Alkenes - These are hydrocarbons in the molecules of which there is ONE double C \u003d C bond.

Alkene nomenclature: suffix appears in the name -EN.

The first member of the homologous series is C2H4 (ethene).

For the simplest alkenes, historically established names are also used:

ethylene (ethene)

propylene (propene),

The following monovalent alkene radicals are often used in the nomenclature:

CH2-CH=CH2

Types of isomerism of alkenes:

1. Isomerism of the carbon skeleton:(starting from C4H8 - butene and 2-methylpropene)

2. Multiple bond position isomerism:(starting with C4H8): butene-1 and butene-2.

3. Interclass isomerism: with cycloalkanes(starting with propene):

C4H8 - butene and cyclobutane.

4. Spatial isomerism of alkenes:

Due to the fact that free rotation around the double bond is impossible, it becomes possible cis-trans- isomerism.

Alkenes having two carbon atoms at each double bond various substitutes, can exist in the form of two isomers that differ in the arrangement of substituents relative to the π-bond plane:

Chemical properties of alkenes.

Alkenes are characterized by:

· double bond addition reactions,

· oxidation reactions,

· substitution reactions in the "side chain".

1. Double bond addition reactions: the weaker π-bond is broken, a saturated compound is formed. These are electrophilic addition reactions - AE.	1) Hydrogenation: CH3-CH=CH2 + H2 à CH3-CH2-CH3 2) Halogenation: CH3-CH=CH2 + Br2 (solution)à CH3-CHBr-CH2Br Discoloration of bromine water is a qualitative reaction to a double bond. 3) Hydrohalogenation: CH3-CH=CH2 + HBr à CH3-CHBr-CH3 (MARKOVNIKOV'S RULE: hydrogen is attached to the most hydrogenated carbon atom). 4) Hydration - water connection: CH3-CH=CH2 + HOH à CH3-CH-CH3 (attachment also occurs according to Markovnikov's rule)
2. Addition of hydrogen bromide to presence of peroxides (Harash effect) - this is a radical addition - AR	CH3-CH=CH2 + HBr -(H2O2)à CH3-CH2-CH2Br (reaction with hydrogen bromide in the presence of peroxide proceeds against Markovnikov's rule )
3. Combustion- complete oxidation of alkenes with oxygen to carbon dioxide and water.	С2Н4 + 3О2 = 2СО2 + 2Н2О
4. Soft oxidation of alkenes - Wagner reaction : reaction with a cold aqueous solution of potassium permanganate.	3CH3- CH=CH2+ 2KMnO4 + 4H2O à 2MnO2 + 2KOH + 3 CH3 - CH - CH2 Oh Oh ( a diol is formed) Discoloration of an aqueous solution of potassium permanganate with alkenes is a qualitative reaction for alkenes.
5. Hard oxidation of alkenes- hot neutral or acidic potassium permanganate solution. Comes with a break in the C=C double bond.	1. Under the action of potassium permanganate in an acidic environment, depending on the structure of the alkene skeleton, the following is formed: Fragment of the carbon chain at the double bond What does it turn into = CH -R R– COOHcarboxylic acid = C – R ketoneR – C – R CH3-C-1 H=C-2Н2 +2 KMn+7O4 + 3H2SO4 a CH3-C+3 Oh + C+4 O2 + 2Mn+2SO4 + K2SO4 + 4H2O 2. If the reaction proceeds in a neutral environment when heated, then, accordingly, potassium salt: Fragment of a chain near a double bond What does it turn into K2CO3 = CH -R R– COOTo- carboxylic acid salt = C – R ketoneR – C – R 3CH3C-1H=With-2Н2 +10 K MnO4 - ta 3 CH3 C+3OO K + + 3K 2C+4O3 + 10MnO2 +4Н2О+ K Oh
6. Oxidation ethylene oxygen in the presence of palladium salts.	CH2=CH2 + O2 –(kat)à CH3CHO (acetaldehyde)
7. Chlorination and bromination to the side chain: if the reaction with chlorine is carried out in the light or at a high temperature, hydrogen is replaced in the side chain.	CH3-CH=CH2 + Cl2 – (light)à CH2-CH=CH2 + HCl
8. Polymerization:	n CH3-CH=CH2 а(-CH–CH2-)n propylene ô polypropylene

ALKENES PRODUCTION

I . Cracking alkanes:	С7Н16 –(t)а CH3-CH=CH2 + C4H10 alkene alkane
II. Dehydrohalogenation of haloalkanes under the action of an alcohol solution of alkali - the reaction ELIMINATING.	Zaitsev's rule: The elimination of a hydrogen atom in elimination reactions occurs predominantly from the least hydrogenated carbon atom.
III. Dehydration of alcohols at an elevated temperature (above 140°C) in the presence of depriving reagents - aluminum oxide or concentrated sulfuric acid - the elimination reaction.	CH3- CH-CH2-CH3 – (H2SO4,t>140o)à à H2O+CH3- CH=CH-CH3 (also obeys the Zaitsev rule)
IV. Dehalogenation of dihaloalkanes having halogen atoms at neighboring carbon atoms, under the action of active metals.	CH2 Br-CH Br-CH3+ mg aCH2=CH-CH3+ MgBr2 Zinc may also be used.
V. Dehydrogenation of alkanes at 500°С:
VI. Incomplete hydrogenation of dienes and alkynes	С2Н2 + Н2 (deficiency) –(kat)à С2Н4

ALKADIENES.

These are hydrocarbons containing two double bonds. The first member of the series is C3H4 (propadiene or allene). The suffix appears in the name - DIEN .

Types of double bonds in dienes:

1.Insulateddouble bonds separated in chain by two or more σ-bonds: CH2=CH–CH2–CH=CH2. Dienes of this type exhibit properties characteristic of alkenes.

2. Cumulativedouble bonds located on one carbon atom: CH2=C=CH2(allen) Such dienes (allenes) belong to a rather rare and unstable type of compounds.

3.Paireddouble bonds separated by one σ-bond: CH2=CH–CH=CH2 Conjugated dienes are characterized by characteristic properties due to the electronic structure of the molecules, namely, a continuous sequence of four sp2 carbon atoms.

Diene isomerism

1. Isomerism double bond positions:

2. Isomerism carbon skeleton:

3. Interclass isomerism with alkynes and cycloalkenes . For example, the following compounds correspond to the formula C4H6:

4. Spatial isomerism Dienes having various substituents at carbon atoms at double bonds, like alkenes, exhibit cis-trans isomerism. (1) Cis isomer (2) Trans isomer

Electronic structure of conjugated dienes.

Molecule of butadiene-1,3 CH2=CH-CH=CH2 contains four carbon atoms sp2 - hybridized state and has a flat structure.

π-electrons of double bonds form a single π-electron cloud (adjoint system ) and are delocalized between all carbon atoms.

The multiplicity of bonds (the number of common electron pairs) between carbon atoms has an intermediate value: there are no purely single and purely double bonds. The structure of butadiene is more accurately reflected by the formula with delocalized "one and a half" bonds.

CHEMICAL PROPERTIES OF CONJUGATED ALKADIENES.

REACTIONS OF ADDITION TO CONJUGATED DIENES. The addition of halogens, hydrogen halides, water and other polar reagents occurs by an electrophilic mechanism (as in alkenes). In addition to addition at one of the two double bonds (1,2-addition), conjugated dienes are characterized by the so-called 1,4-addition, when the entire delocalized system of two double bonds participates in the reaction: The ratio of 1,2- and 1,4-addition products depends on the reaction conditions (with an increase in temperature, the probability of 1,4-addition usually increases).

1. Hydrogenation. CH3-CH2-CH=CH2 (1,2 product) CH2=CH-CH=CH2 + H2 CH3-CH=CH-CH3 (1,4 product) In the presence of a Ni catalyst, a complete hydrogenation product is obtained: CH2=CH-CH=CH2 + 2 H2 –(Ni, t)à CH3-CH2-CH2-CH3

2. Halogenation, hydrohalogenation and hydration 1,4-attachment. 1,2-attachment. With an excess of bromine, one more of its molecule is added at the site of the remaining double bond to form 1,2,3,4-tetrabromobutane.

3. polymerization reaction. The reaction proceeds predominantly by the 1,4-mechanism, with the formation of a polymer with multiple bonds, called rubber : nCH2=CH-CH=CH2 à (-CH2-CH=CH-CH2-)n polymerization of isoprene: nCH2=C–CH=CH2 à(–CH2 –C =CH –CH2 –)n CH3 CH3 (polyisoprene)

OXIDATION REACTIONS - soft, hard, as well as burning. They proceed in the same way as in the case of alkenes - mild oxidation leads to a polyhydric alcohol, and hard oxidation leads to a mixture of various products depending on the structure of the diene: CH2=CH –CH=CH2 + KMnO4 + H2O à CH2 – CH – CH – CH2 + MnO2 + KOH

Alkadienes are burning to carbon dioxide and water. C4H6 + 5.5O2 à 4CO2 + 3H2O

OBTAINING ALKADIENES.

1. catalytic dehydrogenation alkanes (through the stage of formation of alkenes). In this way, divinyl is obtained in industry from butane contained in oil refining gases and associated gases: Isoprene is obtained by catalytic dehydrogenation of isopentane (2-methylbutane):

2. Lebedev's synthesis: (catalyst - a mixture of oxides Al2O3, MgO, ZnO 2 C2H5OH –(Al2O3,MgO, ZnO, 450˚C)à CH2=CH-CH=CH2 + 2H2O + H2

3. Dehydration of dihydric alcohols:

4. Action of an alcoholic solution of alkali on dihaloalkanes (dehydrohalogenation):

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

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

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

6.5. Oxidation of ethylene to acetaldehyde

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

(63)

ethanal (acetaldehyde)

6.6. Ethylene chlorine oxidation

Vinyl chloride is obtained by ethylene chlorine:

6.7. Oxidative ammonolysis

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

acrylonitrile

Hydrocyanic acid is obtained by oxidative ammonolysis of methane:

(66)

7. Hydroformylation of alkenes (Oxosynthesis)

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

Mechanism:

1. Ligand cleavage

2. Addition of ethylene

3. Introduction of ethylene

4. Attachment of a ligand

5. Introduction of CO

6. Oxidative addition of hydrogen

7. Reductive elimination of propanal

8. Addition of carbenes and carbenoids

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

8.1. The structure of carbenes

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

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

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

A. Dichlorocarbene

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

dichlorocarbene

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

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

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

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

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

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

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

(69)

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

dichlorocyclopropane

(70)

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

dichlorocyclopropane

(71)

7,7-dichlorobicycloheptane

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

bicycloheptane

Ex. eleven. Complete reactions:

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

Answer

B. methylene

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

(73)

diazomethane

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

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

B. Simmons reaction-Smith

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

diiodomethane iodomethylzinkiodide

(Simmons-Smith reagent)

(75)

The reaction proceeds according to the following mechanism:

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

Ex. 12. Complete reactions:

Answer

(76)

methylenecyclopentane spiroheptane

(77)

styrene cyclopropylbenzene

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

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

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

Download:

Preview:

To use the preview, create yourself a Google account (account) and log in: https://accounts.google.com

On the topic: methodological developments, presentations and notes

Arrangement of coefficients in chemical equations

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

Redox reactions involving organic 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.

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 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.

The dependence of the redox ability of organic matter on its structure:

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

• multiple bonds(that is why alkenes, alkynes, alkadienes are so easily oxidized);
• certain functional groups, capable of being easily oxidized (--SH, -OH (phenolic and alcohol), - NH 2;
• activated alkyl groups located next to multiple bonds. For example, propene can be oxidized to the unsaturated aldehyde acrolein with atmospheric 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 the oxidation of toluene to benzoic acid with potassium permanganate 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 at a carbon atom containing a functional group.

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

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

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

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 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.

When studying the comparative characteristics of inorganic and organic compounds, we got acquainted with 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 an organic molecule:

-8/3 +1

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

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

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, taken into account with the “+” sign at the carbon atom, and the number of bonds with hydrogen atoms (or another more electropositive element), taken into account with the sign "-" at the carbon atom. In this case, bonds with neighboring 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: -3 + 1 = -2. 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.

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

The processes of oxidation, combustion, halogenation, nitration, dehydrogenation, decomposition are redox processes.

When moving from one class of organic compounds to another andincrease in the degree of branching of the carbon skeleton molecules of compounds within a separate class the oxidation state of the carbon atom responsible for the reducing ability of the compound changes.

Organic substances whose molecules contain carbon atoms with maximum(- and +) CO values(-4, -3, +2, +3), enter into a complete oxidation-combustion reaction, but resistant to mild and medium strength oxidizers.

Substances whose molecules contain carbon atoms in CO -1; 0; +1, oxidize easily, their reduction abilities are close, so their incomplete oxidation can be achieved by one of the known oxidizing agents of low and medium strength. These substances may show dual nature, acting as an oxidizing agent, just as it is inherent in inorganic substances.

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:

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:

Pay attention to the first half of the electronic balance: the carbon atom in the fractional value of s.d. the denominator is 4, so we calculate the transfer of electrons using this coefficient.

Those. going from -2.5 to +4 corresponds to going 2.5 + 4 = 6.5 units. Because 4 carbon atoms are involved, then 6.5 4 \u003d 26 electrons will be given away in total by butane carbon atoms.

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

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

(4 C) -10 …… → (1 C) +4 , taking into account that the number of atoms before the = sign and after must be the same, we equalize (4C) -10 …… →[(1 C) +4 ] 4

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

In other cases, we determine the values ​​of s.d. each carbon atom in the compound, while paying attention to the sequence of substitution of hydrogen atoms at primary, secondary, tertiary carbon atoms:

First, the process of substitution occurs at the tertiary, then at the secondary, and, last of all, at the primary carbon atoms.

Alkenes

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

1. When alkenes are oxidized with a concentrated solution of potassium permanganate KMnO 4 in an acidic environment (hard oxidation) σ- and π-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.

a) 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:

b) 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 when it is oxidized, a ketone is formed, since the transformation of such an atom into an atom of the carboxyl group is impossible without breaking the C–C bond, which is relatively stable under these conditions:

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 the carbon atoms in the double bond contain two carbon radicals, is the formation of two ketones:

2. In neutral or slightly alkaline environments, 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:

During this reaction, the violet color of the aqueous solution of KMnO 4 is discolored. Therefore, it is used as qualitative reaction into alkenes (Wagner reaction).

3. Oxidation of alkenes in the presence of palladium salts (Wacker process) leads to the formation aldehydes and ketones:

2CH 2 \u003d CH 2 + O 2 PdCl2/H2O→ 2 CH 3 -CO-H

Homologues are oxidized at the less hydrogenated carbon atom:

CH 3 -CH 2 -CH \u003d CH 2 + 1 / 2O 2 PdCl2/H2O→ CH 3 - CH 2 -CO-CH 3

Alkynes

The oxidation of acetylene and its homologues proceeds depending on the medium in which the process takes place.

a) 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:

For acetylene:

1) In an acidic environment:

H-C≡C-H KMnO 4, H 2 SO 4 → HOOC-COOH (oxalic acid)

3CH≡CH +8KMnO 4 H 2 O→ 3KOOC-COOK potassium oxalate+8MnO 2 ↓+ 2KOH+ 2H 2 O

Arenas

(benzene and its homologues)

When oxidizing arenes in an acidic medium, one should expect the formation of acids, and in an alkaline medium, salts.

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

5C 6 H 5 -CH 3 + 6KMnO 4 + 9H 2 SO 4 \u003d 5C 6 H 5 COOH + 6MnSO 4 + 3K 2 SO 4 + 14H 2 O,

5C 6 H 5 -C 2 H 5 + 12KMnO 4 + 18H 2 SO 4 \u003d 5C 6 H 5 COOH + 5CO 2 + 12MnSO 4 + 6K 2 SO 4 + 28H 2 O,

C 6 H 5 -CH 3 + 2KMnO 4 \u003d C 6 H 5 COOK + 2MnO 2 + KOH + H 2 O.

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

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+CO2

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 homologues with potassium permanganate 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 with oxygen in the presence of a catalyst (cumene 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 - 18'→ C 6 H 5 COOH + 2CO 2 + 18H + | x5

MnO 4 - + 8H + + 5ē→ Mn +2 + 4H 2 O | x18

Attention should be paid to the fact that at mild oxidation of styrene with potassium permanganate KMnO 4 in a neutral or slightly alkaline medium the π-bond breaks, glycol (dihydric alcohol) is formed. As a result of the reaction, the colored solution of potassium permanganate quickly becomes colorless and a brown precipitate of manganese (IV) oxide precipitates.

Oxidation strong oxidizing agent- potassium permanganate in an acidic environment - leads to a complete rupture 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 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

Alcohols

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 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.

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.

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,

If three or more OH groups are bonded to adjacent carbon atoms, then the middle or middle atoms are converted to formic acid when oxidized with hydrochloric acid.

The oxidation of glycols with potassium permanganate in an acidic medium proceeds similarly to the oxidative cleavage of alkenes and also leads to the formation of acids or ketones, depending on the structure of the initial glycol.

Aldehydes and ketones

Aldehydes are more easily oxidized than alcohols to 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 also under the action of weak ones (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!!! Oxidation of methanal with an ammonia solution of silver oxide leads to the formation of ammonium carbonate, and not formic acid:

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

To compile the equations of redox reactions, both the electron balance method and the half-reaction method (electron-ion method) are used.

For organic chemistry, it is not the oxidation state of the atom that is important, but the shift in electron density, as a result of which partial charges appear on atoms that are in no way consistent with the values ​​of the oxidation states.

Many higher education institutions include tasks on the selection of coefficients in OVR equations by the ion-electronic method (half-reaction method) in tickets for entrance exams. 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 oxidation of sucrose with potassium permanganate in an acidic medium.

The 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 2+ cation, easily oxidized organic compounds are oxidized to CO 2.

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

On the left side, 13 oxygen atoms are missing; to eliminate this contradiction, let's add 13 H 2 O molecules.

The left side now contains 48 hydrogen atoms, they are released as H + cations:

Now we equalize the total charges on the right and left:

The half-reaction scheme is ready. Drawing up a scheme of the second half-reaction usually does not cause difficulties:

Let's combine both schemes:

Task for independent work:

Finish UHR and arrange the coefficients using the electronic balance method or the half-reaction method:

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

CH 3 -CH \u003d CH-CH 3 + KMnO 4 + H 2O →

(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 →

WithH 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 of students should be paid to the behavior of the oxidizing agent - potassium permanganate KMnO 4 in various environments. This is due to the fact that redox reactions in CMMs occur not only in tasks C1 and C2. In the tasks of SZ, representing a chain of transformations of organic substances, oxidation-reduction equations are not uncommon. At school, the oxidizing agent is often written above the arrow as [O]. The requirement for the performance of such tasks at the USE is the mandatory designation of all starting substances and reaction products with the arrangement of the necessary coefficients.

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 by 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 are broken 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