Select the main carbon chain in the molecule. Firstly, it must be the longest. Secondly, if there are two or more chains of equal length, then the most branched one is selected. For example, in a molecule there are 2 chains with the same number (7) of C atoms (highlighted in color):

In case (a) the chain has 1 substituent, and in (b) - 2. Therefore, you should choose option (b).

1. Number the carbon atoms in the main chain so that the C atoms associated with the substituents receive the lowest numbers possible. Therefore, numbering begins from the end of the chain closest to the branch. For example:

Name all radicals (substituents), indicating in front the numbers indicating their location in the main chain. If there are several identical substituents, then for each of them a number (location) is written separated by a comma, and their number is indicated by prefixes di-, three-, tetra-, penta- etc. (For example, 2,2-dimethyl or 2,3,3,5-tetramethyl).

Place the names of all substituents in alphabetical order (as established by the latest IUPAC rules).

Name the main chain of carbon atoms, i.e. the corresponding normal alkane.

Thus, in the name of a branched alkane, the root + suffix is ​​the name of a normal alkane (Greek numeral + suffix “an”), the prefixes are numbers and names of hydrocarbon radicals. Example of title construction:

Chem. Saints of alkanesCracking of alkanes. Cracking is a process of thermal decomposition of hydrocarbons, which is based on the reactions of splitting the carbon chain of large molecules with the formation of compounds with a shorter chain. Isomerization of alkanes Alkanes of normal structure under the influence of catalysts and upon heating are able to transform into branched alkanes without changing the composition of the molecules, i.e. enter into isomerization reactions. These reactions involve alkanes whose molecules contain at least 4 carbon atoms. For example, the isomerization of n-pentane into isopentane (2-methylbutane) occurs at 100°C in the presence of an aluminum chloride catalyst:

The starting material and the product of the isomerization reaction have the same molecular formulas and are structural isomers (carbon skeleton isomerism).

Dehydrogenation of alkanes

When alkanes are heated in the presence of catalysts (Pt, Pd, Ni, Fe, Cr 2 O 3, Fe 2 O 3, ZnO), their catalytic dehydrogenation– abstraction of hydrogen atoms due to the breaking of C-H bonds.

The structure of dehydrogenation products depends on the reaction conditions and the length of the main chain in the starting alkane molecule.

1. Lower alkanes containing from 2 to 4 carbon atoms in the chain, when heated over a Ni catalyst, remove hydrogen from neighboring carbon atoms and turn into alkenes:

Along with butene-2 this reaction produces butene-1 CH 2 =CH-CH 2 -CH 3. In the presence of a Cr 2 O 3 /Al 2 O 3 catalyst at 450-650 °C from n-butane is also obtained butadiene-1,3 CH 2 =CH-CH=CH 2.

2. Alkanes containing more than 4 carbon atoms in the main chain are used to obtain cyclical connections. This happens dehydrocyclization– dehydrogenation reaction, which leads to the closure of the chain into a stable cycle.

If the main chain of an alkane molecule contains 5 (but not more) carbon atoms ( n-pentane and its alkyl derivatives), then when heated over a Pt catalyst, hydrogen atoms are split off from the terminal atoms of the carbon chain, and a five-membered cycle is formed (cyclopentane or its derivatives):

Alkanes with a main chain of 6 or more carbon atoms also undergo dehydrocyclization, but always form a 6-membered ring (cyclohexane and its derivatives). Under reaction conditions, this cycle undergoes further dehydrogenation and turns into the energetically more stable benzene ring of an aromatic hydrocarbon (arene). For example:

These reactions underlie the process reforming– processing of petroleum products to obtain arenes ( aromatization saturated hydrocarbons) and hydrogen. Transformation n- alkanes in the arena leads to an improvement in the detonation resistance of gasoline.

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

What exactly will we be talking about?

Besides complete oxidation(burning), for some classes organic compounds typical reactions incomplete oxidation, and 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 classic oxidizing agents that, so to speak, are universal for many classes.

This is potassium permanganate - KMnO 4. And potassium bichromate (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 oxidizing agents occur quite often, but nowhere is there a comprehensive guide on what principle to choose the products of such reactions.

In practice, there are many factors that influence the course of the reaction (temperature, environment, concentration of reagents, etc.). Often the result is a mixture of products. Therefore, it is almost impossible to predict the product that will form.

But this is not suitable for the Unified State Exam: you cannot write there “maybe either this, or this, or that, or a mixture of products.” There needs to be specifics.

The writers of the assignments put in a certain logic, a certain principle according to which a certain product should be written. Unfortunately, they didn't share it with anyone.

This issue is rather avoided in most manuals: two or three reactions are given as an example.

In this article I present what can be called the results of a research-analysis of Unified State Examination tasks. The logic and principles of composing oxidation reactions with permanganate and dichromate have been solved quite accurately (in accordance with Unified State Examination standards). First things first.

Determination of oxidation state.

First, when we deal 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 determine the products; the reaction must be equalized. For equalization, the electronic balance method is traditionally used. To apply this method, it is necessary to determine the oxidation states of the reducing agents and oxidizing agents before and after the reaction.

U Not organic matter We know oxidation states from the 9th grade:

But they probably didn’t take the organic class in 9th grade. Therefore, before learning how 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 a little differently, differently than in inorganic chemistry.

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

First you need to remember what an oxidation state is.

The oxidation state is the conventional charge that appears on an atom, assuming that the electron pairs are shifted entirely towards the more electronegative atom.

Therefore, the degree of oxidation is determined by the number of displaced electron pairs: if it is displaced towards a given atom, then it acquires an excess minus (-) charge, if from the atom, then it acquires an excess plus (+) charge. In principle, this is all the theory 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 see in which direction the electron pair will shift and what excess charge (+ or -) will arise from this on the carbon atom.

Let's look at specific examples:

At carbon three bonds with hydrogen. Carbon and hydrogen - which is more electronegative? Carbon, which means that 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 connection is with chlorine. Carbon and chlorine - which is more electronegative? Chlorine, which means that along 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 - which is more electronegative? Carbon, which means that 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 connection to another carbon. Carbon and another carbon - their electronegativity is 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 attract an electron pair from carbon, and carbon acquires a charge of +3.

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

Carbon is connected to hydrogen atoms by two bonds. Carbon, being more electronegative, takes away one pair of electrons for each bond with hydrogen and acquires a charge of -2.

A carbon double bond is connected to an oxygen atom. The more electronegative oxygen attracts one electron pair to itself along each bond. Together it turns out that carbon has two electron pairs. Carbon gains a charge of +2.

Together we get +2 -2 = 0.

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

A triple bond with a more electronegative nitrogen gives the carbon a charge of +3; the bond with carbon does not shift the electron pair.

Oxidation with permanganate.

What will happen to the permanganate?

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

Therefore, it can go in three directions:

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

1. Acidic environment.

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

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

1. Alkaline environment.

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

KMnO 4 + KOH → K 2 MnO 4 + H 2 O

1. Neutral environment(and slightly alkaline).

In a neutral environment, in addition to permanganate, water also reacts (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 organic matter?

The first thing you need to understand 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.

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

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

Oxidation secondary alcohol breaks off at the second stage. Since the carbon is in the middle, a ketone is formed, not an aldehyde (the carbon atom in the ketone group can no longer 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 the conditions for this, the reaction continues. It all ends with a product that does not oxidize under the given conditions: a tertiary alcohol, a ketone or an acid.

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

The peculiarity 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 a carboxyl group:

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

H 2 CO 3 → H 2 O + CO 2

Therefore, methanol, formicaldehyde 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 (write 0 above the reaction ° or 20 °) .

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

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

In this case, the π-bond is broken and the hydroxyl group “sits” on the released bonds. The result is a dihydric alcohol:

Let's write the reaction of mild oxidation of ethylene (ethene). Let's write down the starting substances and predict the products. At the same time, we do not write H 2 O and KOH yet: they can appear either on the right side of the equation or on the left. And we immediately determine the oxidation degrees of the substances involved in the redox reaction:

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

Let's set 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 there will be alkali on the right. We put a coefficient in front of it. There is not enough hydrogen on the left, so there is water on the left. We put a coefficient in front of it:

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

Cycloalkene is often slipped in. Don't let it bother you. It is a regular hydrocarbon with a double bond:

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

1. Containing an aldehyde group.

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

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

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

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

In this case, an acid and an alkali appear side by side in one 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 there are 3 moles of acid, and 2 moles of alkali. 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:

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

A similar situation arises during the oxidation of methanal. It, as we remember, 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 an acid salt and a medium salt can be formed. It depends on the ratio between alkali and carbon dioxide:

If alkali has a 2:1 ratio to carbon dioxide, then the average salt will be:

Or there may be significantly more alkali (more than twice). If it is more than doubled, 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 relates to carbon dioxide as 1:1, then there will be an acidic 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 starting substances, products, draw up a balance, put 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 during 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

So 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:

But 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 during 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 go into the formation of an 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

Final equation:

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

1. Containing a triple bondC≡ C.

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

The π bond breaks and the hydroxyl group is attached to the carbon atoms. It's the same principle here. Just remember that in triple bond there are two π bonds. First this happens along the first π bond:

Then via another π-bond:

A structure in which one carbon atom has two hydroxyl groups is extremely unstable. When something is unstable in chemistry, it tends to make something “fall off.” The water falls off, like this:

This results in a carbonyl group.

Let's look at examples:

Ethine (acetylene). Let's consider the stages of oxidation of this substance:

Water elimination:

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

Consider the oxidation of butine-2:

Water elimination:

No acid is formed here, so there is no need to bother with 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 pentine:

Water elimination:

The result is a substance with an interesting structure:

The aldehyde group continues to oxidize:

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

2 moles of alkali should be formed (since the coefficient in front of permanganate is 2), therefore, all the acid is neutralized:

Severe oxidation.

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

IN acidic environment sometimes they heat it up too. But for severe oxidation to occur in a non-acidic environment, heating is a prerequisite.

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

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

1. Alcohols. Aldehydes.

Let's consider the oxidation of ethanol. It gradually oxidizes to acid:

Let's write down the equation. We write down the starting substances, the products of the redox reaction, enter 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 in very gentle conditions (low heat). In this case, we write aldehyde as the product:

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

Let's consider the oxidation of aldehydes using ethanal. It also oxidizes to acid:

Reaction equation:

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

Metanal:

1. Containing multiple bonds.

In this case, the chain breaks at the multiple bond. And the atoms that formed it undergo oxidation (they acquire a bond with oxygen). Oxidize as much as possible.

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

Let's look at 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. Let's determine the oxidation state of the atoms that change it, draw up a balance, equalize the reaction:

When compiling an electronic balance, we mean that there are two reducing agents - two carbon atoms, and 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 an acid, without the intermediate formation of an aldehyde):

Reaction equation:

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

Reaction equation:

1. Double crowned acid.

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

This is oxalic acid. It’s difficult for two crowns to get along side by side. It is certainly stable under normal conditions. But because it has two carboxylic acid groups attached 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:

1. Benzene homologues (and their derivatives).

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

But its homologues are oxidized. In this case, the circuit also breaks, the main thing is to know where exactly. Some principles apply:

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

Let's look at the oxidation of methylbenzene. There, one carbon atom in the radical is oxidized:

Reaction equation:

Let's look at the oxidation of isobutylbenzene:

Reaction equation:

Let's look at the oxidation of sec-butylbenzene:

Reaction equation:

When benzene homologues (and derivatives of homologues) are oxidized 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”:

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

Severe 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 reactions, these turned out to be the most controversial.

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

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

Highly alkaline environment :

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

KMnO 4 + KOH → K 2 MnO 4 .

In a strongly alkaline environment, there is always an excess of alkali, so complete neutralization will take place: if carbon dioxide is formed, there will be carbonate, if an acid is formed, there will be salt (if the acid is polybasic, there will be a medium salt).

For example, propene oxidation:

Oxidation of ethylbenzene:

Slightly alkaline or neutral environment when heated :

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

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

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

We discussed this when we looked at the mild oxidation of aldehydes.

Therefore, if you form acid in a neutral environment, you need to carefully see whether it is enough to neutralize all the acid. Particular attention should be paid to neutralizing 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 environment. And the reaction equation in both neutral and weakly alkaline media will be the same.

For example, let's look at the oxidation of ethylbenzene:

The alkali is quite enough to completely neutralize the resulting acidic compounds, even excess will remain:

3 moles of alkali are consumed - 1 is left.

Final equation:

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

Redox reactions involving potassium dichromate (dichromate).

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

Oxidation with dichromate is usually carried out only in an acidic environment. In this case, chromium is restored to +3. Recovery Products:

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

Let's look at some reactions.

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

Otherwise, the alcohol may be directly oxidized to acid.

The aldehyde produced in the previous reaction can be “trapped” and forced to oxidize to an acid:

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

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

Reaction equation:

Let's consider the oxidation of cyclopentene.

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

Some features of oxidation in the Unified State Examination, with which we do not entirely agree.

We consider those “rules”, principles and reactions that will be discussed in this section to be 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 Unified State Exam in particular.

But nevertheless, we are forced to provide this material exactly in the form required by the Unified State Exam.

We are talking specifically about HARD oxidation.

Remember how benzene homologues and their derivatives are oxidized under harsh conditions? The radicals are all terminated and carboxyl groups are formed. The scraps undergo oxidation “on their own”:

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

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

Let's look at the oxidation of each substance:

First substance:

You need to ignore the fact that there is a benzene ring. From the point of view of the Unified State Examination, this is just secondary alcohol. Secondary alcohols are oxidized to ketones, but ketones are not oxidized further:

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 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, this is how vinylbenzene is oxidized:

And it oxidizes to benzoic acid, you need to keep in mind that according to the logic of the Unified State Examination, it oxidizes this way not because it is a benzene derivative. But because it contains a double bond.

Conclusion.

That's all you need to know about redox reactions involving permanganate and dichromate in organic matter.

Don’t be surprised if you are hearing 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 it receives very little attention.

As you may have seen, two or three reactions cannot explain all the patterns of these reactions. This requires an integrated approach and detailed explanations of all points. Unfortunately, in textbooks and on Internet resources the topic is not fully covered, or not covered at all.

I tried to eliminate these shortcomings and shortcomings and consider this topic as a 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!

In redox reactions organic substances more often they exhibit the properties of reducing agents, and themselves are oxidized. The ease of oxidation of organic compounds depends on the availability of electrons when interacting with the oxidizing agent. All known factors causing an increase in electron density in molecules of organic compounds (for example, positive inductive and mesomeric effects), will increase their ability to oxidize and vice versa.

The tendency of organic compounds to oxidize increases with their nucleophilicity, which corresponds to the following rows:

Increase in nucleophilicity in the series

Let's consider redox reactions representatives of the most important classes organic matter with some inorganic oxidizing agents.

Oxidation of alkenes

During mild oxidation, alkenes are converted to glycols (dihydric alcohols). The reducing atoms in these reactions are carbon atoms linked by a double bond.

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

3C 2 H 4 + 2KMnO 4 + 4H 2 O → 3CH 2 OH–CH 2 OH + 2MnO 2 + 2KOH

Under more severe conditions, oxidation leads to the rupture of the carbon chain at the double bond and the formation of two acids (in a strongly alkaline environment - two salts) or an acid and carbon dioxide (in a strongly alkaline environment - a salt and a 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

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

3) CH 3 CH=CHCH 2 CH 3 + 8KMnO 4 + 10KOH → CH 3 COOK + C 2 H 5 COOK + 6H 2 O + 8K 2 MnO 4

4) CH 3 CH=CH 2 + 10KMnO 4 + 13KOH → CH 3 COOK + K 2 CO 3 + 8H 2 O + 10K 2 MnO 4

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

During the oxidation of alkenes, in which the carbon atoms at the double bond contain two carbon radicals, two ketones are formed:

Alkyne oxidation

Alkynes oxidize under slightly more severe conditions than alkenes, so they usually oxidize by breaking the carbon chain at the triple bond. As in the case of alkenes, the reducing atoms here are carbon atoms connected by a multiple bond. As a result of the reactions, acids and carbon dioxide are formed. Oxidation can be carried out with potassium permanganate or 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

Acetylene can be oxidized with potassium permanganate in a neutral environment to potassium oxalate:

3CH≡CH +8KMnO 4 → 3KOOC –COOK +8MnO 2 +2KOH +2H 2 O

In an acidic environment, oxidation proceeds to oxalic acid or carbon dioxide:

5CH≡CH +8KMnO 4 +12H 2 SO 4 → 5HOOC –COOH +8MnSO 4 +4K 2 SO 4 +12H 2 O
CH≡CH + 2KMnO 4 +3H 2 SO 4 → 2CO 2 + 2MnSO 4 + 4H 2 O + K 2 SO 4

Oxidation of benzene homologues

Benzene does not oxidize even under fairly harsh conditions. Benzene homologues can be oxidized with a solution of potassium permanganate in a neutral environment to potassium benzoate:

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

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

Oxidation of benzene homologues with potassium dichromate or permanganate in an acidic environment leads to the formation of benzoic acid.

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

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

Oxidation of alcohols

The direct oxidation product of primary alcohols is aldehydes, and the oxidation products of secondary alcohols are ketones.

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. When aldehydes evaporate, they 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

With an excess of oxidizing agent (KMnO 4, K 2 Cr 2 O 7) in any environment, primary alcohols are oxidized to carboxylic acids or their salts, and secondary alcohols are oxidized to ketones.

5C 2 H 5 OH + 4KMnO 4 + 6H 2 SO 4 → 5CH 3 COOH + 4MnSO 4 + 2K 2 SO 4 + 11H 2 O

3CH 3 –CH 2 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 → 3CH 3 –COOH + 2K 2 SO 4 + 2Cr 2 (SO 4) 3 + 11H 2 O

Tertiary alcohols do not oxidize under these conditions, but methyl alcohol is oxidized to carbon dioxide.

Dihydric alcohol, ethylene glycol HOCH 2 –CH 2 OH, when heated in an acidic environment with a solution of KMnO 4 or K 2 Cr 2 O 7, is easily oxidized to oxalic acid, and in a neutral environment to potassium oxalate.

5CH 2 (OH) – CH 2 (OH) + 8КMnO 4 +12H 2 SO 4 → 5HOOC –COOH +8MnSO 4 +4К 2 SO 4 +22Н 2 О

3CH 2 (OH) – CH 2 (OH) + 8KMnO 4 → 3KOOC –COOK +8MnO 2 +2KOH +8H 2 O

Oxidation of aldehydes and ketones

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, Cu(OH) 2. All reactions occur 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 + 2KMnO 4 + 3KOH → CH 3 COOK + 2K 2 MnO 4 + 2H 2 O

5CH 3 CHO + 2KMnO 4 + 3H 2 SO 4 → 5CH 3 COOH + 2MnSO 4 + K 2 SO 4 + 3H 2 O

CH 3 CHO + Br 2 + 3NaOH → CH 3 COONa + 2NaBr + 2H 2 O

"silver mirror" reaction

With an ammonia solution of silver oxide, aldehydes are oxidized to carboxylic acids, which in an ammonia solution give ammonium salts (the “silver mirror” reaction):

CH 3 CH=O + 2OH → CH 3 COONH 4 + 2Ag + H 2 O + 3NH 3

CH 3 –CH=O + 2Cu(OH) 2 → CH 3 COOH + Cu 2 O + 2H 2 O

Formic aldehyde (formaldehyde) is usually oxidized to carbon dioxide:

5HCOH + 4KMnO4 (hut) + 6H 2 SO 4 → 4MnSO 4 + 2K 2 SO 4 + 5CO 2 + 11H 2 O

3CH 2 O + 2K 2 Cr 2 O 7 + 8H 2 SO 4 → 3CO 2 +2K 2 SO 4 + 2Cr 2 (SO 4) 3 + 11H 2 O

HCHO + 4OH → (NH 4) 2 CO 3 + 4Ag↓ + 2H 2 O + 6NH 3

HCOH + 4Cu(OH) 2 → CO 2 + 2Cu 2 O↓+ 5H 2 O

Ketones are oxidized under harsh conditions by strong oxidizing agents with the rupture of C-C bonds and give mixtures of acids:

Carboxylic acids. Among the acids, formic and oxalic acids have strong reducing properties, which oxidize to carbon dioxide.

HCOOH + HgCl 2 =CO 2 + Hg + 2HCl

HCOOH+ Cl 2 = CO 2 +2HCl

HOOC-COOH+ Cl 2 =2CO 2 +2HCl

Formic acid, in addition to acidic properties, also exhibits some properties of aldehydes, in particular, reducing ones. At the same time, it is oxidized to carbon dioxide. For example:

2KMnO4 + 5HCOOH + 3H2SO4 → K2SO4 + 2MnSO4 + 5CO2 + 8H2O

When heated with strong dewatering agents (H2SO4 (conc.) or P4O10) it decomposes:

HCOOH →(t)CO + H2O

Catalytic oxidation of alkanes:

Catalytic oxidation of alkenes:

Oxidation of phenols:

Drawing up equations of redox reactions involving organic substances

IN connection with the introduction of final certification of graduates as the only form high school single state exam(USE) and transition high school on specialized training It is becoming increasingly important to prepare high school students to complete the most “expensive” tasks in part “C” in terms of points. Unified State Exam test in chemistry. Despite the fact that the five tasks of part “C” are considered different: chemical properties of inorganic substances, chains of transformations of organic compounds, calculation problems, all of them are, to one degree or another, related to redox reactions (ORR). If you have mastered the basic knowledge of the theory of ODD, then you can correctly complete the first and second tasks in full, and the third - partially. In our opinion, a significant part of the success of completing part “C” lies precisely in this. Experience shows that if, while studying inorganic chemistry, students cope quite well with tasks on writing OVR equations, then similar tasks in organic chemistry cause them great difficulties. Therefore, throughout the entire course of organic chemistry in specialized classes, we try to develop in high school students the skills of composing ORR equations.

When studying comparative characteristics of inorganic and organic compounds, we introduce students to the use of oxidation state (s.o.) (in organic chemistry, primarily carbon) and methods for determining it:

1) calculation of average s.o. carbon in a molecule of organic matter;

2) definition of s.o. each carbon atom.

Let us clarify in what cases it is better to use one method or another.

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P When studying the topic “Alkanes,” we show that the processes of oxidation, combustion, halogenation, nitration, dehydrogenation, and decomposition belong to redox processes. When writing equations for the reactions of combustion and decomposition of organic substances, it is better to use the average value of s.o. carbon. For example:

We pay attention to the first half of the electron balance: the carbon atom has a fractional d.o. the denominator is 4, so we calculate the transfer of electrons using this coefficient.

In other cases, when studying the topic “Alkanes”, we determine the values ​​of d.o. each carbon atom in the compound, while drawing students' attention to the sequence of substitution of hydrogen atoms at primary, secondary, tertiary carbon atoms:

Thus, we lead students to the conclusion that first the process of substitution occurs at tertiary carbon atoms, then at secondary carbon atoms, and, lastly, at primary carbon atoms.

P When studying the topic “Alkenes”, we consider oxidation processes depending on the structure of the alkene and the reaction environment.

When alkenes are oxidized with a concentrated solution of potassium permanganate KMnO 4 in an acidic environment (hard oxidation), - and - bonds are broken to form carboxylic acids, ketones and carbon monoxide (IV). This reaction is used to determine the position of the double bond.

If the double bond is at the end of the molecule (for example, in butene-1), then one of the oxidation products is formic acid, which is easily oxidized to carbon dioxide and water:

We emphasize that if in an alkene molecule the carbon atom at the double bond contains two carbon substituents (for example, in the 2-methylbutene-2 ​​molecule), then during its oxidation a ketone is formed, since the transformation of such an atom into a carboxyl group atom is impossible without breaking C–C bond, relatively stable under these conditions:

We clarify that if the alkene molecule is symmetrical and the double bond is contained in the middle of the molecule, then only one acid is formed during oxidation:

We inform you that a feature of the oxidation of alkenes, in which the carbon atoms at the double bond contain two carbon radicals, is the formation of two ketones:

When considering the oxidation of alkenes in neutral or slightly alkaline media, we focus the attention of high school students on the fact that under such conditions, oxidation is accompanied by the formation of diols (dihydric alcohols), and hydroxyl groups are added to those carbon atoms between which there was a double bond:

IN In a similar way, we consider the oxidation of acetylene and its homologues, depending on the environment in which the process takes place. So, we clarify that in an acidic environment the oxidation process is accompanied by the formation of carboxylic acids:

The reaction is used to determine the structure of alkynes based on their oxidation products:

In neutral and slightly alkaline environments, the oxidation of acetylene is accompanied by the formation of the corresponding oxalates (oxalic acid salts), and the oxidation of homologues is accompanied by the rupture of the triple bond and the formation of carboxylic acid salts:

IN All rules are practiced with students using specific examples, which leads to their better assimilation of theoretical material. Therefore, when studying the oxidation of arenes in various environments, students can independently make assumptions that the formation of acids should be expected in an acidic environment, and salts in an alkaline environment. The teacher will only have to clarify which reaction products are formed depending on the structure of the corresponding arena.

We show with examples that homologs of benzene with one side chain (regardless of its length) are oxidized by a strong oxidizing agent to benzoic acid at the -carbon atom. When heated, benzene homologues are oxidized by potassium permanganate in a neutral environment to form potassium salts of aromatic acids.

5C 6 H 5 –CH 3 + 6KMnO 4 + 9H 2 SO 4 = 5C 6 H 5 COOH + 6MnSO 4 + 3K 2 SO 4 + 14H 2 O,

5C 6 H 5 –C 2 H 5 + 12KMnO 4 + 18H 2 SO 4 = 5C 6 H 5 COOH + 5CO 2 + 12MnSO 4 + 6K 2 SO 4 + 28H 2 O,

C 6 H 5 –CH 3 + 2KMnO 4 = C 6 H 5 COOK + 2MnO 2 + KOH + H 2 O.

We emphasize that if there are several side chains in an arene molecule, then in an acidic environment each of them is oxidized at the a-carbon atom to a carboxyl group, resulting in the formation of polybasic aromatic acids:

P The acquired skills in drawing up ORR equations for hydrocarbons allow them to be used when studying the section “Oxygen-containing compounds”.

Thus, when studying the topic “Alcohols,” students independently compose equations for the oxidation of alcohols using the following rules:

1) primary alcohols are oxidized to aldehydes

3CH 3 –CH 2 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;

2) secondary alcohols are oxidized to ketones

3) oxidation reaction is not typical for tertiary alcohols.

In order to prepare for Unified State Examination for the teacher It is advisable to provide additional information on these properties, which will undoubtedly be useful for students.

When methanol is oxidized with an acidified solution of potassium permanganate or potassium dichromate, CO 2 is formed; primary alcohols during oxidation, depending on the reaction conditions, can form not only aldehydes, but also acids. For example, the oxidation of ethanol with potassium dichromate in the cold ends with the formation acetic acid, and when heated - acetaldehyde:

3CH 3 –CH 2 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 = 3CH 3 –COOH + 2K 2 SO 4 + 2Cr 2 (SO 4) 3 + 11H 2 O,

3CH 3 –CH 2 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.

Let us again remind students of the influence of the environment on the products of alcohol oxidation reactions, namely: a hot neutral solution of KMnO 4 oxidizes methanol to potassium carbonate, and the remaining alcohols to salts of the corresponding carboxylic acids:

When studying the topic “Aldehydes and ketones,” we focus students’ attention on the fact that aldehydes are more easily oxidized than alcohols into the corresponding carboxylic acids not only under the influence of strong oxidizing agents (air oxygen, acidified solutions of KMnO 4 and K 2 Cr 2 O 7), but and under the influence of weak (ammonia solution of silver oxide or copper(II) hydroxide):

5CH 3 –CHO + 2KMnO 4 + 3H 2 SO 4 = 5CH 3 –COOH + 2MnSO 4 + K 2 SO 4 + 3H 2 O,

3CH 3 –CHO + K 2 Cr 2 O 7 + 4H 2 SO 4 = 3CH 3 –COOH + Cr 2 (SO 4) 3 + K 2 SO 4 + 4H 2 O,

CH 3 –CHO + 2OH CH 3 –COONH 4 + 2Ag + 3NH 3 + H 2 O.

We pay special attention to the oxidation of methanal with an ammonia solution of silver oxide, because in this case, ammonium carbonate is formed, not formic acid:

HCHO + 4OH = (NH 4) 2 CO 3 + 4Ag + 6NH 3 + 2H 2 O.

As our many years of experience show, the proposed methodology for teaching high school students how to compose ORR equations involving organic substances increases their final Unified State Exam result in chemistry for a few points.

18. Redox reactions (continued 2)

18.9. OVR involving organic substances

In the ORR of organic substances with inorganic substances, organic substances are most often reducing agents. Thus, when organic matter burns in excess oxygen, carbon dioxide and water are always formed. Reactions are more complicated when using less active oxidizing agents. This section discusses only the reactions of representatives of the most important classes of organic substances with some inorganic oxidizing agents.

Alkenes. During mild oxidation, alkenes are converted to glycols (dihydric alcohols). The reducing atoms in these reactions are carbon atoms linked by a double bond.

The reaction with a solution of potassium permanganate occurs in a neutral or slightly 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 rupture of the carbon chain at the double bond and the formation of two acids (in a strongly alkaline environment - two salts) or an acid and carbon dioxide (in a strongly alkaline environment - a salt and a 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.

Alkynes. Alkynes begin to oxidize under slightly more severe conditions than alkenes, so they usually oxidize by breaking the carbon chain at the triple bond. As in the case of alkanes, the reducing atoms here are carbon atoms, in this case connected by a triple bond. As a result of the reactions, acids and carbon dioxide are formed. Oxidation can be carried out with potassium permanganate or 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

Benzene homologues. Benzene homologues can be oxidized with a solution of potassium permanganate in a neutral environment to potassium benzoate:

C 6 H 5 CH 3 + 2KMnO 4 = C 6 H 5 COOK + 2MnO 2 + KOH + H 2 O (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 environment leads to the formation of benzoic acid.

Alcohols. The direct oxidation product of primary alcohols is aldehydes, and the oxidation products of secondary alcohols are ketones.

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. When aldehydes evaporate, they 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 oxidizing agent (KMnO 4, K 2 Cr 2 O 7) in any environment, primary alcohols are oxidized to carboxylic acids or their salts, and secondary alcohols are oxidized to ketones. Tertiary alcohols do not oxidize under these conditions, but methyl alcohol is oxidized to carbon dioxide. All reactions occur when heated.

Dihydric alcohol, ethylene glycol HOCH 2 –CH 2 OH, when heated in an acidic environment 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, 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 occur 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 atom” 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 agents and reducing agents.

At the same time, based only on the degree of oxidation, it is impossible to unambiguously assess the redox properties of substances. As an example, consider the connections of elements of the VA group. Nitrogen(V) and antimony(V) compounds are more or less strong oxidizing agents, bismuth(V) compounds are very strong oxidizing agents, and phosphorus(V) compounds have virtually no oxidizing properties. In this and other similar cases, what matters is how characteristic a given oxidation state is for a given element, that is, how stable are the compounds containing atoms of a given element in this oxidation state.

Any redox reaction 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 ORR occurring, like any other reaction, can be determined by the sign of the change in the Gibbs energy. In addition, to quantify the redox activity of substances, the electrochemical characteristics of oxidizing agents 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 metal stresses 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 you place ions (oxidizing agents) 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 (top line) increase from left to right, and the reducing properties of metals (bottom line), 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 oxidizing agents. Thus, for reactions in an acidic environment (pH = 0), a “continuation” of the series of metal activities in the direction of increasing oxidative properties is obtained

As in the metal activity series, in this series the oxidizing properties of oxidizing agents (top line) increase from left to right. But, using this series, it is possible to compare the reducing activity of reducing agents (bottom line) only if their oxidized form coincides with that shown in the top line; in this case it intensifies from right to left.

Let's look at a few examples. To find out whether this ORR is possible, we will use a general rule that determines the direction 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 reduce 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 in an acidic environment with a solution of FeCl 3?
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 to obtain a solution of FeCl 3 by blowing oxygen through a solution of FeCl 2 acidified with hydrochloric acid?
It would seem not, since in our series oxygen is to the left of the Fe 3 ions and is a weaker oxidizing agent than these ions. But in 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 environment with potassium permanganate?
In this case, H 2 O 2 is a reducing agent and a stronger reducing agent than Mn 2B ions, and MnO 4 ions are stronger oxidizing agents than the oxygen formed from peroxide. Therefore, it is possible.

A similar series constructed for ORR in an alkaline medium is as follows:

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

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

1. Using the electron-ion balance method, compose equations for the reactions that occur when a) H 2 S (S, more precisely, S 8 ) solution 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; e) K 2 SO 3; e) HNO 2; g) KNO 2; i) KI (I 2 ); j) FeSO 4; l) C 2 H 5 OH (CH 3 COOH); m) CH 3 CHO; n) (COOH) 2 (CO 2 ); n) K 2 C 2 O 4 . Here and below, where necessary, oxidation products are indicated in curly brackets.
2. Write down equations for the reactions that occur when the following gases are passed 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; e) HCHO.
3. The same, but a reducing agent solution 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; e) KI.
4. The same, but a solution of potassium hydroxide is 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).
5. Write down equations for the following reactions occurring in solution: a) KMnO 4 + H 2 S ...;
   b) KMnO 4 + HCl ...;
   c) KMnO 4 + HBr ...;
   d) KMnO 4 + HI ...
6. Make up the following equations for the ORR of manganese dioxide:
7. 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; d) KNO 2; e) KI; f) FeSO 4; g) CH 3 CH 2 CHO; i) H 2 SO 3; j) KHSO 3; k) K 2 SO 3. Write down the equations for the reactions that take place.
8. The same, but the following gases are passed through the solution: a) H 2 S; b) SO 2.
9. Solutions of a) K 2 S (K 2 SO 4 ); b) K 2 SO 3; c) KNO 2; d) KI (KIO 3). Write down the equations for the reactions that take place.
10. A solution of potassium hydroxide was added to the solution of chromium(III) chloride until the initially formed precipitate was dissolved, and then bromine water was added. Write down the equations for the reactions that take place.
11. 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.
12. Write down the equations for the reactions occurring 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.

13. Write down equations for the reactions occurring between solid chromium trioxide and the following substances: a) C; b) CO; c) S (SO 2 ); d) H 2 S; e) NH 3; e) C 2 H 5 OH (CO 2 and H 2 O); g) CH 3 COCH 3 .
14. Write down equations for the reactions that occur 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; l) FeO; m) FeS; m) MgO; n) MgS; p) Fe(OH) 2; c) P 2 O 3; t) As 2 O 3 (H 3 AsO 4 ); y) As 2 S 3 ; f) Fe(NO 3) 2; x) P 4 O 10; v) Cu 2 S.
15. 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.
16. The reactions will proceed the same or differently in the following cases: a) a piece of magnesium was placed in a tall 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 down reaction equations.
17. What is the difference between the reaction of concentrated nitric acid with hydrogen sulfide acid and with gaseous hydrogen sulfide? Write down reaction equations.
18. Will the ORR 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?
19. A mixture of the following substances was treated with concentrated nitric acid: Cu, Fe, Zn, Si and Cr. Write down the equations for the reactions that take place.
20. Write down equations for 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; l) CuS; m) CuO; n) Na 2 S cr; p) Na 2 S p; c) P 4 O 10 .
21. What processes will occur when a) ammonia, b) hydrogen sulfide, c) carbon dioxide are passed through a dilute solution of nitric acid?
22. Write down equations for the reactions that occur when added to concentrated sulfuric acid the following substances: a) Ag; b) Cu; c) graphite; d) HCOOH; e) C 6 H 12 O 6; f) NaCl cr; g) C 2 H 5 OH.
23. 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 down reaction equations. How will the reaction proceed between hot concentrated H 2 SO 4 and hydrogen sulfide?
24. Why is hydrogen chloride obtained by treating crystalline sodium chloride with concentrated sulfuric acid, but hydrogen bromide and hydrogen iodide are not obtained by this method?
25. Write down equations for 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.
26. Write down reaction equations that characterize the redox properties of hydrogen peroxide:

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

27. 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; l) 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; r) LiNO 3; y) HgO; f) Ca(NO 3) 2; x) Fe(OH) 3; v) CuCl 2; h) KClO 3; w) KClO 2; y) CrO 3 ?
28. When hot solutions of ammonium chloride and potassium nitrate are combined, a reaction occurs accompanied by the release of gas. Write an equation for this reaction.
29. Write down equations for the reactions that occur when a) chlorine, b) bromine vapor is passed through a cold solution of sodium hydroxide. The same, but through a hot solution.
30. When interacting with a hot concentrated solution of potassium hydroxide, selenium undergoes dismutation to the nearest stable oxidation states (–II and +IV). Write an equation for this ORR.
31. Under the same conditions, sulfur undergoes a similar dismutation, but excess sulfur reacts with sulfite ions to form thiosulfate ions S 2 O 3 2. Write down the equations for the reactions that take place. ;
32. Write down equations for the electrolysis reactions of a) a solution of copper nitrate with a silver anode, b) a solution of lead nitrate with a copper anode.

Experience 1. Oxidative properties of potassium permanganate in an acidic environment. To 3-4 drops of a solution of potassium permanganate, add an equal volume of a dilute solution of sulfuric acid, and then a solution of sodium sulfite until discolored. Write an equation for the reaction. 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 released as a precipitate? Experience 3. Oxidative 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. Oxidative 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 test tube and a piece of copper tape in the other. Add 8-10 drops of a dilute sulfuric acid solution to both test tubes. Compare the phenomena occurring. CONDUCT THE EXPERIMENT IN A FUME HOOK! 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 release of gaseous reaction products, introduce strips of filter paper moistened with solutions of potassium permanganate and copper sulfate into the test tubes. Explain the occurring phenomena. CONDUCT THE EXPERIMENT IN A FUME HOOK! 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. CONDUCT THE EXPERIMENT IN A FUME HOOK! 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. Heat gently until the metal is completely dissolved. CONDUCT THE EXPERIMENT IN A FUME HOOK! Experience 9. Oxidizing properties of potassium nitrite. To 5-6 drops of potassium nitrite solution add an equal volume of a dilute sulfuric acid solution and 5 drops of potassium iodide solution. What substances are being formed? Experience 10. Reducing 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 until the mixture is completely discolored. Experience 11.Thermal decomposition of copper nitrate. Place one microspatula of copper nitrate trihydrate in a test tube, secure it in a stand and gently heat it with an open flame. Observe dehydration and subsequent decomposition of salt. CONDUCT THE EXPERIMENT IN A FUME HOOK! Experience 12.Thermal decomposition of lead nitrate. Carry out the same procedure as experiment 11, placing lead nitrate in a test tube. CONDUCT THE EXPERIMENT IN A FUME HOOK! What is the difference between the processes that occur during the decomposition of these salts?