Oxidation of formic acid with potassium permanganate equation. Redox reactions involving organic substances
Interaction of formic acid with ammonia solutionsilver hydroxide(reaction of a silver mirror). The formic acid molecule HCOOH has an aldehyde group, so it can be opened in solution by reactions characteristic of aldehydes, for example, by the silver mirror reaction.
An ammonia solution of argentum (Ι) hydroxide is prepared in a test tube. To do this, 1 - 2 drops of a 10% solution of sodium hydroxide are added to 1 - 2 ml of a 1% solution of argentum (Ι) nitrate, the resulting precipitate of argentum (Ι) oxide is dissolved by adding dropwise a 5% solution of ammonia. 0.5 ml of formic acid is added to the resulting clear solution. The test tube with the reaction mixture is heated for several minutes in a water bath (water temperature in the bath is 60 0 -70 0 C). Metallic silver is released as a mirror coating on the walls of the test tube or as a dark precipitate.
HCOOH + 2Ag [(NH 3) 2 ]OH → CO 2 + H 2 O + 2Ag + 4NH 3
b) Oxidation of formic acid with potassium permanganate. Approximately 0.5 g of formic acid or its salt, 0.5 ml of a 10% solution of sulfate acid and 1 ml of a 5% solution of potassium permanganate are placed in a test tube. The tube is closed with a stopper with a gas outlet tube, the end of which is lowered into another tube with 2 ml of lime (or barite) water, and the reaction mixture is heated.
5HCOOH + 2KMnO 4 + 3H 2 SO 4 → 5CO 2 + 8H 2 O + K 2 SO 4 + 2MnSO 4
in) Decomposition of formic acid when heated withconcentrated sulfuric acid. (Thrust!) Add 1 ml of formic acid or 1 g of its salt and 1 ml of concentrated sulfate acid into a dry test tube. The tube is closed with a stopper with a gas outlet tube and gently heated. Formic acid decomposes to form carbon(II) oxide and water. Carbon (II) oxide is ignited at the opening of the gas outlet tube. Pay attention to the nature of the flame.
After completion of work, the test tube with the reaction mixture must be cooled to stop the release of poisonous carbon monoxide.
Experience 12. Interaction of stearic and oleic acids with alkali.
Dissolve approximately 0.5 g of stearin in diethyl ether (without heating) in a dry test tube and add 2 drops of a 1% alcohol solution of phenolphthalein. Then, a 10% solution of sodium hydroxide is added dropwise. The crimson color that appears at the beginning disappears when shaken.
Write the equation for the reaction of stearic acid with sodium hydroxide. (Stearin is a mixture of stearic and palmitic acids.)
C 17 H 35 COOH + NaOH → C 17 H 35 COONa + H 2 O
sodium stearate
Repeat the experiment using 0.5 ml of oleic acid.
C 17 H 33 COOH + NaOH → C 17 H 33 COONa + H 2 O
sodium oleate
Experience13. The ratio of oleic acid to bromine water and potassium permanganate solution.
a) Reaction of oleic acid with bromine water Pour 2 ml of water into a test tube and add about 0.5 g of oleic acid. The mixture is shaken vigorously.
b) Oxidation of oleic acid with potassium permanganate. 1 ml of a 5% potassium permanganate solution, 1 ml of a 10% sodium carbonate solution and 0.5 ml of oleic acid are placed in a test tube. The mixture is vigorously stirred. Note the changes that occur with the reaction mixture.
Experience 14. Sublimation of benzoic acid.
Sublimation of small amounts of benzoic acid is carried out in a porcelain cup, closed with a wide end of a conical funnel (see Fig. 1), the diameter of which is somewhat smaller than the diameter of the cup.
The nose of the funnel is fixed in the foot of the tripod and tightly covered with cotton wool, and in order to prevent sublimation from falling back into the cup, it is covered with a round sheet of filter paper with several holes in it. A porcelain cup with small crystals of benzoic acid (t pl \u003d 122.4 0 C; sublimates below t pl) is carefully slowly heated on a small flame of a gas burner (on an asbestos grid). You can cool the top funnel by applying a piece of filter paper soaked in cold water. After the sublimation stops (after 15-20 minutes), the sublimate is carefully transferred with a spatula into a flask.
Note. For work, benzoic acid can be contaminated with sand.
The test tube in which the emulsion has formed is closed with a stopper under reflux, heated in a water bath until boiling starts and shaken. Does the solubility of oil increase when heated?
The experiment is repeated, but instead of sunflower oil, a small amount of animal fat (pork, beef or mutton fat) is added to test tubes with organic solvents,
b) Determination of the degree of unsaturation of fat by reaction with brominewater. (Thrust!) 0.5 ml of sunflower oil and 3 ml of bromine water are poured into a test tube. The contents of the tube are shaken vigorously. What happens to bromine water?
in) The interaction of vegetable oil with an aqueous solution of potassiumpermanganate (reaction of E. E. Wagner). About 0.5 ml of sunflower oil, 1 ml of a 10% sodium carbonate solution and 1 ml of a 2% potassium permanganate solution are poured into a test tube. Shake the contents of the tube vigorously. The purple color of potassium permanganate disappears.
Discoloration of bromine water and reaction with an aqueous solution of potassium permanganate are qualitative reactions to the presence of a multiple bond (unsaturation) in an organic molecule.
G) Saponification of fat with an alcohol solution of sodium hydroxide In a conical flask with a capacity of 50 - 100 ml, 1.5 - 2 g of solid fat is placed and 6 ml of a 15% alcoholic solution of sodium hydroxide is poured. The flask is stoppered with an air cooler, the reaction mixture is stirred and the flask is heated in a water bath with shaking for 10–12 min (water temperature in the bath is about 80 0 C). To determine the end of the reaction, a few drops of the hydrolyzate are poured into 2-3 ml of hot distilled water: if the hydrolyzate dissolves completely, without the release of fat drops, then the reaction can be considered complete. After saponification is completed, soap is salted out from the hydrolyzate by adding 6–7 ml of a hot saturated sodium chloride solution. The released soap floats, forming a layer on the surface of the solution. After settling, the mixture is cooled with cold water, the hardened soap is separated.
Chemistry of the process on the example of tristearin:
Experience 17. Comparison of the properties of soap and synthetic detergents
a) relation to phenolphthalein. Pour 2-3 ml of a 1% solution of laundry soap into one test tube, and the same amount of a 1% solution of synthetic washing powder into another. Add 2-3 drops of phenolphthalein solution to both tubes. Can these detergents be used to wash alkali-sensitive fabrics?
b) relation to acids. Add a few drops of a 10% solution of acid (chloride or sulfate) to solutions of soap and washing powder in test tubes. Does foam form when shaken? Do the detergent properties of the studied products remain in an acidic environment?
C 17 H 35 COONa+HCl→C 17 H 35 COOH↓+NaCl
in) Attitudetocalcium chloride. To solutions of soap and washing powder in test tubes, add 0.5 ml of a 10% solution of calcium chloride. Shake the contents of the tubes. Does this produce foam? Can these detergents be used in hard water?
C 17 H 35 COONa + CaCl 2 → Ca (C 17 H 35 COO) 2 ↓ + 2NaCl
An experience 18 . Interaction of glucose with an ammonia solution of argentum (Ι) oxide (silver mirror reaction).
0.5 ml of a 1% solution of argentum (Ι) nitrate, 1 ml of a 10% solution of sodium hydroxide are poured into a test tube, and a 5% solution of ammonia is added dropwise until the precipitate of argentum (Ι) hydroxide is dissolved. Then add 1 ml of 1% glucose solution and heat the contents of the tube for 5-10 minutes in a water bath at 70 0 - 80 0 C. Metallic silver is released on the walls of the tube in the form of a mirror coating. During heating, the test tubes must not be shaken, otherwise metallic silver will stand out not on the walls of the test tubes, but in the form of a dark precipitate. To obtain a good mirror, a 10% sodium hydroxide solution is first boiled in test tubes, then they are rinsed with distilled water.
Pour 3 ml of 1% sucrose solution into a test tube and add 1 ml of 10% sulfuric acid solution. The resulting solution is boiled for 5 min, then cooled and neutralized with dry sodium bicarbonate, adding it in small portions with stirring (carefully, the liquid foams from the evolved carbon monoxide (IY)). After neutralization (when the evolution of CO 2 stops), an equal volume of Fehling's reagent is added and the upper part of the liquid is heated until boiling begins.
Does the color of the reaction mixture change?
In another test tube, a mixture of 1.5 ml of a 1% sucrose solution with an equal volume of Fehling's reagent is heated. Compare the results of the experiment - the reaction of sucrose with Fehling's reagent before hydrolysis and after hydrolysis.
C 12 H 22 O 11 + H 2 O C 6 H 12 O 6 + C 6 H 12 O 6
glucose fructose
Note. In a school laboratory, Fehling's reagent can be replaced with cuprum (ΙΙ) hydroxide.
Experience 20. Hydrolysis of cellulose.
In a dry conical flask with a capacity of 50 - 100 ml, put some very finely chopped pieces of filter paper (cellulose) and moisten them with concentrated sulfate acid. Thoroughly mix the contents of the flask with a glass rod until the paper is completely destroyed and a colorless viscous solution is formed. After that, 15 - 20 ml of water is added to it in small portions with stirring (carefully!), The flask is connected to an air reflux condenser and the reaction mixture is boiled for 20 - 30 minutes, stirring it periodically. After hydrolysis is completed, 2–3 ml of liquid is poured, neutralized with dry sodium carbonate, adding it in small portions (the liquid foams), and the presence of reducing sugars is detected by reaction with Fehling's reagent or cuprum (ΙΙ) hydroxide.
(C 6 H 10 O 5)n+nH 2 O→nC 6 H 12 O 6
Cellulose glucose
Experience 21. Interaction of glucose with cuprum (ΙΙ) hydroxide.
a) Place 2 ml of 1% glucose solution and 1 ml of 10% sodium hydroxide into a test tube. Add 1-2 drops of a 5% solution of cuprum (ΙΙ) sulfate to the resulting mixture and shake the contents of the test tube. The bluish precipitate of cuprum (II) hydroxide formed at the beginning instantly dissolves, a blue transparent solution of cuprum (ΙΙ) saccharate is obtained. Process chemistry (simplified):- 
b) The contents of the test tube are heated over the flame of the burner, holding the test tube at an angle so that only the upper part of the solution is heated, and the lower part remains unheated (for control). When gently heated to boiling, the heated part of the blue solution turns orange-yellow due to the formation of cuprum (Ι) hydroxide. With longer heating, a precipitate of cuprum (Ι) oxide may form.
Experience 22. Interaction of sucrose with metal hydroxides. a) Reaction with cuprum (ΙΙ) hydroxide) in an alkaline medium. In a test tube, mix 1.5 ml of a 1% sucrose solution and 1.5 ml of a 10% sodium hydroxide solution. Then, a 5% solution of cuprum (ΙΙ) sulfate is added dropwise. The initially formed pale blue precipitate of cuprum (ΙΙ) hydroxide dissolves upon shaking, the solution acquires a blue-violet color due to the formation of complex cuprum (ΙΙ) saccharate.
b) Obtaining calcium sucrose. Into a small glass (25 - 50 ml) pour 5 - 7 ml of a 20% sucrose solution and add freshly prepared milk of lime dropwise with stirring. Calcium hydroxide dissolves in sucrose solution. The ability of sucrose to give soluble calcium sucrose is used in industry to purify sugar when it is isolated from sugar beets. in) Specific color reactions. 2-5 ml of a 10% sucrose solution and 1 ml of a 5% sodium hydroxide solution are poured into two test tubes. Then a few drops are added to one test tube. 5- percentage solution of cobalt (ΙΙ) sulfate, in another - a few drops 5- percentage solution of nickel (ΙΙ) sulfate. In a test tube with a cobalt salt, a violet color appears, and a green color appears with a nickel salt, Experiment 23. Interaction of starch with iodine. 1 ml of a 1% solution of starch paste is poured into a test tube and then a few drops of iodine strongly diluted with water in potassium iodide are added. The contents of the tube turn blue. The resulting dark blue liquid is heated to a boil. The color disappears, but reappears on cooling. Starch is a heterogeneous compound. It is a mixture of two polysaccharides - amylose (20%) and amylopectin (80%). Amylose is soluble in warm water and gives a blue color with iodine. Amylose consists of almost unbranched chains of glucose residues with a screw or helix structure (approximately 6 glucose residues in one screw). A free channel with a diameter of about 5 microns remains inside the helix, into which iodine molecules are introduced, forming colored complexes. When heated, these complexes are destroyed. Amylopectin is insoluble in warm water, swells in it, forming a starch paste. It consists of branched chains of glucose residues. Amylopectin with iodine gives a reddish-violet color due to the adsorption of iodine molecules on the surface of the side chains. Experience 24. hydrolysis of starch. a) Acid hydrolysis of starch. In a conical flask with a capacity of 50 ml, pour 20 - 25 ml of 1% starch paste and 3 - 5 ml of a 10% solution of sulfate acid. In 7 - 8 tubes pour 1 ml of a very dilute solution of iodine in potassium iodide (light yellow), the tubes are placed in a tripod. 1-3 drops of the starch solution prepared for the experiment are added to the first test tube. Note the resulting color. The flask is then heated on an asbestos grid with a small burner flame. 30 seconds after the start of boiling, a second sample of the solution is taken with a pipette, which is added to the second test tube with an iodine solution, after shaking, the color of the solution is noted. In the future, samples of the solution are taken every 30 seconds and added to subsequent test tubes with iodine solution. Note the gradual change in color of the solutions upon reaction with iodine. The color change occurs in the following order, see table.
After the reaction mixture ceases to give color with iodine, the mixture is boiled for another 2-3 minutes, after which it is cooled and neutralized with a 10% sodium hydroxide solution, adding it dropwise until the medium is alkaline (the appearance of a pink color on phenolphthalein indicator paper). Part of the alkaline solution is poured into a test tube, mixed with an equal volume of Fehling's reagent or a freshly prepared suspension of cuprum (ΙΙ) hydroxide, and the upper part of the liquid is heated until boiling begins.
(
Soluble
Dextrins
C 6 H 10 O 5) n (C 6 H 10 O 5) x (C 6 H 10 O 5) y
maltose
n/2 C 12 H 22 O 11 nC 6 H 12 O 6
b) Enzymatic hydrolysis of starch.
A small piece of black bread is chewed well and placed in a test tube. A few drops of a 5% solution of cuprum (ΙΙ) sulfate and 05 - 1 ml of a 10% solution of sodium hydroxide are added to it. The test tube with the contents is heated. 3. Technique and methodology for demonstration experiments on obtaining and studying the properties of nitrogen-containing organic substances.
Equipment: chemical beakers, glass rod, test tubes, Wurtz flask, dropping funnel, chemical glass, glass vapor tubes, connecting rubber tubes, splinter.
Reagents: aniline, methylamine, litmus and phenolphthalein solutions, concentrated chloride acid, sodium hydroxide solution (10%), bleach solution, concentrated sulfate acid, concentrated nitrate acid, egg white, copper sulfate solution, plumbum (ΙΙ) acetate, phenol solution , formalin.
Experience 1. Getting methylamine. In a Wurtz flask with a volume of 100 - 150 ml, add 5-7 g of methylamine chloride and close the stopper with an addition funnel inserted into it. Connect the gas outlet tube with a rubber tube with a glass tip and lower it into a glass of water. Add potassium hydroxide solution (50%) drop by drop from the funnel. Heat the mixture in the flask gently. Salt decomposes and methylamine is released, which is easily recognizable by its characteristic smell, which resembles the smell of ammonia. Methylamine is collected at the bottom of the glass under a layer of water: + Cl - +KOH → H 3 C - NH 2 + KCl + H 2 O
Experience 2. burning of methylamine. Methylamine burns with a colorless flame in air. Bring a burning splinter to the opening of the gas outlet tube of the device described in the previous experiment and observe the combustion of methylamine: 4H 3 C - NH 2 + 9O 2 → 4CO 2 +10 H 2 O + 2N 2
Experience 3. The ratio of methylamine to indicators. Pass the resulting methylamine into a test tube filled with water and one of the indicators. Litmus turns blue, and phenolphthalein becomes crimson: H 3 C - NH 2 + H - OH → OH This indicates the basic properties of methylamine.
Experience 4. Formation of salts by methylamine. a) A glass rod moistened with concentrated hydrochloric acid is brought to the opening of the test tube from which gaseous methylamine is released. The wand is shrouded in mist.
H 3 C - NH 2 + HCl → + Cl -
b) Pour 1-2 ml into two test tubes: into one - a 3% solution of ferum (III) chloride, into the other a 5% solution of cuprum (ΙΙ) sulfate. Gaseous methylamine is passed into each tube. In a test tube with a solution of ferum (III) chloride, a brown precipitate precipitates, and in a test tube with a solution of cuprum (ΙΙ) sulfate, the blue precipitate formed at the beginning dissolves to form a complex salt, colored bright blue. Process chemistry:
3 + OH - + FeCl 3 → Fe (OH) ↓ + 3 + Cl -
2 + OH - + CuSO 4 →Cu(OH) 2 ↓+ + SO 4 -
4 + OH - + Cu (OH) 2 → (OH) 2 + 4H 2 O
Experience 5. Reaction of aniline with hydrochloric acid. In a test tube with 5 ml of aniline add the same amount of concentrated hydrochloric acid. Cool the tube in cold water. A precipitate of aniline hydrogen chloride precipitates. Pour some water into a test tube with solid hydrogen chloride aniline. After stirring, aniline hydrogen chloride dissolves in water.
C 6 H 5 - NH 2 + HCl → Cl - Experiment 6. Interaction of aniline with bromine water. Add 2-3 drops of aniline to 5 ml of water and shake the mixture vigorously. Add bromine water dropwise to the resulting emulsion. The mixture becomes colorless and a white precipitate of tribromaniline precipitates.
An experience 7. Fabric dyeing with aniline dye. Wool dyeing and silk with acid dyes. Dissolve 0.1 g of methyl orange in 50 ml of water. The solution is poured into 2 glasses. To one of them add 5 ml of a 4N solution of sulfate acid. Then pieces of white woolen (or silk) fabric are lowered into both glasses. Solutions with tissue are boiled for 5 minutes. Then the fabric is taken out, washed with water, squeezed and dried in air, hung on glass rods. Pay attention to the difference in the color intensity of the pieces of fabric. How does the acidity of the environment affect the fabric dyeing process?
Experience 8. Proof of the presence of functional groups in amino acid solutions. a) Detection of the carboxyl group. To 1 ml of a 0.2% solution of sodium hydroxide, colored pink with phenolphthalein, add dropwise a 1% solution of aminoacetate acid (glycine) until the mixture of HOOC - CH 2 - NH 2 + NaOH → NaOOC - CH 2 - NH 2 becomes colorless + H 2 O b) Detection of the amino group. To 1 ml of a 0.2% solution of perchloric acid, colored blue by the Congo indicator (acidic medium), add dropwise a 1% solution of glycine until the color of the mixture changes to pink (neutral medium):
HOOC - CH 2 - NH 2 + HCl → Cl -
Experience 9. Action of amino acids on indicators. Add 0.3 g of glycine to a test tube and add 3 ml of water. Divide the solution into three test tubes. Add 1-2 drops of methyl orange to the first tube, the same amount of phenolphthalein solution to the second, and litmus solution to the third. The color of the indicators does not change, which is explained by the presence of acidic (-COOH) and basic (-NH 2) groups in the glycine molecule, which are mutually neutralized.
Experience 10. Protein precipitation. a) In two test tubes with a protein solution, add dropwise solutions of copper sulphate and plumbum (ΙΙ) acetate. Flocculent precipitates are formed, which dissolve in an excess of salt solutions.
b) Equal volumes of phenol and formalin solutions are added to two test tubes with a protein solution. Observe protein precipitation. c) Heat the protein solution in a burner flame. Observe the turbidity of the solution, which is due to the destruction of the hydration shells near the protein particles and their increase.
Experience 11. Color reactions of proteins. a) Xantoprotein reaction. Add 5-6 drops of concentrated nitrate acid to 1 ml of protein. When heated, the solution and the precipitate turns bright yellow. b) Biuret reaction. To 1 - 2 ml of protein solution add the same amount of diluted copper sulphate solution. The liquid turns red-violet. The biuret reaction makes it possible to identify a peptide bond in a protein molecule. The xantoprotein reaction occurs only if the protein molecules contain residues of aromatic amino acids (phenylalanine, tyrosine, tryptophan).
Experience 12. Reactions with urea. a) Solubility of urea in water. Placed in a test tube 0,5 g of crystalline urea and gradually add water until the urea is completely dissolved. A drop of the resulting solution is applied to red and blue litmus paper. What reaction (acidic, neutral or alkaline) does an aqueous solution of urea have? In aqueous solution, urea is in the form of two tautomeric forms:
b) hydrolysis of urea. Like all acid amides, urea is easily hydrolyzed in both acidic and alkaline media. Pour 1 ml of a 20% urea solution into a test tube and add 2 ml of clear barite water. The solution is boiled until a precipitate of barium carbonate appears in the test tube. Ammonia released from the test tube is detected by the blue color of wet litmus paper.
H 2 N - C - NH 2 + 2H 2 O → 2NH 3 + [HO - C - OH] → CO 2
→H 2 O 
Ba(OH) 2 + CO 2 →BaCO 3 ↓+ H 2 O
c) Biuret formation. Heat in a dry test tube 0,2 g urea. First, urea melts (at 133 C), then, upon further heating, it decomposes with the release of ammonia. Ammonia is detected by smell (carefully!) and by the blue of wet red litmus paper brought to the opening of the test tube. After some time, the melt in the test tube solidifies despite continued heating:
Cool the tube, add 1-2 ml of water and with low heating dissolve the biuret. In addition to biuret, the melt contains a certain amount of cyanuric acid, which is sparingly soluble in water, so the solution is cloudy. When the precipitate settles, pour the biuret solution from it into another test tube, add a few drops of a 10% sodium hydroxide solution (the solution becomes transparent) and 1-2 drops of a 1% solution of cuprum (ΙΙ) sulfate. The solution turns pink-violet. Excess cuprum (ΙΙ) sulfate masks the characteristic color, causing the solution to turn blue, and should therefore be avoided.
Experience 13. Functional analysis of organic substances. 1. Qualitative elemental analysis of organic compounds. The most common elements in organic compounds, in addition to carbon, are hydrogen, oxygen, nitrogen, halogens, sulfur, phosphorus. Conventional qualitative analysis methods are not applicable to the analysis of organic compounds. To detect Carbon, Nitrogen, Sulfur and other elements, organic matter is destroyed by fusion with sodium, while the elements under study are converted into inorganic compounds. For example, Carbon goes into carbon (IV) oxide, Hydrogen - into water, Nitrogen - into sodium cyanide, Sulfur - into sodium sulfide, halogens - into sodium halides. The elements are then discovered by conventional methods of analytical chemistry.
1. Detection of Carbon and Hydrogen by oxidation of substance cuprum(II) oxide.
Device for the simultaneous detection of carbon and hydrogen in organic matter:
1 - dry test tube with a mixture of sucrose and cuprum (II) oxide;
2 - test tube with lime water;
4 - anhydrous cuprum (ΙΙ) sulfate.
The most common, universal method of detection in organic matter. carbon and at the same time hydrogen is the oxidation of cuprum (II) oxide. In this case, carbon is converted into carbon (IU) oxide, and Hydrogen is converted into water. Place 0.2 - 0.3 g of sucrose and 1 - 2 g of cuprum (II) oxide powder. The contents of the test tube are thoroughly mixed, the mixture is covered with a layer of cuprum (II) oxide on top. - about 1 g. A small piece of cotton wool is placed in the upper part of the test tube (under the cork), on which is sprinkled with a little anhydrous copper (II) sulfate. The test tube is closed with a cork with a gas outlet tube and fixed in the leg of the tripod with a slight inclination towards the cork. I lower the free end of the gas outlet tube into a test tube with lime (or barite) water so that the tube almost touches the surface of the liquid. First, the entire test tube is heated, then the part where the reaction mixture is located is strongly heated. Note what happens to the lime water. Why does the color of cuprum (ΙΙ) sulfate change?
Chemistry of processes: C 12 H 22 O 11 + 24CuO → 12CO 2 + 11H 2 O + 24Cu
Ca (OH) 2 + CO 2 → CaCO 3 ↓ + H 2 O
CuSO 4 +5H 2 O → CuSO 4 ∙ 5H 2 O
2. Beilstein test on on halogens. When organic matter is calcined with cuprum (II) oxide, it is oxidized. Carbon turns into carbon (ІУ) oxide, Hydrogen - into water, and halogens (except fluorine) form volatile halides with Cuprum, which color the flame bright green. The response is very sensitive. However, it should be borne in mind that some other cuprum salts, such as cyanides, formed during the calcination of nitrogen-containing organic compounds (urea, pyridine derivatives, quinoline, etc.), also color the flame. The copper wire is held by the plug and its other end (loop) is calcined in the flame of the burner until the coloring of the flame stops and a black coating of cuprum(II) oxide forms on the surface. The cooled loop is moistened with chloroform, poured into a test tube, and again introduced into the flame of the burner. First, the flame becomes luminous (Carbon burns), then an intense green color appears. 2Cu+O 2 →2CuO
2CH - Cl 3 + 5CuO → CuCl 2 + 4CuCl + 2CO 2 + H 2 O
A control experiment should be made using a substance that does not contain halogen (benzene, water, alcohol) instead of chloroform. For cleaning, the wire is moistened with hydrochloric acid and calcined.
II. Opening of functional groups. Based on a preliminary analysis (physical properties, elemental analysis), it is possible to roughly determine the class to which a given test substance belongs. These assumptions are confirmed by qualitative reactions to functional groups.
1. Qualitative reactions to multiple carbon - carbon bonds. a) the addition of bromine. Hydrocarbons containing double and triple bonds easily add bromine:
To a solution of 0.1 g (or 0.1 ml) of the substance in 2 - 3 ml of carbon tetrachloride or chloroform, add dropwise with shaking a 5% solution of bromine in the same solvent. The instant disappearance of the color of bromine indicates the presence of a multiple bond in the substance. But the bromine solution is also decolorized by compounds containing mobile hydrogen (phenols, aromatic amines, tertiary hydrocarbons). However, in this case, a substitution reaction occurs with the release of hydrogen bromide, the presence of which is easily detected using a damp paper of blue litmus or Congo. b) Potassium permanganate test. In a weakly alkaline medium, under the action of potassium permanganate, the substance is oxidized with the breaking of a multiple bond, the solution becomes colorless, and a flocculent precipitate of MnO 2 is formed. - manganese (IU) oxide. To 0.1 g (or 0.1 ml) of a substance dissolved in water or acetone, add dropwise with shaking a 1% solution of potassium permanganate. There is a rapid disappearance of the crimson-violet color, and a brown precipitate of MnO 2 appears. However, potassium permanganate oxidizes substances of other classes: aldehydes, polyhydric alcohols, aromatic amines. In this case, the solutions also become discolored, but the oxidation proceeds for the most part much more slowly.
2. Detection of aromatic systems. Aromatic compounds, unlike aliphatic compounds, are able to easily enter into substitution reactions, often forming colored compounds. Usually, a nitration and alkylation reaction is used for this. Nitration of aromatic compounds. (‘Careful! Thrust!,) Nitration is carried out with nitric acid or a nitrating mixture:
R - H + HNO 3 → RNO 2 + H 2 O
0.1 g (or 0.1 ml) of the substance is placed in a test tube and, with continuous shaking, 3 ml of the nitrating mixture (1 part of concentrated nitrate acid and 1 part of concentrated sulfate acid) is gradually added. The test tube is stoppered with a long glass tube, which serves as a reflux condenser, and heated in a water bath. 5 min at 50 0 C. The mixture is poured into a glass with 10 g of crushed ice. If a solid product or an oil that is insoluble in water and different from the original substance precipitates, then the presence of an aromatic system can be assumed. 3. Qualitative reactions of alcohols. In the analysis for alcohols, substitution reactions are used both for the mobile hydrogen in the hydroxyl group and for the entire hydroxyl group. a) Reaction with metallic sodium. Alcohols readily react with sodium to form alcoholates that are soluble in alcohol:
2 R - OH + 2 Na → 2 RONa + H 2
Place 0.2 - 0.3 ml of anhydrous test substance in a test tube and carefully add a small piece of metallic sodium the size of a millet grain. The evolution of gas upon dissolution of sodium indicates the presence of active hydrogen. (However, acids and CH-acids can also give this reaction.) b) Reaction with cuprum (II) hydroxide. In di-, tri- and polyhydric alcohols, in contrast to monohydric alcohols, freshly prepared cuprum (II) hydroxide dissolves to form a dark blue solution of complex salts of the corresponding derivatives (glycolates, glycerates). Pour a few drops (0.3 - 0.5 ml) of a 3% solution of cuprum (ΙΙ) sulfate, and then 1 ml of a 10% solution of sodium hydroxide. A gelatinous blue precipitate of cuprum (ΙΙ) hydroxide precipitates. The dissolution of the precipitate upon the addition of 0.1 g of the test substance and the change in color of the solution to dark blue confirm the presence of a polyhydric alcohol with hydroxyl groups located at adjacent carbon atoms.
4. Qualitative reactions of phenols. a) Reaction with ferum (III) chloride. Phenols give intensely colored complex salts with ferum (III) chloride. A deep blue or purple color usually appears. Some phenols give a green or red color, which is more pronounced in water and chloroform and worse in alcohol. Place several crystals (or 1 - 2 drops) of the test substance in 2 ml of water or chloroform in a test tube, then add 1 - 2 drops of a 3% ferum (III) chloride solution with shaking. In the presence of phenol, an intense violet or blue color appears. Aliphatic phenols with ferum (ΙΙΙ) chloride in alcohol give a brighter color than in water, and blood-red color is characteristic of phenols. b) Reaction with bromine water. Phenols with free ortho- and pair-positions in the benzene ring easily decolorize bromine water, resulting in a precipitate of 2,4,6- tribromophenol 
A small amount of the test substance is shaken with 1 ml of water, then bromine water is added dropwise. Discoloration of the solution and precipitation of a white precipitate.
5. Qualitative reactions of aldehydes. Unlike ketones, all aldehydes are easily oxidized. The discovery of aldehydes, but not of ketones, is based on this property. a) Silver mirror reaction. All aldehydes easily reduce the ammonia solution of argentum (Ι) oxide. Ketones do not give this reaction:
In a well-washed test tube, mix 1 ml of a silver nitrate solution with 1 ml of a dilute sodium hydroxide solution. The precipitation of argentum (Ι) hydroxide is dissolved by adding a 25% ammonia solution. A few drops of an alcoholic solution of the analyte are added to the resulting solution. The tube is placed in a water bath and heated to 50 0 - 60 0 C. If a shiny deposit of metallic silver is released on the walls of the tube, this indicates the presence of an aldehyde group in the sample. It should be noted that other easily oxidized compounds can also give this reaction: polyhydric phenols, diketones, some aromatic amines. b) Reaction with Fehling's liquid. Fatty aldehydes are capable of reducing divalent cuprum to monovalent:
A test tube with 0.05 g of the substance and 3 ml of Fehling's liquid is heated for 3 - 5 minutes in a boiling water bath. The appearance of a yellow or red precipitate of cuprum (I) oxide confirms the presence of an aldehyde group. b. Qualitative reactions of acids. a) Determination of acidity. Water-alcohol solutions of carboxylic acids show an acid reaction to litmus, congo, or a universal indicator. A drop of a water-alcohol solution of the test substance is applied to a blue wet paper of litmus, congo or a universal indicator. In the presence of acid, the indicator changes its color: litmus becomes pink, Congo blue, and the universal indicator, depending on acidity, from yellow to orange. It should be borne in mind that sulfonic acids, nitrophenols and
some other compounds with a mobile "acidic" hydrogen that do not contain a carboxyl group may also give a color change to the indicator. b) Reaction with sodium bicarbonate. When carboxylic acids interact with sodium bicarbonate, carbon (IY) oxide is released: 1 - 1.5 ml of a saturated solution of sodium bicarbonate is poured into a test tube and 0.1 - 0.2 ml of an aqueous-alcoholic solution of the test substance is added. Isolation of bubbles of carbon(IY) oxide indicates the presence of acid.
RCOOH + NaHCO 3 → RCOONa + CO 2 + H 2 O
7. Qualitative reactions of amines. Amines dissolve in acids. Many amines (especially of the aliphatic series) have a characteristic odor (herring, ammonia, etc.). basicity of amines. Aliphatic amines, as strong bases, are capable of changing the color of indicators such as red litmus, phenolphthalein, and universal indicator paper. A drop of an aqueous solution of the test substance is applied to an indicator paper (litmus, phenolphthalein, universal indicator paper). A change in the color of the indicator indicates the presence of amines. Depending on the structure of the amine, its basicity varies over a wide range. Therefore, it is better to use universal indicator paper. eight. Qualitative reactions of polyfunctional compounds. For qualitative detection of bifunctional compounds (carbohydrates, amino acids), use the complex of the reactions described above.
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:
This material can be difficult to master with self-study, due to the large amount of information, many nuances, all kinds of BUT and IF. Read carefully!
What exactly will be discussed?
In addition to complete oxidation (combustion), some classes of organic compounds are characterized by partial oxidation reactions, while they are converted into other classes.
There are specific oxidizing agents for each class: CuO (for alcohols), Cu (OH) 2 and OH (for aldehydes) and others.
But there are two classic oxidizing agents, which, so to speak, are universal for many classes.
This is potassium permanganate - KMnO 4. And potassium dichromate (dichromate) - K 2 Cr 2 O 7. These substances are strong oxidizing agents due to manganese in the +7 oxidation state, and chromium in the +6 oxidation state, respectively.
Reactions with these oxidizing agents are quite common, but nowhere is there a holistic guide on how to choose the products of such reactions.
In practice, there are a lot of factors that affect the course of the reaction (temperature, medium, concentration of reagents, etc.). Often a mixture of products is obtained. Therefore, it is almost impossible to predict the product that is formed.
But this is not good for the Unified State Examination: there you can’t write “maybe either this, or this, or otherwise, or a mixture of products.” There needs to be specifics.
The compilers of the assignments have invested a certain logic, a certain principle according to which a certain product should be written. Unfortunately, they did not share with anyone.
This question in most manuals is rather slippery bypassed: two or three reactions are given as an example.
I present in this article what can be called the results of a study-analysis of USE tasks. The logic and principles of compiling the oxidation reactions with permanganate and dichromate have been unraveled with fairly high accuracy (in accordance with the USE standards). About everything in order.
Determination of the degree of oxidation.
First, when dealing with redox reactions, there is always an oxidizing agent and a reducing agent.
The oxidizing agent is manganese in permanganate or chromium in dichromate, the reducing agent is atoms in the organic (namely, carbon atoms).
It is not enough to define the products, the reaction must be equalized. For equalization, the electronic balance method is traditionally used. To apply this method, it is necessary to determine the oxidation states of reducing agents and oxidizing agents before and after the reaction.
For inorganic substances, we know the oxidation states from grade 9:
But in organic, probably, in the 9th grade they were not determined. Therefore, before learning how to write OVR in organic chemistry, you need to learn how to determine the degree of oxidation of carbon in organic substances. This is done a little differently than in inorganic chemistry.
Carbon has a maximum oxidation state of +4, a minimum of -4. And it can show any degree of oxidation of this interval: -4, -3, -2, -1, 0, +1, +2, +3, +4.
First you need to remember what an oxidation state is.
The oxidation state is the conditional charge that occurs on an atom, assuming that the electron pairs are shifted completely towards the more electronegative atom.
Therefore, the oxidation state is determined by the number of displaced electron pairs: if it is shifted to a given atom, then it acquires an excess minus (-) charge, if from an atom, then it acquires an excess plus (+) charge. In principle, this is the whole theory that you need to know to determine the oxidation state of a carbon atom.
To determine the degree of oxidation of a particular carbon atom in a compound, we need to consider EACH of its bonds and see in which direction the electron pair will shift and what excess charge (+ or -) will arise from this on the carbon atom.
Let's look at specific examples:
At carbon three hydrogen bonds. Carbon and hydrogen - which is more electronegative? Carbon, then, along these three bonds, the electron pair will shift towards carbon. Carbon takes one negative charge from each hydrogen: it turns out -3
The fourth bond is with chlorine. Carbon and chlorine - which is more electronegative? Chlorine, which means that over this bond, the electron pair will shift towards chlorine. Carbon has one positive +1 charge.
Then, you just need to add: -3 + 1 = -2. The oxidation state of this carbon atom is -2.
Let's determine the oxidation state of each carbon atom:
Carbon has three bonds to hydrogen. Carbon and hydrogen - which is more electronegative? Carbon, then, along these three bonds, the electron pair will shift towards carbon. Carbon takes one negative charge from each hydrogen: it turns out -3
And one more bond with another carbon. Carbon and other carbon - their electronegativity is equal, so there is no displacement of the electron pair (the bond is not polar). 
This atom has two bonds with one oxygen atom, and one more bond with another oxygen atom (as part of the OH group). More electronegative oxygen atoms in three bonds pull an electron pair from carbon, and carbon has a +3 charge.
By the fourth bond, carbon is connected to another carbon, as we have already said, the electron pair does not shift along this bond.
Carbon is bonded to hydrogen atoms by two bonds. Carbon, as more electronegative, pulls one pair of electrons for each bond with hydrogen, acquires a charge of -2.
A carbon double bond is linked to an oxygen atom. The more electronegative oxygen attracts one electron pair for each bond. Together, two electron pairs are pulled from carbon. Carbon acquires a +2 charge.
Together it turns out +2 -2 = 0.
Let's determine the oxidation state of this carbon atom:
A triple bond with a more electronegative nitrogen gives carbon a charge of +3; there is no displacement of the electron pair due to the bond with carbon.
Oxidation with permanganate.
What will happen to permanganate?
The redox reaction with permanganate can proceed in different environments (neutral, alkaline, acidic). And it depends on the medium how exactly the reaction will proceed, and what products are formed in this case.
Therefore, it can go in three directions:
Permanganate, being an oxidizing agent, is reduced. Here are the products of his recovery:
- acid environment.
The medium is acidified with sulfuric acid (H 2 SO 4). Manganese is reduced to the +2 oxidation state. And the recovery products will be:
KMnO 4 + H 2 SO 4 → MnSO 4 + K 2 SO 4 + H 2 O
- Alkaline environment.
To create an alkaline environment, a fairly concentrated alkali (KOH) is added. Manganese is reduced to an oxidation state of +6. Recovery Products
KMnO 4 + KOH → K 2 MnO 4 + H 2 O
- Neutral environment(and slightly alkaline).
In a neutral medium, in addition to permanganate, water also enters into the reaction (which we write on the left side of the equation), manganese will be reduced to +4 (MnO 2), the reduction products will be:
KMnO 4 + H 2 O → MnO 2 + KOH
And in a slightly alkaline environment (in the presence of a low concentration KOH solution):
KMnO 4 + KOH → MnO 2 + H 2 O
What will happen to organics?
The first thing to learn is that it all starts with alcohol! This is the initial stage of oxidation. The carbon to which the hydroxyl group is attached undergoes oxidation.
When oxidized, the carbon atom "acquires" a bond with oxygen. Therefore, when writing down the scheme of the oxidation reaction, they write [O] above the arrow:
primary alcohol oxidized first to an aldehyde, then to a carboxylic acid:
Oxidation secondary alcohol breaks in the second stage. Since the carbon is in the middle, a ketone is formed, not an aldehyde (the carbon atom in the ketone group can no longer physically form a bond with the hydroxyl group):
Ketones, tertiary alcohols and carboxylic acids no longer oxidized
The oxidation process is stepwise - as long as there is where to oxidize and there are all conditions for this - the reaction goes on. Everything ends up with a product that does not oxidize under given conditions: a tertiary alcohol, a ketone, or an acid.
It is worth noting the stages of methanol oxidation. First, it is oxidized to the corresponding aldehyde, then to the corresponding acid:
A feature of this product (formic acid) is that the carbon in the carboxyl group is bonded to hydrogen, and if you look closely, you can see that this is nothing more than an aldehyde group:
And the aldehyde group, as we found out earlier, is oxidized further to the carboxyl:
Did you recognize the resulting substance? Its gross formula is H 2 CO 3 . This is carbonic acid, which breaks down into carbon dioxide and water:
H 2 CO 3 → H 2 O + CO 2
Therefore, methanol, formic aldehyde and formic acid (due to the aldehyde group) are oxidized to carbon dioxide.
mild oxidation.
Mild oxidation is oxidation without strong heating in a neutral or slightly alkaline medium (0 is written above the reaction ° or 20 °) .
It is important to remember that alcohols do not oxidize under mild conditions. Therefore, if they are formed, then oxidation stops on them. What substances will enter into a mild oxidation reaction?
- Containing a C=C double bond (Wagner reaction).
In this case, the π-bond breaks and "sits" on the released bonds along the hydroxyl group. It turns out dihydric alcohol:
Let's write the reaction of mild oxidation of ethylene (ethene). Let's write down the initial substances and predict the products. At the same time, we do not write H 2 O and KOH yet: they can appear both on the right side of the equation and on the left. And we immediately determine the oxidation states of the substances involved in the OVR:
Let's make an electronic balance (we mean that there are two or two carbon atoms of the reducing agent, they are oxidized separately):
Let's set the coefficients:
At the end, add the missing products (H 2 O and KOH). There is not enough potassium on the right - it means that the alkali will be on the right. We put a coefficient in front of it. There is not enough hydrogen on the left, so water is on the left. We put a coefficient in front of it:
Let's do the same with propylene (propene):
Cycloalkene is often slipped. Let him not confuse you. This is a regular hydrocarbon with a double bond:
Wherever this double bond is, the oxidation will proceed in the same way:
- containing an aldehyde group.
The aldehyde group is more reactive (more easily reacts) than the alcohol group. Therefore, the aldehyde will oxidize. Before acid:
Consider the example of acetaldehyde (ethanal). Let's write down the reactants and products and arrange the oxidation states. Let's make a balance and put the coefficients in front of the reducing agent and oxidizing agent:
In a neutral medium and slightly alkaline, the course of the reaction will be slightly different.
In a neutral environment, as we remember, we write water on the left side of the equation, and alkali on the right side of the equation (formed during the reaction):
In this case, in the same mixture, acid and alkali are nearby. Neutralization takes place.
They cannot exist side by side and react, salt is formed:
Moreover, if we look at the coefficients in the equation, we will understand that acids are 3 moles, and alkalis are 2 moles. 2 moles of alkali can only neutralize 2 moles of acid (2 moles of salt are formed). And one mole of acid remains. So the final equation will be:
In a slightly alkaline environment, alkali is in excess - it is added before the reaction, so all the acid is neutralized:
A similar situation arises in the oxidation of methanal. It, as we remember, is oxidized to carbon dioxide:
It must be borne in mind that carbon monoxide (IV) CO 2 is acidic. And will react with alkali. And since carbonic acid is dibasic, both an acid salt and an average salt can be formed. It depends on the ratio between alkali and carbon dioxide:
If alkali is related to carbon dioxide as 2:1, then there will be an average salt:
Or alkali can be significantly more (more than twice). If it is more than twice, then the remainder of the alkali will remain:
3KOH + CO 2 → K 2 CO 3 + H 2 O + KOH
This will occur in an alkaline environment (where there is an excess of alkali, since it was added to the reaction mixture before the reaction) or in a neutral environment, when a lot of alkali is formed.
But if alkali is related to carbon dioxide as 1:1, then there will be an acid salt:
KOH + CO 2 → KHCO 3
If there is more carbon dioxide than needed, then it remains in excess:
KOH + 2CO 2 → KHCO 3 + CO 2
This will be in a neutral environment if little alkali is formed.
We write down the starting substances, products, draw up a balance, put down the oxidation states in front of the oxidizing agent, reducing agent and the products that are formed from them:
In a neutral environment, an alkali (4KOH) will form on the right:
Now we need to understand what will be formed when three moles of CO 2 and four moles of alkali interact.
3CO 2 + 4KOH → 3KHCO 3 + KOH
KHCO 3 + KOH → K 2 CO 3 + H 2 O
So it turns out like this:
3CO 2 + 4KOH → 2KHCO 3 + K 2 CO 3 + H 2 O
Therefore, on the right side of the equation we write two moles of hydrocarbonate and one mole of carbonate:
And in a slightly alkaline environment, there are no such problems: due to the fact that there is an excess of alkali, an average salt will form:
The same will happen with the oxidation of oxalic acid aldehyde:
As in the previous example, a dibasic acid is formed, and according to the equation, 4 moles of alkali should be obtained (since 4 moles of permanganate).
In a neutral environment, again, all the alkali is not enough to completely neutralize all the acid.
Three moles of alkali go to form an acid salt, one mole of alkali remains:
3HOOC–COOH + 4KOH → 3KOOC–COOH + KOH
And this one mole of alkali goes into interaction with one mole of acid salt:
KOOC–COOH + KOH → KOOC–COOK + H2O
It turns out like this:
3HOOC–COOH + 4KOH → 2KOOC–COOH + KOOC–COOK + H2O
Final equation:
In a weakly alkaline medium, an average salt is formed due to an excess of alkali:
- containing a triple bondC≡ C.
Remember what happened during the mild oxidation of double bond compounds? If you do not remember, then scroll back - remember.
The π-bond breaks, attaches to the carbon atoms at the hydroxyl group. Here the same principle. Just remember that there are two pi bonds in a triple bond. First, this happens at the first π-bond:
Then on another π-bond:
A structure in which one carbon atom has two hydroxyl groups is extremely unstable. When something is unstable in chemistry, it tends to “fall off” something. Water falls off, like this:
This results in a carbonyl group.
Consider examples:
Ethine (acetylene). Consider the stages of oxidation of this substance:
Water splitting:
As in the previous example, in one reaction mixture, acid and alkali. Neutralization occurs - salt is formed. As can be seen from the coefficient in front of the alkali permanganate, there will be 8 moles, that is, it is quite enough to neutralize the acid. Final equation:
Consider the oxidation of butyne-2:
Water splitting:
No acid is formed here, so there is no need to fool around with neutralization.
Reaction equation:
These differences (between the oxidation of carbon at the edge and in the middle of the chain) are clearly demonstrated by the example of pentyn:
Water splitting:
It turns out a substance of an interesting structure:
The aldehyde group continues to oxidize:
Let's write down the starting materials, products, determine the degree of oxidation, draw up a balance, put down the coefficients in front of the oxidizing agent and reducing agent:
Alkali should form 2 mol (since the coefficient in front of permanganate is 2), therefore, all acid is neutralized:
Hard oxidation.
Hard oxidation is the oxidation sour, strongly alkaline environment. And also, in neutral (or slightly alkaline), but when heated.
In an acidic environment, they are also sometimes heated. But in order for hard oxidation to proceed not in an acidic environment, heating is a prerequisite.
What substances will undergo severe oxidation? (First, we will analyze only in an acidic environment - and then we will add the nuances that arise during oxidation in a strongly alkaline and neutral or slightly alkaline (when heated) environment).
With hard oxidation, the process goes to the maximum. As long as there is something to oxidize, oxidation continues.
- Alcohols. Aldehydes.
Consider the oxidation of ethanol. Gradually, it oxidizes to an acid:
We write down the equation. We write down the starting substances, OVR products, put down the oxidation states, draw up a balance. Equalize the reaction:
If the reaction is carried out at the boiling point of the aldehyde, when it is formed, it will evaporate (fly away) from the reaction mixture without having time to oxidize further. The same effect can be achieved under very gentle conditions (low heat). In this case, we write aldehyde as a product:
Consider the oxidation of secondary alcohol using the example of propanol-2. As already mentioned, the oxidation terminates at the second stage (the formation of a carbonyl compound). Since a ketone is formed, which is not oxidized. Reaction equation:
Consider the oxidation of aldehydes in terms of ethanal. It also oxidizes to acid:
Reaction equation:
Methanal and methanol, as mentioned earlier, are oxidized to carbon dioxide:
Metanal:
- Containing multiple bonds.
In this case, the chain breaks along the multiple bond. And the atoms that formed it undergo oxidation (acquire a bond with oxygen). Oxidize as much as possible.
When a double bond is broken, carbonyl compounds are formed from fragments (in the scheme below: from one fragment - aldehyde, from the other - ketone)
Let's analyze the oxidation of pentene-2:
Oxidation of "scraps":
It turns out that two acids are formed. Write down the starting materials and products. Let's determine the oxidation states of the atoms that change it, draw up a balance, equalize the reaction: 
When compiling the electronic balance, we mean that there are two or two carbon atoms of the reducing agent, they are oxidized separately:
Acid will not always form. Consider, for example, the oxidation of 2-methylbutene:
Reaction equation:
Absolutely the same principle in the oxidation of compounds with a triple bond (only oxidation occurs immediately with the formation of an acid, without the intermediate formation of an aldehyde):
Reaction equation:
When a multiple bond is located exactly in the middle, then not two products are obtained, but one. Since the "scraps" are the same and they are oxidized to the same products:
Reaction equation:
- Double corona acid.
There is one acid in which carboxyl groups (crowns) are connected to each other:
This is oxalic acid. Two crowns side by side are difficult to get along. It is certainly stable under normal conditions. But due to the fact that it has two carboxyl groups connected to each other, it is less stable than other carboxylic acids.
And therefore, under especially harsh conditions, it can be oxidized. There is a break in the connection between the "two crowns":
Reaction equation:
- Benzene homologues (and their derivatives).
Benzene itself does not oxidize, due to the fact that aromaticity makes this structure very stable. 
But its homologues are oxidized. In this case, the circuit also breaks, the main thing is to know exactly where. Some principles apply:
- The benzene ring itself is not destroyed, and remains intact until the end, the bond is broken in the radical.
- The atom directly bonded to the benzene ring is oxidized. If after it the carbon chain in the radical continues, then the gap will be after it.
Let's analyze the oxidation of methylbenzene. There, one carbon atom in the radical is oxidized: 
Reaction equation:
Let's analyze the oxidation of isobutylbenzene:
Reaction equation:
Let's analyze the oxidation of sec-butylbenzene:
Reaction equation:
During the oxidation of benzene homologues (and derivatives of homologues) with several radicals, two-three- and more basic aromatic acids are formed. For example, the oxidation of 1,2-dimethylbenzene:
Derivatives of benzene homologues (in which the benzene ring has non-hydrocarbon radicals) are oxidized in the same way. Another functional group on the benzene ring does not interfere:
Subtotal. Algorithm "how to write down the reaction of hard oxidation with permanganate in an acidic environment":
- Write down the starting materials (organics + KMnO 4 + H 2 SO 4).
- Write down the products of organic oxidation (compounds containing alcohol, aldehyde groups, multiple bonds, as well as benzene homologues will be oxidized).
- Record the permanganate reduction product (MnSO 4 + K 2 SO 4 + H 2 O).
- Determine the degree of oxidation in OVR participants. Draw up a balance. Put down the coefficients for the oxidizing agent and reducing agent, as well as for the substances that are formed from them.
- Then it is recommended to calculate how many sulfate anions are on the right side of the equation, in accordance with this, put the coefficient in front of sulfuric acid on the left.
- At the end, put the coefficient in front of the water.
Severe oxidation in a strongly alkaline medium and a neutral or slightly alkaline (when heated) medium.
These reactions are much less common. We can say that such reactions are exotic. And as befits any exotic reactions, these were the most controversial.
Hard oxidation is also hard in Africa, so organics are oxidized in the same way as in an acidic environment.
Separately, we will not analyze the reactions for each class, since the general principle has already been stated earlier. We will analyze only the nuances.
Strongly alkaline environment :
In a strongly alkaline environment, permanganate is reduced to an oxidation state of +6 (potassium manganate):
KMnO 4 + KOH → K 2 MnO 4 .
In a strongly alkaline environment, there is always an excess of alkali, therefore, complete neutralization will take place: if carbon dioxide is formed, there will be a carbonate, if an acid is formed, there will be a salt (if the acid is polybasic - an average salt).
For example, the oxidation of propene:
Ethylbenzene oxidation:
Slightly alkaline or neutral when heated :
Here, too, the possibility of neutralization must always be taken into account.
If oxidation proceeds in a neutral environment and an acidic compound (acid or carbon dioxide) is formed, then the resulting alkali will neutralize this acidic compound. But not always alkali is enough to completely neutralize the acid.
When aldehydes are oxidized, for example, it is not enough (oxidation will proceed in the same way as in mild conditions - the temperature will simply speed up the reaction). Therefore, both salt and acid are formed (roughly speaking, remaining in excess).
We discussed this when we discussed the mild oxidation of aldehydes.
Therefore, if you have acid in a neutral environment, you need to carefully see if it is enough to neutralize all the acid. Particular attention should be paid to the neutralization of polybasic acids.
In a weakly alkaline environment, due to a sufficient amount of alkali, only medium salts are formed, as there is an excess of alkali.
As a rule, alkali during oxidation in a neutral environment is quite enough. And the reaction equation that in a neutral, that in a slightly alkaline medium will be the same.
For example, consider the oxidation of ethylbenzene:
Alkali is enough to completely neutralize the resulting acid compounds, even excess will remain:
3 moles of alkali are consumed - 1 remains.
Final equation:
This reaction in a neutral and slightly alkaline medium will proceed in the same way (in a slightly alkaline medium there is no alkali on the left, but this does not mean that it does not exist, it simply does not enter into a reaction).
Redox reactions involving potassium dichromate (bichromate).
Bichromate does not have such a wide variety of organic oxidation reactions in the exam.
Oxidation with dichromate is usually carried out only in an acidic environment. At the same time, chromium is restored to +3. Recovery products:
The oxidation will be tough. The reaction will be very similar to permanganate oxidation. The same substances will be oxidized that are oxidized by permanganate in an acidic environment, the same products will be formed.
Let's take a look at some of the reactions.
Consider the oxidation of alcohol. If the oxidation is carried out at the boiling point of the aldehyde, then it will leave their reaction mixture without being oxidized:
Otherwise, the alcohol can be directly oxidized to an acid.
The aldehyde produced in the previous reaction can be "caught" and made to oxidize to an acid:
Oxidation of cyclohexanol. Cyclohexanol is a secondary alcohol, so a ketone is formed:
If it is difficult to determine the oxidation states of carbon atoms using this formula, you can write on the draft:
Reaction equation:
Consider the oxidation of cyclopentene.
The double bond breaks (the cycle opens), the atoms that formed it are oxidized to the maximum (in this case, to the carboxyl group):
Some features of oxidation in the USE with which we do not entirely agree.
Those "rules", principles and reactions that will be discussed in this section, we consider not entirely correct. They contradict not only the real state of affairs (chemistry as a science), but also the internal logic of the school curriculum and the USE in particular.
But nevertheless, we are forced to give this material in the form that the USE requires.
We are talking about HARD oxidation.
Remember how benzene homologues and their derivatives are oxidized under harsh conditions? All radicals are terminated - carboxyl groups are formed. Scraps are oxidized already "independently":
So, if suddenly a hydroxyl group, or a multiple bond, appears on the radical, you need to forget that there is a benzene ring there. The reaction will go ONLY along this functional group (or multiple bond).
The functional group and multiple bond is more important than the benzene ring.
Let's analyze the oxidation of each substance:
First substance:
It is necessary not to pay attention to the fact that there is a benzene ring. From the point of view of the exam, this is just secondary alcohol. Secondary alcohols are oxidized to ketones, and ketones are not further oxidized:
Let this substance be oxidized with dichromate:
Second substance:
This substance is oxidized, just as a compound with a double bond (we do not pay attention to the benzene ring):
Let it oxidize in neutral permanganate when heated:
The resulting alkali is enough to completely neutralize carbon dioxide:
2KOH + CO 2 → K 2 CO 3 + H 2 O
Final equation:
Oxidation of the third substance:
Let the oxidation proceed with potassium permanganate in an acidic medium:
Oxidation of the fourth substance:
Let it oxidize in a strongly alkaline environment. The reaction equation will be:
And finally, this is how vinylbenzene is oxidized:
And it oxidizes to benzoic acid, it must be borne in mind that, according to the logic of the Unified State Examination, it oxidizes this way not because it is a derivative of benzene. Because it contains a double bond.
Conclusion.
This is all you need to know about redox reactions involving permanganate and dichromate in organics.
Do not be surprised if, some of the points outlined in this article, you hear for the first time. As already mentioned, this topic is very extensive and controversial. And despite this, for some reason, very little attention is paid to it.
As you may have seen, two or three reactions do not explain all the patterns of these reactions. Here you need an integrated approach and a detailed explanation of all points. Unfortunately, in textbooks and on Internet resources, the topic is not fully disclosed, or not disclosed at all.
I tried to eliminate these shortcomings and shortcomings and consider this topic in its entirety, and not in part. I hope I succeeded.
Thank you for your attention, all the best to you! Good luck in mastering chemical science and passing exams!
C 6 H 5 -CHO + O 2 ® C 6 H 5 -CO-O-OH
The resulting perbenzoic acid oxidizes the second molecule of benzoic aldehyde to benzoic acid:
C 6 H 5 -CHO + C 6 H 5 -CO-O-OH ® 2C 6 H 5 -COOH
Experiment No. 34. Oxidation of benzoic aldehyde with potassium permanganate
Reagents:
benzoic aldehyde
Potassium permanganate solution
Ethanol
Progress:
Place ~3 drops of benzaldehyde in a test tube, add ~2 ml of potassium permanganate solution and heat on a water bath with shaking until the smell of aldehyde disappears. If the solution does not discolor, then the color is destroyed with a few drops of alcohol. The solution is cooled. Crystals of benzoic acid fall out:
C 6 H 5 -CHO + [O] ® C 6 H 5 -COOH
Experiment No. 35. The oxidation-reduction reaction of benzaldehyde (Cannizzaro reaction)
Reagents:
benzoic aldehyde
Alcoholic solution of potassium hydroxide
Progress:
Add ~5 ml of a 10% alcoholic solution of potassium hydroxide to ~1 ml of benzoic aldehyde in a test tube and shake vigorously. In this case, heat is released and the liquid solidifies.
The redox reaction of benzoic aldehyde in the presence of alkali proceeds according to the following scheme:
2C 6 H 5 -CHO + KOH ® C 6 H 5 -COOK + C 6 H 5 -CH 2 -OH
The potassium salt of benzoic acid (a product of the oxidation of benzoic aldehyde) and benzyl alcohol (a product of the reduction of benzoic aldehyde) are formed.
The resulting crystals are filtered off and dissolved in a minimum amount of water. When ~1 ml of a 10% hydrochloric acid solution is added to a solution, free benzoic acid precipitates:
C 6 H 5 -COOK + HCl ® C 6 H 5 -COOH¯ + KCl
Benzyl alcohol is in the solution remaining after separation of the crystals of the potassium salt of benzoic acid (the solution has the smell of benzyl alcohol).
VII. CARBOXY ACIDS AND THEIR DERIVATIVES
Experience No. 36. Oxidation of formic acid
Reagents:
Formic acid
10% sulfuric acid solution
Potassium permanganate solution
Barite or lime water
Progress:
~0.5-1 ml of formic acid, ~1 ml of 10% sulfuric acid solution and ~4-5 ml of potassium permanganate solution are poured into a test tube with a gas outlet tube. The gas outlet tube is immersed in a test tube with a solution of lime or barite water. The reaction mixture is gently heated by placing boiling stones in a test tube for uniform boiling. The solution first turns brown, then discolors, carbon dioxide is released:
5H-COOH + 2KMnO 4 + 3H 2 SO 4 ® 5HO-CO-OH + K 2 SO 4 + 2MnSO 4 + 3H 2 O
HO-CO-OH ® CO 2 + H 2 O
Experience No. 37. Recovery of an ammonia solution of silver hydroxide with formic acid
Reagents:
Ammonia silver hydroxide solution (Tollens' reagent)
Formic acid
This substance can be considered not only as an acid, but also as an aldehyde. The aldehyde group is circled in brown.
Therefore, formic acid exhibits the reducing properties typical of aldehydes:
1. Silver mirror reaction:
2Ag (NH3)2OH ® NH4HCO3 + 3NH3 + 2Ag + H2O.
2. Reaction with copper hydroxide when heated:
НСООНa + 2Cu (OH)2 + NaOH ® Na2CO3 + Cu2O¯ + 3H2O.
3. Oxidation with chlorine to carbon dioxide:
HCOOH + Cl2 ® CO2 + 2HCl.
Concentrated sulfuric acid removes water from formic acid. This produces carbon monoxide:
In the molecule of acetic acid there is a methyl group, the rest of the saturated hydrocarbon - methane.
Therefore, acetic acid (and other saturated acids) will enter into radical substitution reactions characteristic of alkanes, for example:
CH3COOH + Cl2 + HCl
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presentation source - http://prezentacii.com/po_himii/13798-schavelevaya-kislota.html
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