1.What hydrocarbons are called unsaturated and how are they divided? Write the general formulas of unsaturated hydrocarbons.
Unsaturated hydrocarbons are those containing one or more multiple bonds. The main classes of unsaturated hydrocarbons are alkenes, alkynes and alkadienes.
2. Alkenes include:
3)
3. Based on modern ideas about electronic orbitals, characterize the nature of chemical bonds in the ethylene molecule. 
4. The number of σ-bonds in an ethylene molecule is equal to:
2) 5
5. What types of isomerism are observed in saturated and unsaturated hydrocarbons? 
6. Draw abbreviated structural formulas of all hydrocarbons whose molecular formula is . Name these connections. 
7. Why is the number of isomers in hydrocarbons of the ethylene series greater than in saturated hydrocarbons? To prove this, give hydrocarbon isomers with molecular formulas and. 
8. The isomer of pentene-1 is
3) 2-methylbutene-1
9. A homologue of propene is
1) butene-1
10. How are ethylene and hydrocarbons of the ethylene series produced? Write the equations for the corresponding reactions. 
11. What chemical properties do ethylene and its homologues have? Confirm your answer with equations of the corresponding reactions. 
12. Write the equations for the combustion reactions of ethylene and propylene and their interaction with bromine water and a solution of potassium permanganate. 
13. How do propylene and butenes react with hydrogen halides? Using these examples, explain the essence of Markovnikov's rule. 
14. For what purposes are ethylene and propylene used?
Ethylene is used to produce ethylene glycol, polyethylene, and ethyl alcohol. In addition, ethylene accelerates the ripening of fruits. Isopropyl alcohol and polypropylene are obtained from propylene.
15. What reactions are called polymerization reactions? Write the equation for the polymerization reaction of propylene. 
16. Write equations chemical reactions, confirming the genetic relationship between classes of organic compounds in Scheme 5. 
Problem 1. Calculate:
a) what volume and what mass of ethylene can be obtained from 160 ml of ethyl alcohol, the density of which is 0.8 g/cm3; b) what volume of absolute (anhydrous) ethyl alcohol (ρ=0.8 g/cm3) can be obtained from 100 m3 of ethylene (n.s.) 
Problem 2. When ethylene was passed through bromine water, the mass of the solution increased by 7 g. What volume of gas reacted and what mass of 1,2-dibromoethane was formed (no.)? 
Problem 3. What volume of ethylene (no.) will be required to produce 126 kg of ethylene oxide if the mass fraction of ethylene production losses is 0.1? 
Problem 4. What volume of air is required to burn 50 liters of propylene (n.o.)? 
1. Write the structural formulas of isomers of carbonyl compounds with the general formula c5h10o and name them using rational nomenclature, Iupak nomenclature.
Solution:
For carbonyl compounds the formation structural isomers is associated with a change in the structure of the carbon chain, a different arrangement of the carbonyl group in the carbon chain:
The functional group in the structural formulas of aldehydes is often written in the form of an abbreviated formula as CHO (COH - incorrect!), in ketones the oxygen atom is placed in parentheses (O).
2. Write the reaction equations for the oxidation of butanal and 2-methylpentanone-3 with a chromium mixture.
Solution:
When aldehydes are oxidized, carboxylic acids are formed without changing the number of carbon atoms in their composition. The oxidation of ketones is accompanied by the destruction of the carbon chain, resulting in the formation of a mixture of carboxylic acids and (or) ketones with a smaller number of carbon atoms (compared to the original ketone) in their composition:
3. Write the reaction schemes for acetone and acetaldehyde with hydrocyanic acid and hydroxylamine. Compare the reactivity of carbonyl compounds in these reactions.
Solution:
The interactions of acetone and acetaldehyde with these reagents refer to nucleophilic addition reactions (AN). The reaction schemes look like:
a) interaction with hydrocyanic acid
Hydrolysis of the resulting oxynitriles is accompanied by the formation of α-hydroxy acids, and the number of carbon atoms in the acid composition increases by one: 2-hydroxypropanoic acid
(lactic acid)
b) interaction with hydroxylamine:
Nucleophilic addition reactions with nitrogen-containing nucleophiles (NH 3 , NH 2 OH, NH 2 -NH 2 , etc.) are accompanied by the subsequent elimination of a water molecule from the reaction product.
The reactivity of aldehydes in these reactions is higher than that of ketones, because the magnitude of the positive charge on the carbon atom of the carbonyl group in aldehydes compared to ketones is greater:
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4. Obtain acetone using all possible methods.
Solution:
Carbonyl compounds can be obtained as a result of the following transformations:
1. hydrolysis of 33 emm-dihalogenated hydrocarbons:
2. oxidation of alcohols, in which aldehydes are formed from primary alcohols, and ketones from secondary alcohols:
3. ozonolysis of alkenes followed by hydrolysis:
4. hydration of alkynes (Kucherov reaction):
5. pyrolysis of calcium (barium) salts of carboxylic acids:
3.3. Carboxylic acids and their derivatives
Literature
1. Ivanov, V.G. Organic chemistry: textbook. manual for universities / V.G. Ivanov, V.A. Gorlenko. – M.: Publishing Center “Academy”, 2009. -P. 222-254,469-479.
2. Grandberg, I.I. Organic chemistry: textbook. for universities / I.I. Grandberg. - M.: Bustard, 2002. -S. 357-392, 410-426, 393-399.
3. Artemenko, A.I. Organic chemistry: textbook. for universities / A.I. Artemenko. - M.: graduate School, 2009. -P.170-199, 251-253, 259-272, 402-408.
4. Petrov, A.A. Organic chemistry: textbook. for universities / A.A. Petrov. - St. Petersburg: Ivan Fedorov, 2002. -S. 189-2003, 207-220, 257-285, 421-434.
The recommended textbooks present a significantly larger volume of material than is necessary for the program. All textbooks discuss in detail the electronic structure, chemical properties of carboxylic acids, the preparation and use of acids and their derivatives. In the recommended textbook, the material is presented systematically and briefly, so it is better to independently study the topic “Carboxylic acids and their derivatives” using the textbook.
Hydrocarbon derivatives containing a carboxyl group are called carboxylic acids.
The study of carboxylic acids should begin with a consideration of the carboxyl group and the general classification of acids. Depending on the nature of the radical and the number of carboxylic groups, there are several types of carboxylic acids: aliphatic and aromatic carboxylic acids, mono- and dicarboxylic acids, as well as with mixed functions - hydroxy-, oxo-, and amino acids.
When studying the nomenclature, it should be noted that the most common are the trivial names of acids; some names need to be remembered: acetic (ethane), butyric (butane), oleic, stearic, linoleic, tartaric, oxalic, etc. The principles of compiling the names of acids according to the IUPAC nomenclature were discussed earlier (see “Nomenclature of organic compounds”).
When studying physical properties carboxylic acids, attention should be paid to the formation of intermolecular hydrogen bonds.
When studying chemical properties acids, it is necessary to establish similarities with inorganic acids: the ability to dissociate, the formation of salts. The presence of a mobile hydrogen atom in the carboxyl group -COO N defines acid properties carboxylic acids. The strength of acids depends on a number of factors: the structure and composition of the radical, the presence of electron-withdrawing or electron-donating groups in the radical, the number of carboxyl groups, etc. As a result of interaction with metals, metal oxides, bases and some salts, acids form salts. You should consider the properties of sodium and calcium salts (solubility, relation to hydrolysis, heating), pay attention to salts of higher carboxylic acids (soaps).
Carboxylic acids are characterized by nucleophilic substitution reactions. Substitution of the hydroxyl group in the carboxyl group leads to the production of acid derivatives: anhydrides, acid halides, amides, esters. The student should pay special attention to the esterification reaction (production of esters), its conditions and mechanism. When studying the chemistry of fats, it is necessary to remember the formulas of five basic acids (2 saturated and 3 unsaturated) and trihydric alcohol - glycerol, which are part of triglycerides. You should learn to write reaction equations for the formation of fats, and learn the most important chemical properties of fats: hydrolysis, hydrogenation.
A common property of all acid derivatives is their hydrolysis (acidic and alkaline), which results in the formation of a carboxylic acid or its salt. Acid derivatives are widely used in organic synthesis for the acylation of alcohols, phenols, amines and other compounds.
When studying dicarboxylic acids, it is necessary to note both the similarity and difference with monocarboxylic acids, due to the large number of carboxyl groups and their mutual influence.
When studying the structure and chemical properties of unsaturated acids, it is necessary to repeat the previously studied material on unsaturated hydrocarbons (see “Alkenes”).
When studying hydroxy acids, one should pay attention to the optical isomerism characteristic of these compounds, consider the properties of acids in the carboxyl group (formation of salts, esters) and alcohol group (substitution of the OH group, formation of alcoholates, ethers and esters). It is necessary to consider the relationship of hydroxy acids to heating and the elimination of water with the formation of products of various structures.
Keto acids can be considered using the example of pyruvic and acetoacetic acids. Pyruvic acid is an important intermediate product in the conversion of carbohydrates and proteins in organisms; it is necessary to take into account the existence of enol and ketone forms of the acid.
The main methods for producing carboxylic acids: hydrolysis of their derivatives (nitriles, esters, etc.), oxidation of alkenes, alcohols and aldehydes, fermentation of carbohydrates, etc.
Test questions and exercises:
1. Write the structural formulas of all isomeric acids of the composition C 5 H 10 O 2. Name the compounds using trivial and substitutive nomenclatures. Compare the acidic properties of the isomers.
2. Write for propionic (propanoic) acid the equations for possible reactions with metallic sodium, copper (II) hydroxide, sodium carbonate, ethyl alcohol, ammonia, followed by heating. Name the resulting compounds using IUPAC nomenclature.
3. Write the structural formulas of isomeric dicarboxylic acids with the composition C 4 H 6 O 4. Name the resulting isomers.
4. Write the structural formulas of isomeric unsaturated acids with the composition C 4 H 6 O 2. Which of them will have geometric isomers? Name the structural and spatial isomers.
5. Write the structural formula of the triglyceride - dipalmitoolein. Write the equations for bromination, saponification with sodium hydroxide, oxidation with potassium permanganate of the indicated fat.
6. Compare the acidic properties of acetic, oxoacetic, oxalic and hydroxyacetic acids.
EXAMPLES OF SOLUTIONS TO TYPICAL PROBLEMS
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The valency of carbon is ... (number). Therefore, when writing structural formulas, four lines should depart from carbon, depicting chemical bonds.
The form of writing the composition of an organic molecule in which each C atom is shown separately with bonds is called c………. f…….. . Chemically bonded carbon atoms represent carbon skeleton molecules of matter.
Three types of structural formulas
1. The most full form writing a hydrocarbon formula is when each atom of the molecule is shown separately:
This type of recording is bulky, takes up a lot of space and is rarely used.
2.
A form of recording in which the total number of hydrogen atoms for each C atom is indicated, and dashes are placed between adjacent carbons,
meaning x……… with…. :
CH 3 –CH 2 –CH 3, Cl–CH 2 –CH 2 –Br.
3. A structural formula in which dashes between atoms located in a record on one line are not indicated, while atoms appearing on other lines are connected by dashes to a straight chain:
Sometimes carbon chains are represented by broken lines, geometric shapes(triangle, square, cube). In this case, at each break in the chain, as well as at the beginning and end of the chain, a C atom is implied. For example, in the images
correspond to the structural formulas
Below are some properties of individual saturated hydrocarbons and their recording forms (Table 1).
Table 1
Names of saturated hydrocarbons (alkanes) of linear structure
| Name alkane | Molecular formula | Structural formula | Aggregate state | Temperature boiling point, °C |
|---|---|---|---|---|
| Methane | CH 4 | CH 4 | Gas | –161,6 |
| Ethane | C 2 H 6 | CH 3 CH 3 | Gas | –88,6 |
| Propane | C 3 H 8 | CH 3 CH 2 CH 3 | Gas | –42,1 |
| Butane | C 4 H 10 | CH 3 CH 2 CH 2 CH 3 | Gas | –0,5 |
| Pentane | C 5 H 12 | CH 3 (CH 2) 3 CH 3 | Liquid | 36,1 |
| Hexane | C 6 H 14 | CH 3(CH 2) 4 CH 3 | Liquid | 68,7 |
| Heptane | C 7 H 16 | CH 3 (CH 2) 5 CH 3 | Liquid | 98,5 |
| Octane | C 8 H 18 | CH 3 (CH 2) 6 CH 3 | Liquid | 125,6 |
| Nonan | C 9 H 20 | CH 3 (CH 2) 7 CH 3 | Liquid | 150,7 |
| Dean | C 10 H 22 | CH 3 (CH 2) 8 CH 3 | Liquid | 174,0 |
Compiling the names of branched and substituted alkanes
1. Select the main carbon chain and number it in such a way (left or right) that the incoming substituents receive the smallest numbers.
2. The name begins with a digital locant - the number of the carbon at which the substituent is located. After the number, the name of the deputy is written through a dash. The different substituents are listed sequentially. If identical substituents are repeated twice, then the prefix “di” is written in the name after the digital locators indicating the position of these substituents. Accordingly, with three identical substituents the prefix is “three”, with four – “tetra”, with five substituents – “penta”, etc.
Names of alternates
3. The name of the hydrocarbon numbered as the main carbon chain is written together with the prefix and substituent:
a) 2-methylbutane; b) 2,3-dimethylpentane; c) 2-chloro-4-methylpentane.
The names of cycloalkanes are formed in a similar way, only the prefix “cyclo” is added to the name of the hydrocarbon - according to the number of carbon atoms in the cycle:
Substances that are similar in structure, but differ in one or more groups -CH 2 - are known as g....... .
Examples homologues:
CH 3 –CH 3, CH 3 –CH 2 –CH 3, CH 3 –CH 2 –CH 2 –CH 3.
Element of similarity – linear chain alkanes:
The similarity of the three formulas of the substances in the last example - in each case, at the second C atom of the main carbon chain there is the same substituent - the CH 3 group.
The phenomenon of the existence of substances that are different in structure and properties, but have the same qualitative and quantitative composition, is called and....... .
Substances that have the same m……….. formula, but different s………. formulas are......
(Table 2).
table 2
Examples of isomeric hydrocarbons
| Molecular formula | Structural formulas |
|---|---|
| C 4 H 10 | CH 3 CH 2 CH 2 CH 3, |
| C 5 H 12 | CH 3 CH 2 CH 2 CH 2 CH 3, |
| C 6 H 14 | CH 3 (CH 2) 4 CH 3, 
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Rules for compiling isomers using the example of the compound C 5 H 11 Cl.
1. Write down the linear carbon chain C5:
S–S–S–S–S.
2. Determine which class of hydrocarbons this compound belongs to. The determination is made using general formulas for hydrocarbons of different classes (C n H 2 n+2 , C n H 2 n and so on.). The substance C 5 H 11 Cl is a chloroalkane, i.e. is a derivative of an alkane of the form C n H 2 n+2 (n= 5), in which one H atom is replaced by Cl. This means that all bonds in the molecule are single and there are no cycles.
3. Number the C atoms of the carbon chain (carbon skeleton) and place a Cl heteroatom at C-1:
4. Write down the required number of hydrogen atoms at each carbon of the chain, taking into account that the valence of C is IV. The result is isomer a):
5. The chlorine atom is moved along the main C5 chain, connecting it in series with the C-2 and C-3 atoms. This is how isomers b) and c are obtained:
A further shift of chlorine to the right along the chain does not produce new isomers. Thus, isomer a*) is identical to isomer a), isomer b*) is identical to isomer b). It’s just that in isomers a*) and b*) the direction of numbering of C atoms changes, counting is done from right to left (without asterisks it was from left to right):
6. Based on the carbon skeleton (see point 3), the outermost (fifth) C atom is removed and placed as a substituent on the internal carbon of the chain (first to C-2, then to C-3). Main chains C4 with a carbon substituent at C-2 and C-3 are obtained:
Write down the structural formulas of the new isomers:
7. By placing chlorine at the internal C atoms of the main carbon chain C4, two additional isomers are obtained:
8. A substance of the formula C 5 H 11 Cl may have a three-carbon main chain C 3:
Thus, for a substance with the molecular formula C 5 H 11 Cl, eight structural formulas of isomers a)–h) differing in structure can be compiled.
