Compilation of equations of redox reactions involving organic substances

IN connections with the introduction as the only form of the final certification of high school graduates of the Unified state exam (USE) and transition senior school The preparation of high school students is becoming increasingly relevant to the preparation of high school students to fulfill the most "expensive" parties of part "C" test EGE in chemistry. Despite the fact that five tasks of part "C" are considered different: chemical properties inorganic substances, chains of conversion of organic compounds, calculated tasks, are all of them to one way or another due to redox reactions (OSR). If the main knowledge of the theory of the OSR is learned, then you can correctly perform the first and second tasks completely, and the third is partially. In our opinion, a significant part of success when performing a part "C" is precisely in this. Experience shows that if, studying inorganic chemistry, students are well coped with the tasks of writing the OSR equations, the similar tasks for organic chemistry cause them great difficulties. Therefore, throughout the study of the entire course of organic chemistry in profile classes, we try to form the skills of the compilation of the OPR equations in high school students.

When studying comparative characteristics Inorganic and organic compounds, we introduce students using the degree of oxidation (C.O.) (in organic chemistry, first of all carbon) and methods for its determination:

1) Calculation of Medium S.O. carbon in the organic matter molecule;

2) Definition of S.O. Each carbon atom.

We specify in what cases it is better to use one or another way.

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P The study of the topic "Alkana" show that the processes of oxidation, combustion, halogenation, nitration, dehydration, decomposition relate to redox processes. When writing the equations of combustion reactions and decomposition of organic substances, it is better to use the average value of C.O. Carbon. For example:

We draw attention to the first half of the electronic balance: at the carbon atom in the fractional value of C.O. The denominator is 4, so the calculation of electron transmission is carried out according to this coefficient.

In other cases, when studying the topic "Alkans", we determine the values \u200b\u200bof S.O. Each carbon atom in the compound, drawing the attention of students to the sequence of replacement of hydrogen atoms in primary, secondary, tertiary carbon atoms:

Thus, we are summing up students in the conclusion that the process of substitution in tertiary, then in secondary, and, last time - in primary carbon atoms.

P The study of the topic "Alkenes" consider oxidation processes depending on the structure of alkenet and the reaction medium.

With the oxidation of alkenes with a concentrated solution of potassium permanganate KMNO 4 in an acidic environment (rigid oxidation), a break occurs - and - connections to form carboxylic acids, ketones and carbon oxide (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, at Bouthena-1), one of the oxidation products is formic acid, easily oxidized to carbon dioxide and water:

We emphasize that if in the alkin molecule, the carbon atom at a double bond comprises two carbon substituents (for example, in a 2-methylbutene-2 \u200b\u200bmolecule), then when it is oxidation, the ketone formation occurs, because the conversion of such an atom into an atom of the carboxyl group is impossible without breaking C-C-communication, relatively stable under these conditions:

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

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

Considering the oxidation of alkenes in neutral or weakly alkaline media, focusing the attention of high school students on the fact that in such conditions the oxidation is accompanied by the formation of diols (ductomic alcohols), and the hydroxyl groups are joined by the carbon atoms, between which there was a double bond:

IN Similarly, we consider the oxidation of acetylene and its homologs, depending on which the environment proceeds. So, we specify 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 alkins on oxidation products:

In neutral and low-alkaline media, the oxidation of acetylene is accompanied by the formation of appropriate oxalate (salts of oxalic acid), and the oxidation of homologues is a breakdown of triple bonds and the formation of carboxylic acid salts:

IN The CE rules are carried out with students on specific examples, which leads to a better absorption of them theoretical material. Therefore, when studying the oxidation of arena in various environments, students can independently suggest that acid formation should be expected in an acidic medium, and in alkaline salts. The teacher will only be clarified which reaction products are formed depending on the structure of the relevant arena.

Showing on the examples that gomologists of benzene with one side chain (regardless of its length) are oxidized by a strong oxidizing agent to benzoic acid by carbon atom. Gomezol homologs 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 the molecule, then in an acidic medium, each of them is oxidized according to A-carbon atom to the carboxyl group, as a result of which polybent aromatic acids are formed:

P Ollinized skills to compile the equations of the OSR for hydrocarbons allow you to use them when studying the section "Oxygen-containing compounds".

Thus, when studying the theme "Alcohol", students independently make up the equations of alcohol oxidation 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 3CHO 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 characteristic.

In order to prepare for Ege teacher It is advisable to give additional information to the specified properties, which will undoubtedly be useful for students.

With the oxidation of methanol with an acidified solution of potassium permanganate or potassium dichromate, CO 2 is formed, primary alcohols during oxidation, depending on the conditions of the reaction, can form not only aldehydes, but also acids. For example, the oxidation of ethanol with dichromate potassium on cold ends with an acetic acid belt, 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 3CHO 3 -CHO + K 2 SO 4 + CR 2 (SO 4) 3 + 7H 2 O.

We again recall the student on the effect of the medium on the products of alcohol oxidation reactions, namely: the 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", emphasize the attention of students on the fact that aldehydes are easier than alcohols, oxidized into the appropriate carboxylic acids not only under the action of strong oxidizing agents (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 hydroxide (II)):

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.

Special attention is paid to the oxidation of methanal with ammonia solution of silver oxide, since In this case, ammonium carbonate is formed, and not formic acid:

Hchex + 4OH \u003d (NH 4) 2 CO 3 + 4AG + 6NH 3 + 2H 2 O.

As our many years of experience shows, the proposed methodology for learning high school students to compile the AURO equations involving organic substances increases their final the result of the ege in chemistry for several points.

Redox reactions in organic chemistry are the greatest interest, because The transition from one degree of oxidation to another strongly depends on right choice Reagent and reaction conditions. OSR studies at a mandatory chemistry course not enough, but in control and measuring eME Materials It is found not only in the tasks C1 and C2, but also the tasks of the NW representing the chain of conversion of organic substances.

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Redox reactions in organic chemistry

"Think easily, it is difficult to act, and turn the idea is the most difficult thing in the world." I. Guete oxidative and restoration reactions in organic chemistry are the greatest interest, since The selectivity of the transition from one oxidation degree to another strongly depends on the correct choice of reagent and the conditions for carrying out reactions. But the OSR is studying at the obligatory course of chemistry is not fully fully. Particular attention should be paid to the oxidative and recovery processes derived from the participation of organic substances. This is due to the fact that the redox reactions in the control and measuring materials of the EGE are found not only in the tasks C1 and C2, but also the tasks of the NW representing the chain of the conversion of organic substances. In school textbooks, the oxidizer is often recorded above the arrow as [o]. The requirement to fulfill such tasks on the USE is the mandatory designation of all starting materials and reaction products with the arrangement of the necessary coefficients. Redox reactions are traditionally important, and at the same time studying in the 10th grade, in the course "Organic Chemistry" causes certain difficulties in students.

C3. The tasks of this block check the knowledge of organic chemistry in chains of conversion of organic substances in the overwhelming majority of tasks are ORV. The expert has the right to accrue the score only if the equation is recorded, and not the reaction scheme, i.e. Factors are true. In reactions involving inorganic oxidants (potassium permanganate, chromium compounds (VI), hydrogen peroxide, etc.) This is not easy, without an electronic balance.

Determination of the degree of oxidation of atoms in organic compound molecules The rule: CO (atom) \u003d the number of bonds with more EO atoms minus the number of connections with less EO atoms.

Changes in the degree of oxidation of carbon atoms in organic compound molecules. Class of organic compounds The degree of oxidation of carbon atom -4 / -3 -2 -1 0 +1 +2 +3 +4 Alkanes CH 4 CH 3 -CH 3 CH 3 -CH 2 -CH 3 CH 3 | C H 3 -C H-CH 3 CH 3 | C H 3 -C -CH 3 | CH 3 - - - - alkenes - CH 2 \u003d CH 2 CH 3 -CH \u003d CH 2 - - - - alkins - - CH \u003d CH CH 3 -C \u003d CH - - - - alcohols _ _ H 3 C-CH 2 - He H 3 CC H-CH 3 | Oh CH 3 | H 3 C - C - CH 3 | OH - - - - Halogens - - H 3 C-CH 2 - CI H 3 C - C H - CH 3 | CI CH 3 | H 3 C - C - CH 3 | Ci - - - aldehydes and ketones - - - - H 3 C-CH \u003d O H 3 C-C OCH 3 - - carboxylic acids - - - - - - H 3 C-C OOH - Full oxidation products - - - - - - - CO 2

The propensity of organic compounds to oxidation is associated with the presence: multiple connections (alkenes, alkins, alkadiennes are easily oxidized); Functional groups capable of easy to oxidize (-OH, - SNO, - NH 2); activated alkyl groups located adjacent to multiple bonds or benzene ring (for example, proposed can be oxidized to non-precious aldehyde acheroin, oxidation of toluene to benzoic acid permanganate potassium in an acidic environment); The presence of hydrogen atoms with a carbon atom containing a functional group.

1. The might oxidation of organic compounds for the soft oxidation of organic compounds (alcohols, aldegs, unshakable compounds) are used chromium compounds (VI) - chromium oxide (VI), CRO 3, potassium dichromate to 2 s R 2 O 7, etc.. As a rule, Oxidation is carried out in an acidic medium, recovery products are chromium salts (III), for example: 3CH 3 -CHO + K 2 CR 2 O 7 + 4H 2 SO 4 → 3CH 3 -COOH + 4K 2 SO 4 + Cr 2 (SO 4) 3 + 4H 2 O T 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 with oxidation of alcohols by dichromate Potassium on cold oxidation can be stopped at the aldehyde formation stage, while heated carboxylic acids are formed: 3CH 3 -CH 2 OH + K 2 CR 2 O 7 + 4H 2 SO 4 → 3CH 3 -C H O + K 2 SO 4 + CR 2 (SO 4) 3 + 7h 2 O

Alk EN + KMNO4 -1 Kon H 2SO4 Diol Salt Carbonova K-you + Carbonate Carbonic K-TA + CO 2 Alk EN + KMNO4 -2 Kon H 2SO4 2 Salts Carbonovaya K-You 2 Carbon K-Ta Diol 2. Extremely stronger The oxidant is permanganate potassium neutors. Neutre.

C 2 H 2 + 2KMNO 4 + 3H 2 SO 4 \u003d 2CO 2 + 2MNSO 4 + 4H 2 O + K 2 SO 4 Alk In + Kmno4 -1 Kon H 2SO4 Salt Carbonova K-You + Carbonate Carbonovaya K-TA + CO 2 Alk In + KmnO4 -2 Kon N 2SO4 2 Salts Carb. K-you 2 carbon k-you 5CH 3 C \u003d CH + 8KMNO 4 + 12H 2 SO 4 \u003d 5CH 3 COOH + 5CO 2 + 8MNSO 4 + 4K 2 SO 4 + 12H 2 O

5C 6 H 5 -CH 3 +6 KMNO 4 + H 2 SO 4  5C 6 H 5 COOH + 6MNSO 4 + K 2 SO 4 + 14H 2 OC 6 H 5 CH 3 + 2KMNO 4  C 6 H 5 Cook + 2MNO 2 + KOH + H 2 OC 6 H 5 CH 2 CH 3 + 4KMNO 4  C 6 H 5 COOK + K 2 CO 3 + 2H 2 O + 4MNO 2 + KOH 2 O + 4MNO 2 + KOH Benzole Homologists + KMNO4 Kon H 2SO4 Benzoic Acid Neutre. Benzoate

The redox properties of oxygen-containing compounds of alcohol oxidizers are most often the oxide of copper (II) or permanganate potassium, and oxidizing agents of aldehydes and ketones - copper hydroxide (II), ammonia solution of silver oxide and other oxidizing agents

Ol + kmno4 -1 Kon H 2SO4 Aldehyde Ol + KMNO4 -2 Kon H 2SO4 ketone Ol + to MnO4 (Balls) -1 Kon H 2SO4 Neutron Carboxylic Acid Salt Carboxylic Acid Salt Carboxylic Acid

Al dehyd + KMNO4 Kon H 2SO4 Carboxylic acid + carboxylic acid salt Carboxylic acid carboxylic acid neutre. 3ch 3 cho + 2kmno 4 \u003d CH 3 COOH + 2CH 3 Cook + 2mno 2 + H 2 O

Aldehydes are quite strong reducing agents, and therefore easily oxidized by various oxidizing agents CH 3 CHO + 2OH  CH 3 COONH 4 + 2AG + H 2 O + 3NH 3

The selection algorithm because in the C3 task in the preparation of the OPR equations does not require the writing of the electronic balance equations, select the coefficients conveniently by the method of a substrate balance - a simplified method of electronic balance. one . A diagram of the OSR is drawn up. For example, for the oxidation of toluene to benzoic acid by acidified solution of potassium permanganate, the reaction scheme is as follows: from 6 H 5 -CH 3 + KMNO 4 + H 2 SO 4  C 6 H 5 -C OO H + K 2 SO 4 + MNSO 4 + H 2 O 2. indicate from. Atoms. S.O. The carbon atom is determined according to the above method. C 6 H 5 -C -3 H 3 + KMN +7 O 4 + H 2 SO 4  C 6 H 5 -C +3 OO H + K 2 SO 4 + Mn +2 SO 4 + H 2 O 3. Number electrons given by carbon atom (6) is written as a coefficient in front of the oxidant formula (potassium permanganate): C 6 H 5 -C -3 H 3 + 6 KMN +7 O 4 + H 2 SO 4  C 6 H 5 -C + 3 oO H + K 2 SO 4 + Mn +2 SO 4 + H 2 O 4. The number of electrons adopted by the manganese atom (5) is written as a coefficient in front of the reducing agent formula (toluene): 5 s 6 H 5 -C -3 N 3 + 6 kmn +7 o 4 + H 2 SO 4  C 6 H 5 -C +3 OO H + K 2 SO 4 + Mn +2 SO 4 + H 2 O 5. The most important coefficients in place. Further selection is not labor: 5 C 6 H 5 -CH 3 + 6 KMNO 4 + 9 H 2 SO 4  5 C 6 H 5 -C OO H + 3 K 2 SO 4 + 6 MNSO 4 + 14 H 2 O

An example of a test task (C3) 1. Write the reaction equations with which the following transformations can be carried out: HG 2+, H + KMNO 4, H + C L 2 (Equimol.), H  C 2 H 2  x 1  CH 3 SOON  X 2  CH 4  x 3 1. Kucherov's reaction. HG 2+, H + CH  CH + H 2 O  CH 3 CHO 2.Lehydes are easily oxidized to carboxylic acids, including such a strong oxidizing agent as potassium permanganate in an acidic environment. CH 3 CHO + KMNO 4 + H 2 SO 4  CH 3 COOH + K 2 SO 4 + MNSO 4 + H 2 O CH 3 C +1 N O + KMN +7 O 4 + H 2 SO 4  CH 3 -C +3 oO H + K 2 SO 4 + Mn +2 SO 4 + H 2 O 5 CH 3 CHO + 2 KMNO 4 + 3 H 2 SO 4  5 CH 3 COOH + K 2 SO 4 + 2 MNSO 4 + 3 H 2 o 3. For the implementation of the next chain link, it is necessary to estimate the substance X 2 from two positions: first, it is formed in one step from acetic acid, secondly, it is possible to obtain methane. This substance is acetate alkali metal. The equations of the third and fourth reactions are recorded. CH 3 COH + NaOH  CH 3 COONA + H 2 O Splice 4. CH 3 Coona + NaOH  CH 4 + Na 2 CO 3 5. The flow of the next reaction (light) is uniquely indicated by its radical character. Taking into account the specified ratio of reagents (Equomolar), the equation of the last reaction is recorded: H  CH 4 + CL 2  CH 3 CL + HCl

Sites simulators: http://reshuege.ru/ (RTUM EGE) http://4ege.ru/himiya/4181-demoversiya-ep-po-himii-2014.html ( Ege Portal) http://www.alleng.ru/edu/chem3.htm (Internet educational resources - chemistry) http://ege.yandex.ru/ (online tests)

Redox reactions involving organic substances

The propensity of organic compounds to oxidation is associated with the presence multiple bonds, functional groups, hydrogen atoms with carbon atom containing a functional group.

The sequential oxidation of organic substances can be represented as the following chain of transformations:

Saturated hydrocarbon → Unsaturated hydrocarbon → Alcohol → Aldehyde (ketone) → Carboxylic acid → CO 2 + H 2 O

The genetic relationship between classes of organic compounds seems to be here as a series of redox reactions that provide the transition from one class of organic compounds to another. Complete its products of complete oxidation (combustion) of any of the representatives of the classes of organic compounds.

The dependence of the oxidation and reductive capacity of the organic matter from its structure:

The increased tendency of organic compounds to oxidation is due to the presence in the molecule of substances:

• multiple connections (which is why alkenes, alkins, alkadins are so easily oxidized;
• defined functional groupscapable of easy to oxidize (--sh, -oh (phenolic and alcohol), - NH 2;
• activated alkyl groupslocated adjacent to multiple connections. For example, propane can be oxidized to an unspecified aldehyde acherin oxygen in the presence of water vapor on bismuth-molybdenum catalysts.

H 2 C-CH-CH 3 → H 2 C --CH-COH

As well as oxidation of toluene to benzoic acid permanganate potassium in an acidic environment.

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

• the presence of hydrogen atoms under carbon atom containing a functional group.

An example is the reactivity in the reactions of oxidation of primary, secondary and tertiary alcohols for oxidation reactivity.

Despite the fact that during any oxidation reaction reactions occurs both oxidation and recovery, the reaction is classified depending on what is happening directly with the organic compound (if it is oxidized, they say about the oxidation process, if restored - about the recovery process) .

So, in the reaction of ethylene with permanganate potassium ethylene will be oxidized, and potassium permanganate is restored. The reaction is called ethylene oxidation.

The application of the concept of "degree of oxidation" (CO) in organic chemistry is very limited and is implemented primarily in the preparation of equations of redox reactions. However, given that more or less constant composition of the reaction products is possible only with full oxidation (combustion) of organic substances, the expediency of the placement of coefficients in the reactions of incomplete oxidation disappears. For this reason, it is usually limited to the preparation of the circuit of the conversion of organic compounds.

When studying the comparative characteristics of inorganic and organic compounds, we acquainted using the degree of oxidation (C.O.) (in organic chemistry, especially carbon) and methods for its determination:

1) calculation of Medium S.O. Carbon in the organic matter molecule:

-8/3 +1

This approach is justified if all the reactions in the organic substance destroys all chemical ties (burning, full decomposition).

2) definition of S.O. Each carbon atom:

In this case, the degree of oxidation of any carbon atom in the organic compound is equal to the algebraic sum of the numbers of all bonds with atoms of more electronegative elements, taken into account with the "+" sign at the carbon atom, and the number of bonds with hydrogen atoms (or other more electropositive element) taken into account with the sign "-" at the carbon atom. In this case, connections with neighboring carbon atoms are not taken into account.

As the simplest example, we define the degree of carbon oxidation in the methanol molecule.

The carbon atom is associated with three atoms of hydrogen (these relationships are taken into account with the sign "-"), one bond with an oxygen atom (it is taken into account with the "+" sign). We obtain: -3 + 1 \u003d -2. In order, the degree of carbon oxidation in methanol is -2.

The calculated degree of carbon oxidation is, although the conditional value, but it indicates the nature of the electron density displacement in the molecule, and its change as a result of the reaction indicates a place of an oxidation-reducing process.

We specify in what cases it is better to use one or another way.

The processes of oxidation, combustion, halogenation, nice, dehydrogenation, decomposition relate to redox processes.

When moving from one class of organic compounds to another andincrease the degree of branching of the carbon skeleton Connection molecules inside a separate class the degree of oxidation of the carbon atom responsible for the restoring ability of the compound is changed.

Organic substances in the molecules of which contain carbon atoms with maximum (- and +) values \u200b\u200bof s (-4, -3, +2, +3), react complete oxidation-burning, but resistant to the effects of soft oxidizing agents and oxidizing agents.

Substances in the molecules of which contain carbon atoms in CO -1; 0; +1, oxidized easily, their restoration abilities are close, the poet of their incomplete oxidation can be achieved at the expense of one of the known small and medium oxidizers. These substances may exercise dual nature, speaking and as an oxidizing agentJust like this is inherent inorganic substances.

When writing the equations of combustion reactions and decomposition of organic substances, it is better to use the average value of C.O. Carbon.

For example:

Make a complete equation chemical reaction Balance method.

The average value of the degree of carbon oxidation in H-Bhutan:

The degree of carbon oxidation in carbon oxide (IV) is +4.

We will make an electronic balance sheet scheme:

Pay attention to the first half of the electronic balance: at the carbon atom in the fractional value of S.O. The denominator is 4, so the calculation of electron transmission is carried out according to this coefficient.

Those. The transition from -2.5 to +4 corresponds to the transition 2.5 + 4 \u003d 6.5 units. Because 4 carbon atoms are involved, 6.5 · 4 \u003d 26 electrons will be given to the total carbon atoms of Bhutan.

In view of the coefficients found, the equation of the chemical combustion of n-butane burning will look like this:

You can use the method of determining the total charge of carbon atoms in the molecule:

(4 C.) -10 …… → (1 C.) +4, given that the number of atoms to the sign \u003d and after it should be equally equalized (4C.) -10 …… →[(1 C.) +4] · 4

Consequently, the transition from -10 to +16 is associated with a loss of 26 electrons.

In other cases, we define the values \u200b\u200bof C.O. Each carbon atom in the compound, drawing attention to the sequence of replacement of hydrogen atoms in primary, secondary, tertiary carbon atoms:

Initially, the process of substitution in tertiary, then at the secondary, and, last of all - in the primary carbon atoms.

Alkenes

Oxidation processes depend on the structure of alkenet and the reaction medium.

1. In the oxidation of alkenes with a concentrated solution of potassium permanganate KMNO 4 in an acidic environment (hard oxidation) There is a rupture of σ- and π-bonds to form carboxylic acids, ketones and carbon oxide (IV). This reaction is used to determine the position of the double bond.

a) if the double bond is located at the end of the molecule (for example, at Bouthen-1), then one of the oxidation products is formic acid, easily oxidized to carbon dioxide and water:

b) if in the alkin molecule, the carbon atom with a double bond comprises two carbon substituents (for example, in a 2-methylbutene-2 \u200b\u200bmolecule), then when it is oxidation, the formation of ketone, T. K. The conversion of such an atom into an atom of the carboxyl group is impossible without breaking C-C-C-communication, relatively stable under these conditions:

c) 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:

A feature of the oxidation of alkenes, in which carbon atoms at a double bond comprise two carbon radicals, is the formation of two ketones:

2. With neutral or weakly alkaline media, oxidation is accompanied by the formation of diols (ductomic alcohols) , wherein the hydroxyl groups are joined by carbon atoms, between which there was a double bond:

During this reaction, the purple color of the KMNO 4 aqueous solution is discolored. So it is used as quality reaction on alkenes (Wagner Reaction).

3. The oxidation of alkenes in the presence of palladium salts (vacuker process) leads to education aldehydes and ketones:

2ch 2 \u003d CH 2 + O 2 PDCl2 / H2O. → 2 CH 3 -CO-H

Homologists are oxidized to a less hydrogenated carbon atom:

CH 3 -CH 2 -CH \u003d CH 2 + 1 / 2O 2 PDCl2 / H2O. → CH 3 - CH 2 -CO-CH 3

Alkina

The oxidation of acetylene and its homologues flows depending on which the process proceeds in which environment.

but) In an acidic environment, the oxidation process is accompanied by the formation of carboxylic acids:

The reaction is used to determine the structure of alkins on oxidation products:

In neutral and low-alkaline media, the oxidation of acetylene is accompanied by the formation of appropriate oxalate (salts of oxalic acid), and the oxidation of homologues is a breakdown of triple bonds and the formation of carboxylic acid salts:

For acetylene:

1) in an acidic environment:

H-C≡C-H KMNO. 4, H. 2 SO. 4 → HOOC-COOH (sorvelic acid)

3ch≡ch + 8kmno 4 H. 2 O.→ 3kooc Cook Oxalat potassium + 8mno 2 ↓ + 2KOH + 2H 2 O

Arena

(Benzene and his homologues)

In the oxidation of arena in an acidic medium, an acid formation should be expected, and in alkaline - salts.

Gomezol homologs with one side chain (regardless of its length) are oxidized by a strong oxidizing agent to benzoic acid by α-carbon atom. Gomezol homologs 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 the molecule, then in an acidic medium, each of them is oxidized according to A-carbon atom to the carboxyl group, as a result of which polybent aromatic acids are formed:

1) in an acidic environment:

C 6 H 5 -CH 2 -R KMNO. 4, H. 2 SO. 4 → C 6 H 5 -COOH benzoic acid+ Co 2.

2) in a neutral or alkaline environment:

C 6 H 5 -CH 2 -R KMNO4, H2O / (OH)→ C 6 H 5 -Cook + CO 2

3) oxidation of benzene homologists permanganate potassium or potassium bichromate when heated:

C 6 H 5 -CH 2 -R KMNO. 4, H. 2 SO. 4, T. ˚ C.→ C 6 H 5 -COOH benzoic acid+ R-COOH

4) oxidation of cumene oxygen in the presence of a catalyst (a cumorole method for producing phenol):

C 6 H 5 CH (CH 3) 2 O2, H2SO4→ C 6 H 5 -OH phenol + CH 3 -CO-CH 3 acetone

5C 6 H 5 CH (CH 3) 2 + 18KMNO 4 + 27H 2 SO 4 → 5C 6 H 5 COOH + 42H 2 O + 18MNSO 4 + 10CO 2 + K 2 SO 4

C 6 H 5 CH (CH 3) 2 + 6H 2 O - 18C→ C 6 H 5 COOH + 2CO 2 + 18H + | x 5.

MNO 4 - + 8H + + 5C→ Mn +2 + 4H 2 O | x 18.

Attention should be paid that when soft oxidation of styrene permanganate potassium KMNO 4 in neutral or slightly alkaline mediumthere is a gap π-meansy, glycol is formed (ductomic alcohol). As a result of the reaction, the painted solution of potassium permanganate is quickly discharged and a brown precipitate of manganese oxide (IV) falls.

Oxidation of the same strong oxidizer - Potassium permanganate in an acidic medium - leads to a complete break of the double bond and the formation of carbon dioxide and benzoic acid, the solution is bleached.

C 6 H 5 -CH mon 2 + 2 kmno 4 + 3 H 2 SO 4 → C 6 H 5 -COOH + CO 2 + K 2 SO 4 + 2 MNSO 4 +4 H 2 O

Alcohol

It should be remembered that:

1) Primary alcohols are oxidized to aldehydes:

3CH 3 -CH 2 OH + K 2 CR 2 O 7 + 4H 2 SO 4 \u003d 3CHO 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 characteristic.

Tertiary alcohols, in the molecules of which there is no hydrogen atom with a carbon atom containing a group, it is not oxidized under normal conditions. In harsh conditions (under the action of strong oxidizing agents and at high temperatures), they can be oxidized to a mixture of low molecular weight carboxylic acids, i.e. The destruction of the carbon skeleton occurs.

When the methanol is oxidized by the acidified solution of potassium permanganate or potassium dichromate, CO 2 is formed.

Primary alcohols When oxidation, depending on the conditions of the reaction, not only aldehydes can be formed, but also acids.

For example, the oxidation of ethanol with dichromate potassium on cold ends with an acetic acid belt, 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,

If three or more HE-groups are associated with neighboring carbon atoms, then with the oxidation of iodine acid, medium or average atoms are converted into formic acid

The oxidation of glycols permanganate potassium in an acidic medium extends similarly to the oxidative splitting of alkenes and also leads to the formation of acids or ketones depending on the structure of the original glycol.

Aldehydes and Ketones

Aldehydes are easier than alcohols, oxidized into the appropriate carboxylic acids not only under the action of strong oxidizers (air oxygen, acidified solutions of KMNO 4 and K 2 Cr 2 O 7), but also under the action of weak (ammonia solution of silver oxide or copper hydroxide (II) ):

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

Special attention!!! The oxidation of methanal with ammonia solution of silver oxide leads to the formation of ammonium carbonate, and not formic acid:

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

To compile the equations of oxidation-reducing reactions, both the electronic balance method and the half-formation method (electron-ion method) are used.

For organic chemistry, the degree of oxidation of the atom is important, and the offset of electron density, as a result of which partial charges appear on atoms, in no way consistent with the values \u200b\u200bof oxidation degrees.

Many universities include in tickets for the entrance examinations of the task for the selection of coefficients in the OPR equations by ion-electronic method (by half-formation). If at school and pays at least some attention to this method, then, mainly in the oxidation of inorganic substances.

Let's try to apply the half-formation method for oxidation of sucrose permanganate potassium in an acidic environment.

The advantage of this method is that there is no need to immediately guess and record reaction products. They are fairly easily determined during the equation. Oxidizer in an acidic environment most fully manifests its oxidative properties, for example, an Anion MNO - turns into a Mn 2+ cation, easily oxidizing organic compounds are oxidized to CO 2.

We write in the molecular form of the conversion of sucrose:

In the left part there are lack of 13 oxygen atoms to eliminate this contradiction, add 13 molecules H 2 O.

The left part now contains 48 hydrogen atoms, they are released in the form of n + cations:

Now equal the total charges on the right and left:

Scheme of semi-reaction is ready. Drawing up the second half-formation scheme usually does not cause difficulties:

Combine both schemes:

Task for independent work:

Finish UHR and spread the coefficients by the method of electronic balance or by half-formation:

CH 3 -CH \u003d CH-CH 3 + KMNO 4 + H 2 SO 4 →

CH 3 -CH \u003d CH-CH 3 + KMNO 4 + H 2ABOUT →

(CH 3) 2 C \u003d C-CH 3 + KMNO 4 + H 2 SO 4 →

CH 3 -CH 2 -CH \u003d CH 2 + KMNO 4 + H 2 SO 4 →

FROMH 3 -CH 2 -C≡C-CH 3 + KMNO 4 + H 2 SO 4 →

C 6 H 5 -CH 3 + KMNO 4 + H2O →

C 6 H 5 -C 2 H 5 + KMNO 4 + H 2 SO 4 →

C. 6 H. 5 - Ch 3 + KMNO. 4 + H. 2 SO. 4 →

My notes:

Particular attention to students should be paid to the behavior of the oxidant - potassium permanganate KMNO 4 in various environments. This is due to the fact that the redox reactions in the kima are found not only in the tasks C1 and C2. In the tasks of the NW, representing the chain of the conversion of organic substances, often the equation of the reduction oxidation. At school, the oxidizer is often recorded above the arrow as [O]. The requirement to fulfill such tasks on the USE is the mandatory designation of all starting materials and the remedy products with the arrangement of the necessary coefficients.

Description of the presentation Oxidation-Restore reactions involving organic substances by slides

Redox reactions with the participation of organic substances Kocholeva L. R., Chemistry Teacher Mobu "Lyceum No. 9" of Orenburg

In organic chemistry, oxidation is defined as a process in which the connection of the functional group converts from one category to higher: alkin alcohite alpine (ketone) carboxylic acid. Most oxidation reactions include the introduction of an oxygen atom into the molecule or the formation of a double bond with an already existing oxygen atom due to the loss of hydrogen atoms.

Oxidifiers for oxidation of organic substances typically use transition metal compounds, oxygen, ozone, peroxides and sulfur compounds, selenium, iodine, nitrogen and others. From oxidizing agents based on transition metal, compounds of chromium (VI) and manganese (VII), (VI) and (IV) are predominantly used. The most common chromium compounds (VI) is a solution of potassium dichromate K 2 Cr 2 O 7 in sulfuric acid, a solution of chromium trioxide CR. O 3 in dilute sulfuric acid.

Oxidifiers in the oxidation of organic substances Chrom (VI) in any medium is restored to chromium (III), however, the oxidation in an alkaline medium in organic chemistry does not find practical application. Permanganate potassium KMN. O 4 In different environments, various oxidative properties show, and the power of the oxidant increases in an acidic environment. Caline Manganat K 2 Mn. O 4 and manganese oxide (IV) Mn. O 2 exhibit oxidative properties only in an acidic environment

Alkenes, depending on the nature of the oxidizing agent and the reaction conditions, various products are formed: dioxide alcohols, aldehydes, ketones, carboxylic acids during the oxidation of KMN aqueous solution. O 4 At room temperature, there is a rupture of π-bonds and two-heeded alcohols (Wagner reaction) are formed: discoloration of potassium permanganate solution - high-quality response to multiple communication

Alkenes oxidation of alkenes with a concentrated solution of potassium permanganate KMN. O 4 or potassium dichromate K 2 Cr 2 O 7 in an acidic medium is accompanied by a breaking of not only π-, but also σ-bonds reaction products - carboxylic acids and ketones (depending on the structure of alkene) using this reaction on alkene oxidation products can be determined The position of the dual connection in its molecule:

Alkenes 5 CH 3 -CH \u003d CH-CH 3 +8 KMN. O 4 +12 H 2 SO 4 → 10 CH 3 COOH +8 MN. SO 4 + 4 K 2 SO 4 + 12 H 2 O 5 CH 3 -CH \u003d CH-CH 2 -CH 3 +8 KMN. O 4 +12 H 2 SO 4 → 5 CH 3 COOH +5 CH 3 CH 2 COOH +8 MN. SO 4 +4 K 2 SO 4 +12 H 2 O CH 3 -CH 2 -CH \u003d CH 2 +2 KMN. O 4 +3 H 2 SO 4 → CH 3 CH 2 COOH + CO 2 +2 Mn. SO 4 + K 2 SO 4 +4 H 2 O

Alkenes alkenes of branched structure containing a hydrocarbon radical at the carbon atom connected by a double bond, a mixture of carboxylic acid and ketone is formed during oxidation:

Alkenes 5 CH 3 -CH \u003d C-CH 3 + 6 KMN. O 4 +9 H 2 SO 4 → │ CH 3 5 CH 3 COOH + 5 O \u003d C-CH 3 + 6 MN. SO 4 + 3 K 2 SO 4+ │ CH 3 9 H 2 O

Alkenes alkenes of branched structure containing hydrocarbon radicals in both carbon atoms connected by a double bond, a mixture of ketones formed during oxidation:

Alkenes 5 CH 3 -C \u003d C-CH 3 + 4 KMN. O 4 +6 H 2 SO 4 → │ │ CH 3 10 O \u003d C-CH 3 + 4 Mn. SO 4 + 2 K 2 SO 4 + 6 H 2 O │ CH

Alkenes As a result of catalytic oxidation of alkens of air oxygen, epoxides are obtained: in harsh conditions, in the air, alkenes are burning, like other hydrocarbons, burn with the formation of carbon dioxide and water: C 2 H 4 + 3 O 2 → 2 CO 2 + 2 H 2

Alkadienin CH 2 \u003d CH-CH \u003d CH 2 in an oxidized molecule Two terminal double bonds, therefore, two carbon dioxide molecules are formed. The carbon skeleton is not branched, so when oxidizing 2 -to and 3 -to carbon atoms, carboxyl groups CH 2 \u003d CH-CH \u003d CH 2 + 4 KMN are formed. O 4 + 6 H 2 SO 4 → 2 CO 2 + NSOO-Soon + 4 Mn. SO 4 +2 k 2 SO 4 + 8 H 2

Alkina Alkina are easily oxidized by potassium permanganate and potassium dichromate at the place of multiple communication under action on alkina aqueous solution KMN. O 4 It occurs its discoloration (high-quality reaction to a multiple bond) in the interaction of acetylene with an aqueous solution of potassium permanganate, salt of oxalic acid (potassium oxalate) is formed:

Alkina Acetylene can be oxidized by potassium permanganate in a neutral medium to potassium oxalate: 3 CH≡CH +8 KMN. O 4 → 3 KOOC - Cook +8 Mn. O 2 +2 Kon +2 H 2 O in an acidic oxidation oxidation goes to oxalic acid or carbon dioxide: 5 CH≡CH +8 KMN. O 4 +12 H 2 SO 4 → 5 HOOC - COOH +8 MN. SO 4 +4 K 2 SO 4 +12 N 2 About Ch≡ch + 2 KMN. O 4 +3 H 2 SO 4 \u003d 2 CO 2 + 2 Mn. SO 4 + 4 H 2 O + K 2 SO

Alkina oxidation of potassium permanganagats in an acidic medium during heating is accompanied by a gap of a carbon chain at the site of the triple bond and leads to acid formation: the oxidation of alkins containing a triple bond from the extreme carbon atom is accompanied under these conditions by the formation of carboxylic acid and CO 2:

Alkina CH 3 C≡CCH 2 CH 3 + K 2 CR 2 O 7 + 4 H 2 SO 4 → CH 3 COOH + CH 3 CH 2 COOH + CR 2 (SO 4) 3 + K 2 SO 4 + 3 H 2 O 3 CH 3 C≡CH + 4 K 2 CR 2 O 7 +16 H 2 SO 4 → CH 3 COOH + 3 CO 2 ++ 4 CR 2 (SO 4) 3 + 4 K 2 SO 4 +16 H 2 O CH 3 C≡CH + 8 KMN. O 4 + 11 KOH → CH 3 Cook + K 2 CO 3 + 8 K 2 Mn. O 4 +6 H 2 O

Cycloalkanes and cycloalkanes under the action of strong oxidizing agents (KMN. O 4, K 2 Cr 2 O 7 et al.) Cycloalkanes and cycloalkens form two-headed carboxylic acids with the same number of carbon atoms: 5 C 6 H 12 + 8 KMN. O 4 + 12 H 2 SO 4 → 5 HOOC (CH 2) 4 COOH + 4 K 2 SO 4 + 8 Mn. SO 4 +12 H 2 O

Arena benzene resistant to oxidizers at room temperature does not react with aqueous solutions of potassium permanganate, potassium dichromate and other oxidizing agents can be oxidized with the formation of dialdehyde:

Arena gomologists of benzene are oxidized relatively easily. The oxidation is subjected to a side chain, in toluene - a methyl group. Soft oxidizers (Mn. O 2) oxidize the methyl group to the aldehyde group: C 6 H 5 CH 3 + 2 Mn. O 2 + H 2 SO 4 → C 6 H 5 CHO + 2 MN. SO 4 + 3 H 2 O

Arena stronger oxidizers - KMN. O 4 In an acidic medium or a chromium mixture, heating is oxidized by a methyl group to carboxyl: in a neutral or weakly alkaline medium, a benzoic acid is formed, and its salt - potassium benzoate:

Arena in an acidic medium 5 s 6 H 5 CH 3 +6 KMN. O 4 +9 H 2 SO 4 → 5 C 6 H 5 Soam + 6 Mn. SO 4 +3 K 2 SO 4 + 14 H 2 O in neutral medium C 6 H 5 CH 3 +2 KMN. O 4 \u003d C 6 H 5 Cook + 2 Mn. O 2 + KOH + H 2 O in an alkaline medium C 6 H 5 CH 2 CH 3 + 4 KMN. O 4 \u003d C 6 H 5 Cook + K 2 CO 3 + 2 H 2 O + 4 Mn. O 2 + KOH

Arena under the action of strong oxidizing agents (KMN. O 4 in an acidic medium or chromium mixture) side chains are oxidized regardless of the structure: carbon atom directly related to the benzene core, to the carboxyl group, the remaining carbon atoms in the side chain - up to CO 2 oxidation of any homologue Benzene with one side chain under the action of KMN. O 4 in an acidic medium or chromium mixture leads to the formation of benzoic acid:

Arena gomologists of benzene containing several side chains, for the oxidation form appropriate multi-axis aromatic acids:

Arena in a neutral or weakly alkaline medium during oxidation permanganate potassium formed a carboxylic acid salt and potassium carbonate:

Arena 5 C 6 H 5 -C 2 H 5 + 12 KMN. O 4 + 18 H 2 SO 4 -\u003e 5 C 6 H 5 -COOH + 5 CO 2 + 12 Mn. SO 4 + 6 K 2 SO 4 + 28 H 2 O C 6 H 5 -C 2 H 5 +4 KMN. O 4 → C 6 H 5 -COOK + K 2 CO 3 + KON +4 MN. O 2 +2 H 2 O 5 C 6 H 5 -CH (CH 3) 2 + 18 KMN. O 4 + 27 H 2 SO 4 -\u003e 5 C 6 H 5 -COOH + 10 CO 2 + 18 Mn. SO 4 + 9 K 2 SO 4 + 42 H 2 O 5 CH 3 -C 6 H 4 -CH 3 +12 KMN. O 4 +18 H 2 SO 4 → 5 C 6 H 4 (COO) 2 +12 Mn. SO 4 +6 k 2 SO 4 + 28 H 2 O CH 3 -C 6 H 4 -CH 3 + 4 KMN. O 4 → C 6 H 4 (Cook) 2 +4 Mn. O 2 +2 KOH + 2 H 2 O

Styrene oxidation of styrene (vinylbenzene) solution of potassium permanganate in an acidic and neutral medium: 3 C 6 H 5 -CH mon 2 + 2 KMN. O 4 + 4 H 2 O → 3 C 6 H 5 -CH-CH 2 + 2 Mn. O 2 + 2 KOH ı ı Oh Oh oxidation with a strong oxidizing agent - potassium permanganate in an acidic medium - leads to a complete break of the double bond and the combination of carbon dioxide and benzoic acid, the solution is bleached. C 6 H 5 -CH mon 2 + 2 KMN. O 4 + 3 H 2 SO 4 → C 6 H 5 -COOH + CO 2 + K 2 SO 4 + 2 Mn. SO 4 +4 H 2 O

Alcohols are the most suitable oxidizing agents for primary and secondary alcohols: KMN. O 4, chrome mix. Primary alcohols, except methanol, are oxidized to aldehydes or carboxylic acids:

Alcohols methanol is oxidized to CO 2: ethanol under the action of CL 2 is oxidized to acetic aldehyde: secondary alcohols are oxidized to ketones:

Alcohols Double-spectacular alcohol, ethylene glycol HOCH 2 -CH 2 OH, when heated in an acidic medium with a solution of KMN. O 4 or k 2 Cr 2 O 7 is easily oxidized to oxalic acid, and in neutral - to potassium oxalate. 5 CH 2 (OH) - CH 2 (OH) + 8 KMN. O 4 +12 H 2 SO 4 → 5 HOOC - COOH +8 MN. SO 4 +4 K 2 SO 4 +22 H 2 O 3 CH 2 (OH) - CH 2 (OH) + 8 KMN. O 4 → 3 KOOC - Cook +8 Mn. O 2 +2 Kon +8 H 2 O

Phenols are oxidized easily due to the presence of hydroxochroup connected to the phenol benoline ring oxidizes hydrogen peroxide in the presence of a catalyst to a dioxide phenol pyrochetchin, during the oxidation of the chromium mixture - to the pair -Benzochinone:

Aldehydes and ketones aldehydes are easily oxidized, while the aldehyde group is oxidized to carboxyl: 3 CH 3 CHO + 2 KMN. O 4 + 3 H 2 O → 2 CH 3 Cook + CH 3 COOH + 2 MN. O 2 + H 2 O 3 CH 3 CH \u003d O + K 2 CR 2 O 7 + 4 H 2 SO 4 \u003d 3 CH 3 COOH + CR 2 (SO 4) 3 + 7 H 2 O methanal oxidizes to CO 2:

Aldehydes and ketones Qualitative reactions to aldehydes: oxidation of copper hydroxide (II) The reaction of the "silver mirror" salt, and not acid!

Aldehydes and ketones are oxidized with difficulty, weak oxidizers on them do not act under the action of strong oxidizers, there is a break of C - with bonds on both sides of the carbonyl group to form a mixture of acids (or ketones) with a smaller number of carbon atoms than in the source connection:

Aldehydes and ketones In the case of an asymmetric structure of ketone, oxidation is mainly carried out by a less hydrogenated carbon atom in the carbonyl group (Popov Rule - Wagner) on the ketone oxidation products, it is possible to establish its structure:

Formic acid among marginal monoxide acids is easily oxidized only formic acid. This is due to the fact that in formic acid, in addition to the carboxyl group, an aldehyde group can be distinguished. 5 NSON + 2 KMN. O 4 + 3 H 2 SO 4 → 2 Mn. SO 4 + K 2 SO 4 + 5 CO 2 + 8 H 2 O Formic acid reacts with ammonia silver oxide solution and HCOOH + 2OH hydroxide (II) HCOOH + 2OH → 2 AG + (NH 4) 2 CO 3 + 2 NH 3 + H 2 o HCOOH + 2 Cu (OH) 2 CO 2 + Cu 2 O ↓ + 3 H 2 O In addition, formic acid is oxidized by chlorine: Nson + CL 2 → CO 2 + 2 HCl

Unforeseen carboxylic acids are easily oxidized by an aqueous solution of KMN. O 4 In a weakly alkaline medium with the formation of dihydroxyxic acids and their salts: in an acidic medium, a carbon skeleton is ruptied at a dual connection site C \u003d C with the formation of a mixture of acids:

Sorrelic acid is easily oxidized under the action of KMN. O 4 in an acidic medium when heated to CO 2 (permanganateometry method): When heated is subjected to decarboxylation (disproportionation reaction): in the presence of concentrated H 2 SO 4, when heated oxalic acid and its salts (oxalates) are disproportionated:

We write down the reaction equations: 1) CH 3 CH 2 CH 2 CH 3 2) 3) 4) 5) 16, 32% (36, 68%, 23, 82%) PT, TO X 3 x 2 PT, TO. KMN. O 4 KOH X 4 Heptane Koh, to benzene. X 1 FE, HCl. HNO 3 H 2 SO 4 CH 3 + 4 H 2 CH 3 + 6 KMN. O 4 + 7 Kohcook + 6 K 2 Mn. O 4 + 5 H 2 O COOK + KOH + K 2 CO 3 TO NO 2 + H 2 O + HNO 3 H 2 SO 4 NH 3 C L + 3 F E C L 2 + 2 H 2 ON 2 + 3 F E + 7 HC L.

Oxidative and recovery processes have long been interested in chemists and even alchemists. Among the chemical reactions occurring in nature, everyday life, a huge set are redox: combustion of fuel, oxidation of nutrients, tissue breathing, photosynthesis, spanking food, etc. In such reactions, both inorganic substances and organic can participate in such reactions. However, if in the school year of inorganic chemistry, sections devoted to redox reactions occupy a significant place, then there is not enough attention to the course of organic chemistry.

What are restorative and oxidative processes?

All chemical reactions can be divided into two types. The first includes reactions flowing without changing the degree of oxidation of atoms that are part of the reacting substances.

The second type includes all reactions going with a change in the degree of oxidation of atoms that are part of the reacting substances.

Reactions flowing with a change in the degree of oxidation of atoms included in the reactant substances are called redox.

From a modern point of view, the change in the degree of oxidation is associated with pulling or moving electrons. Therefore, along with the above, it is possible to give such a definition of restorative and oxidative reactions: these are such reactions in which the transition of electrons from one atoms, molecules or ions to the other.

Consider the main provisions relating to the theory of redox reactions.

1. The oxidation is called the process of returning by an electron atom, a molecule or an electron ion, the degree of oxidation increases.

2. Recovery is the process of connecting electrons by an atom, molecule or ion, the degree of oxidation decreases.

3. Atoms, molecules or ions that give electrons are called reducing agents. During the reaction, they are oxidized. Atoms, molecules or ions that attach electrons are called oxidizing agents. During the reaction, they are restored.

4. Oxidation is always accompanied by restoration; Recovery is always associated with oxidation, which can be expressed by equations.

Therefore, oxidative reaction reactions are the unity of two opposite processes - oxidation and recovery. In these reactions, the number of electrons given by the reducing agent is equal to the number of electrons connected by the oxidizing agent. At the same time, no matter whether electrons are transmitted from one atom to another completely or only partially delayed to one of the atoms, it is conventionally indicated on the return and addition of electrons.

Redox reactions of organic substances are the most important property combining these substances. The propensity of organic compounds to oxidation is associated with the presence of multiple bonds, functional groups, hydrogen atoms with a carbon atom containing a functional group.

The application of the concept of "degree of oxidation" (CO) in organic chemistry is very limited and is implemented primarily in the preparation of equations of redox reactions. However, given that more or less constant composition of the reaction products is possible only with full oxidation (combustion) of organic substances, the expediency of the placement of coefficients in the reactions of incomplete oxidation disappears. For this reason, it is usually limited to the preparation of the circuit of the conversion of organic compounds.

It seems to us important to indicate the value from the carbon atom when studying the entire set of properties of organic compounds. Systematization of information about oxidants, the establishment of the relationship between the structure of organic substances and they will help to teach students:

Choose laboratory and industrial oxidizers;

Find the dependence of the oxidation and reduction capacity of the organic matter from its structure;

Establish a connection between the class of organic substances and the oxidizing agent of the desired force, the aggregate state and the mechanism of action;

Predict the reaction conditions and expected oxidation products.

Determination of the degree of oxidation of atoms in organic substances

The degree of oxidation of any carbon atom in the organic matter is equal to the algebraic sum of all its bonds with more electronegative elements (CL, O, S, N, etc.), taken into account with the "+" sign, and bonds with hydrogen atoms (or other more electrical element ), accounted for with the sign "-". In this case, connections with neighboring carbon atoms are not taken into account.

We define the degree of oxidation of carbon atoms in the limiting hydrocarbon molecules of propane and alcohol of ethanol:

The sequential oxidation of organic substances can be represented as the following chain of transformations:

Saturated hydrocarbon unsaturated hydrocarbon alcohol aldehyde (ketone) CO + CO + H O. carboxylic acid

The genetic relationship between the classes of organic compounds seems to be here as a number of redox reactions that provide the transition from one class of organic compounds to another. Complete its products of complete oxidation (combustion) of any of the representatives of the classes of organic compounds.

Application . Table number 1.

Changes in carbon atoms in carbon molecule in organic compound molecules are shown in the table. From the data of the table, it can be seen that during the transition from one class of organic compounds to another and increasing the degree of branching of the carbon skeleton of molecules of compounds inside a separate class, the degree of oxidation of the carbon atom responsible for the reducing connection ability changes. Organic substances, in the molecules of which contain carbon atoms with maximum (- and +) values \u200b\u200bof CO (-4, -3, +2, +3), react complete oxidation-burning, but resistant to the effects of soft oxidizing agents and oxidizing agents . Substances in the molecules of which contain carbon atoms in CO -1; 0; +1, oxidized easily, the restoration abilities of them are close, the poet of their incomplete oxidation can be achieved at the expense of one of the famous oxidizing agents of small and medium power. These substances can show a dual nature, speaking and as an oxidizing agent, just as inherent inorganic substances.

Oxidation and restoration of organic substances

The increased tendency of organic compounds to oxidation is due to the presence in the molecule of substances:

• hydrogen atoms with carbon atom containing a functional group.

Compare primary, secondary and tertiary alcohols on oxidation reactivity:

Primary and secondary alcohols having hydrogen atoms with a carbon atom carrying a functional group; Oxisions easily: the first to the aldehydes, the second to ketones. At the same time, the structure of the carbon skeleton of the source alcohol is preserved. Tertiary alcohols, in the molecules of which there is no hydrogen atom with a carbon atom containing a group, it is not oxidized under normal conditions. In harsh conditions (under the action of strong oxidizing agents and at high temperatures), they can be oxidized to a mixture of low molecular weight carboxylic acids, i.e. The destruction of the carbon skeleton occurs.

There are two approaches to the determination of degrees of oxidation of elements in organic substances.

1. Calculate the average degree of oxidation of the carbon atom in the organic compound molecule, for example propane.

This approach is justified if all chemical bonds (combustion, full decomposition) are destroyed during the reaction in the organic matter.

Note that formally fractional oxidation degrees calculated in this way may also be in the case of inorganic substances. For example, in conjunction of KO (Potassium Superoxide), the degree of oxidation of oxygen is equal to 1/2.

2. Determine the degree of oxidation of each carbon atom, for example in Bhutan.

In this case, the degree of oxidation of any carbon atom in the organic compound is equal to the algebraic sum of the numbers of all bonds with atoms of more electronegative elements, taken into account with the "+" sign, and the number of bonds with hydrogen atoms (or other more electropositive element), taken into account with the "-" sign . In this case, the bonds with carbon atoms are not taken into account.

As the simplest example, we define the degree of carbon oxidation in the methanol molecule.

The carbon atom is associated with three atoms of hydrogen (these relationships are taken into account with the sign "-"), one bond with an oxygen atom (it is taken into account with the "+" sign). We get:

Thus, the degree of carbon oxidation in methanol is -2.

The calculated degree of carbon oxidation is, although the conditional value, but it indicates the nature of the electron density displacement in the molecule, and its change as a result of the reaction indicates a place of an oxidation-reducing process.

Consider the chain of transformations of substances:

For catalytic dehydrogenation ethane is ethylene; Product of ethylene hydration - ethanol; Its oxidation will lead to ethanne, and then to acetic acid; With its combustion, carbon dioxide and water is formed.

We define the degree of oxidation of each carbon atom in the molecules of these substances.

It can be noted that in the course of each of these transformations, the degree of oxidation of one of the carbon atoms is constantly changing. In the direction of the ethane to carbon oxide (IV) there is an increase in the degree of oxidation of the carbon atom.

Despite the fact that in the course of any oxidation reaction reactions occurs both oxidation and recovery, they are classified depending on this, which is directly related to the organic compound (if it is oxidized, they say about the oxidation process, if restored - about the recovery process).

So, in the reaction of ethanol with permanganate potassium ethanol will be oxidized, and potassium permanganate is restored. The reaction is called oxidation of ethanol.

Compilation of oxidative - reducing equations

To compile the equations of oxidation-restorative reactions, both the electronic balance method and the semi-resource method (electronically ion method) are used. Consider several examples of oxidation-restorative reactions involving organic substances.

1. The burning of H-Bhutan.

The reaction scheme has the form:

Let's make a complete chemical reaction equation by the balance sheet.

The average value of the degree of carbon oxidation in H-Bhutan:

The degree of carbon oxidation in carbon oxide (IV) is +4.

We will make an electronic balance sheet scheme:

In view of the coefficients found, the equation of the chemical combustion of n-butane burning will look like this:

The coefficients for this equation can be found in another method that has already been mentioned. Calculating the degree of oxidation of each of the carbon atoms, we see that they differ:

In this case, the electronic balance diagram will look like this:

Since all chemical bonds are destroyed in its molecules in its molecules, in this case, the first approach is fully justified, especially since the electronic balance circuit, compiled by the second way, is somewhat more complicated.

2. The reaction of ethylene oxidation with potassium permanganate solution in the neutral medium on cold (Wagner reaction).

We put the coefficients in the reaction equation by the electronic balance.

The complete chemical reaction equation will look like this:

To determine the coefficients, it is possible to use the half-formation method. Ethylene is oxidized in this reaction to ethylene glycol, and permanganate - ions are restored to the formation of manganese dioxide.

Schemes of the respective semoretosis:

Total electron-ion equation:

3. The reaction of oxidation of glucose permanganate potassium in an acidic environment.

A. Electronic Balance Method.

First option

Second option

Calculate the degrees of oxidation of each of the carbon atoms in the glucose molecule:

The electronic balance sheet is compared with the previous examples:

B. The half-formation method in this case looks like this:

Total ion equation:

Molecular Glucose Reaction Equation permanganate potassium:

In organic chemistry it is advisable to use the determination of oxidation as an increase in oxygen content or reduce the content of hydrogen. The reduction in this case is defined as a decrease in oxygen content or an increase in hydrogen content. With this definition, the sequential oxidation of organic substances can be submitted to the following scheme:

Practice shows that the selection of coefficients in the oxidation reactions of organic substances causes certain difficulties, as it is necessary to deal with very unusual degrees of oxidation .. Some students due to the lack of experience continue to identify the degree of oxidation with valence and, as a result, the degree of carbon oxidation is incorrectly determined. in organic compounds. Carbon valence in these compounds is always equal to four, and the degree of oxidation can take different values \u200b\u200b(from -3 to +4, including fractional values). An unusual moment when oxidizing organic substances is the zero degree of oxidation of the carbon atom in some complex compounds. If overcome the psychological barrier, the compilation of such equations does not represent complexity, for example:

The degree of oxidation of the carbon atom in sucrose is zero. Rewrite the reaction scheme indicating the degrees of oxidation of atoms that change them:

We compile electronic equations and find coefficients during the oxidizing and reducing agent and the products of their oxidation and recovery:

We substitute the resulting coefficients in the reaction scheme:

The remaining coefficients are selected in such a sequence: K SO, H SO, H O. The final equation has the form:

Many universities include on-tickets for entrance examinations for the selection of coefficients in the ORP equations by electronic method (by half-formation). If at school and pays at least some attention to this method, then, mainly in the oxidation of inorganic substances. Let's try to apply the half-formation method for the above example of the oxidation of sucrose permanganate potassium in an acidic environment.

The first advantage of this method is that there is no need to immediately guess and record reaction products. They are fairly easily determined during the equation. The oxidizing agent in an acidic environment most fully manifests its oxidative properties, for example, an Anion MNO turns into a MN cation, easily oxidizing organic oxidation to CO.

We write in the molecular form of the conversion of sucrose:

In the left part there are lack of 13 oxygen atoms to eliminate this contradiction, add 13 molecules H O. CH

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