Drawing up equations of redox reactions involving organic substances

AT 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 connected with redox reactions (ORD). 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 C–C bond, relatively stable under these conditions:

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

We 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:

AT 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:

AT 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 environment, the formation of acids should be expected, and in an alkaline environment, 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 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 allow them to be used 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 students' attention 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 (air oxygen, acidified solutions of KMnO 4 and K 2 Cr 2 O 7), but and under the influence of weak (ammonia solution of silver oxide or copper(II) hydroxide):

5CH 3 -CHO + 2KMnO 4 + 3H 2 SO 4 \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.

Physical Properties

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

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

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

Table. Physical properties of some arenas

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

Benzene - low-boiling ( tkip= 80.1°C), colorless liquid, insoluble in water

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

Most aromatic hydrocarbons are life threatening and toxic.

Obtaining arenes (benzene and its homologues)

In the laboratory

1. Fusion of salts of benzoic acid with solid alkalis

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

sodium benzoate

2. Wurtz-Fitting reaction: (here G is halogen)

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

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

In industry

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

1. Dehydrocyclization of alkanes with more than 6 carbon atoms:

C 6 H 14 t , kat→C 6 H 6 + 4H 2

2. Trimerization of acetylene(only for benzene) – R. Zelinsky:

3C 2 H2 600°C, Act. coal→C 6 H 6

3. Dehydrogenation cyclohexane and its homologues:

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

C 6 H 12 t, cat→C 6 H 6 + 3H 2

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

methylcyclohexanetoluene

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

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

chloroethane ethylbenzene

Chemical properties of arenes

I. OXIDATION REACTIONS

1. Combustion (smoky flame):

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

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

3. Benzene homologues are oxidized by potassium permanganate (discolor potassium permanganate):

A) in an acidic environment to benzoic acid

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

Homologues of benzene with one side chain give benzoic acid:

Homologues containing two side chains give dibasic acids:

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

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

Simplified :

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

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

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

II. ADDITION REACTIONS (harder than alkenes)

1. Halogenation

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

2. Hydrogenation

C 6 H 6 + 3H 2 t , PtorNi→C 6 H 12 (cyclohexane)

3. Polymerization

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

1. Halogenation -

a ) benzene

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

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

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

b) benzene homologues upon irradiation or heating

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

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

c) benzene homologues in the presence of a catalyst

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

2. Nitration (with nitric acid)

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

nitrobenzene - smell almond!

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

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

The use of benzene and its homologues

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

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

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

Isopropylbenzene C 6 H 5 -CH (CH 3) 2 serves to obtain phenol and acetone.

Chlorine derivatives of benzene used for plant protection. So, the product of substitution of H atoms in benzene with chlorine atoms is hexachlorobenzene C 6 Cl 6 - a fungicide; it is used for dry seed dressing of wheat and rye against hard smut. The product of the addition of chlorine to benzene is hexachlorocyclohexane (hexachloran) C 6 H 6 Cl 6 - an insecticide; it is used to control harmful insects. These substances refer to pesticides - chemical means of combating microorganisms, plants and animals.

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

VIDEO EXPERIENCES

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

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

The growth of nucleophilicity in the series

Consider redox reactions representatives of the most important classes organic matter with some inorganic oxidizing agents.

Alkene oxidation

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

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

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

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

1) 5CH 3 CH=CHCH 2 CH 3 + 8KMnO 4 + 12H 2 SO 4 → 5CH 3 COOH + 5C 2 H 5 COOH + 8MnSO 4 + 4K 2 SO 4 + 17H 2 O

2) 5CH 3 CH=CH 2 + 10KMnO 4 + 15H 2 SO 4 → 5CH 3 COOH + 5CO 2 + 10MnSO 4 + 5K 2 SO 4 + 20H 2 O

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

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

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

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

Alkyne oxidation

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

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

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

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

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

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

Oxidation of benzene homologues

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

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

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

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

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

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

Alcohol oxidation

The direct products of the oxidation of primary alcohols are aldehydes, while those of secondary alcohols are ketones.

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

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

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

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

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

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

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

5CH 2 (OH) - CH 2 (OH) + 8KMnO 4 + 12H 2 SO 4 → 5HOOC -COOH + 8MnSO 4 + 4K 2 SO 4 + 22H 2 O

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

Oxidation of aldehydes and ketones

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

3CH 3 CHO + 2KMnO 4 → CH 3 COOH + 2CH 3 COOK + 2MnO 2 + H 2 O

3CH 3 CHO + K 2 Cr 2 O 7 + 4H 2 SO 4 → 3CH 3 COOH + Cr 2 (SO 4) 3 + 7H 2 O

CH 3 CHO + 2KMnO 4 + 3KOH → CH 3 COOK + 2K 2 MnO 4 + 2H 2 O

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

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

silver mirror reaction

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

CH 3 CH \u003d O + 2OH → CH 3 COONH 4 + 2Ag + H 2 O + 3NH 3

CH 3 -CH \u003d O + 2Cu (OH) 2 → CH 3 COOH + Cu 2 O + 2H 2 O

Formic aldehyde (formaldehyde) is oxidized, as a rule, to carbon dioxide:

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

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

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

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

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

carboxylic acids. Among the acids, formic and oxalic acids have strong reducing properties, which are oxidized to carbon dioxide.

HCOOH + HgCl 2 \u003d CO 2 + Hg + 2HCl

HCOOH + Cl 2 \u003d CO 2 + 2HCl

HOOC-COOH + Cl 2 \u003d 2CO 2 + 2HCl

Formic acid, in addition to acidic properties, also exhibits some properties of aldehydes, in particular, reducing. It is then oxidized to carbon dioxide. For example:

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

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

HCOOH →(t)CO + H2O

Catalytic oxidation of alkanes:

Catalytic oxidation of alkenes:

Phenol oxidation:

Redox processes have long been of interest to chemists and even alchemists. Among the chemical reactions that occur in nature, everyday life and technology, a great many are redox reactions: fuel combustion, nutrient oxidation, tissue respiration, photosynthesis, food spoilage, etc. Both inorganic and organic substances can participate in such reactions. However, if the sections devoted to redox reactions occupy a significant place in the school course of inorganic chemistry, then insufficient attention is paid to this issue in the course of organic chemistry.

What are redox processes?

All chemical reactions can be divided into two types. The first includes reactions that proceed without changing the oxidation state of the atoms that make up the reactants.

The second type includes all reactions that occur with a change in the oxidation state of the atoms that make up the reactants.

Reactions that occur with a change in the oxidation state of the atoms that make up the reactants are called redox reactions.

From a modern point of view, a change in the oxidation state is associated with the withdrawal or movement of electrons. Therefore, along with the above, one can also give such a definition of redox reactions: these are reactions in which electrons transfer from one atoms, molecules or ions to others.

Let us consider the main provisions related to the theory of redox reactions.

1. Oxidation is the process of giving away an electron by an atom, molecule or ion of electrons, while the oxidation states increase.

2. Recovery is the process of adding electrons to an atom, molecule or ion, while the oxidation state decreases.

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

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

Therefore, redox reactions are a unity of two opposite processes - oxidation and reduction. In these reactions, the number of electrons donated by the reducing agent is equal to the number of electrons added by the oxidizing agent. In this case, regardless of whether the electrons pass from one atom to another completely or are only partially drawn to one of the atoms, we conditionally speak only of the return and attachment of electrons.

Redox reactions of organic substances are the most important property that unites these substances. The tendency of organic compounds to oxidize is associated with the presence of multiple bonds, functional groups, hydrogen atoms at the carbon atom containing the functional group.

The use of the concept of "oxidation state" (CO) in organic chemistry is very limited and is realized, first of all, in the formulation of equations for redox reactions. However, taking into account that a more or less constant composition of the reaction products is possible only with the complete oxidation (combustion) of organic substances, the expediency of arranging the coefficients in the reactions of incomplete oxidation disappears. For this reason, they usually confine themselves to drawing up a scheme for the transformations of organic compounds.

It seems to us important to indicate the value of the CO of the carbon atom in the study of the totality of the properties of organic compounds. Systematization of information about oxidizing agents, establishing a relationship between the structure of organic substances and their CO will help teach students:

Choose laboratory and industrial oxidizers;

Find the dependence of the redox ability of organic matter on its structure;

Establish a relationship between a class of organic substances and an oxidizing agent of the required strength, state of aggregation and mechanism of action;

Predict reaction conditions and expected oxidation products.

Determination of the oxidation state of atoms in organic substances

The oxidation state of any carbon atom in 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 another more electropositive element ), taken into account with the "-" sign. In this case, bonds with neighboring carbon atoms are not taken into account.

Let us determine the oxidation states of carbon atoms in the molecules of the saturated hydrocarbon propane and ethanol alcohol:

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

Saturated hydrocarbon Unsaturated hydrocarbon Alcohol Aldehyde (ketone) Carboxylic acid CO + HO.

The genetic link between classes of organic compounds is presented here as a series of redox reactions that ensure the transition from one class of organic compounds to another. It is completed by the products of complete oxidation (combustion) of any of the representatives of the classes of organic compounds.

Appendix . Table number 1.

Changes in CO at carbon atoms in a carbon molecule in molecules of organic compounds are shown in the table. It can be seen from the table data that when moving from one class of organic compounds to another and increasing the degree of branching of the carbon skeleton of compound molecules within a separate class, the degree of oxidation of the carbon atom responsible for the reducing ability of the compound changes. Organic substances, the molecules of which contain carbon atoms with maximum (- and +) CO values ​​(-4, -3, +2, +3), enter into a complete oxidation-burning reaction, but are resistant to mild and medium-strength oxidizers . Substances whose molecules contain carbon atoms in CO -1; 0; +1, are easily oxidized, their reduction abilities are close, so their incomplete oxidation can be achieved by using one of the known oxidizing agents of low and medium strength. These substances can exhibit a dual nature, acting as an oxidizing agent, just as it is inherent in inorganic substances.

Oxidation and reduction of organic substances

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

• hydrogen atoms at the carbon atom containing the functional group.

Let's compare primary, secondary and tertiary alcohols in terms of reactivity to oxidation:

Primary and secondary alcohols having hydrogen atoms at the carbon atom bearing the functional group; easily oxidized: the former to aldehydes, the latter to ketones. At the same time, the structure of the carbon skeleton of the initial alcohol is preserved. Tertiary alcohols, in the molecules of which there is no hydrogen atom at the carbon atom containing the OH group, do not oxidize under normal conditions. Under 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. destruction of the carbon skeleton.

There are two approaches to determining the oxidation states of elements in organic substances.

1. Calculate the average oxidation state of a carbon atom in a molecule of an organic compound, such as propane.

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

Note that, formally, the fractional oxidation states calculated in this way can also be in the case of inorganic substances. For example, in the compound KO (potassium superoxide), the oxidation state of oxygen is -1/2.

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

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

As the simplest example, let's determine the oxidation state of carbon in a methanol molecule.

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

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

The calculated degree of oxidation of carbon, although a conditional value, but it indicates the nature of the shift in the electron density in the molecule, and its change as a result of the reaction indicates an ongoing redox process.

Consider the chain of transformations of substances:

The catalytic dehydrogenation of ethane produces ethylene; the product of ethylene hydration is ethanol; its oxidation will lead to ethanal and then to acetic acid; When it burns, it produces carbon dioxide and water.

Let us determine the oxidation states of each carbon atom in the molecules of the listed substances.

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

Despite the fact that in the course of any redox reactions both oxidation and reduction occur, they are classified depending on what happens directly to the organic compound (if it is oxidized, they speak of an oxidation process, if it is reduced, it is a reduction process).

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

Drawing up redox equations

To compile the equations of redox reactions, both the electron balance method and the half-reaction method (electron - ion method) are used. Consider a few examples of redox reactions involving organic substances.

1. Combustion of n-butane.

The reaction scheme looks like:

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

The average value of the oxidation state of carbon in n-butane:

The oxidation state of carbon in carbon monoxide (IV) is +4.

Let's make an electronic balance diagram:

Taking into account the found coefficients, the equation for the chemical reaction of n-butane combustion will look like this:

The coefficients for this equation can also be found by another method, which has already been mentioned. Having calculated the oxidation states of each of the carbon atoms, we see that they differ:

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

Since all chemical bonds in its molecules are destroyed during the combustion of n-butane, in this case the first approach is quite justified, especially since the electronic balance scheme compiled by the second method is somewhat more complicated.

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

Let's arrange the coefficients in the reaction equation using the electron balance method.

The complete equation for a chemical reaction would look like this:

To determine the coefficients, you can also use the method of half-reactions. Ethylene is oxidized in this reaction to ethylene glycol, and permanganate ions are reduced to form manganese dioxide.

Schemes of the corresponding half-reactions:

The total electron-ion equation:

3. Oxidation reactions of potassium permanganate glucose in an acidic medium.

A. Electronic balance method.

First option

Second option

Let's calculate the oxidation states of each of the carbon atoms in the glucose molecule:

The electronic balance scheme becomes more complicated compared to the previous examples:

B. The half-reaction method in this case is as follows:

Total ionic equation:

Molecular equation for the reaction of glucose with potassium permanganate:

In organic chemistry, it is appropriate to use the definition of oxidation as an increase in oxygen content or a decrease in hydrogen content. Recovery is then defined as a decrease in oxygen content or an increase in hydrogen content. With this definition, the sequential oxidation of organic substances can be represented by the following scheme:

Practice shows that the selection of coefficients in the oxidation reactions of organic substances causes certain difficulties, since one has to deal with very unusual oxidation states. Some students, due to lack of experience, continue to identify the oxidation state with valence and, as a result, incorrectly determine the oxidation state of carbon in organic compounds. The valency of carbon in these compounds is always four, and the degree of oxidation can take on various values ​​(from -3 to +4, including fractional values). An unusual moment in the oxidation of organic substances is the zero degree of oxidation of the carbon atom in some complex compounds. If you overcome the psychological barrier, the compilation of such equations is not difficult, for example:

The oxidation state of the carbon atom in sucrose is zero. We rewrite the reaction scheme indicating the oxidation states of the atoms that change them:

We compose electronic equations and find the coefficients for the oxidizing agent and reducing agent and the products of their oxidation and reduction:

We substitute the obtained coefficients into the reaction scheme:

We select the remaining coefficients in the following sequence: K SO , H SO , HO. The final equation looks like:

Many universities include in tickets for entrance exams tasks on the selection of coefficients in the OVR equations by the electronic method (half-reaction method). If the school pays at least some attention to this method, it is mainly in the oxidation of inorganic substances. Let's try to apply the half-reaction method for the above example of the oxidation of sucrose with potassium permanganate in an acidic medium.

The first advantage of this method is that there is no need to immediately guess and write down the reaction products. They are fairly easy to determine in the course of the equation. An oxidizing agent in an acidic environment most fully manifests its oxidizing properties, for example, the MnO anion turns into a Mn cation, easily oxidized organic ones are oxidized to CO.

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

On the left side, 13 oxygen atoms are missing, to eliminate this contradiction, we add 13 HO molecules. CH

2. Kartsova A.A., Levkin A.N. Redox reactions in organic chemistry // Chemistry at school. - 2004. - No. 2. - P.55-61.

3. Khomchenko G.P., Savostyanova K.I. Redox reactions: A guide for students. M.-: Enlightenment, 1980.

4. Sharafutdinov V. Redox reactions in organic chemistry // Bashkortostan ukytyusyhy. - 2002. - No. 5. - P.79 -81.

Alkynes (otherwise acetylenic hydrocarbons) are hydrocarbons containing a triple bond between carbon atoms, with the general formula CnH2n-2. The carbon atoms in the triple bond are in a state of sp - hybridization.

Reaction of acetylene with bromine water

The acetylene molecule contains a triple bond, bromine destroys it and joins the acetylene. Terabromomethane is formed. Bromine is consumed in the formation of tetrabromoethane. Bromine water (yellow) - discolors.

This reaction proceeds at a lower rate than in the series of ethylene hydrocarbons. The reaction also proceeds in steps:

HC ≡ CH + Br 2 → CHBr = CHBr + Br 2 → CHBr 2 - CHBr 2

acetylene → 1,2-dibromoethane → 1,1,2,2-tetrabromoethane

The decolorization of bromine water proves the unsaturation of acetylene.

The reaction of acetylene with a solution of potassium permanganate

In a solution of potassium permanganate, acetylene is oxidized, and the molecule breaks at the site of the triple bond, the solution quickly becomes colorless.

3HC ≡ CH + 10KMnO 4 + 2H 2 O → 6CO 2 + 10KOH + 10MnO 2

This reaction is a qualitative reaction for double and triple bonds.

The reaction of acetylene with an ammonia solution of silver oxide

If acetylene is passed through an ammonia solution of silver oxide, the hydrogen atoms in the acetylene molecule are easily replaced by metals, since they have high mobility. In this experiment, hydrogen atoms are replaced by silver atoms. Silver acetylenide is formed - a yellow precipitate (explosive).

CH ≡ CH + OH → AgC≡CAg↓ + NH 3 + H 2 O

This reaction is a qualitative reaction for a triple bond.