chlorine - who is more electronegative? Chlorine, therefore, due to this connection, the electron pair will shift towards chlorine. Carbon has one positive charge +1.

Then, you just need to add: -3 + 1 \u003d -2. The oxidation state of this carbon atom is -2.

We determine the oxidation state of each carbon atom:

Carbon has three bonds with hydrogen. Carbon and hydrogen - who is more electronegative? Carbon, therefore, along these three bonds, the electron pair will shift towards carbon. Carbon takes away from each hydrogen one negative charge: it turns out -3

And another bond with another carbon. Carbon and other carbon - their electronegativity is equal, therefore, the displacement of the electron pair does not occur (the bond is not polar).

This atom has two bonds with one oxygen atom, and another bond with another oxygen atom (as part of the OH group). More electronegative oxygen atoms in three bonds pull the electron pair on carbon, the charge +3 appears on carbon.

By the fourth bond, carbon is bonded to another carbon, as we have already said, the electron pair does not shift through this bond.

By two bonds, carbon is bonded to hydrogen atoms. Carbon, as more electronegative, draws one pair of electrons for each bond with hydrogen, acquires a charge of -2.

The double bond of carbon is bonded to an oxygen atom. More electronegative oxygen pulls one electron pair over each bond. Together, it turns out that carbon pulls two electron pairs. Carbon gains a charge of +2.

Together it turns out +2 -2 \u003d 0.

Let's determine the oxidation state of this carbon atom:

Triple bond with more electronegative nitrogen - gives carbon a charge of +3, due to carbon bond, the electron pair does not shift.

Permanganate oxidation.

What will happen to permanaganate?

The redox reaction with permanganate can occur in different environments (neutral, alkaline, acidic). And it depends on the environment how the reaction will proceed and what products are formed in this way.

Therefore, it can go in three directions:

Permanganate, being an oxidizing agent, is restored. Here are the products of its recovery:

Acidic environment.

The medium is acidified with sulfuric acid (H 2 SO 4). Manganese is reduced to oxidation state +2. And recovery products will be:

KMnO 4 + H 2 SO 4 → MnSO 4 + K 2 SO 4 + H 2 O

Alkaline environment.

A fairly concentrated alkali (KOH) is added to create an alkaline environment. Manganese is reduced to oxidation state +6. Recovery Products

KMnO 4 + KOH → K 2 MnO 4 + H 2 O

Neutral environment (and slightly alkaline).

In a neutral environment, in addition to permanganate, water also enters the reaction (which we write on the left side of the equation), manganese will be reduced to +4 (MnO 2), the reduction products will be:

KMnO 4 + H 2 O → MnO 2 + KOH

And in a slightly alkaline environment (in the presence of a low concentration KOH solution):

KMnO 4 + KOH → MnO 2 + H 2 O

What will happen to organics?

The first thing to learn - it all starts with alcohol! This is the initial stage of oxidation. The carbon to which the hydroxyl group is attached is oxidized.

During oxidation, a carbon atom “acquires” a bond with oxygen. Therefore, when the oxidation reaction scheme is written, [O] is written above the arrow:

Primary alcohol oxidized first to aldehyde, then to carboxylic acid:

Oxidation secondary alcohol breaks off in the second stage. Since carbon is in the middle, a ketone is formed, not an aldehyde (the carbon atom in the ketone group cannot physically form a bond with the hydroxyl group):

Ketones, tertiary alcohols and carboxylic acids no longer oxidize:

The oxidation process is stepwise - as long as there is room for oxidation and there are all conditions for this, the reaction is ongoing. Everything ends with a product that under these conditions does not oxidize: tertiary alcohol, ketone or acid.

It is worth noting the stages of methanol oxidation. First, it oxidizes to the corresponding aldehyde, then to the corresponding acid:

A feature of this product (formic acid) is that the carbon in the carboxyl group is bound to hydrogen, and if you look closely, you can see that this is nothing more than an aldehyde group:

And the aldehyde group, as we found out earlier, is oxidized further to the carboxyl group:

Did you recognize the substance obtained? Its gross formula is H 2 CO 3. This is carbonic acid, which breaks down into carbon dioxide and water:

H 2 CO 3 → H 2 O + CO 2

Therefore, methanol, formic aldehyde and formic acid (due to the aldehyde group) are oxidized to carbon dioxide.

Mild oxidation.

Mild oxidation is oxidation without strong heating in a neutral or slightly alkaline environment (0 is written over the reaction ° or 20 °) .

It is important to remember that alcohols do not oxidize under mild conditions. Therefore, if they form, then oxidation stops on them. What substances will undergo a mild oxidation reaction?

Containing a double bond C \u003d C (Wagner reaction).

In this case, the π-bond breaks and “sits down” to the vacant bonds on the hydroxyl group. It turns out dihydric alcohol:

We write the reaction of mild oxidation of ethylene (ethene). We write down the starting materials and predict the products. At the same time, we are not writing H 2 O and KOH: they may appear both on the right side of the equation and on the left. And immediately determine the degree of oxidation of the substances involved in the OVR:

We compose an electronic balance (we mean that there are two or two carbon atoms of a reducing agent, they are oxidized separately):

Let's put the coefficients:

In the end, the missing products (H 2 O and KOH) must be added. There is not enough potassium on the right - then the alkali will be on the right. We put the coefficient in front of her. There is not enough hydrogen on the left, which means water on the left. We put before it the coefficient:

Let's do the same with propylene (propene):

Often slip cycloalkene. Let him not confuse you. This is a common double bond hydrocarbon:

Wherever this double bond is, the oxidation will go the same:

Aldehyde-containing.

The aldehyde group is more reactive (reacts more easily) than alcohol. Therefore, the aldehyde will be oxidized. To acid:

Consider the example of acetaldehyde (ethanal). We write down the reagents and products and arrange the degrees of oxidation. We draw up a balance and put the coefficients in front of the reducing agent and oxidizing agent:

In a neutral environment and slightly alkaline course of the reaction will be slightly different.

In a neutral environment, as we recall, we write water on the left side of the equation, and alkali on the right side of the equation (formed during the reaction):

At the same time, acid and alkali are found in the same mixture. There is a neutralization.

They cannot exist side by side and react, salt is formed:

Moreover, if we look at the coefficients in the equation, we will understand that acids are 3 moles and alkali 2 moles. 2 moles of alkali can neutralize only 2 moles of acid (2 moles of salt is formed). And one mole of acid remains. Therefore, the final equation will be like this:

In a slightly alkaline environment, alkali is in excess - it is added before the reaction, so all acid is neutralized:

A similar situation occurs during the oxidation of methanal. He, as we recall, is oxidized to carbon dioxide:

It must be borne in mind that carbon monoxide (IV) CO 2 is acidic. And it will react with alkali. And since carbonic acid is dibasic, both acidic and medium salts can form. It depends on the ratio between alkali and carbon dioxide:

If alkali refers to carbon dioxide as 2: 1then there will be an average salt:

Or alkali can be much larger (more than twice). If it is more than twice, then the alkali residue will remain:

3KOH + CO 2 → K 2 CO 3 + H 2 O + KOH

This will occur in an alkaline environment (where there is an excess of alkali, since it is added to the reaction mixture before the reaction) or in a neutral environment, when a lot of alkali is formed.

But if the alkali refers to carbon dioxide as 1: 1then there will be acid salt:

KOH + CO 2 → KHCO 3

If there is more carbon dioxide than necessary, then it remains in excess:

KOH + 2CO 2 → KHCO 3 + CO 2

This will happen in a neutral environment if little alkali is formed.

We write the starting materials, products, balance, put down the degree of oxidation in front of the oxidizing agent, reducing agent and products that are formed from them:

In a neutral environment, an alkali (4KOH) will form on the right:

Now we need to understand what will be formed by the interaction of three moles of CO 2 and four moles of alkali.

3CO 2 + 4KOH → 3KHCO 3 + KOH

KHCO 3 + KOH → K 2 CO 3 + H 2 O

Therefore, it turns out like this:

3CO 2 + 4KOH → 2KHCO 3 + K 2 CO 3 + H 2 O

Therefore, on the right side of the equation, we write two moles of bicarbonate and one mole of carbonate:

And in a slightly alkaline environment there are no such troubles: due to the fact that there is an excess of alkali, an average salt will form:

The same will happen with the oxidation of oxalic acid aldehyde:

As in the previous example, a dibasic acid is formed, and according to the equation 4 moles of alkali should be obtained (since 4 moles of permanganate).

In a neutral environment, again, all the alkali is not enough to completely neutralize all the acid.

Three moles of alkali goes to the formation of acid salt, one mole of alkali remains:

3HOOC – COOH + 4KOH → 3KOOC – COOH + KOH

And this one mole of alkali goes into interaction with one mole of acid salt:

KOOC – COOH + KOH → KOOC – COOK + H 2 O

It turns out like this:

3HOOC – COOH + 4KOH → 2KOOC – COOH + KOOC – COOK + H 2 O

The final equation:

In a slightly alkaline medium, an average salt is formed due to an excess of alkali:

Containing triple bondC≡ C.

Remember what happened with the mild oxidation of double bond compounds? If you do not remember, then scroll back - remember.

The π bond breaks, attaches to the carbon atoms at the hydroxyl group. Here is the same principle. It is only worth remembering that in a triple bond there are two π-bonds. First, this occurs in the first π-bond:

Then through another π-connection:

A structure in which one carbon atom has two hydroxyl groups is extremely unstable. When something is not stable in chemistry, it strives for something to “fall off”. Water falls off, like this:

It turns out the carbonyl group.

Consider the following examples:

Ethine (acetylene). Consider the stages of oxidation of this substance:

Cleavage of water:

As in the previous example, in one reaction mixture, acid and alkali. Neutralization occurs - salt is formed. As can be seen from the coefficient before alkali permanganate, there will be 8 moles, that is, it is enough to neutralize the acid. The final equation:

Consider the oxidation of butin-2:

Cleavage of water:

Here acid does not form, so no fooling over neutralization.

Reaction equation:

These differences (between the oxidation of carbon at the edge and in the middle of the chain) are clearly demonstrated by the example of pentin:

Cleavage of water:

It turns out a substance of interesting structure:

The aldehyde group continues to oxidize:

We write the starting materials, products, determine the degree of oxidation, balance, put down the coefficients in front of the oxidizing agent and reducing agent:

Alkalis should form 2 moles (since the coefficient before permanganate is 2), therefore, all acid is neutralized:

Hard oxidation.

Hard oxidation is oxidation in sour, strongly alkaline environment. And also, in neutral (or slightly alkaline), but when heated.

In an acidic environment, they are also sometimes heated. But in order for hard oxidation not to proceed in an acidic environment, heating is a prerequisite.

What substances will be subjected to severe oxidation? (First, we will analyze it only in an acidic environment - and then we will supplement it with the nuances that arise during oxidation in a strongly alkaline and neutral or slightly alkaline (when heated) medium).

With severe oxidation, the process goes to the maximum. While there is something to oxidize, oxidation is on.

Alcohols. Aldehydes.

Consider the oxidation of ethanol. It is gradually oxidized to acid:

We write the equation. We write down the starting materials, the OVR products, put down the oxidation state, and draw up a balance. We equalize the reaction:

If the reaction is carried out at the boiling temperature of the aldehyde, when it is formed, it will evaporate (fly away) from the reaction mixture, not having time to oxidize further. The same effect can be achieved under very gentle conditions (weak heating). In this case, we write aldehyde as a product:

Consider the oxidation of secondary alcohol by the example of propanol-2. As already mentioned, oxidation terminates in the second stage (the formation of a carbonyl compound). Since a ketone is formed that does not oxidize. Reaction equation:

We will consider the oxidation of aldehydes based on ethanal. It also oxidizes to acid:

Reaction equation:

Methanal and methanol, as mentioned earlier, are oxidized to carbon dioxide:

Metanal:

Multiple Links.

In this case, the circuit breaks in a multiple connection. And the atoms that formed it undergo oxidation (acquire a bond with oxygen). Oxidized as possible.

When a double bond is broken, carbonyl compounds are formed from the fragments (in the scheme below: from one fragment — aldehyde, from the other — ketone)

We analyze the oxidation of pentene-2:

Oxidation of "scraps":

It turns out that two acids are formed. We write the starting materials and products. We determine the oxidation states of the atoms that change it, make up the balance, balance the reaction:

When compiling the electronic balance, we mean that there are two or two carbon atoms of the reducing agent; they are oxidized separately:

Acid will not always form. Let us examine, for example, the oxidation of 2-methylbutene:

Reaction equation:

Absolutely the same principle for the oxidation of compounds with a triple bond (only oxidation occurs immediately with the formation of acid, without intermediate formation of aldehyde):

Reaction equation:

When the multiple bond is located exactly in the middle, it turns out not two products, but one. Since the "scraps" are the same and they are oxidized to the same products:

Reaction equation:

Double Crowned Acid.

There is one acid in which carboxyl groups (crowns) are connected to each other:

This is oxalic acid. Two crowns nearby are difficult to get along. It is of course stable under normal conditions. But due to the fact that in it two carboxyl groups are connected to each other, it is less stable than other carboxylic acids.

And therefore, under particularly harsh conditions, it can be oxidized. There is a break in the connection between the "two crowns":

Reaction equation:

Benzene homologues (and their derivatives).

Benzene itself does not oxidize, due to the fact that aromaticity makes this structure very stable.

But his homologs are oxidized. At the same time, a circuit break occurs, the main thing is to know exactly where. Some principles apply:

The benzene ring itself does not collapse, and remains intact until the end, bond breaking occurs in the radical.
An atom is oxidized that is directly bonded to the benzene ring. If after it the carbon chain in the radical continues, then the gap will be after it.

Let us examine the oxidation of methylbenzene. There, one carbon atom in the radical is oxidized:

Reaction equation:

Let's analyze the oxidation of isobutylbenzene:

Reaction equation:

We analyze the oxidation of sec-butylbenzene:

Reaction equation:

When oxidizing benzene homologs (and derivatives of homologs) with several radicals, two to three or more basic aromatic acids are formed. For example, the oxidation of 1,2-dimethylbenzene:

Derivatives of the benzene homologs (in which the benzene ring has non-hydrocarbon radicals) are oxidized in the same way. Another functional group on the benzene ring does not interfere:

Subtotal. The algorithm "how to record the reaction of hard oxidation with permanganate in an acidic environment":

Record the starting materials (organics + KMnO 4 + H 2 SO 4).
Write down the products of organic oxidation (compounds containing alcohol, aldehyde groups, multiple bonds, and also benzene homologs will be oxidized).
Record the reduction product of permanganate (MnSO 4 + K 2 SO 4 + H 2 O).
Determine the degree of oxidation in participants of the RIA. Make a balance. Put the coefficients of the oxidizing agent and reducing agent, as well as the substances that are formed from them.
Then it is recommended to calculate how many sulfate anions are on the right side of the equation, in accordance with this, put the coefficient in front of sulfuric acid on the left.
At the end, put the coefficient in front of the water.

Hard oxidation in a highly alkaline environment and a neutral or slightly alkaline (when heated) environment.

These reactions are much less common. We can say that such reactions are exotic. And as befits any exotic reactions, these turned out to be the most controversial.

Rigid oxidation is also tough in Africa, so organics are oxidized in the same way as in an acidic environment.

We will not analyze the reactions for each class separately, since the general principle has already been stated earlier. We will analyze only the nuances.

Highly alkaline environment :

In a highly alkaline environment, permanganate is reduced to an oxidation state of +6 (potassium manganate):

KMnO 4 + KOH → K 2 MnO 4.

In a highly alkaline environment, alkali is always in excess, therefore complete neutralization will take place: if carbon dioxide is formed, there will be carbonate, if acid is formed, there will be salt (if polybasic acid - medium salt).

For example, the oxidation of propene:

Ethylbenzene Oxidation:

Slightly alkaline or neutral medium when heated :

It is also necessary to always consider the possibility of neutralization.

If oxidation takes place in a neutral environment and an acid compound (acid or carbon dioxide) is formed, then the alkali that forms will neutralize this acid compound. But alkali is not always enough to completely neutralize the acid.

During the oxidation of aldehydes, for example, it is not enough (oxidation will proceed in the same way as under mild conditions - the temperature will simply accelerate the reaction). Therefore, both salt and acid are formed (remaining roughly speaking in excess).

We discussed this when we examined the mild oxidation of aldehydes.

Therefore, if you form an acid in a neutral environment, you need to carefully see if it is enough to neutralize all acid. Particular attention should be paid to the neutralization of polybasic acids.

In a slightly alkaline environment, due to a sufficient amount of alkali, only medium salts are formed, since there is an excess of alkali.

As a rule, alkali during oxidation in a neutral environment is quite enough. And the reaction equation is that in a neutral, that in a slightly alkaline environment will be the same.

For example, we analyze the oxidation of ethylbenzene:

Alkali is enough to completely neutralize the acidic compounds obtained, even the excess will remain:

Consumes 3 moles of alkali - 1 remains.

The final equation:

This reaction in the neutral and slightly alkaline environment will proceed the same way (in the weakly alkaline medium there is no alkali on the left, but this does not mean that it is not there, it just does not enter the reaction).

Redox reactions involving potassium dichromate (bichromate).

Dichromate does not have such a wide variety of organic oxidation reactions in the Unified State Examination.

Bichromate oxidation is usually carried out only in an acidic environment. In this case, chrome is restored to +3. Recovery Products:

The oxidation will be tough. The reaction will be very similar to permanganate oxidation. The same substances that are oxidized by permanganate in an acidic environment will be oxidized, the same products will be formed.

Let's analyze some reactions.

Consider the oxidation of alcohol. If the oxidation is carried out at the boiling temperature of the aldehyde, then it will leave their reaction mixture without being subjected to oxidation:

Otherwise, the alcohol can be directly oxidized to acid.

The aldehyde obtained in the previous reaction can be “caught” and made to oxidize to acid:

Oxidation of cyclohexanol. Cyclohexanol is a secondary alcohol, so ketone is formed:

If it is difficult to determine the degree of oxidation of carbon atoms using this formula, you can write in a draft:

Reaction equation:

Consider the oxidation of cyclopentene.

The double bond breaks (the cycle opens), the atoms that formed it are oxidized to the maximum (in this case, to the carboxyl group):

Some features of oxidation in the exam, with which we do not quite agree.

Those “rules”, principles and reactions that will be discussed in this section, we believe are not entirely correct. They contradict not only the real state of affairs (chemistry as a science), but also the internal logic of the school curriculum and the USE in particular.

But nevertheless, we are forced to give this material in the form that the exam requires.

It will be a question of HARD oxidation.

Remember how benzene homologs and their derivatives are oxidized under harsh conditions? Radicals all break off - carboxyl groups are formed. Scraps undergo oxidation already "on their own":

So, if suddenly a radical appears hydroxyl group, or a multiple bond, you need to forget that there is a benzene ring. The reaction will go ONLY on this functional group (or multiple link).

The functional group and multiple bonds are more important than the benzene ring.

Let's analyze the oxidation of each substance:

The first substance:

It is necessary not to pay attention to the fact that there is a benzene ring. From the point of view of the exam, this is just secondary alcohol. Secondary alcohols are oxidized to ketones, and ketones are not further oxidized:

Let this substance be oxidized with bichromate in us:

The second substance:

This substance is oxidized, simply as a double bond compound (we do not pay attention to the benzene ring):

Let it oxidize in neutral permanganate when heated:

The alkali formed is enough to completely neutralize carbon dioxide:

2KOH + CO 2 → K 2 CO 3 + H 2 O

The final equation:

Oxidation of the third substance:

Let the oxidation proceed with potassium permanganate in an acidic environment:

Oxidation of the fourth substance:

Let it oxidize in a highly alkaline environment. The reaction equation will be:

And finally, that's how vinylbenzene is oxidized:

But it oxidizes to benzoic acid, it must be borne in mind that, according to the logic of the Unified State Examination, it is oxidized not because it is a benzene derivative. But because it contains a double bond.

Conclusion.

This is all you need to know about redox reactions involving permanganate and dichromate in organic matter.

Do not be surprised if, for some of the points outlined in this article, you hear for the first time. As already mentioned, this topic is very extensive and controversial. And despite this, for some reason, very little attention is paid to it.

As you may have seen, two or three reactions cannot explain all the laws of these reactions. An integrated approach and a detailed explanation of all points are needed here. Unfortunately, in textbooks and on Internet resources, the topic is not fully disclosed, or not disclosed at all.

I tried to eliminate these shortcomings and shortcomings and consider this topic in its entirety, and not in part. I hope I succeeded.

Thank you for your attention, all the best to you! Good luck in mastering chemical science and passing exams!

Equalization of redox reactions with the participation of organic substances by the electronic balance method.

Organic oxidation reactions are often found in basic chemistry courses. Moreover, their recording is usually presented in the form of simple schemes, some of which give only a general idea about the transformations of substances of different classes into each other, not taking into account the specific conditions of the process (for example, the reaction of the medium) that affect the composition of the reaction products. Meanwhile, the requirements of the exam in chemistry in part C are such that it becomes necessary to write down the reaction equation with a certain set of coefficients. This paper provides recommendations on the methodology for compiling such equations.

Two methods are used to describe redox reactions: the electron-ion equation method and the electronic balance method. Without dwelling on the first, we note that the electronic balance method is studied in the chemistry course of the main school and therefore is quite applicable to continue the study of the subject.

The equations of electronic balance primarily describe the processes of oxidation and reduction of atoms. In addition, special factors indicate the coefficients in front of the formulas of substances containing atoms that participated in the processes of oxidation and reduction. This, in turn, allows you to find the remaining coefficients.

Example 1. Oxidation of toluene with potassium permanganate in an acidic environment.

C 6 H 5 -CH 3 + KMnO 4 + H 2 SO 4 \u003d ...

It is known that the side methyl radicals of arenes are usually oxidized to carboxyl, therefore, in this case, benzoic acid is formed. Potassium permanganate in an acidic medium is reduced to doubly charged manganese cations. Given the presence of a sulfuric acid environment, the products will be manganese (II) sulfate and potassium sulfate. In addition, water is formed during oxidation in an acidic environment. Now the reaction scheme looks like this:

C 6 H 5 -CH 3 + KMnO 4 + H 2 SO 4 \u003d C 6 H 5 COOH + MnSO 4 + K 2 SO 4 + H 2 O

The diagram shows that the state of the carbon atom in the methyl radical, as well as the manganese atom, changes. The degree of oxidation of manganese is determined by the general rules of calculation: in potassium permanganate +7, in manganese sulfate +2. The oxidation states of the carbon atom can be easily determined from the structural formulas of the methyl radical and carboxyl. For this, it is necessary to consider the shift of the electron density on the basis that, due to electronegativity, carbon occupies an intermediate position between hydrogen and oxygen, and the C – C bond is formally considered nonpolar. In a methyl radical, a carbon atom attracts three electrons from three hydrogen atoms, so its oxidation state is -3. In a carboxyl, a carbon atom gives two electrons to the carbonyl oxygen atom and one electron to the oxygen atom of the hydroxyl group, therefore the oxidation state of the carbon atom is +3.

Equation of electronic balance:

Mn +7 + 5e \u003d Mn +2 6

C -3 - 6e \u003d C +3 5

Before formulas of substances containing manganese, a coefficient of 6 is required, and before the formulas of toluene and benzoic acid, a factor of 5.

5C 6 H 5 -CH 3 +6 KMnO 4 + H 2 SO 4 \u003d 5C 6 H 5 COOH + 6MnSO 4 + K 2 SO 4 + H 2 O

5C 6 H 5 -CH 3 +6 KMnO 4 + H 2 SO 4 \u003d 5C 6 H 5 COOH + 6MnSO 4 + 3K 2 SO 4 + H 2 O

And the number of sulfur atoms:

5C 6 H 5 -CH 3 +6 KMnO 4 + 9H 2 SO 4 \u003d 5C 6 H 5 COOH + 6MnSO 4 + 3K 2 SO 4 + H 2 O

At the final stage, a coefficient is needed in front of the water formula, which can be derived by selecting the number of hydrogen or oxygen atoms:

5C 6 H 5 -CH 3 +6 KMnO 4 + H 2 SO 4 \u003d 5C 6 H 5 COOH + 6MnSO 4 + K 2 SO 4 + 14H 2 O

Example 2. The reaction of the silver mirror.

Most literature indicates that aldehydes in these reactions are oxidized to the corresponding carboxylic acids. The oxidizing agent is an ammonia solution of silver oxide (I) - Ag2 O amm.r. In fact, the reaction proceeds in an alkaline ammonia medium; therefore, an ammonium salt or CO should form2 in case of formaldehyde oxidation.

Consider the oxidation of acetic aldehyde with Tollens reagent:

CH 3 CHO + Ag (NH 3) 2 OH \u003d ...

In this case, the product of oxidation will be ammonium acetate, and the product of recovery is silver:

CH 3 CHO + Ag (NH 3) 2 OH \u003d CH 3 COONH 4 + Ag + ...

The carbon atom of the carbonyl group undergoes oxidation. According to the structure of carbonyl, a carbon atom gives two electrons to an oxygen atom and takes one electron from a hydrogen atom, i.e. carbon oxidation state +1. In the carboxyl group of ammonium acetate, a carbon atom gives three electrons to oxygen atoms and has an oxidation state of +3. Equation of electronic balance:

C +1 - 2e \u003d C +3 1

Ag +1 + 1e \u003d Ag 0 2

We put the coefficients in front of the formulas of substances containing carbon and silver atoms:

CH 3 CHO + 2Ag (NH 3) 2 OH \u003d CH 3 COONH 4 + 2Ag + ...

Of the four ammonia molecules on the left side of the equation, one will participate in salt formation, and the remaining three will be released in free form. Also in the composition of the reaction products will be water, the coefficient before the formula of which can be found by selection (1):

CH 3 CHO + 2Ag (NH 3) 2 OH \u003d CH 3 COONH 4 + 2Ag + H 2 O

In conclusion, we note that an alternative way to describe OVR - the method of electron-ion equations - with its advantages, requires additional training time for study and development, which, as a rule, is extremely limited. However, the well-known method of electronic balance with its proper use leads to the desired results.

Physical properties

Benzene and its closest homologues are colorless liquids with a specific odor. Aromatic hydrocarbons are lighter than water and do not dissolve in it, however, they are easily dissolved in organic solvents - alcohol, ether, acetone.

Benzene and its homologues themselves are good solvents for many organic substances. All arenas burn with a smoky flame due to the high carbon content of their molecules.

The physical properties of some arenas are presented in the table.

Table. Physical properties of some arenas

Title	Formula	t ° .pl., ° C	t °. ° C
Benzene	C 6 H 6	5,5	80,1
Toluene (methylbenzene)	C 6 H 5 CH 3	95,0	110,6
Ethylbenzene	C 6 H 5 C 2 H 5	95,0	136,2
Xylene (dimethylbenzene)	C 6 H 4 (CH 3) 2
ortho-		25,18	144,41
meta-		47,87	139,10
couple-		13,26	138,35
Propylbenzene	C 6 H 5 (CH 2) 2 CH 3	99,0	159,20
Cumene (isopropylbenzene)	C 6 H 5 CH (CH 3) 2	96,0	152,39
Styrene (vinylbenzene)	C 6 H 5 CH \u003d CH 2	30,6	145,2

Benzene - low boiling ( t bale\u003d 80.1 ° C), colorless liquid, insoluble in water

Attention! Benzene - poison, acts on the kidneys, changes the blood formula (with prolonged exposure), can disrupt the structure of chromosomes.

Most aromatic hydrocarbons are life-threatening, toxic.

Obtaining arenes (benzene and its homologs)

In the laboratory

1. The fusion of salts of benzoic acid with solid alkalis

C 6 H 5 -COONa + NaOH t →C 6 H 6 + Na 2 CO 3

sodium benzoate

2. Würz-Fitting Reaction: (here G is halogen)

C 6H 5 -G + 2Na + R-G →C 6 H 5 - R + 2 NaG

FROM 6 H 5 -Cl + 2Na + CH 3 -Cl → C 6 H 5 -CH 3 + 2NaCl

In industry

isolated from oil and coal by fractional distillation, reforming;
from coal tar and coke oven gas

1. Dehydrocyclization of alkanes with the number of carbon atoms greater than 6:

C 6 H 14 t , kat→ C 6 H 6 + 4H 2

2. Acetylene Trimerization (only for benzene) - r. Zelinsky:

3C 2 H 2 600 ° C , Act. coal → C 6 H 6

3. Dehydrogenation cyclohexane and its homologues:

Soviet academician Nikolai Dmitrievich Zelinsky established that benzene is formed from cyclohexane (dehydrogenation of cycloalkanes

C 6 H 12 t, kat→ C 6 H 6 + 3H 2

C 6 H 11 -CH 3 t , kat→ C 6 H 5 -CH 3 + 3H 2

methylcyclohexantholuene

4. Alkylation of benzene (obtaining homologues of benzene) - r Friedel-Crafts.

C 6 H 6 + C 2 H 5 -Cl t, AlCl3→ C 6 H 5 -C 2 H 5 + HCl

chloroethane ethylbenzene

Chemical properties of arenas

I. OXIDATION REACTIONS

1. Burning (smoking flame):

2C 6 H 6 + 15O 2 t → 12CO 2 + 6H 2 O + Q

2. Under normal conditions, benzene does not discolor bromine water and an aqueous solution of potassium permanganate

3. Homologues of benzene are oxidized by potassium permanganate (bleach potassium permanganate):

A) in an acidic environment to benzoic acid

Under the action of potassium permanganate and other strong oxidizing agents on benzene homologs, the side chains are oxidized. No matter how complex the substituent chain, it breaks down, with the exception of the a-carbon atom, which is oxidized to the carboxyl group.

Homologs of benzene with one side chain give benzoic acid:

Homologists containing two side chains produce dibasic acids:

5C 6 H 5 -C 2 H 5 + 12KMnO 4 + 18H 2 SO 4 → 5C 6 H 5 COOH + 5CO 2 + 6K 2 SO 4 + 12MnSO 4 + 28H 2 O

5C 6 H 5 -CH 3 + 6KMnO 4 + 9H 2 SO 4 → 5C 6 H 5 COOH + 3K 2 SO 4 + 6MnSO 4 + 14H 2 O

Simplified :

C 6 H 5 -CH 3 + 3O KMnO4→ C 6 H 5 COOH + H 2 O

B) in neutral and slightly alkaline to salts of benzoic acid

C 6 H 5 -CH 3 + 2KMnO 4 → C 6 H 5 COOK + K OH + 2MnO 2 + H 2 O

II. ADDITION REACTIONS (harder than alkenes)

1. Halogenation

C 6 H 6 + 3Cl 2 h ν → C 6 H 6 Cl 6 (hexachlorocyclohexane - hexachloran)

2. Hydrogenation

C 6 H 6 + 3H 2 t , Pt or Ni → C 6 H 12 (cyclohexane)

3. Polymerization

III. SUBSTITUTION REACTIONS - ionic mechanism (lighter than alkanes)

1. Halogenation -

a ) benzene

C 6 H 6 + Cl 2 AlCl 3 → C 6 H 5 -Cl + HCl (chlorobenzene)

C 6 H 6 + 6Cl 2 t, AlCl3→ C 6 Cl 6 + 6HCl( hexachlorobenzene)

C 6 H 6 + Br 2 t, FeCl3→ C 6 H 5 -Br + HBr( bromobenzene)

b) benzene homologs when irradiated or heated

In chemical properties, alkyl radicals are similar to alkanes. The hydrogen atoms in them are replaced by halogen by the free-radical mechanism. Therefore, in the absence of a catalyst, a radical substitution reaction occurs in the side chain during heating or UV irradiation. The effect of the benzene ring on alkyl substituents leads to the fact that always replaced by a hydrogen atom at a carbon atom directly bonded to the benzene ring (a-carbon atom).

1) C 6 H 5 -CH 3 + Cl 2 h ν → C 6 H 5 -CH 2 -Cl + HCl

c) benzene homologs in the presence of a catalyst

C 6 H 5 -CH 3 + Cl 2 AlCl 3 → (mixture of orth, a pair of derivatives) + HCl

2. Nitration (with nitric acid)

C 6 H 6 + HO-NO 2 t, H2SO4→ C 6 H 5 -NO 2 + H 2 O

nitrobenzene - smell almonds!

C 6 H 5 -CH 3 + 3HO-NO 2 t, H2SO4→ FROM H 3 -C 6 H 2 (NO 2) 3 + 3H 2 O

2,4,6-trinitrotoluene (tol, trotyl)

The use of benzene and its homologues

Benzene C 6 H 6 is a good solvent. Benzene as an additive improves the quality of motor fuel. It serves as a raw material for the production of many aromatic organic compounds - nitrobenzene C 6 H 5 NO 2 (solvent, from which aniline is obtained), chlorobenzene C 6 H 5 Cl, phenol C 6 H 5 OH, styrene, etc.

Toluene C 6 H 5 –CH 3 is a solvent used in the manufacture of dyes, drugs, and explosives (TNT (thol), or 2,4,6-trinitrotoluene TNT).

XylenesC 6 H 4 (CH 3) 2. Technical xylene is a mixture of three isomers ( ortho-, meta- and couple-xylene) - is used as a solvent and starting material for the synthesis of many organic compounds.

Isopropylbenzene C 6 H 5 –CH (CH 3) 2 is used to produce phenol and acetone.

Chlorine derivatives of benzene used to protect plants. So, the product of substitution of H atoms in benzene by chlorine atoms - hexachlorobenzene С 6 Сl 6 - fungicide; it is used for dry dressing of wheat and rye seeds against smut. The product of the addition of chlorine to benzene is hexachlorocyclohexane (hexachloran) С 6 Н 6 Сl 6 - insecticide; It is used to combat harmful insects. Mentioned substances relate to pesticides - chemical agents against microorganisms, plants and animals.

Styrene C 6 H 5 - CH \u003d CH 2 polymerizes very easily, forming polystyrene, and butadiene styrene rubbers copolymerizing with butadiene.

VIDEO EXPERIENCE

18. Redox reactions (continued 2)

18.9. Organic organic matter

In the OVR of organic substances with inorganic organic substances are most often reducing agents. So, during the combustion of organic matter in excess of oxygen, carbon dioxide and water are always formed. The reactions are more difficult when using less active oxidizing agents. This section considers only the reactions of representatives of the most important classes of organic substances with some inorganic oxidizing agents.

Alkenes. In mild oxidation, alkenes are converted to glycols (dihydric alcohols). Reducing atoms in these reactions are carbon atoms bound by a double bond.

The reaction with a solution of potassium permanganate proceeds in a neutral or slightly alkaline environment as follows:

C 2 H 4 + 2KMnO 4 + 2H 2 O CH 2 OH – CH 2 OH + 2MnO 2 + 2KOH (cooling)

Under more severe conditions, oxidation leads to the breaking of the carbon chain in a double bond and the formation of two acids (in a highly alkaline medium - two salts) or acid and carbon dioxide (in a highly alkaline medium - salt and carbonate):

1) 5CH 3 CH \u003d CHCH 2 CH 3 + 8KMnO 4 + 12H 2 SO 4 5CH 3 COOH + 5C 2 H 5 COOH + 8MnSO 4 + 4K 2 SO 4 + 17H 2 O (heating)

2) 5CH 3 CH \u003d CH 2 + 10KMnO 4 + 15H 2 SO 4 5CH 3 COOH + 5CO 2 + 10MnSO 4 + 5K 2 SO 4 + 20H 2 O (heating)

3) CH 3 CH \u003d CHCH 2 CH 3 + 6KMnO 4 + 10KOH CH 3 COOK + C 2 H 5 COOK + 6H 2 O + 6K 2 MnO 4 (heating)

4) CH 3 CH \u003d CH 2 + 10KMnO 4 + 13KOH CH 3 COOK + K 2 CO 3 + 8H 2 O + 10K 2 MnO 4 (heating)

Potassium dichromate in a sulfuric acid environment oxidizes alkenes similarly to reactions 1 and 2.

Alkines. Alkines begin to oxidize under somewhat more stringent conditions than alkenes, so they are usually oxidized with a carbon chain breaking through a triple bond. As in the case of alkanes, the reducing atoms here are carbon atoms bonded in this case by a triple bond. As a result of the reactions, acids and carbon dioxide are formed. The oxidation can be carried out with permanganate or potassium dichromate in an acidic medium, for example:

5CH 3 C CH + 8KMnO 4 + 12H 2 SO 4 5CH 3 COOH + 5CO 2 + 8MnSO 4 + 4K 2 SO 4 + 12H 2 O (heating)

Sometimes it is possible to isolate intermediate oxidation products. Depending on the position of the triple bond in the molecule, these are either diketones (R 1 –CO – CO – R 2) or aldoketones (R – CO – CHO).

Acetylene can be oxidized with potassium permanganate in a slightly alkaline medium to potassium oxalate:

3C 2 H 2 + 8KMnO 4 \u003d 3K 2 C 2 O 4 + 2H 2 O + 8MnO 2 + 2KOH

In an acidic environment, oxidation goes to carbon dioxide:

C 2 H 2 + 2KMnO 4 + 3H 2 SO 4 \u003d 2CO 2 + 2MnSO 4 + 4H 2 O + K 2 SO 4

Homologists of benzene. Homologs of benzene can be oxidized with a solution of potassium permanganate in a neutral medium to potassium benzoate:

C 6 H 5 CH 3 + 2KMnO 4 \u003d C 6 H 5 COOK + 2MnO 2 + KOH + H 2 O (when boiled)

C 6 H 5 CH 2 CH 3 + 4KMnO 4 \u003d C 6 H 5 COOK + K 2 CO 3 + 2H 2 O + 4MnO 2 + KOH (when heated)

Oxidation of these substances with potassium dichromate or potassium permanganate in an acidic medium leads to the formation of benzoic acid.

Alcohols. The direct oxidation products of primary alcohols are aldehydes, and secondary ones are ketones.

The aldehydes formed during the oxidation of alcohols are easily oxidized to acids; therefore, aldehydes from primary alcohols are obtained by oxidation with potassium dichromate in an acidic medium at the boiling temperature of the aldehyde. Evaporating, aldehydes do not have time to oxidize.

3C 2 H 5 OH + K 2 Cr 2 O 7 + 4H 2 SO 4 \u003d 3CH 3 CHO + K 2 SO 4 + Cr 2 (SO 4) 3 + 7H 2 O (heating)

With an excess of oxidizing agent (KMnO 4, K 2 Cr 2 O 7) in any medium, primary alcohols are oxidized to carboxylic acids or their salts, and secondary alcohols to ketones. Tertiary alcohols do not oxidize under these conditions, and methyl alcohol oxidizes to carbon dioxide. All reactions occur when heated.

The dihydric alcohol, ethylene glycol HOCH 2 –CH 2 OH, when heated in an acidic medium with a solution of KMnO 4 or K 2 Cr 2 O 7 is easily oxidized to carbon dioxide and water, but sometimes it is possible to isolate intermediate products (HOCH 2 –COOH, HOOC– COOH et al.).

Aldehydes. Aldehydes are quite powerful reducing agents, and therefore are easily oxidized by various oxidizing agents, for example: KMnO 4, K 2 Cr 2 O 7, OH. All reactions occur when heated:

3CH 3 CHO + 2KMnO 4 \u003d CH 3 COOH + 2CH 3 COOK + 2MnO 2 + H 2 O
3CH 3 CHO + K 2 Cr 2 O 7 + 4H 2 SO 4 \u003d 3CH 3 COOH + Cr 2 (SO 4) 3 + 7H 2 O
CH 3 CHO + 2OH \u003d CH 3 COONH 4 + 2Ag + H 2 O + 3NH 3

Formaldehyde with an excess of oxidizing agent is oxidized to carbon dioxide.

10/18. Comparison of the redox activity of various substances

From the definitions of the concepts of "atom-oxidizer" and "atom-reducing agent" it follows that only the oxidizing properties of atoms in the highest degree of oxidation. On the contrary, atoms with a lower oxidation state possess only reducing properties. Atoms in intermediate oxidation states can be either oxidizing agents or reducing agents.

However, based only on the degree of oxidation, it is impossible to unambiguously assess the redox properties of substances. As an example, consider compounds of elements of the VA group. The nitrogen (V) and antimony (V) compounds are more or less strong oxidizing agents, bismuth (V) compounds are very strong oxidizing agents, and the phosphorus (V) compounds practically do not have oxidizing properties. In this and other similar cases, it matters how much a given oxidation state is characteristic of a given element, that is, how stable are compounds containing atoms of a given element in this oxidation state.

Any OVR proceeds in the direction of the formation of a weaker oxidizing agent and a weaker reducing agent. In the general case, the possibility of any OVR occurring, like any other reaction, can be determined by the sign of the Gibbs energy change. In addition, for the quantitative assessment of the redox activity of substances, the electrochemical characteristics of oxidizing agents and reducing agents are used (standard potentials of redox pairs). Based on these quantitative characteristics, it is possible to construct series of redox activity of various substances. The series of metal stresses known to you is constructed in this way. This series makes it possible to compare the reducing properties of metals in aqueous solutions under standard conditions ( from \u003d 1 mol / l, T \u003d 298.15 K), as well as the oxidizing properties of simple aquacations. If ions (oxidizing agents) are placed in the top row of this row, and metal atoms (reducing agents) in the bottom row, then the left side of this row (before hydrogen) will look like this:

In this series, the oxidizing properties of ions (upper row) are enhanced from left to right, and the reducing properties of metals (lower row), on the contrary, from right to left.

Given the differences in redox activity in different environments, it is possible to build similar series for oxidizing agents. So, for reactions in an acidic environment (pH \u003d 0), a “continuation” of a series of metal activity is obtained in the direction of enhancing oxidative properties

As in the activity series of metals, in this series the oxidizing properties of oxidizing agents (top line) are enhanced from left to right. But, using this series, it is possible to compare the reducing activity of the reducing agents (bottom line) only when their oxidized form coincides with that given in the top line; in this case, it is amplified from right to left.

Let's look at a few examples. To find out whether this OVR is possible, we will use the general rule that determines the direction of the occurrence of redox reactions (reactions proceed in the direction of the formation of a weaker oxidizing agent and a weaker reducing agent).

1. Is it possible to recover cobalt from magnesium with a CoSO 4 solution with magnesium?
Magnesium is a stronger reducing agent than cobalt, and Co 2 ions are stronger oxidizing agents than Mg 2 ions, therefore, it is possible.
2. Can FeCl 3 be used to oxidize copper to CuCl 2 in an acidic environment?
Since Fe 3B ions are stronger oxidizing agents than Cu 2 ions, and copper is a stronger reducing agent than Fe 2 ions, it is possible.
3. Is it possible, by blowing oxygen through a solution of FeCl 2 acidified with hydrochloric acid, to obtain a solution of FeCl 3?
It would seem not, since in our series oxygen is to the left of Fe 3 ions and is a weaker oxidizing agent than these ions. But in an aqueous solution, oxygen is almost never reduced to H 2 O 2, in this case it is reduced to H 2 O and takes place between Br 2 and MnO 2. Therefore, such a reaction is possible, although it proceeds rather slowly (why?).
4. Is it possible to oxidize H 2 O 2 in an acidic medium with potassium permanganate?
In this case, the H 2 O 2 reducing agent and reducing agent are stronger than the Mn 2B ions, and the MnO 4 oxidizing ions are stronger than the oxygen formed from peroxide. Therefore, it is possible.

A similar series, constructed for OVR in an alkaline environment, is as follows:

Unlike the "acid" series, this series cannot be used in conjunction with a series of metal activity.

The electron-ion balance method (half-reaction method), intermolecular OVR, intramolecular OVR, OVM dismutation (disproportionation, self-oxidation-self-healing), OVR switching, passivation.

Using the electron-ion balance method, make up the equations of the reactions that occur when a solution of a) H 2 S (S, more precisely, S 8) is added to a solution of potassium permanganate acidified with sulfuric acid; b) KHS; c) K 2 S; g) H 2 SO 3; d) KHSO 3; e) K 2 SO 3; g) HNO 2; g) KNO 2; i) KI (I 2); k) FeSO 4; k) C 2 H 5 OH (CH 3 COOH); m) CH 3 CHO; m) (COOH) 2 (CO 2); o) K 2 C 2 O 4. Hereinafter, if necessary, oxidation products are indicated in curly brackets.
Draw up the equations of the reactions that occur when the following gases are passed through a potassium permanganate solution acidified with sulfuric acid: a) C 2 H 2 (CO 2); b) C 2 H 4 (CO 2); c) C 3 H 4 (propyne) (CO 2 and CH 3 COOH); g) C 3 H 6; d) CH 4; e) HCHO.
The same, but the reducing agent solution was added to the neutral potassium permanganate solution: a) KHS; b) K 2 S; c) KHSO 3; g) K 2 SO 3; d) KNO 2; e) KI.
The same, but a solution of potassium hydroxide was previously added to the potassium permanganate solution: a) K 2 S (K 2 SO 4); b) K 2 SO 3; c) KNO 2; d) KI (KIO 3).
Make the equations of the following reactions in solution: a) KMnO 4 + H 2 S ...;
   b) KMnO 4 + HCl ...;
   c) KMnO 4 + HBr ...;
   d) KMnO 4 + HI ...
Make the following equations for the OVR of manganese dioxide:
Solutions of the following substances were added to a solution of potassium dichromate acidified with sulfuric acid: a) KHS; b) K 2 S; c) HNO 2; g) KNO 2; d) KI; e) FeSO 4; g) CH 3 CH 2 CHO; i) H 2 SO 3; j) KHSO 3; l) K 2 SO 3. Make the equations of the proceeding reactions.
The same, but the following gases were passed through the solution: a) H 2 S; b) SO 2.
To a potassium chromate solution containing potassium hydroxide, a) K 2 S (K 2 SO 4) solutions were added; b) K 2 SO 3; c) KNO 2; d) KI (KIO 3). Make the equations of the proceeding reactions.
A solution of potassium hydroxide was added to a solution of chromium (III) chloride to dissolve the initially formed precipitate, and then bromine water. Make the equations of the proceeding reactions.
The same, but at the last stage, a solution of potassium peroxodisulfate K 2 S 2 O 8 was added, which was reduced to sulfate during the reaction.
Make up the equations of the reactions proceeding in solution:

a) CrCl 2 + FeCl 3; b) CrSO 4 + FeCl 3; c) CrSO 4 + H 2 SO 4 + O 2;

d) CrSO 4 + H 2 SO 4 + MnO 2; d) CrSO 4 + H 2 SO 4 + KMnO 4.

Write the equations of the reactions between solid chromium trioxide and the following substances: a) C; b) CO; c) S (SO 2); g) H 2 S; d) NH 3; e) C 2 H 5 OH (CO 2 and H 2 O); g) CH 3 COCH 3.
Make the equations of the reactions proceeding when the following substances are added to concentrated nitric acid: a) S (H 2 SO 4); b) P 4 ((HPO 3) 4); c) graphite; g) Se; d) I 2 (HIO 3); e) Ag; g) Cu; i) Pb; k) KF; l) FeO; m) FeS; m) MgO; p) MgS; p) Fe (OH) 2; c) P 2 O 3; r) As 2 O 3 (H 3 AsO 4); y) As 2 S 3; f) Fe (NO 3) 2; x) P 4 O 10; c) Cu 2 S.
The same, but when passing the following gases: a) CO; b) H 2 S; c) N 2 O; g) NH 3; d) NO; e) H 2 Se; g) HI.
The reactions will occur in the same or different ways in the following cases: a) a piece of magnesium was placed in a two-thirds high test tube filled with concentrated nitric acid; b) a drop of concentrated nitric acid was placed on the surface of a magnesium plate? Make up the reaction equations.
What is the difference between the reaction of concentrated nitric acid with hydrogen sulfide and gaseous hydrogen sulfide? Make up the reaction equations.
Will OVD occur the same way when anhydrous crystalline sodium sulfide and its 0.1 M solution are added to the concentrated nitric acid solution?
Concentrated nitric acid treated a mixture of the following substances: Cu, Fe, Zn, Si and Cr. Make the equations of the proceeding reactions.
Make the equations of the reactions proceeding when the following substances are added to diluted nitric acid: a) I 2; b) Mg; c) Al; g) Fe; d) FeO; e) FeS; g) Fe (OH) 2; i) Fe (OH) 3; j) MnS; l) Cu 2 S; m) CuS; m) CuO; o) Na 2 S cr; p) Na 2 S p; c) P 4 O 10.
What processes will occur when passing through a dilute solution of nitric acid a) ammonia, b) hydrogen sulfide, c) carbon dioxide?
Make the equations of the reactions occurring when the following substances are added to concentrated sulfuric acid: a) Ag; b) Cu; c) graphite; g) HCOOH; d) C 6 H 12 O 6; e) NaCl cr; g) C 2 H 5 OH.
When hydrogen sulfide is passed through cold concentrated sulfuric acid, S and SO 2 are formed; hot concentrated H 2 SO 4 oxidizes sulfur to SO 2. Make up the reaction equations. How will the reaction between hot concentrated H 2 SO 4 and hydrogen sulfide proceed?
Why is hydrogen chloride obtained by treating crystalline sodium chloride with concentrated sulfuric acid, but hydrogen bromide and hydrogen iodide are not obtained in this way?
Make up the equations of the reactions that occur during the interaction of dilute sulfuric acid with a) Zn, b) Al, c) Fe, d) chromium in the absence of oxygen, e) chromium in air.
Make the equations of reactions characterizing the redox properties of hydrogen peroxide:

In which of these reactions is hydrogen peroxide an oxidizing agent, and in which a reducing agent?

What reactions proceed when the following substances are heated: a) (NH 4) 2 CrO 4; b) NaNO 3; c) CaCO 3; d) Al (NO 3) 3; d) Pb (NO 3) 3; e) AgNO 3; g) Hg (NO 3) 2; i) Cu (NO 3) 2; k) CuO; l) NaClO 4; m) Ca (ClO 4) 2; m) Fe (NO 3) 2; o) PCl 5; p) MnCl 4; c) H 2 C 2 O 4; r) LiNO 3; y) HgO; f) Ca (NO 3) 2; x) Fe (OH) 3; c) CuCl 2; h) KClO 3; w) KClO 2; u) CrO 3?
When merging hot solutions of ammonium chloride and potassium nitrate, a reaction occurs, accompanied by evolution of gas. Equate this reaction.
Draw up the equations of the reactions that occur when passing through a cold solution of sodium hydroxide a) chlorine, b) bromine vapor. The same, but through a hot solution.
When interacting with a hot concentrated potassium hydroxide solution, selenium undergoes dismutation to the nearest stable oxidation states (–II and + IV). Make the equation of this OVR.
Under the same conditions, sulfur undergoes a similar dismutation, but an excess of sulfur reacts with sulfite ions to form thiosulfate ions S 2 O 3 2. Make the equations of the proceeding reactions. ;
Make the equations of the electrolysis reactions a) a solution of copper nitrate with a silver anode, b) a solution of lead nitrate with a copper anode.

Experience 1. Oxidizing properties of potassium permanganate in an acidic environment. K 3-4 drops of a solution of potassium permanganate pour an equal volume of a dilute solution of sulfuric acid, and then a solution of sodium sulfite until discoloration. Compose a reaction equation.

Experience 2.Oxidizing properties of potassium permanganate in a neutral environment. To 3-4 drops of a solution of potassium permanganate add 5-6 drops of a solution of sodium sulfite. What substance was precipitated?

Experience 3. Oxidizing properties of potassium permanganate in an alkaline environment. To 3-4 drops of potassium permanganate solution add 10 drops of concentrated sodium hydroxide solution and 2 drops of sodium sulfite solution. The solution should turn green.

Experience 4. Oxidizing properties of potassium dichromate in an acidic environment. Acidify 6 drops of potassium dichromate solution with four drops of a dilute sulfuric acid solution and add sodium sulfite solution until the color of the mixture changes.

Experience 5. Oxidizing properties of dilute sulfuric acid. Place a zinc granule in one tube and a piece of copper tape in another. Add 8-10 drops of dilute sulfuric acid solution to both tubes. Compare occurring events. EXPERIENCE IN A EXHAUST CABINET!

Experience 6. Oxidizing properties of concentrated sulfuric acid. Similar to experiment 5, but add a concentrated solution of sulfuric acid. A minute after the start of the evolution of gaseous reaction products, add test strips of filter paper moistened with solutions of potassium permanganate and copper sulfate to the tubes. Explain the occurring phenomena. EXPERIENCE IN A EXHAUST CABINET!

Experience 7. Oxidizing properties of diluted nitric acid. Similar to experiment 5, but add a dilute solution of nitric acid. Observe the color change of the gaseous reaction products. EXPERIENCE IN A EXHAUST CABINET!

Experience 8. Oxidizing properties of concentrated nitric acid. Place a piece of copper tape in a test tube and add 10 drops of a concentrated solution of nitric acid. Gently heat until metal is completely dissolved. EXPERIENCE IN A EXHAUST CABINET!

Experience 9. Oxidizing properties of potassium nitrite.To 5-6 drops of a solution of potassium nitrite add an equal volume of a dilute solution of sulfuric acid and 5 drops of a solution of potassium iodide. What substances are observed?

Experience 10. Restorative properties of potassium nitrite. To 5-6 drops of a solution of potassium permanganate add an equal volume of a dilute solution of sulfuric acid and a solution of potassium nitrite to completely discolor the mixture.

Experience 11. Thermal decomposition of copper nitrate. Place one micro spatula of copper nitrate trihydrate in a test tube, fix it in a tripod and carefully heat it with an open flame. Observe dehydration and subsequent salt decomposition. EXPERIENCE IN A EXHAUST CABINET!

Experience 12. Thermal decomposition of lead nitrate. Conduct similarly to experiment 11, placing lead nitrate in a test tube. EXPERIENCE IN A EXHAUST CABINET! What is the difference between the processes occurring during the decomposition of these salts?