Description of the presentation REDOX REACTIONS INVOLVING ORGANIC SUBSTANCES on slides

REDOX REACTIONS WITH THE PARTICIPATION OF ORGANIC SUBSTANCES Kochuleva L.R., Chemistry teacher, Lyceum No. 9, Orenburg

In organic chemistry, oxidation is defined as a process in which, as a result of the transformation of a functional group, a compound passes from one category to a higher one: alkene alcohol aldehyde (ketone) carboxylic acid. Most oxidation reactions involve 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.

OXIDIZERS For the oxidation of organic substances, compounds of transition metals, oxygen, ozone, peroxides, and compounds of sulfur, selenium, iodine, nitrogen, and others are usually used. Of the oxidizing agents based on transition metals, chromium (VI) and manganese (VII), (VI) and (IV) compounds are mainly used. The most common chromium (VI) compounds are 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.

OXIDIZERS During the oxidation of organic substances, chromium (VI) in any medium is reduced to chromium (III), however, oxidation in an alkaline medium in organic chemistry does not find practical application. Potassium permanganate KMn. O 4 in different environments exhibits different oxidizing properties, while the strength of the oxidizing agent increases in an acidic environment. Potassium manganate K 2 Mn. O 4 and manganese (IV) oxide Mn. O 2 exhibit oxidizing properties only in an acidic environment

ALKENES Depending on the nature of the oxidizing agent and the reaction conditions, various products are formed: dihydric alcohols, aldehydes, ketones, carboxylic acids When oxidized with an aqueous solution of KMn. O 4 at room temperature, the π-bond breaks and dihydric alcohols are formed (Wagner reaction): Discoloration of the potassium permanganate solution - a qualitative reaction for a multiple bond

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 rupture of not only π-, but also σ-bonds Reaction products - carboxylic acids and ketones (depending on the structure of the alkene) Using this reaction, the products of alkene oxidation can be determined the position of the double bond 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=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 Branched alkenes containing a hydrocarbon radical at the carbon atom connected by a double bond form a mixture of carboxylic acid and ketone upon 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 Branched alkenes containing hydrocarbon radicals at both carbon atoms connected by a double bond form a mixture of ketones upon oxidation:

ALKENES 5 CH 3 -C=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 the catalytic oxidation of alkenes with atmospheric oxygen, epoxides are obtained: Under harsh conditions, when burned in air, alkenes, like other hydrocarbons, burn to form carbon dioxide and water: C 2 H 4 + 3 O 2 → 2 CO 2 + 2 H 2 O

ALKADIENES CH 2 =CH−CH=CH 2 There are two terminal double bonds in the oxidized molecule, therefore, two molecules of carbon dioxide are formed. The carbon skeleton is not branched, therefore, when the 2nd and 3rd carbon atoms are oxidized, carboxyl groups CH 2 \u003d CH - CH \u003d CH 2 + 4 KMn are formed. O 4 + 6 H 2 SO 4 → 2 CO 2 + HCOO−COOH + 4 Mn. SO 4 +2 K 2 SO 4 + 8 H 2 O

ALKYNES Alkynes are easily oxidized by potassium permanganate and potassium dichromate at the site of a multiple bond When alkynes are treated with an aqueous solution of KMn. O 4 it becomes discolored (qualitative reaction to a multiple bond) When acetylene reacts with an aqueous solution of potassium permanganate, a salt of oxalic acid (potassium oxalate) is formed:

ALKYNES Acetylene can be oxidized with potassium permanganate in a neutral medium to potassium oxalate: 3 CH≡CH +8 KMn. O 4 → 3 KOOC–COOK +8 Mn. O 2 +2 KOH +2 H 2 O In an acidic environment, 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 H 2 O 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

ALKYNES Oxidation with potassium permanganates in an acidic medium when heated is accompanied by a break in the carbon chain at the site of the triple bond and leads to the formation of acids: Oxidation of alkynes containing a triple bond at the extreme carbon atom is accompanied under these conditions by the formation of carboxylic acid and CO 2:

ALKYNES 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 3C≡CH+8KMn. O 4+11 KOH →CH 3 COOK + K 2 CO 3 + 8 K 2 Mn. O 4 +6 H 2 O

Cycloalkanes and Cycloalkenes Under the action of strong oxidizing agents (KMn. O 4 , K 2 Cr 2 O 7 , etc.), cycloalkanes and cycloalkenes form dibasic 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

ARENES Benzene Resistant to oxidizing agents at room temperature Does not react with aqueous solutions of potassium permanganate, potassium dichromate and other oxidizing agents Can be oxidized with ozone to form dialdehyde:

ARENES Benzene homologues Oxidize relatively easily. The side chain undergoes oxidation, in toluene - the methyl group. Mild oxidizing agents (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, when heated, oxidizes the methyl group to a carboxyl group: In a neutral or slightly alkaline medium, not benzoic acid itself is formed, but its salt, potassium benzoate:

ARENE In acid medium 5 C 6 H 5 CH 3 +6 KMn. O 4 +9 H 2 SO 4 → 5 C 6 H 5 COOH + 6 Mn. SO 4 +3 K 2 SO 4 + 14 H 2 O In a neutral environment 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 environment 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. O2 + KOH

ARENES Under the action of strong oxidizing agents (KMn. O 4 in an acid medium or a chromium mixture), the side chains are oxidized regardless of the structure: the carbon atom directly attached to the benzene ring to a carboxyl group, the remaining carbon atoms in the side chain to CO 2 Oxidation of any homologue benzene with one side chain under the action of KMn. O 4 in an acidic environment or a chromium mixture leads to the formation of benzoic acid:

ARENES Benzene homologues containing several side chains form the corresponding polybasic aromatic acids upon oxidation:

ARENES In a neutral or slightly alkaline medium, oxidation with potassium permanganate produces 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 -> 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 + KOH +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 ---> 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 (COOH) 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) with a solution of potassium permanganate in an acidic and neutral medium: 3 C 6 H 5 -CH═CH 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—results in the complete breaking of the double bond and the formation of carbon dioxide and benzoic acid; the solution becomes colorless. C 6 H 5 -CH═CH 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 The most suitable oxidizing agents for primary and secondary alcohols are: KMn. O 4 chromium mixture. 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 acetaldehyde: Secondary alcohols are oxidized to ketones:

ALCOHOLS Dihydric 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 KOH +8 H 2 O

PHENOLS They are easily oxidized due to the presence of a hydroxo group connected to the benzene ring. Phenol is oxidized with hydrogen peroxide in the presence of a catalyst to diatomic phenol pyrocatechol, and when oxidized with a chromium mixture, to para-benzoquinone:

ALDEHYDES AND KETONES Aldehydes are easily oxidized, while the aldehyde group is oxidized to a carboxyl group: 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 is oxidized to CO 2:

ALDEHYDES AND KETONES Qualitative reactions to aldehydes: oxidation with copper (II) hydroxide "silver mirror" reaction Salt, not acid!

ALDEHYDES AND KETONES Ketones are oxidized with difficulty, weak oxidizing agents do not act on them Under the action of strong oxidizing agents, C-C bonds break 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 original compound:

ALDEHYDES AND KETONES In the case of an asymmetric ketone structure, oxidation is predominantly carried out from the side of the less hydrogenated carbon atom at the carbonyl group (Popov-Wagner rule). Based on the oxidation products of the ketone, its structure can be established:

FORMIC ACID Among the saturated monobasic acids, only formic acid is easily oxidized. This is due to the fact that in formic acid, in addition to the carboxyl group, an aldehyde group can also be isolated. 5 NUN + 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 an ammonia solution of silver oxide and copper (II) hydroxide 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: HCOOH + Cl 2 → CO 2 + 2 HCl

UNSATURATED CARBOXIC ACIDS Easily oxidized with an aqueous solution of KMn. O 4 in a weakly alkaline medium with the formation of dihydroxy acids and their salts: In an acidic medium, the carbon skeleton breaks at the site of the C=C double bond with the formation of a mixture of acids:

OXALIC ACID Easily oxidized by KMn. O 4 in an acidic environment when heated to CO 2 (permanganatometry method): When heated, it undergoes decarboxylation (disproportionation reaction): In the presence of concentrated H 2 SO 4, when heated, oxalic acid and its salts (oxalates) disproportionate:

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 N H 3 C l + 3 F e C l 2 + 2 H 2 ON O 2 + 3 F e + 7 H C l

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 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 implemented, 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.

Application . 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 electronic 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 higher education institutions 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.

Redox reactions in organic chemistry are of the greatest interest, because. the transition from one oxidation state to another strongly depends on the correct choice of the reagent and the reaction conditions. OVR is not studied fully enough in the compulsory course of chemistry, but in the USE control and measuring materials they are found not only in tasks C1 and C2, but also in tasks SZ, representing a chain of transformations of organic substances.

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REDOX REACTIONS IN ORGANIC CHEMISTRY

“Thinking is easy, acting is difficult, and turning a thought into action is the most difficult thing in the world” I. Goethe Redox reactions in organic chemistry are of the greatest interest, because. the selectivity of the transition from one oxidation state to another strongly depends on the correct choice of the reagent and the reaction conditions. But OVR is not fully studied in the compulsory course of chemistry. Students should pay special attention to the redox processes that occur with the participation of organic substances. This is due to the fact that redox reactions in the USE control and measuring materials are found not only in tasks C1 and C2, but also in tasks SZ, representing a chain of transformations of organic substances. In school textbooks, the oxidizing agent is often written above the arrow as [O]. The requirement for completing such assignments for the USE is the mandatory designation of all starting substances 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 for students.

C3. The tasks of this block test knowledge of organic chemistry In the chains of transformations of organic substances, in the vast majority of tasks, OVRs are encountered. The expert has the right to award a point only if the equation is written, and not the reaction scheme, i.e. coefficients are correct. In reactions involving inorganic oxidants (potassium permanganate, chromium (VI) compounds, hydrogen peroxide, etc.), this can be difficult to do, without electronic balance.

Determination of the oxidation state of atoms in molecules of organic compounds RULE: CO (atom) = the number of bonds with more EO atoms minus the number of bonds with less EO atoms.

Change in the degree of oxidation of carbon atoms in the molecules of organic compounds. Class of organic compounds The degree of oxidation of the 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 - - - - Alkynes - - CH \u003d CH CH 3 -C \u003d CH - - - - Alcohols _ _ H 3 C-CH 2 - OH H 3 C-C H-CH 3 | OH CH 3 | H 3 C - C - CH 3 | OH - - - Halogenalkanes - - 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 - Complete oxidation products - - - - - - - CO 2

The tendency of organic compounds to oxidize is associated with the presence of: multiple bonds (alkenes, alkynes, alkadienes are easily oxidized); functional groups that can be easily oxidized (–OH, - CHO, - NH 2); activated alkyl groups located adjacent to multiple bonds or a benzene ring (for example, propene can be oxidized to unsaturated acrolein aldehyde, oxidation of toluene to benzoic acid with potassium permanganate in an acidic environment); the presence of hydrogen atoms at the carbon atom containing the functional group.

1. SOFT OXIDATION OF ORGANIC COMPOUNDS For mild oxidation of organic compounds (alcohols, aldehydes, unsaturated compounds), chromium (VI) compounds are used - chromium oxide (VI), CrO 3, potassium dichromate K 2 С r 2 O 7, etc. As a rule, oxidation is carried out in an acidic environment, the reduction products are chromium (III) salts, 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 When alcohols are oxidized with dichromate potassium in the cold, oxidation can be stopped at the stage of aldehyde formation, but when heated, carboxylic acids are formed: 2 (SO 4) 3 + 7H 2 O

ALC EN + KMnO4 -1 KOH H 2SO4 Diol Carbonic acid salt + carbonate Carbonic acid + CO 2 ALC EN + KMnO4 -2 KOH N 2SO4 2 carboxylic salts 2 carboxylic acids Diol 2. Significantly stronger The oxidizing agent is potassium permanganate NEUTRE. NEUTRAL

C 2 H 2 + 2KMnO 4 + 3H 2 SO 4 \u003d 2CO 2 + 2MnSO 4 + 4H 2 O + K 2 SO 4 ALA IN + KMnO4 -1 KOH H 2SO4 Salt of carboxylic acid + carbonate Carbonic acid + CO 2 ALK IN + KMnO4 -2 KOH H 2SO4 2 carb. to-you 2 carboxylic to-you 5CH 3 C = CH + 8KMnO 4 + 12H 2 SO 4 = 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 O 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 Benzene homologs + KMnO4 KOH H 2SO4 benzoic acid NEUT. Benzoate

Redox properties of oxygen-containing compounds Oxidizing alcohols are most often copper (II) oxide or potassium permanganate, and oxidizing agents for aldehydes and ketones - copper (II) hydroxide, ammonia solution of silver oxide and other oxidizing agents

OL + KMnO4 -1 KOH H 2SO4 ALDEHYDE OL + KMnO4 -2 KOH H 2SO4 ketone OL + K MnO4 (eq) -1 KOH H 2SO4 NEUTRAL Carboxylic acid salt Carboxylic acid salt Carboxylic acid

Aldehyde + KMnO4 KOH H 2SO4 Carboxylic acid + Carboxylic acid salt Carboxylic acid salt Carboxylic acid NEUT. 3CH 3 CHO + 2KMnO 4 \u003d CH 3 COOH + 2CH 3 COOK + 2MnO 2 + H 2 O

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

Algorithm for selection of coefficients Since in task C3, when compiling the equations of the OVR, it is not required to write the equations of the electronic balance, it is convenient to select the coefficients by the sublinear balance method - a simplified method of electronic balance. one . An OVR scheme is being drawn up. For example, for the oxidation of toluene to benzoic acid with an acidified solution of potassium permanganate, the reaction scheme is as follows: C 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. The d.d. atoms. S.o. carbon atom is determined by 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 donated by the carbon atom (6) is written as a coefficient in front of the formula of the oxidizing agent (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 accepted by the manganese atom (5) is written as a coefficient in front of the formula of the reducing agent (toluene): 5 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 5. The most important coefficients are in place. Further selection is not difficult: 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 you can carry out the following transformations: Hg 2+, H + KMnO 4, H + C l 2 (equimol.), h  C 2 H 2  X 1  CH 3 COOH  X 2  CH 4  X 3 1. Kucherov reaction. Hg 2+, H + CH  CH + H 2 O  CH 3 CHO 2. Aldehydes 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 H 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. To perform the next link in the chain, it is necessary to evaluate the substance X 2 from two positions: firstly, it is formed from acetic acid in one stage, and secondly, methane can be obtained from it. This substance is an alkali metal acetate. The equations of the third and fourth reactions are written down. CH 3 COOH + NaOH  CH 3 COONa + H 2 O fusion 4. CH 3 COONa + NaOH  CH 4 + Na 2 CO 3 5. The conditions for the next reaction (light) clearly indicate its radical character. Taking into account the indicated ratio of reagents (equimolar), the equation of the last reaction is written: h  CH 4 + Cl 2  CH 3 Cl + HCl

Simulator sites: http://reshuege.ru/ (I will solve the USE) http://4ege.ru/himiya/4181-demoversiya-ege-po-himii-2014.html (USE portal) http://www.alleng. ru/edu/chem3.htm (Internet educational resources - Chemistry) http://ege.yandex.ru/ (online tests)

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.