Chemistry. Thematic tests to prepare for the exam. Tasks high level complexity (C1-C5). Ed. Doronkina V.N.

3rd ed. - R. n / D: 2012. - 234 p. R. n / D: 2011. - 128 p.

The proposed manual has been compiled in accordance with the requirements of the new USE specification and is intended to prepare for a unified state exam in chemistry. The book includes tasks of a high level of complexity (С1-С5). Each section contains the necessary theoretical information, parsed (demo) examples of tasks that allow you to master the methodology for completing tasks of part C, and groups of training tasks by topic. The book is addressed to students in grades 10-11 educational institutions preparing for the exam and planning to receive high score on the exam, as well as teachers and methodologists who organize the process of preparing for the exam in chemistry. The manual is part of the educational and methodological complex “Chemistry. Preparation for the Unified State Exam”, which includes such manuals as “Chemistry. Preparation for the USE-2013”, “Chemistry. 10-11 grades. Thematic tests to prepare for the exam. Basic and elevated levels" and etc.

Format: pdf (2012 , 3rd ed., rev. and add., 234s.)

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CONTENT
Introduction 3
Question C1. Redox reactions. Corrosion of metals and methods of protection against it 4
Question questions C1 12
Question C2. Reactions confirming the relationship of various classes not organic matter 17
Question questions C2 28
SZ question. Reactions confirming the relationship of hydrocarbons and oxygen-containing organic compounds 54
Asking questions HH 55
Question C4. Calculations: mass (volume, amount of substance) of the reaction products, if one of the substances is given in excess (has impurities), if one of the substances is given as a solution with a certain mass fraction solute 68
Questions C4 73
Question C5. Finding molecular formula substances 83
Question questions C5 85
Answers 97
Appendix. The relationship of various classes of inorganic substances. Additional tasks 207
Tasks 209
Problem solving 218
Literature 234

INTRODUCTION
This book is designed to prepare you for high-level tasks in general, inorganic, and organic chemistry(tasks of part C).
For each of the questions C1 - C5, a large number of assignments (more than 500 in total), which will allow graduates to test their knowledge, improve their existing skills, and, if necessary, learn the factual material included in the test tasks of part C.
The content of the manual reflects the features USE options offered in last years, and complies with the current specification. Questions and answers correspond to the wording of the USE tests.
Part C assignments have varying degrees difficulties. Maximum score correctly completed task is from 3 to 5 points (depending on the degree of complexity of the task). Checking the tasks of this part is carried out on the basis of a comparison of the graduate's answer with an element-by-element analysis of the given sample answer, each correctly completed element is estimated at 1 point. For example, in the SZ task, you need to write 5 equations for reactions between organic substances that describe the sequential transformation of substances, and you can only write 2 (suppose the second and fifth equations). Be sure to write them down on the answer sheet, you will get 2 points for the task of SZ and significantly increase your score on the exam.
We hope that this book will help you successfully pass the exam.

The proposed manual presents a system of tasks of increased and high levels of complexity, similar to control tasks. measuring materials USE in chemistry. Particular attention is paid to the analysis of tasks that caused the greatest difficulties. For training and self-preparation for the exam, tasks of various levels of complexity are offered in key sections of the chemistry course.
The manual is addressed to high school students, teachers and parents. It will help schoolchildren to test their knowledge and skills in the subject, and teachers to assess the degree of achievement of the requirements. educational standards individual students and ensure their targeted preparation for the exam.

Examples.
When zinc oxide was reduced with carbon monoxide, a metal was formed. The metal reacted with a concentrated potassium hydroxide solution to form a complex salt. An excess of hydrogen sulfide was passed through the salt solution, and a precipitate formed. When this precipitate is heated with concentrated nitric acid brown gas was released. Write equations of four described reactions.

A mixture of nitric oxide (IV) and oxygen was passed through a solution of potassium hydroxide. The resulting salt was dried and calcined. The residue obtained after calcination of the salt was dissolved in water and mixed with a solution of potassium iodide and sulfuric acid. The simple substance formed during this reaction reacted with aluminum. Write the equations for the four described reactions.

CONTENT
INTRODUCTION
PART 1
Section 1. Redox reactions, patterns of their course. Electrolysis of solutions and molten salts
1.1. Material for repetition and systematization of knowledge
1.2. Tasks with comments and solutions
1.3. Tasks for independent work
Section 2. Classification of inorganic substances. Characteristic chemical properties of inorganic substances of various classes. The relationship of inorganic substances
2.1. Material for repetition and systematization of knowledge
2.2. Tasks with comments and solutions
2.3. Tasks for independent work
Section 3. Classification of organic substances. Characteristic properties organic substances of various classes. The relationship of organic substances
3.1. Material for repetition and systematization of knowledge
3.2. Tasks with comments and solutions
3.3. Tasks for independent work
Section 4 Calculation problems
PART 2. TRAINING OPTIONS, INCLUDING TASKS OF INCREASED AND HIGH LEVEL OF COMPLEXITY
Option 1
Option 2
Option 3
Option 4
Option 5
PART 3. ANSWERS TO TASKS FOR INDEPENDENT WORK AND TASKS OF TRAINING OPTIONS.

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In the course of studying organic chemistry, tasks are often used to complete chains of transformations. They are used in the 9th grade at the first stage of the study of organic substances, in the 10th grade when studying the factual material in this course and in the 11th grade at final stage learning. This question is included in the tasks of the third part of the chemistry exam material in the form and USE materials. Tasks for the implementation of transformations are widely used in general lessons

Target.

· To promote the formation of students of a higher level of complexity of training on the issue of the genetic relationship between organic compounds;

Create conditions for systematization and deepening of students' knowledge about the relationship of organic substances according to the scheme: composition - structure - properties of substances

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Theory and practice of solving problems of a high level of complexity in the process of teaching chemistry (based on the topic " genetic connection organic compounds") Completed by: Vidershpan I.P., teacher of chemistry, MBOU "Klyuchevskaya secondary school No. 1", 2015 MBOU "Klyuchevskaya secondary comprehensive school No. 1 "Klyuchevsky district, Altai Territory

In the course of studying organic chemistry, tasks are often used to complete chains of transformations. They are used in the 9th grade at the first stage of studying organic substances, in the 10th grade when studying the factual material in this course in the 11th grade at the final stage of education. this question is included in the tasks of the third part of the exam material in chemistry in the form and materials of the exam. Tasks for the implementation of transformations are widely used in general lessons.

Target. Help students develop a higher level of complexity general training on the issue of the genetic relationship between organic compounds; create conditions for the systematization and deepening of students' knowledge about the relationship of organic substances according to the scheme: composition - structure - properties of substances Tasks. develop in students logical thinking(by establishing a genetic link between different classes hydrocarbons and derivatives of hydrocarbons, hypotheses about chemical properties ah unfamiliar organic matter); develop students' ability to compare (using the example of comparing the chemical properties of organic compounds); develop information and cognitive competence of students.

What does the concept of "genetic connection" mean? The genetic bond is the relationship between substances. different classes compounds based on their mutual transformations and reflecting their origin. The genetic relationship can be reflected in the genetic series. genetic series consists of substances, which reflects the transformation of substances of one class of compounds into substances of other classes containing the same number of carbon atoms. CH 4 → CH 3 NO 2 → CH 3 NH 2 → CH 3 NH 3 Cl → CH 3 NH 2 → N 2

In order to successfully complete tasks that show genetic relationships between classes of organic substances, knowledge of the nomenclature and classification of substances is worked out at chemistry lessons, the chemical properties of compounds and methods for their preparation are studied. This approach can be traced at all stages of the study of this issue, only each time theoretical material deepens and expands.

Nomenclature and classification In the issue of studying the nomenclature and classification of organic substances, you can create a map of compound formulas and use it for conversation, work in groups, in pairs and individually. This will allow you to quickly correct the knowledge of students. They, having formulas in front of them, can quickly and efficiently navigate the task, reason, answer the questions posed, while acquiring new knowledge. The use of this map of compound formulas in chemistry lessons affected the quality of the knowledge gained on the issue of nomenclature and classification of organic substances.

Formula Map chemical compounds(the map of formulas is printed in abbreviation) 1) H - COOH methanoic acid, formic acid 2) CH 3 - COOH ethanoic acid, acetic acid 3) C 17 H 35 - COOH stearic acid 4) CH 3 - CH 2 - OH ethyl alcohol, ethanol C 2 H 5 OH 5) CH 3 - OH methyl alcohol, methanol 6 ) H is COH methanal, formic aldehyde, formaldehyde 7) CH 3 - COH; ethanal, acetaldehyde, acetaldehyde 8) CH 3 - COONa; sodium acetate 9) CH 3 - O - CH 3 dimethyl ether 10) CH 3 - COOCH 3 methyl acetate, methyl ether acetic acid. 1 1) CH 3 - COOC 2 H 5 ethyl acetate, acetic acid ethyl ester 1 2) HCOOCH 3 methyl formate, formic acid methyl ester.

The study of chemical properties and methods for obtaining organic compounds. When studying these questions in the classroom, you can apply reference circuits. Reference schemes allow short form give a lot of information. Students enjoy using them. These schemes help them organize their knowledge and develop the logic of everyone's thinking. Thus, students are prepared to perform transformation exercises. Students perform outline diagrams when studying each topic in stages. Repeated reference to the outline scheme allows you to lay a solid foundation in the knowledge gained.

C 6 H 12 O 6 → C 2 H 5 OH → CH 3 - CHO → CH 3 - COOH → Cl - CH 2 - COOH → H 2 N - CH 2 - COOH To fix educational material and revitalization learning activities recommended to pay Special attention on the solution of problems of the type associated with the mutual transformations of substances (chains of transformations).

Success in completing tasks will depend on the number of transformations carried out 1) CaC 2 → X 1 → X 2 → nitrobenzene [H] X 3 + HCl X 4 2) C 2 H 6 + Cl 2, hN X 1 → X 2 → CH 2 \u003d CH - CH \u003d CH 2 + H 2 (1.4 addition) → X 3 + Cl 2 X 4 3) CH 4 1500 C X 1 +2 H 2 X 2 + Br 2, hv X 3 → ethylbenzene + KMnO 4+ H 2 SO 4 X 4 4) Starch hydrolysis X 1 → C 2 H 5 OH → X 2 → X 3 C act. Benzene 5) 1-chloropropane + Na X 1 -4 H 2 X 2 → chlorobenzene → X 3 + nH 2 CO X 4 6) 1-chlorobutane + NaOH, H 2 O X 1 → butene-1 + HCl X 2 conc. alcohol X 3 KMnO 4, H 2 O X 4 7) ethylene + Br 2 (ad) X 1 + KOH, alcohol X 2 + H 2 O, Hg X 3 → X 4 → methyl acetate 8) ethyl acetate → sodium acetate NaOH (fusion X 1 1500 C X 2 400 C X 3 C 2 H 5 Cl, AlCl 3 X 4 9) 1-bromopropane → hexane → benzene CH 3 Cl X 1 KMnO 4, H 2 SO 4 X 2 CH 3 OH, H X 3 10) butanol- 2 HCl X 1 KOH, C 2 H 5 OH X 2 KMnO 4, H 2 SO 4 X 3 CH 3 OH, H X 4 → potassium acetate. 11) C 6 H 6 → C 6 H 5 - CH (CH 3) 2 KMnO4 X 1 HNO3 (1 mol.) X 2 Fe + HCl X 3 NaOH (ex) X 4 12) CH 3 - CH 2 - CHO Ag2 O X 1 + Cl2, hv X 2 NaOH X 3 CH3OH, H X 4 polymerization X 5 13) CH 3 - CH 2 - CH 2 - CH 2 OH H2SO4, t X 1 HBr X 2 NH3 X 3 14) cyclohexene t, Kat X 1 →C 6 H 5 NO 2 +H2,t X 2 HCl X 3 AgNO3 X 4 15) CaC 2 H2O X 1 H2O, HgSO4 X 2 C4(OH)2, t X 3 CH3 OH, H2 SO4 X 4 16) CH 3 - CH 2 -CH (CH 3) - CH 3 Br2, light. X 1 con. alcohol X 2 HBr X 1 Na X 3 → CO 2 17) CH 4 1000C X 1 C act, t X 2 CH3 Cl, AlCl3 X 3 KMnO4, t X 4 18) methane → X 1 → benzene CH3Cl, AlCl3 X 2 → benzoic acid CH3OH, H X 3 19) CH 4 → CH 3 NO 2 → CH 3 NH 2 → CH 3 NH 3 Cl → CH 3 NH 2 → N 2 20) C 6 H 5 CH 3 KMnO4, H2SO4, t X 1 HNO3 (1 mol) X 2 Fe+HCl ex. X 3 NaOH ex. x4

Most often, the essence of the task is consistent solution the following tasks: construction (lengthening or shortening) of the carbon skeleton; introduction of functional groups into aliphatic and aromatic compounds; substitution of one functional group for another; removal of functional groups; change in the nature of functional groups. The sequence of operations may be different, depending on the structure and nature of the initial and resulting compounds.

Checklist Present the facts and their relationships in visual form. Write down, in as much detail as possible, the essence of the problem in the form of a diagram. Look at the problem as broadly as possible, take into account even solutions that seem unthinkable. In the end, it is they who may turn out to be right and lead you to the right decision. Use the trial and error method. If there is a limited set of options, try them all.

What reactions can be used to carry out transformations according to the scheme: CH 3 COO Na → CH 3 - CH 3 → CH 2 \u003d CH 2 → CH 2 Br - CH 2 Br → CH ≡ CH → KOOC - COOK Solution. 1. To obtain ethane from sodium acetate, we use the Kolbe synthesis: el-z 2CH 3 COO Na + 2H 2 O → CH 3 - CH 3 + H 2 +2 NaHCO 3 2. To convert ethane to ethene, we carry out the dehydrogenation reaction: t, Ni CH 3 - CH 3 → CH 2 \u003d CH 2 + H 2 3. To obtain a dihaloalkane from an alkene, we use the bromination reaction: CH 2 \u003d CH 2 + Br 2 → CH 2 Br - CH 2 Br 4. To obtain ethine from dibromoethane, it is necessary to carry out dehydrohalogenation reaction, for this, an alcohol solution of KOH is used: CH 2 Br - CH 2 Br +2 KOH alcohol. solution → CH ≡ CH +2 KV r +2H 2 O Problem 1

5. Ethine decolorizes an aqueous solution of KMnO 4: 3 CH ≡ CH +8 KMnO 4 → 3 KOOC - COOK + + 8 MnO 2 +2 KOH +2 H 2 O The introduction of four oxygen atoms into the molecule corresponds to the loss of 8 ē, therefore, before MnO 2 c we set the coefficient 8. M manganese changes the oxidation state from +7 to + 4, which corresponds to the acquisition of 3 ē, therefore we put the coefficient 3 in front of the organic matter. Pay attention to the equations of reactions 1 and 5: Kolbe synthesis and oxidation of alkynes aqueous solution potassium permanganate. AT acidic environment permanganate ion is reduced to Mn 2+, and ethyne is oxidized to oxalic acid: 5 CH ≡ CH + 8 KMnO 4 +12 H 2 SO 4 → 5 HOOC - COOH + 8 MnSO 4 + 4 K 2 SO 4 + 12 H 2 O

Determination of the oxidation state of carbon in alkanes -4 -3 -3 CH 4 CH 3 -CH 3 CH 3 -CH 2 -CH 3 -3 -2 -3 CH 3 -CH-CH 3 | CH 3 -3 -1 -3 CH 3 0 | CH 3 -C-CH 3 | CH 3

Determination of the oxidation state of carbon in alcohols -2 -3 -1 0 CH 3 -OH CH 3 -CH 2 -OH CH 3 -CH-CH 3 | OH -1 -1 -3 0 0 -3 CH 2 - CH 2 CH 3 - CH - CH - CH 3 | | | | OH OH OH OH

K2Cr2O7 Oxidizers KMnO 4 Mn 2+ MnO 2 MnO 4 2+ Cr 3+ Cr 3

The processes of oxidation of an alkene depend on its structure and the reaction medium. When alkenes are oxidized to a concentrated solution of potassium permanganate KMnO4 in an acidic medium (hard oxidation), σ and π bonds are broken with the formation carboxylic acids, ketones and carbon monoxide (IV) . This reaction is used to determine the position of the double bond.

If in the alkene molecule the carbon atom at the double bond contains two carbon substituents (for example, in the molecule of 2-methylbutene-2), then during its oxidation a ketone is formed: CH 3 -C=CH -C H 3 | CH 3 + KMnO 4 + H 2 SO 4 = MnSO 4 + K 2 SO 4 + H 2 O + + CH 3 - C - CH 3 || O + CH 3 -COOH -3 0 -1 +2 +3 + 7 +2 0 +2 C - 2ē→C -1 +3 C - 4ē→C +7 +2 Mn +5ē→Mn 6 3 0 5 6 oxidation process 0 - 1 C , C - reducing agents reduction process +7 Mn - oxidizing agent 5 5 5 6 6 3 9 9

Ways of arranging the coefficients in the OVR with the participation of organic substances. C 6 H 5 CH 3 + KMnO 4 + H 2 SO 4  C 6 H 5 COOH + K 2 SO 4 + MnSO 4 + H 2 O We select the coefficients using the electronic balance method: C -3 - 6e - → C +3 6 │ 5 Mn +7 + 5e - → Mn +2 5│6 5C 6 H 5 CH 3 + 6K MnO 4 - + 9 H 2 SO 4  5C 6 H 5 COOH + + 6 MnSO 4 + 3 K 2 SO 4 + 14 H2O

The coefficients can also be selected by the electron-ion balance method (half-reaction method). C 6 H 5 CH 3 + KMnO 4 + H 2 SO 4 → C 6 H 5 COOH + K 2 SO 4 + MnSO 4 + H 2 O C 6 H 5 CH 3 +2 H 2 O - 6e - → C 6 H 5 COOH + 6H + │ 5 reducing agent MnO 4 - + 8H + + 5e - → Mn 2+ +4 H 2 O │ 6 oxidizing agent 5C 6 H 5 CH 3 + 10 H 2 O +6 MnO 4 - + 48H + → 5C 6 H 5 COOH+ 30H + + + 6 Mn +2 + 24 H 2 O 6 K + 18 H + 6SO 4 2- 14H 2 O 9SO 4 2- 6 K + + 3SO 4 2- 5C 6 H 5 CH 3 + 6K MnO 4 - +9 H 2 SO 4 → 5C 6 H 5 COOH + 6 MnSO 4 + + 3 K 2 SO 4 + 14 H 2 O

Write the equation for the reaction between propylene and potassium permanganate in a neutral medium. Write the equation for the reaction between butene-2 ​​and potassium permanganate in an acidic medium. Compare the ratio of all isomeric alcohols of the composition C 4 H 10 O to oxidizing agents. For butanol-1 and butanol-2, write the reaction equations with a solution of potassium dichromate in an acidic medium. Write the equation for the reaction between ethyl alcohol and a solution of potassium dichromate in an acidic medium. Write the equation for the reaction between ethylbenzene and potassium permanganate in an acidic medium. Write the equation for the reaction between styrene and potassium permanganate in a neutral medium. Write the equation for the reduction of 1,3-dimethylnitrobenzene with ammonium sulfide in a neutral medium (Zinin reaction). Tasks for independent solution.

Conclusion Students' knowledge of the questions On the relationship of organic substances according to the scheme: composition - structure- properties Compilation of redox reaction equations in organic chemistry Final certification of graduates in the form and materials of the Unified State Examination in 2012: 6 students successfully completed the tasks of parts A and B, and in task C 3 two people received 5 each, the remaining four people 3 - 4 points out of five maximum possible. in 2013: 4 students successfully coped with the tasks of parts A, B of the examination material, and task C 3 three people received 5 each, one person received 4 points in 2014 one person took the exam and completely coped with these topics in the tasks of part A , B and C 3 exam material.

1. O.S. Gabrielyan, I.G. Ostroumov Chemistry. Toolkit. Grade 10. Bustard, 2001 2. O.S.Gabrielyan, I.G.Ostroumov Chemistry. Toolkit. Grade 11. Bustard, 2004 3 I.G. Norenko. Pedagogical advice. Formation experience educational space schools. Issue 6. Educational - methodical manual. Volgograd. Teacher 2008 4. L.I. Salyakhov. Pedagogical advice. Technology of preparation and practical developments. Teaching aid. The globe. 2006 5. cnit. ssau. ru › Title › chem 2/ u 9. htm 6. http:// otvet . mail. ru / question /52521459 7. L.R. Kochuleva. Methodical manual on organic chemistry. Preparation for the exam. Orenburg 2011 8. N.E. Kuznetsova, A.N. Levkin Task book in chemistry: Grade 10: - M .: Ventana-Graf, 2011. - 144 p. literature

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MBOU "Klyuchevskaya secondary school No. 1"
Klyuchevskoy district, Altai Territory

Theory and practice of solving problems of a high level of complexity in the process of teaching chemistry

(on the material of the topic "Genetic relationship of organic compounds")

Completed by: Widershpan Irina Petrovna

Chemistry teacher

MBOU "Klyuchevskaya secondary school No. 1"

with. Keys 2015

1. Introduction (relevance, importance, significance of the issue under consideration, purpose, tasks)……………………………………………………..3
2. Main part

1. Genetic connection…………………………………………………..3
2. Work on the study of the nomenclature and classification of organic substances………………………………………………………………..5
3. Work on the study of the study of the chemical properties of organic

compounds and methods for their preparation………………………………5

1. Work organizationon deciphering the chains of transformations ...... .6
2. The sequence of actions when performing chains………..7
3. Memo……………………………………………………………… .8
4. Ways of arranging the coefficients in the OVR with the participation of organic substances………………………………………………………………………10

1. Conclusion ……………………………………………………………14
2. Literature……………………………………………………………...14
3. Applications ………………………………………………………………15

1. Introduction

Importance: In the course of studying organic chemistry, tasks for completing chains of transformations are often used. They are used in the 9th grade at the first stage of studying organic substances, in the 10th grade when studying the factual material in this course, and in the 11th grade at the final stage of training. This question is included in the tasks of the third part of the exam material in chemistry in the form and materials of the exam. Tasks for the implementation of transformations are widely used in general lessons.

Relevance: Therefore, performing tasks with genetic links can create problem situation at students. They are on the lookout for a variety of options. Children have creativity to this issue, as they offer their chains of transformations, while showing knowledge of the factual material and the logic of their thinking.

Target.

• Help students develop a higher level

the difficulty of training on the genetic link between

organic compounds;

• create conditions for the systematization and deepening of students' knowledge about the relationship of organic substances according to the scheme: composition - structure - properties of substances

Tasks .

• develop students' logical thinking (by establishing a genetic relationship between different classes of hydrocarbons, putting forward hypotheses about the chemical properties of unfamiliar organic substances);
• develop students' ability to compare (using the example of comparing the chemical properties of organic compounds);
• develop information and cognitive competence of students

2. Main body

2.1 Genetic connection

The material world in which we live and of which we are a tiny part is one and at the same time infinitely diverse. Unity and Diversity chemical substances of this world is most clearly manifested in the genetic connection of substances, which is reflected in the so-called genetic series.

What does the concept mean"genetic connection"?

genetic connectioncalled the connection between substances of different classes of compounds, based on their mutual transformations and reflecting their origin. The genetic link may be reflected ingenetic lines.The genetic series consists of substances that reflectthe transformation of substances of one class of compounds into substances of other classes containing the same number of carbon atoms.

Genetic relationships between substances should be understood as the genetic relationship of substances based on their structure and properties, showing the unity and interconnection of all organic compounds.
Genetic connections reflect the dialectics of nature, show how the process of complication, the development of substances, their composition, structure, the appearance of formations capable of the birth of life proceeded.
AT in practical terms genetic relationships show from which substances and in what ways you can get the right substances. Each transition is both an expression of the chemical properties of matter and possible ways its practical use. 1

In order to successfully complete tasks that show genetic relationships between classes of organic substances, knowledge of the nomenclature and classification of substances is worked out at chemistry lessons, the chemical properties of compounds and methods for their preparation are studied. This approach can be traced at all stages of the study of this issue, only each time the theoretical material deepens and expands.

2.2 Work on the study of the nomenclature and classification of organic substances

In the issue of studying the nomenclature and classification of organic substances, I created and tested a map of compound formulas (see Appendix 1). It is used for frontal conversation, for work in groups, in pairs and individual. This allows you to quickly correct the knowledge of students. They, having formulas in front of them, can quickly and efficiently navigate the task, reason, answer the questions posed, while acquiring new knowledge. The use of this map of compound formulas in chemistry lessons affected the quality of the knowledge gained on the issue of nomenclature and classification of organic substances.

2.3 Work on the study of the study of the chemical properties of organic compounds and methods for their preparation.

The next stage is the study of the chemical properties of organic compounds and methods for their preparation. In the study of these issues in the classroom, I use reference schemes (see Appendix 2).

The study of the structure, chemical properties and methods for producing hydrocarbons various groups shows that they are all genetically related, i.e. transformations of some hydrocarbons into others are possible.

Reference schemes allow you to give a large amount of information in a concise form. Students enjoy using them. These schemes help them organize their knowledge and develop the logic of everyone's thinking. Thus, students are prepared to perform transformation exercises.

Students perform outline diagrams when studying each topic in stages. Repeated reference to the outline scheme allows you to lay a solid foundation in the knowledge gained.

2.4 Organization of workby deciphering the chains of transformations

To consolidate the educational material and intensify educational activities, it is recommended to pay special attention to solving problems of the type associated with the mutual transformations of substances (chains of transformations).

Exercises on deciphering the chains of transformations help to better understand the genetic relationships between organic compounds. For example, after studying the topic "Hydrocarbons", you should pay attention to the solution of this particular type of problem (see.example ).

Task 1. Name the intermediates in the following transformation scheme: H 2 SO 4 (conc.), t HBr Na Cr 2 O 3 , Al 2 O 3

Ethyl ether → X → Y → Z → butadiene-1,3
Decision. In this chain of transformations, including 4 reactions, from ethyl alcohol C 2 H 5 OH butadiene-1,3 should be obtained

CH 2 \u003d CH–CH \u003d CH 2.

1. When heating alcohols with concentrated sulfuric acid
   H 2 SO 4 (water remover) occurs themdehydration with

the formation of an alkene. The splitting of water from ethyl alcohol leads to the formation of ethylene: H 2 SO 4 (conc), t

CH 3 - CH 2 -OH → CH 2 \u003d CH 2 + H 2 O

Thus, the substance X, which is formed when dehydration ethyl alcohol and capable of reacting with HBr is ethylene CH 2 \u003d CH 2.

1. Ethylene is a representative of alkenes. Being an unsaturated compound, it is able to enter into addition reactions. As a resulthydrobromination ethylene:

CH 2 \u003d CH 2 + HBr → CH 3 - CH 2 -Br

bromoethane CH is formed 3 -CH 2 -Br (substance Y):

1. When heated bromoethane in the presence of metallic sodium(wurtz reaction ), n-butane is formed (substance Z):

2CH 3 - CH 2 -Br + 2Na → CH 3 - CH 2 - CH 2 - CH 3 + 2NaBr

1. Dehydrogenation n-butane in the presence of a catalyst is one of the methods for obtaining butadiene-1,3 CH 2 \u003d CH - CH \u003d CH 2 2

Success in completing tasks will depend on the number of transformations carried out. The student handout contains individual tasks(respectively for each class and according to the level of the studied material). Developed and tested Handout for students in grade 10 (see Appendix 3), for grade 11 (see Appendix 4), to prepare for exams in the form and materials of the Unified State Examination (see Appendix 5, 6).

In order to create a "success situation" for the student in the classroom, individual approach. Enough tasks have been selected so that there are no repetitions of them. Therefore, everyone has their own ways of solving problems, and one goal has been achieved, a transformation has been carried out, showing the genetic relationship between individual classes of organic compounds.

2.5 The sequence of actions when performing chains.

One of the most common types of tasks in organic chemistry are those in which it is required to carry out transformations according to the proposed scheme. At the same time, in some cases it is necessary to indicate specific reagents and conditions for the occurrence of reactions leading to substances that make up the chain of transformations. In others, on the contrary, it is necessary to determine which substances are formed under the action of these reagents on the starting compounds.

Usually in such cases it is not required to indicate the fine technical details of the synthesis, the exact concentration of reagents, specific solvents,

purification and isolation methods, etc. however, exemplary reaction conditions must be specified.

Most often, the essence of the task lies in the sequential solution of the following tasks:

• construction (lengthening or shortening) of the carbon skeleton;

2 cnit.ssau.ru › Title › chem2/u9.htm

• introduction of functional groups into aliphatic and aromatic compounds;
• substitution of one functional group for another;
• removal of functional groups;
• change in the nature of functional groups.

The sequence of operations may be different, depending on the structure and nature of the initial and resulting compounds.

2.6 Reminder

Present facts and their relationships visually. Write down, in as much detail as possible, the essence of the problem in the form of a diagram.

Look at the problem as broadly as possible, take into account even solutions that seem unthinkable. In the end, they may be the right ones and lead you to the right decision.

Use the trial and error method. If there is a limited set of options, try them all.

Task 2. CH 4 → CH 3 Br → C 2 H 6 → C 2 H 5 Cl → C 2 H 5 OH →

→ CH 3 SON → CH 3 COOH → CH 3 COOS 2 H 5

Decision.

1. To introduce a halogen atom into a hydrocarbon molecule, you can use the radical chlorination reaction:

CH 4 + Br 2 → CH 3 Br + HBr

2. One of the options leading from a halogen derivative to saturated hydrocarbon with a large number carbon atoms, reaction with metallic sodium (Wurtz reaction):

2CH 3 Br + 2Na → H 3 C - CH 3 + 2NaBr

Light

1. C 2 H 6 + Cl 2 → CH 3 CH 2 Cl + HCl

4. To convert a halogen derivative into an alcohol, it is necessary to replace the halogen atom in the molecule with hydroxyl group, which can be done by carrying out the reaction of nucleophilic substitution (hydrolysis in an alkaline medium): H 2 O

C 2 H 5 Cl + KOH → C 2 H 5 OH + KCl

5. In order to convert alcohol to aldehyde, you need to increase the oxidation state of the carbon atom at functional group, i.e. act as a mild (non-destructive molecule) oxidizing agent:

C 2 H 5 OH + CuO → CH 3 - CHO + Cu + H 2 O

6. Further oxidation will lead to the transformation of the aldehyde group into a carboxyl group: t

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

7. Reactions of carboxylic acids with alcohols lead to the formation esters: H+

CH 3 - COOH + HO - CH 2 - CH 3 → CH 3 - COO - CH 2 - CH 3 + H 2 O

Task 3. Which reactions can be used to carry out transformations according to the scheme: CH 3 COOHa → CH 3 - CH 3 → CH 2 \u003d CH 2 → CH 2 Br - CH 2 Br →

→CH≡CH→KOOC – COOK

Decision.

1. To obtain ethane from sodium acetate, we use the Kolbe synthesis:

El-z

2CH 3 COOHa + 2H 2 O → CH 3 - CH 3 + H 2 + 2NaHCO 3

2. To convert ethane to ethene, we carry out the dehydrogenation reaction:

t, Ni

CH 3 - CH 3 → CH 2 \u003d CH 2 + H 2

3. To obtain a dihaloalkane from an alkene, we use the reaction

Bromination:

CH 2 \u003d CH 2 + Br 2 → CH 2 Br– CH 2 Br

4. To obtain ethine from dibromoethane, it is necessary to carry out the reaction

Dehydrohalogenation, for this, an alcohol solution of KOH is used:

CH 2 Br– CH 2 Br +2 KOH alcohol. rr → CH≡CH +2 KBr +2H 2 O

5. Ethine discolors aqueous KMnO solution 4 :

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

The introduction of four oxygen atoms into the molecule corresponds to the loss of 8 electrons; therefore, before MnO 2 we set the coefficient 8. Mn changes the oxidation state from +7 to +4, which corresponds to the acquisition of 3 electrons, therefore we set the coefficient 3 in front of the organic substance.

Pay attention to reaction equations 1 and 5: Kolbe synthesis and oxidation of alkynes with an aqueous solution of potassium permanganate.

Note : In an acidic environment, the permanganate ion is reduced to Mn 2+ , and ethyne is oxidized to oxalic acid:

5CH≡CH +8KMnO 4 +12H 2 SO 4 →5HOOC – COOH +8MnSO 4 +4К 2 SO 4 +12Н 2 О 2.7 Methods for arranging coefficients in the OVR with the participationorganic substances.

To arrange the coefficients, the electron balance method is usually used, which requires knowledge of the oxidation state of the carbon atom, which can take values ​​from -4 to +4 and depends on the relative electronegativity of the atoms of its immediate environment.

The oxidation state of carbon atoms is determined by the number electron pairs, displaced towards the carbon atom if its electronegativity is higher than that of the neighboring atom, or displaced from the carbon atom if its electronegativity is lower.

For example, the degree of oxidation of the carbon atom in the methyl radical of the toluene molecule is "-3", because the electron density shifts from three hydrogen atoms to the more electronegative carbon atom across the three σ bonds. In the carboxyl group of the molecule benzoic acid the oxidation state of the carbon atom is "+ 3", because. electron density shifts

from the carbon atom to the oxygen atoms along two σ- and one π-bond (only three bonds):

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

C +3

H→C -3 ←H OH

toluene benzoic acid

We select the coefficients using the electronic balance method:

C -3 - 6e - → C +3 5

Mn +7 + 5е - →Mn +2 6

The coefficients can also be selected by the electron-ion balance method (half-reaction method).

Toluene C 6 H 5 CH 3 oxidized to benzoic acid 6 H 5 COOH, permanganate ion MnO 4 - reduced to manganese cation Mn+2 .

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

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

MnO 4 - + 8H + + 5e - → Mn 2+ +4 H 2 O 6

5C 6 H 5 CH 3 + 10H 2 O + 6MnO 4 - + 48H + → 5C 6 H 5 COOH + 30H + + 6Mn +2 + 24H 2 O

6 K + 18H + 6SO 4 2- 14H 2 O

9SO 4 2- 6 K + + 3SO 4 2-

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

In most cases, redox reactions

organic compounds are accompanied by the removal or addition of hydrogen and oxygen atoms.

In oxidation: the introduction of an oxygen atom into a molecule of an organic compound is equivalent to the loss of two electrons, and the elimination of a hydrogen atom is equivalent to the loss of one electron. During reduction: splitting off an oxygen atom - the acquisition of two electrons, the addition of a hydrogen atom - the acquisition of one electron.

This principle is based on the method of arranging the coefficients, which does not require determining the degree of oxidation of carbon. Let us determine the number of hydrogen atoms lost by toluene and the number of oxygen atoms introduced into the molecule. Lost two hydrogen atoms (-2e¯), introduced two oxygen atoms (-4e¯). Total given 6e¯.

Let us consider the application of various methods of arranging the coefficients using the example of the reaction between n-methylcumene and potassium permanganate in an acidic medium.

1. Oxidation of any alkyl substituents in benzene derivatives occurs to carboxyl groups:

CH 3 - CH - CH 3 COOH

KMnO 4 + H 2 SO 4 → + K 2 SO 4 + MnSO 4 + H 2 O + CO 2

CH 3 COOH

The two carbon atoms of the isopropyl radical are oxidized to carbon dioxide.

2. Let's draw up an electronic balance diagram without determining the oxidation states of carbon atoms. Let us determine the number of hydrogen atoms lost by n-methylcumene and the number of oxygen atoms introduced into the molecule. Lost eight hydrogen atoms (-8e¯), introduced four oxygen atoms (-8e¯). In addition, four oxygen atoms were included in the composition of carbon dioxide (-8e¯). Total given 24e¯. Manganese with an oxidation state of +7 was reduced to +2, then the electronic balance scheme can be written:

CH 3 - CH - CH 3 COOH

– 24e¯ → – 24e¯ 5

CH 3 COOH 120

Mn +7 +5 e¯ → Mn +2 +5 e¯ 24

3. Let's draw up an electronic balance diagram by determining the oxidation states

carbon atoms. The oxidation state of carbon atoms in methyl radicals CH 3 is equal to "+3", in the methine group CH - "+1", in carboxyl groups - "-3".Note that only α-carbons (directly bonded to benzene ring) are oxidized to carboxyl groups, the remaining carbon atoms are oxidized to carbon dioxide.

C -1 - 4e - → C +3

C -3 - 6e - → C +3 - 24e - 5

2 C -3 - 14e - → 2C +4

Mn +7 + 5e - →Mn +2 + 5e - 24

CH 3 - CH - CH 3 COOH

5 +24KMnO 4 +36H 2 SO 4 →5 +12K 2 SO 4 +24MnSO 4 +56 H 2 O+10CO 2

CH 3 COOH

4. We select the coefficients by the method of electron-ion balance.

In this case, there is no need to show structural formulas organic substances, so the formulas of n-methylcumene and tere phthalic acid, we write in molecular form:

C 10 H 14 + 8H 2 O - 24e - → C 8 H 6 O 4 + 2CO 2 + 24H + 5

MnO 4 - + 8H + + 5e - → Mn 2+ + 4H 2 O 24

5C 9 H 18 + 40H 2 O + 24MnO - 4 + 192H + → 5C 7 H 6 O 2 + 10CO 2 + 120H + + 24Mn 2 + + 96H 2 O

5C 9 H 18 + 24MnO - 4 + 72H + → 5C 7 H 6 O 2 + 10CO 2 + 24Mn 2 + + 56H 2 O

24K + 36SO 4 2- 24SO 4 2-

24K + + 12SO 4 2-

5C 10 H 14 + 24KMnO 4 + 36H 2 SO 4 → 5C 7 H 6 O 2 +10CO2 +12K2 SO4 + 24MnSO4 + 56H2 O

I offer handouts for students (see Appendix 7).

In preparation for final certification in USE form I use the exercises from the problem book in chemistry grade 10 N.E. Kuznetsova, A.N. Levkin and offer exercises for students (see Appendix 8, 9,10)

13

3. Conclusion

This approach to learning this issue gave a positive result in the final certification of graduates in the form and materials of the Unified State Examination in 2012: 6 students successfully coped with the topics described above in the tasks of parts A and B of the examination material, and in task C3 two people got 5, the other four people got 3 - 4 points out of five maximum possible. In 2014, one person took the exam and completely coped with these topics in the tasks of parts A, B and C3 examination material.

LITERATURE

1. O.S. Gabrielyan, I.G. Ostroumov Chemistry. Toolkit. Grade 10.

Bustard, 2001

2. O.S.Gabrielyan, I.G.Ostroumov Chemistry. Toolkit. Grade 11.

Bustard, 2004

3 I.G. Norenko. Pedagogical advice. Formation experience

educational space of the school. Issue 6. Educational and methodical

allowance. Volgograd. Teacher 2008

4. L.I. Salyakhov. Pedagogical advice. Technology of preparation and

practical developments. Teaching aid. The globe. 2006

5. cnit.ssau.ru › Title › chem2/u9.htm

6. http://answer.mail.ru/question/52521459

7. L.R.Kochuleva. Methodical manual on organic chemistry. Training

to the exam. Orenburg 2011

8. N.E. Kuznetsova, A.N. Levkin Task book in chemistry: Grade 10: − M.: Ventana-

Graf, p.2011. – 144

14

Appendix 1

CHEMICAL FORMULA CHART

(the formula map is printed in abbreviation)

1)CH3 – COOH; 2)CH2 =CH2 ; 3)CH3 –CH2 – OH; 4)CH CH; 5)CH3 – COH;

6)CH3 – COONa; 7)CH3 –CH3 ; 8) CH2 = CH – CH = CH2 ; 9) CH3 – OCH3 ; 10)CH3 OH; 11) CH C-CH3 ; 12) CH3 – COOKH3 ; 13) CH3 – COOC2 H5 ;

14) HCOOCH3 ; 15) HCOOH; 16) CH3 –CH2 Cl; 17) CH3 –CH2 –CH2 –CH3 ;

18) CH3 –CH2 – COOH; 19) CH3 –CH2 – COH; 21) C2 H5 OH;

21) CH3 –CH2 Cl; 20)C6 H6 ; 22) C6 H5 CH3 ; 23) C6 H5 COH; 24) C6 H5 COOH;

25)CH2 CH—COOH; 26) (CH3 – COO)2 Mg; 27) CH3 – NO2 ; 28) CH3 – NH2 ;

29) CH3 – NH – CH3 ; 30) C2 H5 – NO2 ; 31) CH3 – NH – C2 H5 ; 32) CH2 =CHCl

Annex 2

REFERENCE DIAGRAM

CHEMICAL PROPERTIES OF LIMITED HYDROCARBONS (ALKANES)

15

Appendix 3

GRADE 10.

1)CH2 =CH-CH3 → X→ CH3 – CHOH – CH3

2. Methane→ acetylene→ benzene→ chlorobenzene→ phenol.

3) C2 H6 → X→ C2 H5 Oh

4) Ethylene→ chloroethane→ butane→ 2-chlorobutane→ butanol - 2.

5) Ethan→ ethene→ ethanol→ bromoethane→ butane.

6)C6 H6 → X→ C6 H5 Oh

7) Acetylene→ benzene→ bromobenzene→ phenol→ picric acid.

8)C2 H2 → X→ CH3 COOH

9)C2 H5 Oh+CuO,tX1 +Ag2O,tX2 +CH3OH,H2SO4X3

10)CH2 =CH-CH3 → X→ CH3 COCH3

11)C2 H2 H2O,HgX1 +H2 , Ni , PtX2 +H2 SO4 , t≤ 140CX3

12)C2 H5 Oh→ X→ CH3 COOH

13) CH3 COONa+H2SO4X1 +C2 H5 OH , H2 SO4X2 +H2O, NaOHX3

14) CH3 COOC2 H5 → X→ C2 H4

15) CH2 Br–CH2 –CH3 +NaOHX1 +CuOX2 +Ag2 O , tX3

16) C2 H5 Cl+NaOHX1 CuOX2

17) CaC2 +H2OX1 +H2O , HgX2

18) Sodium acetate→ methane→ chloromethane→ methanol→ dimethyl ether.

19) C2 H6 +Br2X1 +NaOHX2

20) Ethylene→ ethanol→ ethanal→ ethanoic acid→ sodium acetate.

21) C2 H4 +Cl2X1 +NaOHX2

22) Starch→ ethanol→ ethylene→ 1, 2 - dichloroethane→ ethylene glycol.

23) Methanol→ X→ methane acid.

16

Appendix 4

A SET OF SCHEMES USED TO PERFORM TRANSFORMATIONS IN ORGANIC CHEMISTRY

GRADE 11.

1) CaC2 → X1 → X2 → nitrobenzene[h]X3 +HClX4

2) C2 H6 +Cl2 , hNX1 → X2 → CH2 =CH-CH=CH2 +H2 (1.4 connection)→ X3 +Cl2X4

3)CH4 1500CX1 +2 H2X2 +Br2 , hvX3 → ethylbenzene+KMnO4+H2SO4X4

4) StarchhydrolysisX1 →C2 H5 OH→X2 → X3 C act.Benzene

5) 1-chloropropane+ NaX1 -4H2X2 → chlorobenzene → X3 +nH2COX4

6) 1-chlorobutane+NaOH,H2OX1 → butene-1+HClX2 conc. alcoholX3 KMnO4, H2OX4

7) stylen+Br2(hell)X1 +KOH, alcoholX2 +H2O, HgX3 → X4 → methyl acetate

8) ethyl acetate → sodium acetateNaOH(fusionX1 1500CX2400CX3 C2H5Cl,AlCl3X4

9) 1-bromopropane → hexane → benzeneCH3 ClX1 KMnO4 , H2SO4X2 CH3OH, HX3

10) butanol-2HClX1 KOH,C2H5OHX2 KMnO4 , H2SO4X3 CH3OH,HX4 → potassium acetate.

11)C6 H6 → C6 H5 –CH(CH3 ) 2 KMnO4X1 HNO3 (1 mol.)X2 Fe+HClX3 NaOH(g)X4

12) CH3 -CH2 – CHOAg2OX1 +Cl2, hvX2 NaOHX3 CH3OH, HX4 polymerizationX5

13) CH3 –CH2 –CH2 –CH2 OhH2SO4, tX1 HBrX2 NH3X3

14) cyclohexenet,KatX1 →C6 H5 NO2 +H2,tX2 HClX3 AgNO3X4

15) CaC2 H2OX1 H2O, HgSO4X2 C4(OH)2, tX3 CH3 OH, H2 SO4X4

16) CH3 – CH2 –CH(CH3 )-CH3 Br2, light.X1 con. alcoholX2 HBrX1 NaX3 → CO2

17) CH4 1000CX1 Act,tX2 CH3 Cl , AlCl3X3 KMnO4 , tX4

18) methane → X1 → benzeneCH3Cl,AlCl3X2 → benzoic acidCH3OH,HX3

19) CH4 → CH3 NO2 → CH3 NH2 → CH3 NH3 Cl→CH3 NH2 → N2

20)C6 H5 CH3 KMnO4, H2SO4, tX1 HNO3 (1 mol)X2 Fe+HClex.X3 NaOH ex.X4

17

Annex 5

A SET OF SCHEMES USED TO PREPARE FOR THE PERFORMANCE OF THE TASK WITH3 EXAMINATION PART IN THE FORM AND ON THE USE MATERIALS

1) Ethine → benzeneCH3Cl , AlCl 3X1 Cl2 , UVX2 KOHaq.X3 HCOOH,HX4

2) 1,3-dibromobutaneZnX1 → 2-,bromobutaneNaX2 → 1,2dimethylbenzeneKMnO4 ,H2SO4,tX3

3) C2 H5 Cl → C3 H8 t, NiX1 KMnO4,H2OX2 ex.HBrX3 KOH (alcohol)X4

4) CH3 OhHBrX1 NH3X2 HBrX3 KOHX2 → N2

5)C2 H5 OhHBrX1 KOH (alcohol) ,tX2 C6 H6 , HX3 Br2 , lightX4 KOH (alcohol) ,tX5

6)C2 H5 OhHCOOH,HX1 KOH , t , H2OX2 → HCOOHH2 SO4 (conc)X3 H2 , t, catX4

7) cellulose → glucose → ethanolCH3COOH,t,HX1 → CH3 COONaelectrolysisX2

8) CH3 CHCl2 → CH3 CHOH2, cat.X1 NH3, 300X2 CO2 + H2OX3 tX4

9)C3 H7 OhAl2 O3 , 400X1 KMnO4 ,H2OX2 HBr(e)X3 KOH (alcohol)X4 C act, tX5

10)C2 H5 OhAl2 O3 , 400X1 KMnO4 , H2OX2 HBr(e)X3 KOH (alcohol)X4 →C2 H4 O

11) CH4 → HCHOH2 , cat.X1 NaX2 HClX1 KMnO4 , H2 SO4X3

12) CH4 → ethyne → vinylacetyleneizb. H2 ,catX1 → ethanoic acidNH3X2

13) CaC2 H2 OX1 KMnO4H2 C2 O4 conc. H2SO4X2 → HCOOKconc. H3PO4X3

14) H2 C2 O4 →CO→CH3 OH→CH3 COOKH3 NaOH + H2O, tX→CH4

18

Appendix 6

1. C6 H12 O6 →C2 H5 OH → CH3 – CHO → CH3 – COOH→Cl-CH2 – COOH →

→ H2 N–CH2 – COOH

Br2 , light KOH, alcohol HBr Na

2.CH3 – CH2 – CH(CH3 ) – CH3 → X1 → X2 → X1 → X3 → CO2

Cu(OH)2 +Cl2 , hν NaOHsp. CH3 OH, H+ polymerization

3.SN3 – CH2 – SON X1 X2 X3 X4 X5

H2 SO4 , 200 °C cat., t° OH HCl KMnO4 , H2 O

4. ethanol X1 X2 Ag2 C2 X2 X3

H2 O, CON., t° 1200 °C cat. Izb. Br2

5. propyl acetate X1 CH4 X2 vinylacetylene X3

+Cl electrolysis2 , light +NaOH, H2 O H2 SO4 (conc.), t

6.CH3 COOH X1 With2 H6 X2 X3 X4

C, 400 °C Cl2 , cat. CO2 + H2 About Br2 + H2 O

7.C2 H2 X1 X2 C6 H5 OK X3 X4

19

Annex 7

Tasks for independent solution.

• Write the equation for the reaction between propylene and potassium permanganate in a neutral medium.
• Write the equation for the reaction between butene-2 ​​and potassium permanganate in an acidic medium.
• Compare the ratio to oxidizing agents of all isomeric alcohols of composition C4 H10 A. For butanol-1 and butanol-2, write the reaction equations with a solution of potassium dichromate in an acidic medium.
• Write the equation for the reaction between ethyl alcohol and a solution of potassium dichromate in an acidic medium.
• Write the equation for the reaction between ethylbenzene and potassium permanganate in an acidic medium.
• Write the equation for the reaction between styrene and potassium permanganate in a neutral medium.
• Write the equation for the reduction of 1,3-dimethylnitrobenzene with ammonium sulfide in a neutral medium (Zinin reaction).
• When glucose is oxidized bromine water gluconic acid is formed, and when oxidized with concentrated nitric acid, glucaric acid is formed. Write down the equations of the corresponding reactions.

20

Annex 8

Annex 9

21

Annex 10

Appendix 11

22|

Author: Doronkin Vladimir Nikolaevich, Sazhneva Tatyana Vladimirovna, Berezhnaya Alexandra Grigoryevna
Publisher: Legion, 2015
Series: Getting ready for the exam
Genre: Unified State Examination in Chemistry, etc.

Annotation to the book "Chemistry. Unified State Examination. Tasks of a high level of complexity (questions 36-40)"

The book is addressed to students in grades 10-11 of educational institutions who are preparing for the Unified State Examination and planning to get a good result on the exam, as well as teachers and methodologists who organize the process of preparing for the exam in chemistry. The manual is an important addition to the educational and methodological complex "Chemistry. Preparation for the Unified State Examination", which includes such manuals as "Chemistry. Preparation for the Unified State Exam-2015. Books ...
Read completely
The proposed manual has been compiled in accordance with the requirements of the new specifications approved at the time of the publication of the book and USE demos and is intended to prepare for the unified state exam in chemistry. The book includes tasks of a high level of complexity (questions 36-40 on the exam). Each of its sections contains the necessary theoretical information, analyzed (demonstration) examples of tasks that allow you to master the methodology for performing tasks of a high level of complexity, and groups training tasks by topics.
The book is addressed to students in grades 10-11 of educational institutions who are preparing for the Unified State Examination and planning to get a good result on the exam, as well as teachers and methodologists who organize the process of preparing for the exam in chemistry. The manual is an important addition to the educational and methodological complex "Chemistry. Preparation for the Unified State Examination", which includes such manuals as "Chemistry. Preparation for the Unified State Examination-2015. Books 1 and 2", "Chemistry. 10-11 grades. Thematic tests to prepare for Unified State Exam. Vasa and advanced levels", "Chemistry. Grades 9-11. Pocket reference book", training notebooks in general, inorganic and organic chemistry, etc.
You can download Chemistry. USE. Tasks of a high level of complexity (questions 36-40) - Doronkin, Sazhneva, Berezhnaya.