Toluene with potassium permanganate in an alkaline medium. Oxidation reactions of organic substances

Toluene is a colorless liquid with a specific odor. Toluene is lighter than water and does not dissolve in it, but it easily dissolves 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 in its molecule.

The physical properties of toluene are presented in the table.

Table. Physical properties of toluene.

Chemical properties of toluene

I. Oxidation reaction.

1. Combustion (smoky 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 medium to benzoic acid

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

B) in neutral and slightly alkaline to benzoic acid salts

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

II... ADDITIONAL REACTIONS

1. Halogenation

WITH 6 N 5 CH 3 + Bg 2  WITH 6 N 5 CH 2 Br + HBg

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

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 ort, 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, TNT)

The use of toluene.

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

2.2. Being in nature

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

2.3. Anthropogenic sources of toluene input 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 found in urban smoke.

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

The background level of toluene in the atmosphere is 0.75 µg / m3 (0.00075 mg / m3).

Also, the main sources of toluene intake in environment is the chemical production of explosives, epoxy resins, varnishes and paints, etc.

This material can be difficult to master during self-study, due to the large amount of information, many nuances, all kinds of BUTs and IFs. Read carefully!

What exactly will be discussed?

In addition to complete oxidation (combustion), some classes of organic compounds are characterized by incomplete oxidation reactions, while they are converted 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 classic oxidizing agents, which, so to speak, are universal for many classes.

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

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

In practice, there are many factors that affect the course of the reaction (temperature, medium, concentration of reagents, etc.). A mixture of products is often obtained. Therefore, it is almost impossible to predict the product that will be formed.

But for the Unified State Exam, this is not suitable: there you cannot write “maybe this way, or this way, or another, or a mixture of products”. Specificity is needed there.

The authors of the assignments have put in a certain logic, a certain principle according to which a certain product should be written. Unfortunately, they haven't shared with anyone.

In most textbooks, this question is rather slippery to avoid: two or three reactions are given as an example.

I present in this article what can be called the results of the research-analysis of the USE tasks. The logic and principles of composing the oxidation reactions with permanganate and dichromate have been solved quite with high precision(in accordance with the USE standards). Everything in order.

Determination of the oxidation state.

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, the reducing agent is atoms in organic matter (namely, carbon atoms).

It is not enough to define the products; the reaction must be equalized. The electronic balance method is traditionally used for equalization. To apply this method, it is necessary to determine the oxidation states of reducing agents and oxidizing agents before and after the reaction.

Have not organic matter We are able to oxidize from grade 9:

But in organics, probably, in the 9th grade, they were not determined. Therefore, before learning how to write OVR in organic chemistry, you need to learn how to determine the oxidation state of carbon in organic matter. This is done a little differently, differently than in inorganic chemistry.

Carbon has a maximum oxidation state of +4, a minimum -4. And it can show any oxidation state of this interval: -4, -3, -2, -1, 0, +1, +2, +3, +4.

First you need to remember what the oxidation state is.

The oxidation state is the conditional charge that occurs on an atom, assuming that the electron pairs are shifted entirely towards the 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 oxidation state of a carbon atom.

To determine the oxidation state of a particular carbon atom in a compound, we need to consider EACH of its bonds and look in which direction the electron pair will move 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 one negative charge from each hydrogen: it turns out -3

The fourth bond is with chlorine. Carbon and chlorine - who is more electronegative? Chlorine means that through this bond the electron pair will shift towards chlorine. Carbon has one positive charge +1.

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

Let's 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 one negative charge from each hydrogen: it turns out -3

And one more bond with another carbon. Carbon and other carbon - their electronegativities are equal, so there is no displacement of the electron pair (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 off an electron pair from carbon, and carbon gets a charge of +3.

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

Carbon is linked to hydrogen atoms by two bonds. Carbon, being more electronegative, draws off one pair of electrons for each bond with hydrogen, acquires a charge of -2.

A carbon double bond is bonded to an oxygen atom. More electronegative oxygen attracts one electron pair for each bond. Together, two electron pairs are pulled from carbon. Carbon gains a +2 charge.

Together it turns out +2 -2 = 0.

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

A triple bond with a more electronegative nitrogen - gives carbon a charge of +3, there is no displacement of the electron pair due to the bond with carbon.

Oxidation with permanganate.

What will happen to the permanaganate?

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

Therefore, it can go in three directions:

Permanganate, being an oxidizing agent, is reduced. Here are the products of its restoration:

Acidic environment.

The medium is acidified with sulfuric acid (H 2 SO 4). Manganese is reduced to the +2 oxidation state. And the 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 the +6 oxidation state. Recovery products

KMnO 4 + KOH → K 2 MnO 4 + H 2 O

Neutral environment(and slightly alkaline).

In a neutral medium, in addition to permanganate, water also enters into 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 medium (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 is that it all starts with alcohol! This is the initial stage of oxidation. The carbon to which the hydroxyl group is attached undergoes oxidation.

When oxidized, a carbon atom "acquires" a bond with oxygen. Therefore, when the oxidation reaction scheme is written, [O] is written above the arrow:

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

Oxidation secondary alcohol breaks off at the second stage. Since carbon is in the middle, a ketone is formed, not an aldehyde (a carbon atom in a ketone group can no longer physically form a bond with a hydroxyl group):

Ketones, tertiary alcohols and carboxylic acids no longer oxidize further:

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

It is worth noting the stages of methanol oxidation. First, it is oxidized 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 bonded to hydrogen, and if you look closely, you will notice that this is nothing more than an aldehyde group:

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

Did you recognize the substance you received? Its gross formula is H 2 CO 3. It is carbonic acid that decomposes 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 medium (write above the reaction 0 ° or 20 °) .

It is important to remember that alcohols in soft conditions do not oxidize. Therefore, if they are formed, then oxidation stops at them. What substances will undergo a mild oxidation reaction?

Containing a double bond C = C (Wagner reaction).

In this case, the π-bond is broken and "sits" on the vacated bonds at the hydroxyl group. It turns out dihydric alcohol:

Let's write the reaction of mild oxidation of ethylene (ethene). Let's write down the initial substances and predict the products. At the same time, we do not write H 2 O and KOH: they can appear both on the right side of the equation and on the left. And we immediately determine the oxidation states of the substances participating in the OVR:

Let's compose an electronic balance (we mean that there are two reducing agents - two carbon atoms, they are oxidized separately):

Let's arrange the coefficients:

At the end, you need to add the missing products (H 2 O and KOH). There is not enough potassium on the right, which means the alkali will be on the right. We put the coefficient in front of it. There is not enough hydrogen on the left, which means there is water on the left. We put the coefficient in front of it:

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

Cycloalkene is often slipped in. Don't let him embarrass you. This is a common double bond hydrocarbon:

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

Containing an aldehyde group.

The aldehyde group is more reactive (reacts more easily) than the alcohol group. Therefore, the aldehyde will oxidize. Before acid:

Consider acetaldehyde (ethanal) as an example. Let's write down the reagents and products and arrange the oxidation states. Let's draw up a balance and put the coefficients in front of the reducing agent and the oxidizing agent:

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

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

In this case, acid and alkali appear side by side in the same mixture. Neutralization occurs.

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 alkalis are 2 moles. 2 moles of alkali can neutralize only 2 moles of acid (2 moles of salt are formed). And one mole of acid remains. Therefore, the final equation will be like this:

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

A similar situation occurs during the oxidation of methanal. As we remember, it is oxidized to carbon dioxide:

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

If alkali is 2: 1 to carbon dioxide, then there will be a medium salt:

Or the alkali can be much more (more than twice). If it is more than twice, then the remainder of the alkali 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 alkali refers to carbon dioxide as 1: 1, then there will be sour salt:

KOH + CO 2 → KHCO 3

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

KOH + 2CO 2 → KHCO 3 + CO 2

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

Let's write down the initial substances, products, draw up a balance, put down the oxidation states in front of the oxidizing agent, reducing agent and the 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 when three moles of CO 2 and four moles of alkali interact.

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 hydrocarbonate and one mole of carbonate:

And in a weakly alkaline environment, there are no such problems: 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 the equation should result in 4 moles of alkali (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 are spent on the formation of acidic salt, one mole of alkali remains:

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

And this one mole of alkali goes to interact with one mole of acidic salt:

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

It turns out like this:

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

Final equation:

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

Containing a triple bondC≡ C.

Remember what happened during the mild oxidation of compounds with a double bond? If you don’t remember, then scroll back - remember.

The π-bond breaks, attaches to the carbon atoms at the hydroxyl group. The principle is the same here. Just remember that there are two π-bonds in the triple bond. First, this happens along the first π-bond:

Then on another π-bond:

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

It turns out a carbonyl group.

Let's consider some examples:

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

Splitting off water:

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

Consider the oxidation of butyne-2:

Splitting off water:

Acid is not formed here, so there is no need to fool around with neutralization.

Reaction equation:

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

Splitting off water:

It turns out a substance of an interesting structure:

The aldehyde group continues to oxidize:

Let's write down the initial substances, products, determine the oxidation states, draw up a balance, put down the coefficients in front of the oxidizing agent and the reducing agent:

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

Harsh 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 also sometimes heat up. But for the harsh oxidation to go not in an acidic environment, heating is a prerequisite.

What substances will undergo severe oxidation? (At first, we will analyze only in an acidic environment - and then we will supplement the nuances that arise during oxidation in a strongly alkaline and neutral or slightly alkaline (when heated) environment).

With severe oxidation, the process goes to the maximum. As long as there is something to oxidize, oxidation is in progress.

Alcohols. Aldehydes.

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

We write down the equation. We write down the initial substances, OVR products, put down the oxidation states, and draw up a balance. Let's equalize the reaction:

If the reaction is carried out at the boiling point of the aldehyde, when it is formed, it will evaporate (fly away) from the reaction mixture, without having time to oxidize further. The same effect can be achieved under very gentle conditions (low heat). In this case, we write aldehyde as a product:

Let us consider the oxidation of secondary alcohol using propanol-2 as an example. As already mentioned, oxidation is terminated at the second stage (formation of a carbonyl compound). Since a ketone is formed, which does not oxidize. Reaction equation:

Let us consider the oxidation of aldehydes on the basis of ethanal. It also oxidizes to acid:

Reaction equation:

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

Methanal:

Containing multiple links.

In this case, a chain break occurs at a multiple connection. And the atoms that formed it undergo oxidation (acquire a bond with oxygen). Oxidized as much as possible.

When a double bond is broken, carbonyl compounds are formed from scraps (in the scheme below: from one scraps - aldehyde, from another - ketone)

Let's analyze the oxidation of pentene-2:

Oxidation of "scraps":

It turns out that two acids are formed. Let's write down the starting materials and products. Determine the oxidation state of the atoms that change it, draw up a balance, equalize the reaction:

Composing the electronic balance, we mean that there are two reducing agents - two carbon atoms, they are oxidized separately:

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

Reaction equation:

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

Reaction equation:

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

Reaction equation:

Double-corona acid.

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

This is oxalic acid. Two crowns are hard to get along side by side. She is of course stable in normal conditions... But due to the fact that it has two carboxyl groups connected to each other, it is less stable than other carboxylic acids.

And therefore, under especially 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 its homologues are oxidized. In this case, the chain also breaks, 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 cleavage occurs in the radical.
An atom directly bound to the benzene ring is oxidized. If after it the carbon chain in the radical continues, then the gap will be after it.

Let us examine the oxidation of methylbenzene. One carbon atom in the radical is oxidized there:

Reaction equation:

Let us analyze the oxidation of isobutylbenzene:

Reaction equation:

Let's analyze the oxidation of sec-butylbenzene:

Reaction equation:

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

Derivatives of benzene homologues (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. Algorithm "how to write the reaction of hard oxidation with permanganate in an acidic medium":

Write down the starting materials (organic + KMnO 4 + H 2 SO 4).
Write down the oxidation products of organic matter (compounds containing alcohol, aldehyde groups, multiple bonds, and benzene homologues will be oxidized).
Write down the product of the reduction of permanganate (MnSO 4 + K 2 SO 4 + H 2 O).
Determine the oxidation state of the OVR participants. Draw up a balance. Put down the coefficients for the oxidizing agent and the reducing agent, as well as for 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.

Rigid oxidation in a strongly 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 reaction, these turned out to be the most controversial.

Hard oxidation is also hard in Africa, so organic matter is oxidized in the same way as in an acidic environment.

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

Strongly alkaline environment :

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

KMnO 4 + KOH → K 2 MnO 4.

In a strongly 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 a salt (if the acid is polybasic, there will be a medium salt).

For example, the oxidation of propene:

Oxidation of ethylbenzene:

Weakly alkaline or neutral environment when heated :

Here, too, the possibility of neutralization must always be taken into account.

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

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

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

Therefore, if you form acid in a neutral environment, you need to carefully see if it is enough to neutralize all the acid. Special attention you need to pay attention to the neutralization of polybasic acids.

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

As a rule, alkali is sufficient for oxidation in a neutral medium. And the reaction equation, whether in a neutral or slightly alkaline medium, will be the same.

For example, let's analyze the oxidation of ethylbenzene:

Alkali is quite enough for the complete neutralization of the acid compounds obtained, even excess will remain:

3 moles of alkali are consumed - 1 remains.

Final equation:

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

Redox reactions involving potassium dichromate (dichromate).

Dichromate does not have such a wide variety of organic oxidation reactions in the USE.

Oxidation with dichromate is usually carried out only in an acidic medium. With this, chromium is restored to +3. Recovery products:

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

Let's look at some of the reactions.

Consider the oxidation of alcohol. If the oxidation is carried out at the boiling point of the aldehyde, then it will leave the reaction mixture without undergoing oxidation:

Otherwise, alcohol can be directly oxidized to acid.

The aldehyde obtained during the previous reaction can be "caught" and made to oxidize to an acid:

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

If it is difficult to determine the oxidation states of carbon atoms using this formula, you can paint on a draft:

Reaction equation:

Consider the oxidation of cyclopentene.

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

Some features of oxidation in the USE, with which we do not entirely agree.

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

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

It's about HARD oxidation.

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

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

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

Let's analyze the oxidation of each substance:

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 USE, this is just a secondary alcohol. Secondary alcohols are oxidized to ketones, and ketones are not further oxidized:

Let this substance be oxidized with dichromate:

Second substance:

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

Let it oxidize in neutral permanganate when heated:

The resulting alkali is enough to completely neutralize carbon dioxide:

2KOH + CO 2 → K 2 CO 3 + H 2 O

Final equation:

Oxidation of the third substance:

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

Oxidation of the fourth substance:

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

And finally, this is how vinyl benzene is oxidized:

And it is oxidized to benzoic acid, it must be borne in mind that, according to the logic of the USE, it is so 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 you hear some of the points outlined in this article for the first time. As already mentioned, this topic is very broad and controversial. And despite this, for some reason, very little attention is paid to her.

As you may have seen, two or three reactions cannot explain all the patterns of these reactions. Needed here A complex approach and detailed explanation all the moments. 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 whole, 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.

Oxidation reactions of organic substances are often found in basic course chemistry. Moreover, their recording is usually presented in the form of simple schemes, some of which give only general idea on the transformation of substances of various classes into each other, without taking into account the specific conditions of the process (for example, the reaction of the medium), which 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 precisely the reaction equation with a certain set of coefficients. This paper provides recommendations on how to compose such equations.

To describe redox reactions, two methods are used: the method of electron-ion equations and the method of electronic balance. Without dwelling on the first, we note that the electronic balance method is studied in the course of chemistry at the basic school and therefore is quite applicable for continuing the study of the subject.

The electronic balance equations 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 oxidation and reduction processes. This, in turn, allows you to find the rest of the coefficients.

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

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

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, when oxidized in an acidic environment, water is formed. Now the reaction diagram looks like this:

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

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

Electronic balance equation:

Mn +7 + 5e = Mn +2 6

C -3 - 6e = C +3 5

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

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

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

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

Example 2. The reaction of the "silver mirror".

Most literary sources indicate that aldehydes in these reactions are oxidized to the corresponding carboxylic acids. In this case, the oxidizing agent is an ammonia solution of silver (I) oxide - Ag 2 O amm.r-r. In fact, the reaction takes place in an alkaline ammonia environment, therefore, an ammonium salt or CO should be formed. 2 in case of oxidation of formaldehyde.

Consider the oxidation of acetaldehyde with the Tollens reagent:

CH 3 CHO + Ag (NH 3) 2 OH = ...

In this case, the oxidation product will be ammonium acetate, and the reduction product will be silver:

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

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

C +1 - 2e = C +3 1

Ag +1 + 1e = Ag 0 2

Let's put the coefficients in front of the formulas of substances containing carbon and silver atoms:

CH 3 CHO + 2Ag (NH 3) 2 OH = 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, the composition of the reaction products will contain water, the coefficient in front of the formula of which can be found by selection (1):

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

In conclusion, note that alternative way descriptions of OVR - the method of electron-ion equations - with its advantages, requires additional study time for study and development, which, as a rule, is extremely limited. However, the well-known method of electronic balance, when used correctly, leads to the required 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, but they easily dissolve in organic solvents - alcohol, ether, acetone.

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

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

Table. Physical properties of some arenas

Name	Formula	t ° .pl., ° C	t °. boil., ° 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
pair-		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 (vinyl benzene)	C 6 H 5 CH = CH 2	30,6	145,2

Benzene - low-boiling ( tbale= 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 and toxic.

Obtaining arenes (benzene and its homologues)

In the laboratory

1. 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)

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

WITH 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 more than 6 carbon atoms:

C 6 H 14 t , kat→ C 6 H 6 + 4H 2

2. Acetylene trimerization(for benzene only) - 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 found 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) - p 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 arenes

I... OXIDATION REACTIONS

1. Combustion (smoky flame):

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

2. Benzene under normal conditions does not discolor bromine water and an aqueous solution of potassium permanganate

3. Homologues of benzene are oxidized with potassium permanganate (decolorize potassium permanganate):

A) in an acidic medium to benzoic acid

When benzene homologues are exposed to potassium permanganate and other strong oxidants, the side chains are oxidized. No matter how complex the substituent chain is, it is destroyed, with the exception of the a-carbon atom, which is oxidized to a carboxyl group.

Homologues of benzene with one side chain give benzoic acid:

Homologues containing two side chains give 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 benzoic acid salts

C 6 H 5 -CH 3 + 2KMnO 4 → C 6 H 5 COOК + K ОН + 2MnO 2 + H 2 O

II... ADDITIONAL 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 , PtorNi→ 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 homologues upon irradiation or heating

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

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

c) homologues of benzene in the presence of a catalyst

C 6 H 5 -CH 3 + Cl 2 AlCl 3 → (mixture of ort, 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→ WITH H 3 -C 6 H 2 (NO 2) 3 + 3H 2 O

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

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. Serves as a raw material for the production of many aromatic organic compounds - nitrobenzene C 6 H 5 NO 2 (solvent, aniline is obtained from it), 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 production of dyes, drugs and explosives (TNT (tol), or 2,4,6-trinitrotoluene TNT).

Xylenes C 6 H 4 (CH 3) 2. Technical xylene is a mixture of three isomers ( ortho-, meta- and pair-xylenes) - is used as a solvent and initial product for the synthesis of many organic compounds.

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

Chlorine derivatives of benzene used for plant protection. So, the product of substitution of H atoms in benzene by chlorine atoms is hexachlorobenzene С 6 Сl 6 - fungicide; it is used for dry dressing of wheat and rye seeds against hard smut. The product of chlorine addition to benzene is hexachlorocyclohexane (hexachloran) С 6 Н 6 Сl 6 - insecticide; it is used to control harmful insects. The substances mentioned are pesticides - chemicals fight against microorganisms, plants and animals.

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

VIDEO EXPERIENCES

18. Redox reactions (continued 2)

18.9. ORP with the participation of organic substances

In OVR organic substances with inorganic organic substances are most often reducing agents. So, when organic matter is burned in an excess of oxygen, carbon dioxide and water are always formed. The reactions are more complicated when less active oxidants are used. In this section, only the reactions of representatives of the most important classes of organic substances with some inorganic oxidants are considered.

Alkenes. Mild oxidation converts alkenes to glycols (dihydric alcohols). The reducing atoms in these reactions are double-bonded carbon atoms.

The reaction with a solution of potassium permanganate proceeds in a neutral or weakly alkaline medium 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 at the double bond and the formation of two acids (in a strongly alkaline medium - two salts) or an acid and carbon dioxide (in a highly alkaline medium - salt and carbonate):

1) 5CH 3 CH = 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 = 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 = 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 = 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 medium oxidizes alkenes similarly to reactions 1 and 2.

Alkyne. Alkynes begin to oxidize under somewhat harsher conditions than alkenes, so they usually oxidize with a triple bond breaking of the carbon chain. As in the case of alkanes, the reducing atoms here are carbon atoms linked in this case by a triple bond. The reactions produce acids and carbon dioxide. Oxidation can be carried out with potassium permanganate or potassium dichromate in an acidic environment, 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 = 3K 2 C 2 O 4 + 2H 2 O + 8MnO 2 + 2KOH

In an acidic environment, oxidation proceeds to carbon dioxide:

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

Homologues of benzene. Homologues 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 = C 6 H 5 COOK + 2MnO 2 + KOH + H 2 O (when boiling)

C 6 H 5 CH 2 CH 3 + 4KMnO 4 = 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 permanganate in an acidic medium leads to the formation of benzoic acid.

Alcohols. Aldehydes are the direct oxidation product of primary alcohols, and ketones are secondary ones.

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 point of the aldehyde. Evaporation, aldehydes do not have time to oxidize.

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

With an excess of an 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 are not oxidized under these conditions, and methyl alcohol is oxidized to carbon dioxide. All reactions take place when heated.

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 intermediate products (HOCH 2 –COOH, HOOC– COOH, etc.).

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

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

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

18.10. Comparison of the redox activity of various substances

From the definitions of the concepts "oxidizing atom" and "reducing agent" it follows that atoms in the highest oxidation state have only oxidizing properties. On the contrary, atoms in the lowest oxidation state have only reducing properties. Atoms in intermediate oxidation states can be both oxidizing and reducing agents.

At the same time, based only on the oxidation state, it is impossible to unambiguously assess the redox properties of substances. As an example, consider the compounds of group VA elements. Compounds of nitrogen (V) and antimony (V) are more or less strong oxidizing agents, compounds of bismuth (V) are very strong oxidizing agents, and compounds of phosphorus (V) have practically no 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 compounds containing atoms of a given element in this oxidation state are.

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

In this row, the oxidizing properties of ions (top row) increase from left to right, and the reducing properties of metals (bottom row), on the contrary, from right to left.

Taking into account the differences in redox activity in different environments, it is possible to construct similar series for oxidants. So, for reactions in an acidic environment (pH = 0), a "continuation" of a series of metal activities in the direction of enhancing oxidizing properties is obtained

As in the row of metal activities, in this row the oxidizing properties of oxidants (top row) increase from left to right. But, using this series, it is possible to compare the reductive activity of reducing agents (bottom line) only if 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 ORR is possible, we will use the general rule that determines the direction of the redox reactions (the 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 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. Is it possible to oxidize copper to CuCl 2 with a FeCl 3 solution 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 the Fe 3 ions and is a weaker oxidant than these ions. But in aqueous solution oxygen is practically never reduced to H 2 O 2, in which case it is reduced to H 2 O and takes a place between Br 2 and MnO 2. Therefore, such a reaction is possible, however, it proceeds rather slowly (why?).
4. Is it possible to oxidize H 2 O 2 in an acidic environment 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 ions are oxidizing agents that are stronger than the oxygen formed from the peroxide. Therefore, it is possible.

A similar series, built for ORP in an alkaline medium, is as follows:

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

Electron-ion balance method (half-reaction method), intermolecular ORR, intramolecular ORR, ORR dismutation (disproportionation, self-oxidation-self-healing), ORR contamination, passivation.

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

Make up the equations for the reactions between solid chromium trioxide and the following substances: a) C; b) CO; c) S (SO 2); d) H 2 S; e) NH 3; f) C 2 H 5 OH (CO 2 and H 2 O); g) CH 3 COCH 3.
Make up the equations of the reactions occurring when the following substances are added to concentrated nitric acid: a) S (H 2 SO 4); b) P 4 ((HPO 3) 4); c) graphite; d) Se; e) I 2 (HIO 3); f) Ag; g) Cu; i) Pb; j) KF; k) FeO; m) FeS; m) MgO; n) MgS; p) Fe (OH) 2; c) P 2 O 3; m) As 2 O 3 (H 3 AsO 4); y) As 2 S 3; t) 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; d) NH 3; e) NO; f) H 2 Se; g) HI.
The reactions will proceed in the same or different way in the following cases: a) a piece of magnesium was placed in a high test tube two-thirds filled with concentrated nitric acid; b) a drop of concentrated nitric acid was placed on the surface of the magnesium plate? Write the reaction equations.
What is the difference between the reaction of concentrated nitric acid with hydrogen sulfide acid and with gaseous hydrogen sulfide? Write the reaction equations.
Will ORS proceed in the same way when anhydrous crystalline sodium sulfide and its 0.1 M solution are added to a concentrated solution of nitric acid?
A mixture of the following substances was treated with concentrated nitric acid: Cu, Fe, Zn, Si and Cr. Make up the equations of the ongoing reactions.
Make up the equations of the reactions that occur when the following substances are added to dilute nitric acid: a) I 2; b) Mg; c) Al; d) Fe; e) FeO; f) FeS; g) Fe (OH) 2; i) Fe (OH) 3; j) MnS; l) Cu 2 S; m) CuS; m) CuO; n) 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 up the equations of the reactions that occur when the following substances are added to concentrated sulfuric acid: a) Ag; b) Cu; c) graphite; d) HCOOH; e) C 6 H 12 O 6; f) 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. Write the reaction equations. How will the reaction proceed between hot concentrated H 2 SO 4 and hydrogen sulfide?
Why is hydrogen chloride obtained by treating crystalline sodium chloride with concentrated sulfuric acid, while hydrogen bromide and hydrogen iodide are not obtained by this method?
Make up the equations for the reactions occurring 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 up 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 occur when the following substances are heated: a) (NH 4) 2 CrO 4; b) NaNO 3; c) CaCO 3; d) Al (NO 3) 3; e) Pb (NO 3) 3; f) AgNO 3; g) Hg (NO 3) 2; i) Cu (NO 3) 2; j) CuO; k) NaClO 4; m) Ca (ClO 4) 2; m) Fe (NO 3) 2; n) PCl 5; p) MnCl 4; c) H 2 C 2 O 4; m) LiNO 3; y) HgO; t) Ca (NO 3) 2; x) Fe (OH) 3; v) CuCl 2; h) KClO 3; w) KClO 2; y) CrO 3?
When hot solutions of ammonium chloride and potassium nitrate are poured, a reaction occurs, accompanied by the evolution of gas. Write the equation for this reaction.
Make up the equations of the reactions proceeding 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 solution of potassium hydroxide, selenium undergoes dismutation to the nearest stable oxidation states (–II and + IV). Equate this RVR.
Under the same conditions, sulfur undergoes a similar dismutation, but the excess sulfur reacts with sulfite ions to form thiosulfate ions S 2 O 3 2. Make up the equations of the ongoing reactions. ;
Make up the equations of 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. Add an equal volume of diluted sulfuric acid solution to 3-4 drops of potassium permanganate solution, and then sodium sulfite solution until discoloration. Make a reaction equation.

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

Test 3. Oxidizing properties of potassium permanganate in an alkaline medium. Add 10 drops of concentrated sodium hydroxide solution and 2 drops of sodium sulfite solution to 3-4 drops of potassium permanganate solution. The solution should turn green.

Test 4. Oxidizing properties of potassium dichromate in an acidic environment. Acidify 6 drops of potassium dichromate solution with four drops of 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 the other. Add 8-10 drops of diluted sulfuric acid solution to both tubes. Compare the events taking place. EXPERIENCE IN A DRAIN CABINET!

Experience 6. Oxidizing properties of concentrated sulfuric acid. Similar to experiment 5, but add concentrated sulfuric acid solution. One minute after the onset of discharge gaseous products the reaction, introduce strips of filter paper moistened with solutions of potassium permanganate and copper sulfate into test tubes. Explain what is happening. EXPERIENCE IN A DRAIN CABINET!

Experience 7. Oxidizing properties of dilute 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 DRAIN CABINET!

Test 8. Oxidizing properties of concentrated nitric acid. Place a piece of copper tape in a test tube and add 10 drops of concentrated nitric acid solution. Heat gently until the metal is completely dissolved. EXPERIENCE IN A DRAIN CABINET!

Test 9. Oxidizing properties of potassium nitrite. Add an equal volume of diluted sulfuric acid solution and 5 drops of potassium iodide solution to 5-6 drops of potassium nitrite solution. What substances are formed?

Test 10. The reducing properties of potassium nitrite. Add an equal volume of diluted sulfuric acid solution and potassium nitrite solution to 5-6 drops of potassium permanganate solution until the mixture is completely discolored.

Experience 11.Thermal decomposition of copper nitrate. Place one microspatula of copper nitrate trihydrate in a test tube, fix it in a rack and gently heat with an open flame. Observe dehydration and subsequent salt decomposition. EXPERIENCE IN A DRAIN CABINET!

Test 12.Thermal decomposition of lead nitrate. Carry out analogously to experiment 11, placing lead nitrate in a test tube. EXPERIENCE IN A DRAIN CABINET! What is the difference between the processes taking place during the decomposition of these salts?