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

What exactly will be discussed?

In addition to complete oxidation (combustion), some classes of organic compounds are characterized by partial 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 oxidizing agents are quite common, but nowhere is there a holistic guide on how to choose the products of such reactions.

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

But this is not good for the Unified State Examination: there you can’t write “maybe either this, or this, or otherwise, or a mixture of products.” There needs to be specifics.

The compilers of the assignments have invested a certain logic, a certain principle according to which a certain product should be written. Unfortunately, they did not share with anyone.

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

I present in this article what can be called the results of a study-analysis of USE tasks. The logic and principles of compiling the oxidation reactions with permanganate and dichromate have been unraveled with fairly high accuracy (in accordance with the USE standards). About everything in order.

Determination of the degree of oxidation.

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 the organic (namely, carbon atoms).

It is not enough to define 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 reducing agents and oxidizing agents before and after the reaction.

For inorganic substances, we know the oxidation states from grade 9:

But in organic, 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 degree of oxidation of carbon in organic substances. This is done a little differently than in inorganic chemistry.

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

First you need to remember what an oxidation state is.

The oxidation state is the conditional charge that occurs on an atom, assuming that the electron pairs are shifted completely towards the more electronegative atom.

Therefore, the oxidation state is determined by the number of displaced electron pairs: if it is shifted 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 degree of oxidation 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 hydrogen bonds. Carbon and hydrogen - which is more electronegative? Carbon, then, 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 - which is more electronegative? Chlorine, which means that over this bond, the electron pair will shift towards chlorine. Carbon has one positive +1 charge.

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 to hydrogen. Carbon and hydrogen - which is more electronegative? Carbon, then, 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 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 one more bond with another oxygen atom (as part of the OH group). More electronegative oxygen atoms in three bonds pull an electron pair from carbon, and carbon has a +3 charge.

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

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

A carbon double bond is linked to an oxygen atom. The more electronegative oxygen attracts one electron pair for each bond. Together, two electron pairs are pulled from carbon. Carbon acquires 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 permanganate?

The redox reaction with permanganate can proceed in different environments (neutral, alkaline, acidic). And it depends on the medium how exactly 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 his recovery:

1. acid 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

1. Alkaline environment.

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

KMnO 4 + KOH → K 2 MnO 4 + H 2 O

1. 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 environment (in the presence of a low concentration KOH solution):

KMnO 4 + KOH → MnO 2 + H 2 O

What will happen to organics?

The first thing to learn 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, the carbon atom "acquires" a bond with oxygen. Therefore, when writing down the scheme of the oxidation reaction, they write [O] above the arrow:

primary alcohol oxidized first to an aldehyde, then to a carboxylic acid:

Oxidation secondary alcohol breaks in 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 oxidized

The oxidation process is stepwise - as long as there is where to oxidize and there are all conditions for this - the reaction goes on. Everything ends up with a product that does not oxidize under 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:

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 can see that this is nothing more than an aldehyde group:

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

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

H 2 CO 3 → H 2 O + CO 2

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

mild oxidation.

Mild oxidation is oxidation without strong heating in a neutral or slightly alkaline medium (0 is written 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 enter into a mild oxidation reaction?

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

In this case, the π-bond breaks and "sits" on the released bonds along 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 yet: 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 involved in the OVR:

Let's make an electronic balance (we mean that there are two or two carbon atoms of the reducing agent, they are oxidized separately):

Let's set the coefficients:

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

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

Cycloalkene is often slipped. Let him not confuse you. This is a regular hydrocarbon with a double bond:

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

1. containing an aldehyde group.

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

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

In a neutral medium and slightly alkaline, 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, in the same mixture, acid and alkali are nearby. Neutralization takes place.

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 only neutralize 2 moles of acid (2 moles of salt are formed). And one mole of acid remains. So the final equation will be:

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

A similar situation arises in 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 will react with alkali. And since carbonic acid is dibasic, both an acid salt and an average salt can be formed. It depends on the ratio between alkali and carbon dioxide:

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

Or alkali can be significantly 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 was added to the reaction mixture before the reaction) or in a neutral environment, when a lot of alkali is formed.

But if alkali is related to carbon dioxide as 1:1, then there will be an acid 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 be in a neutral environment if little alkali is formed.

We write down the starting 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

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 hydrocarbonate and one mole of carbonate:

And in a slightly 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 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 to form 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 + H2O

It turns out like this:

3HOOC–COOH + 4KOH → 2KOOC–COOH + KOOC–COOK + H2O

Final equation:

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

1. containing a triple bondC≡ C.

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

The π-bond breaks, attaches to the carbon atoms at the hydroxyl group. Here the same principle. Just remember that there are two pi bonds in a triple bond. First, this happens at 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 unstable in chemistry, it tends to “fall off” something. Water falls off, like this:

This results in a carbonyl group.

Consider examples:

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

Water splitting:

As in the previous example, in one reaction mixture, acid and alkali. Neutralization occurs - salt is formed. As can be seen from the coefficient in front of 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:

Water splitting:

No acid is 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 clearly demonstrated by the example of pentyn:

Water splitting:

It turns out a substance of an interesting structure:

The aldehyde group continues to oxidize:

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

Alkali should form 2 mol (since the coefficient in front of permanganate is 2), therefore, all acid is neutralized:

Hard oxidation.

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

In an acidic environment, they are also sometimes heated. But in order for hard oxidation to proceed not in an 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 the nuances that arise during oxidation in a strongly alkaline and neutral or slightly alkaline (when heated) environment).

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

1. Alcohols. Aldehydes.

Consider the oxidation of ethanol. Gradually, it oxidizes to an acid:

We write down the equation. We write down the starting substances, OVR products, put down the oxidation states, draw up a balance. 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:

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

Consider the oxidation of aldehydes in terms of 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 along the multiple bond. And the atoms that formed it undergo oxidation (acquire a bond with oxygen). Oxidize as much as possible.

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

Let's analyze the oxidation of pentene-2:

Oxidation of "scraps":

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

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

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

Reaction equation:

Absolutely the same principle in 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 a 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:

1. Double corona acid.

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

This is oxalic acid. Two crowns side by side are difficult to get along. It is certainly stable under 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:

1. 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 circuit also breaks, the main thing is to know exactly where. Some principles apply:

1. The benzene ring itself is not destroyed, and remains intact until the end, the bond is broken 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 gap will be after it.

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

Reaction equation:

Let's analyze the oxidation of isobutylbenzene:

Reaction equation:

Let's analyze the oxidation of sec-butylbenzene:

Reaction equation:

During the oxidation of benzene homologues (and derivatives of homologues) with several radicals, two-three- and more basic aromatic acids are formed. For example, the 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 down the reaction of hard oxidation with permanganate in an acidic environment":

1. Write down the starting materials (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. Record the permanganate reduction product (MnSO 4 + K 2 SO 4 + H 2 O).
4. Determine the degree of oxidation in OVR participants. Draw up a balance. Put down the coefficients for the oxidizing agent and 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, in accordance with this, put the 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 medium and a neutral or slightly alkaline (when heated) medium.

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

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

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

Strongly alkaline environment :

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

KMnO 4 + KOH → K 2 MnO 4 .

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

For example, the oxidation of propene:

Ethylbenzene oxidation:

Slightly alkaline or neutral when heated :

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

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

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

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

Therefore, if you have acid in a neutral environment, you need to carefully see if it is enough to neutralize all the acid. Particular attention should be paid 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 during oxidation in a neutral environment is quite enough. And the reaction equation that in a neutral, that in a slightly alkaline medium will be the same.

For example, consider the oxidation of ethylbenzene:

Alkali is enough to completely neutralize the resulting acid compounds, even excess will remain:

3 moles of alkali are consumed - 1 remains.

Final equation:

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

Redox reactions involving potassium dichromate (bichromate).

Bichromate does not have such a wide variety of organic oxidation reactions in the exam.

Oxidation with dichromate is usually carried out only in an acidic environment. At the same time, chromium is restored to +3. Recovery products:

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

Let's take a 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 their reaction mixture without being oxidized:

Otherwise, the alcohol can be directly oxidized to an acid.

The aldehyde produced in the previous reaction can be "caught" and made 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:

Consider the oxidation of cyclopentene.

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

Some features of oxidation in the 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 of the school curriculum and the USE in particular.

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

We are talking about HARD oxidation.

Remember how benzene homologues and their derivatives are oxidized under harsh conditions? All radicals are terminated - carboxyl groups are formed. Scraps are oxidized already "independently":

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

The functional group and multiple bond is 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 exam, this is just 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, just 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 strongly alkaline environment. The reaction equation will be:

And finally, this is how vinylbenzene is oxidized:

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

Conclusion.

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

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

As you may have seen, two or three reactions do not explain all the patterns of these reactions. Here you need an integrated approach and a detailed explanation of all points. Unfortunately, in textbooks and on Internet resources, the topic is not fully disclosed, or not disclosed at all.

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

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

Drawing up equations of redox reactions involving organic substances

IN In connection with the introduction of the Unified State Examination (USE) as the only form of final certification of secondary school graduates and the transition of high school to specialized education, the preparation of high school students for the most “expensive” tasks in terms of points of part “C” of the USE test in chemistry is becoming increasingly important. Despite the fact that the five tasks of part “C” are considered different: the chemical properties of inorganic substances, the chains of transformations of organic compounds, computational tasks, all of them are to some extent related to redox reactions (ORRs). If the basic knowledge of the OVR theory is mastered, then it is possible to correctly complete the first and second tasks in full, and the third - partially. In our opinion, a significant part of the success in the implementation of part "C" lies precisely in this. Experience shows that if, studying inorganic chemistry, students cope well enough with the tasks of writing OVR equations, then similar tasks in organic chemistry cause great difficulties for them. Therefore, throughout the study of the entire course of organic chemistry in specialized classes, we try to develop in high school students the skills of compiling OVR equations.

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

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

2) definition of s.d. every carbon atom.

We clarify in which cases it is better to use one or another method.

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

We pay attention to the first half of the electronic balance: at the carbon atom in the fractional value of s.d. 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 s.d. 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 bring students to the conclusion that at the beginning the process of substitution takes place at the tertiary, then at the secondary, and, last of all, at the primary carbon atoms.

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

When alkenes are oxidized with a concentrated solution of potassium permanganate KMnO 4 in an acidic medium (hard oxidation), - and - bonds break with the formation of 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 the alkene molecule the carbon atom at the double bond contains two carbon substituents (for example, in the molecule of 2-methylbutene-2), then during its oxidation a ketone is formed, since the transformation of such an atom into an atom of the carboxyl group is impossible without breaking the C–C bond, which is 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 report that a feature of the oxidation of alkenes, in which the carbon atoms in the double bond contain two carbon radicals, is the formation of two ketones:

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 attached 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 medium 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 by oxidation products:

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

IN All rules are worked out with students on specific examples, which leads to a better assimilation of theoretical material. Therefore, when studying the oxidation of arenes in various media, students can independently make assumptions that in an acidic medium one should expect the formation of acids, and in an alkaline medium, salts. The teacher will only have to clarify which reaction products are formed depending on the structure of the corresponding arena.

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

5C 6 H 5 -CH 3 + 6KMnO 4 + 9H 2 SO 4 \u003d 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 \u003d 5C 6 H 5 COOH + 5CO 2 + 12MnSO 4 + 6K 2 SO 4 + 28H 2 O,

C 6 H 5 -CH 3 + 2KMnO 4 \u003d 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 medium each of them is oxidized at an a-carbon atom to a carboxyl group, resulting in the formation of polybasic aromatic acids:

P The acquired skills in compiling OVR equations for hydrocarbons make it possible to use them in the study of the “Oxygen-containing compounds” section.

So, when studying the topic “Alcohols”, students independently compose the 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 \u003d 3CH 3 -CHO + K 2 SO 4 + Cr 2 (SO 4) 3 + 7H 2 O;

2) secondary alcohols are oxidized to ketones

3) for tertiary alcohols, the oxidation reaction is not typical.

In order to prepare for the exam, it is advisable for the teacher to give additional information to 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 of acetic acid, and when heated, acetaldehyde:

3CH 3 -CH 2 OH + 2K 2 Cr 2 O 7 + 8H 2 SO 4 \u003d 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 remind students again about 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 the attention of students on the fact that aldehydes are more easily oxidized than alcohols into the corresponding carboxylic acids not only under the action of strong oxidizing agents (oxygen in the air, acidified solutions of KMnO 4 and K 2 Cr 2 O 7), but also under the influence of weak ones (ammonia solution of silver oxide or copper (II) hydroxide):

5CH 3 -CHO + 2KMnO 4 + 3H 2 SO 4 \u003d 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 \u003d 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, since in this case, ammonium carbonate is formed, and not formic acid:

HCHO + 4OH \u003d (NH 4) 2 CO 3 + 4Ag + 6NH 3 + 2H 2 O.

As our long-term experience shows, the proposed method of teaching high school students how to write OVR equations with the participation of organic substances increases their final USE result in chemistry by several points.

Oxidation of alkenes with potassium permanganate in an alkaline medium when heated (harsh conditions) leads to the destruction of their carbon skeleton at the site of the double bond. In this case, depending on the number of alkyl groups associated with the vinyl fragment, two carboxylic acids can be obtained, an acid and a ketone or two ketones:

Exercise 11. What product is formed during the oxidation of cyclohexene (a) with a dilute solution of potassium permanganate in the cold and (b) with a concentrated solution of potassium permanganate, followed by acidification.

Exercise 12. What products are formed from 1,2-dimethylcyclohexene during its (a) catalytic hydrogenation, (b) oxidation with a dilute solution of potassium permanganate in the cold, (c) ozonation followed by reductive cleavage.

6.5. Oxidation of ethylene to acetaldehyde

Oxidation of ethylene with atmospheric oxygen in the presence of palladium(II) and copper(II) chlorides results in the formation of acetaldehyde ( Wacker process):

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ethanal (acetaldehyde)

6.6. Ethylene chlorine oxidation

Vinyl chloride is obtained by ethylene chlorine:

6.7. Oxidative ammonolysis

The oxidation of hydrocarbons with atmospheric oxygen in the presence of ammonia leads to the transformation of the methyl group into a cyano group. This oxidation is called oxidative ammonolysis. Acrylonitrile is obtained by oxidative ammonolysis of propylene.

acrylonitrile

Hydrocyanic acid is obtained by oxidative ammonolysis of methane:

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7. Hydroformylation of alkenes (Oxosynthesis)

At a temperature of 30 to 250 about C and a pressure of 100-400 atm. in the presence of dicobaltoctacarbonyl, alkenes add hydrogen and carbon monoxide to form aldehydes. Usually a mixture of isomers is obtained:

Mechanism:

1. Ligand cleavage

2. Addition of ethylene

3. Introduction of ethylene

4. Attachment of a ligand

5. Introduction of CO

6. Oxidative addition of hydrogen

7. Reductive elimination of propanal

8. Addition of carbenes and carbenoids

In recent years, much attention in organic chemistry has been paid to compounds of divalent carbon - carbenes. Most of the carbenes are unstable and react immediately after their formation with other compounds.

8.1. The structure of carbenes

The unsubstituted carbene:CH 2 , also called methylene, can be in singlet or triplet form. In the singlet form of a carbene, two unbonding electrons with paired spins are in the same orbital, while in the triplet form, two unpaired electrons with parallel spins are in two orbitals of the same energy. Different electronic configurations of singlet and triplet carbenes are reflected both in different geometry of these particles and in different chemical activity. The divalent carbon atom of the singlet carbene is in the sp 2 -hybrid state, both electrons are located in the sp 2 -hybrid orbital (HOMO), and the p-orbital (LUMO) is free. The triplet carbene is characterized by the sp hybridization of divalent carbon; in this case, two unpaired electrons are located on two p-orbitals, i.e., the triplet carbene is a biradical. The H-C-H angle for singlet methylene, according to spectral data, is equal to 102-105 0 , and for triplet methylene this angle increases to 135140 o . This corresponds to the higher stability of triplet methylene. According to quantum mechanical calculations, triplet methylene is indeed 10 kcal/mol more stable than singlet methylene.

Substituents, however, cause a change in the relative stability of these two forms of carbenes. For dialkylcarbenes, the triplet form is also more stable than the singlet form, but for dihalocarbenes : CHal 2 , and other carbenes with substituents containing a lone pair of electrons, the ground state is singlet. The C1-C-C1 bond angle of 106° for dichlorocarbene is in good agreement with the singlet form. The higher stability of the singlet form of dihalocarbenes compared to the triplet form is apparently due to its stabilization due to the lone pair of electrons of the heteroatom

Such stabilization of the triplet form of dihalocarbenes is impossible. According to the data of quantum mechanical calculation, the energy of the singlet-triplet transition for dichlorocarbene is 13.5 Kcal/mol.

A. Dichlorocarbene

To generate dihalocarbenes, methods have been developed based on the reaction of α-elimination of hydrogen halide from trihalomethanes under the action of strong bases. This method was historically the first to generate the first of the carbenes, dichlorocarbene, as an intermediate (J. Hine 1950). When interacting with strong bases, an anion is formed from chloroform (pKa of chloroform is ~16), bromoform (pKa = 9) and other trihalomethanes, which is stabilized by elimination of the halide ion to form dihalocarbene. By the action of strong bases on chloroform, dichlorocarbene is obtained:

dichlorocarbene

Organolithium compounds in an indifferent aprotic medium can also be used as a base. Then, below -100 0 С, the formation of trichloromethyllithium as an intermediate can be recorded.

Strong bases such as RLi can be used to generate carbenes from 1,1-dihalogen derivatives

In recent years, to generate dihalocarbenes instead of nα-butyllithium is widely used as the base sodium bis(trimethylsilyl)amide.

This releases a chemically inert silazane [bis(trimethylsilyl)amide]. Sodium bis(trimethylsilyl)amide, in contrast to n-butyllithium, can be isolated in an inert atmosphere in dry form. In practice, its ether solutions are more often used, which can be stored at room temperature for a long time.

Dichlorocarbene can also be generated by thermal decarboxylation of dry sodium trichloroacetate:

One of the most accessible modern methods for the generation of dichlorocarbene from chloroform under the action of sodium hydroxide under conditions of interfacial catalysis will be considered in detail later.

Dichlorocarbene adds to alkenes to give dichlorocyclopropanes. The addition occurs stereospecifically - the configuration of the initial alkene is also preserved in the reaction product - cyclopropane:

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trance-2-butene trance-1,2-dimethyl-3,3-

dichlorocyclopropane

(70)

cis-2-butene qiWith-1,2-dimethyl-3,3-

dichlorocyclopropane

(71)

7,7-dichlorobicycloheptane

During the reduction of 1,1-dihalocyclopropanes under the action of lithium in mpem-butyl alcohol, zinc in acetic acid or sodium in liquid ammonia, both halogen atoms are replaced by hydrogen. This is one of the common methods for the preparation of cyclopropane derivatives.

bicycloheptane

Ex. eleven. Complete reactions:

(Z)-3-methyl-2-pentene methylenecyclohexane

Answer

B. methylene

Methylene can be obtained by decomposition of diazomethane. Diazomethane is a relatively unstable substance that decomposes into nitrogen and methylene upon irradiation.

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diazomethane

Methylene:CH 2 during the photolysis of diazomethane is formed in a less stable singlet form. Under the reaction conditions, singlet methylene rapidly loses energy as a result of collisions with diazomethane or nitrogen molecules and turns into more stable triplet methylene.

A singlet carbene is characterized by synchronous addition to the double bond of the alkene with complete preservation of the geometry at the double bond (α-cycloaddition reaction). The addition of the singlet form of the carbene to the double bond thus occurs strictly stereospecifically.

B. Simmons reaction-Smith

An efficient and experimentally very simple method for the conversion of alkenes to cyclopropane derivatives is based on the reaction of alkenes with methylene iodide and a zinc-copper alloy. This reaction was discovered in 1958 by Simmons and Smith and immediately gained wide popularity in the synthesis of cyclopropane derivatives. The active species in this reaction is not a carbene : CH 2, and the carbenoid is iodomethylzinc iodide IZnCH 2 I, formed by the interaction of methylene iodide and a zinc-copper pair.

diiodomethane iodomethylzinkiodide

(Simmons-Smith reagent)

(75)

The reaction proceeds according to the following mechanism:

The Simmons-Smith reaction is a very convenient method for converting alkenes to cyclopropanes.

Ex. 12. Complete reactions:

Answer

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methylenecyclopentane spiroheptane

(77)

styrene cyclopropylbenzene

In tasks of the C3 category of the Unified State Examination, the reactions of oxidation of organic substances with potassium permanganate KMnO 4 in an acidic environment, occurring with a break in the carbon chain, cause particular difficulties. For example, the propene oxidation reaction proceeding according to the equation:

CH 3 – CH = CH 2 + KMnO4 + H 2 SO 4 → CH 3 COOH + CO 2 + MnSO 4 + K 2 SO 4 + H 2 Oh

To factor in complex redox equations like this one, the standard technique suggests an electronic balance, but after another attempt, it becomes obvious that this is not enough. The root of the problem here lies in the fact that the coefficient in front of the oxidizer, taken from the electronic balance, must be replaced. This article offers two methods that allow you to choose the right factor in front of the oxidizer, in order to finally equalize all the elements. Substitution method to replace the coefficient in front of the oxidizing agent, it is more suitable for those who are able to count for a long time and painstakingly, since the arrangement of the coefficients in this way can be lengthy (in this example, it took 4 attempts). The substitution method is used in conjunction with the "TABLE" method, which is also discussed in detail in this article. Method "algebraic" allows you to replace the coefficient in front of the oxidizing agent no less simply and reliably, but much faster KMnO 4 compared to the substitution method, but has a narrower scope. The "algebraic" method can only be used to replace the coefficient in front of the oxidizer KMnO 4 in the equations of oxidation reactions of organic substances proceeding with a break in the carbon chain.

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On the topic: methodological developments, presentations and notes

Arrangement of coefficients in chemical equations

The teacher, being the main character in the organization of the cognitive activity of students, is constantly in search of ways to improve the effectiveness of learning. Organization of effective training...

St. Petersburg State Technological Institute

(Technical University)

Department of Organic Chemistry Faculty 4

Group 476

Course work

Alkene oxidation

Student……………………………………… Rytina A.I.

Lecturer………………………………... Piterskaya Yu.L.

Saint Petersburg

Introduction

1. Epoxidation (reaction by N.A. Prilezhaev, 1909)

2. Hydroxylation

2.1anti-Hydroxylation

2.2syn-Hydroxylation

3. Oxidative cleavage of alkenes

4.Ozonolysis

5. Oxidation of alkenes in the presence of palladium salts

Conclusion

List of sources used

Introduction

Oxidation is one of the most important and widespread transformations of organic compounds.

In organic chemistry, oxidation is understood as processes that lead to the depletion of a compound in hydrogen or its enrichment in oxygen. In this case, electrons are removed from the molecule. Accordingly, reduction is understood as the detachment from an organic oxygen molecule or the addition of hydrogen to it.

In redox reactions, oxidizing agents are compounds with a high electron affinity (electrophiles), and reducing agents are compounds that have a tendency to donate electrons (nucleophiles). The ease of oxidation of a compound increases with the growth of its nucleophilicity.

During the oxidation of organic compounds, as a rule, complete transfer of electrons and, accordingly, a change in the valence of carbon atoms does not occur. Therefore, the concept of the degree of oxidation - the conditional charge of an atom in a molecule, calculated on the basis of the assumption that the molecule consists only of ions - is only conditional, formal.

When compiling the equations of redox reactions, it is necessary to determine the reducing agent, oxidizing agent and the number of given and received electrons. As a rule, the coefficients are selected using the electron-ion balance method (half-reaction method).

This method considers the transition of electrons from one atom or ion to another, taking into account the nature of the medium (acidic, alkaline or neutral) in which the reaction takes place. To equalize the number of oxygen and hydrogen atoms, either water molecules and protons (if the medium is acidic) or water molecules and hydroxide ions (if the medium is alkaline) are introduced.

Thus, when writing the reduction and oxidation half-reactions, one must proceed from the composition of the ions actually present in the solution. Substances that are poorly dissociated, poorly soluble or evolved as a gas should be written in molecular form.

As an example, consider the process of ethylene oxidation with a dilute aqueous solution of potassium permanganate (Wagner reaction). During this reaction, ethylene is oxidized to ethylene glycol, and potassium permanganate is reduced to manganese dioxide. Two hydroxyls are added at the site of the double bond:

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

Reduction half-reaction: MnO 4 ¯ + 2H 2 O + 3 e→ MnO 2 + 4OH ¯ 2

Oxidation half-reaction: C 2 H 4 + 2OH − − 2 e → C 2 H 6 O 2 3

Finally, we have in ionic form:

2MnO 4 ¯ + 4H 2 O + 3C 2 H 4 + 6OH ¯ → 2MnO 2 + 8OH ¯ + 3C 2 H 6 O 2

After carrying out the necessary reductions of similar terms, we write the equation in molecular form:

3C 2 H 4 + 2KMnO 4 + 4 H 2 O \u003d 3C 2 H 6 O 2 + 2MnO 2 + 2KOH.

Characterization of some oxidizers

Oxygen

Air oxygen is widely used in technological processes, as it is the cheapest oxidizing agent. But oxidation with air oxygen is fraught with difficulties associated with the control of the process, which proceeds in different directions. The oxidation is usually carried out at high temperature in the presence of catalysts.

Ozone

Ozone O 3 is used to obtain aldehydes and ketones, if it is difficult to obtain them in other ways. Most often, ozone is used to establish the structure of unsaturated compounds. Ozone is produced by the action of a quiet electric discharge on oxygen. One of the significant advantages of ozonation, compared with chlorination, is the absence of toxins after treatment.

Potassium permanganate

Potassium permanganate is the most commonly used oxidizing agent. The reagent is soluble in water (6.0% at 20ºC), as well as in methanol, acetone and acetic acid. For oxidation, aqueous (sometimes acetone) solutions of KMnO 4 are used in a neutral, acidic or alkaline medium. When carrying out the process in a neutral environment, salts of magnesium, aluminum are added to the reaction mass or carbon dioxide is passed through to neutralize the potassium hydroxide released during the reaction. The oxidation reaction of KMnO 4 in an acidic environment is most often carried out in the presence of sulfuric acid. The alkaline environment during oxidation is created by the KOH formed during the reaction, or it is initially added to the reaction mass. In slightly alkaline and neutral media, KMnO 4 oxidizes according to the equation:

KMnO4+ 3 e+ 2H 2 O \u003d K + + MnO 2 + 4OH ¯

in an acidic environment:

KMnO4+ 5 e+ 8H + = K + + Mn 2+ + 4H 2 O

Potassium permanganate is used to obtain 1,2-diols from alkenes, in the oxidation of primary alcohols, aldehydes and alkylarenes to carboxylic acids, and also for the oxidative cleavage of the carbon skeleton at multiple bonds.

In practice, a fairly large excess (more than 100%) of KMnO 4 is usually used. This is due to the fact that under normal conditions KMnO 4 partially decomposes into manganese dioxide with the release of O 2 . Explosively decomposes with concentrated H 2 SO 4 when heated in the presence of reducing agents; mixtures of potassium permanganate with organics are also explosive.

Peracids

Peracetic and performic acids are obtained by reacting 25-90% hydrogen peroxide with the corresponding carboxylic acid according to the following reaction:

RCOOH + H 2 O 2 \u003d RCOOOH + H 2 O

In the case of acetic acid, this equilibrium is established relatively slowly, and sulfuric acid is usually added as a catalyst to accelerate the formation of peracid. Formic acid is strong enough on its own to provide a quick equilibrium.

Pertrifluoroacetic acid, obtained in a mixture with trifluoroacetic acid by the reaction of trifluoroacetic anhydride with 90% hydrogen peroxide, is an even stronger oxidizing agent. Similarly, peracetic acid can be obtained from acetic anhydride and hydrogen peroxide.

Solid m-chloroperbenzoic acid, because it is relatively safe to handle, quite stable and can be stored for a long time.

Oxidation occurs due to the released oxygen atom:

RCOOOH = RCOOH + [O]

Peracids are used to obtain epoxides from alkenes, as well as lactones from alicyclic ketones.

Hydrogen peroxide

Hydrogen peroxide is a colorless liquid, miscible with water, ethanol and diethyl ether. A 30% solution of H 2 O 2 is called perhydrol. A highly concentrated preparation may react explosively with organic substances. On storage, it decomposes into oxygen and water. The persistence of hydrogen peroxide increases with dilution. For oxidation, aqueous solutions of various concentrations (from 3 to 90%) are used in neutral, acidic or alkaline media.

H 2 O 2 \u003d H 2 O + [O]

By the action of this reagent on α,β-unsaturated carbonyl compounds in an alkaline medium, the corresponding epoxyaldehydes and ketones are obtained, peracids are synthesized by the oxidation of carboxylic acids in an acidic medium. A 30% solution of H 2 O 2 in acetic acid oxidizes alkenes to 1,2-diols. Hydrogen peroxide is used: to obtain organic and inorganic peroxides, Na perborate and percarbonate; as an oxidizing agent in rocket fuels; upon receipt of epoxides, hydroquinone, pyrocatechol, ethylene glycol, glycerin, vulcanization accelerators of the thiuram group, etc.; for bleaching oils, fats, fur, leather, textile materials, paper; for cleaning germanium and silicon semiconductor materials; as a disinfectant for the neutralization of domestic and industrial wastewater; in medicine; as a source of O 2 in submarines; H 2 O 2 is part of Fenton's reagent (Fe 2 + + H 2 O 2), which is used as a source of OH free radicals in organic synthesis.

Ruthenium and osmium tetroxides

Osmium tetroxide OsO 4 is a white to pale yellow powder with mp. 40.6ºС; t. kip. 131.2ºС. Sublimates already at room temperature, soluble in water (7.47 g in 100 ml at 25ºС), СCl 4 (250 g in 100 g of solvent at 20ºС). In the presence of organic compounds, it turns black due to reduction to OsO 2 .

RuO 4 is a golden yellow prism with so pl. 25.4ºС, noticeably sublimates at room temperature. Sparingly soluble in water (2.03 g in 100 ml at 20ºС), very soluble in CCl 4 . A stronger oxidizing agent than OsO 4 . Above 100ºС explodes. Like osmium tetroxide, it has high toxicity and high cost.

These oxidizing agents are used for the oxidation of alkenes to α-glycols under mild conditions.