Chemistry. Subject tests to prepare for the Unified State 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 is compiled in accordance with the requirements of the new Unified State Exam specification and is intended to prepare for the unified state exam in chemistry. The book includes tasks of a high level of complexity (C1-C5). Each section contains the necessary theoretical information, disassembled (demonstration) examples of completing tasks, which allow you to master the methodology for completing tasks in Part C, and groups of training tasks by topic. The book is addressed to students in grades 10-11 of general education institutions who are preparing for the Unified State Exam and planning to receive high result on the exam, as well as teachers and methodologists who organize the process of preparing for the chemistry exam. 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 Unified State Examination 2013", "Chemistry. 10-11 grades. Thematic tests for preparing for the Unified State Exam. Basic and elevated levels" and etc.

Format: pdf (2012 , 3rd ed., rev. and additional, 234 pp.)

Size: 2.9 MB

Watch, download: 14 .12.2018, links removed at the request of the Legion publishing house (see note)

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

INTRODUCTION
This book is intended to prepare you for completing tasks of a high level of complexity in general, inorganic and organic chemistry(Part C tasks).
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 existing skills, and, if necessary, learn factual material included in test tasks parts C.
The contents of the manual reflect the features Unified State Exam options, offered in last years, and complies with the current specification. The questions and answers correspond to the wording of the Unified State Examination tests.
Part C tasks have varying degrees of difficulty. Maximum score A correctly completed task will score from 3 to 5 points (depending on the degree of difficulty of the task). Testing of tasks in this part is carried out on the basis of comparing the graduate’s answer with an element-by-element analysis of the given sample answer; each correctly completed element is scored 1 point. For example, in the SZ task you need to create 5 equations for reactions between organic substances, describing the sequential transformation of substances, but you can only create 2 (let’s say the second and fifth equations). Be sure to write them down in the answer form, you will receive 2 points for the SZ task and will significantly improve your result in the exam.
We hope that this book will help you successfully pass the Unified State Exam.

Chemistry. Thematic tests for preparing for the Unified State Exam. Tasks of a high level of complexity (C1-C5). Ed. Doronkina V.N.

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

The proposed manual is compiled in accordance with the requirements of the new Unified State Examination specification and is intended for preparation for the unified state exam in chemistry. The book includes tasks of a high level of complexity (C1-C5). Each section contains the necessary theoretical information, analyzed (demonstration) examples of completing tasks, which allow you to master the methodology for completing tasks in Part C, and groups of training tasks by topic. The book is addressed to students in grades 10-11 of general education institutions who are preparing for the Unified State Exam and planning to get a high result in the exam, as well as teachers and methodologists who organize the process of preparing for the chemistry exam. 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 Unified State Examination 2013", "Chemistry. 10-11 grades. Thematic tests for preparing for the Unified State Exam. Basic and advanced levels”, etc.

Format: pdf (2012 , 3rd ed., rev. and additional, 234 pp.)

Size: 2.9 MB

Watch, download: 14 .12.2018, links removed at the request of the Legion publishing house (see note)

CONTENT
Introduction 3
Question C1. Redox reactions. Metal corrosion and methods of protection against it 4
Asking question C1 12
Question C2. Reactions confirming the relationship between various classes of inorganic substances 17
Asking question C2 28
SZ question. Reactions confirming the relationship between hydrocarbons and oxygen-containing organic compounds 54
Asking question SZ 55
Question C4. Calculations: masses (volume, amount of substance) of reaction products, if one of the substances is given in excess (has impurities), if one of the substances is given in the form of a solution with a certain mass fraction of the dissolved substance 68
Asking question C4 73
Question C5. Finding the molecular formula of a substance 83
Asking question C5 85
Answers 97
Application. Interrelation of various classes of inorganic substances. Additional tasks 207
Tasks 209
Solving problems 218
Literature 234

INTRODUCTION
This book is intended to prepare you for completing tasks of a high level of complexity in general, inorganic and organic chemistry (part C tasks).
For each of the questions C1 - C5, a large number of tasks are given (more than 500 in total), which will allow graduates to test their knowledge, improve 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 of the Unified State Exam variants offered in recent years and corresponds to the current specifications. The questions and answers correspond to the wording of the Unified State Examination tests.
Part C tasks have varying degrees of difficulty. The maximum score for a correctly completed task is from 3 to 5 points (depending on the degree of complexity of the task). Testing of tasks in this part is carried out on the basis of comparing the graduate’s answer with an element-by-element analysis of the given sample answer; each correctly completed element is scored 1 point. For example, in the SZ task you need to create 5 equations for reactions between organic substances, describing the sequential transformation of substances, but you can only create 2 (let’s say the second and fifth equations). Be sure to write them down in the answer form, you will receive 2 points for the SZ task and will significantly improve your result in the exam.
We hope that this book will help you successfully pass the Unified State Exam.

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

In the course of studying organic chemistry, tasks on performing chains of transformations are often used. They are used in grade 9 at the first stage of studying organic substances, in grade 10 when studying factual material in this course and in grade 11 at final stage training. This question is included in the tasks of the third part of the exam material in chemistry in the form and on the materials of the Unified State Examination. Tasks for carrying out transformations are widely used in general lessons

Target.

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

· Create conditions for systematizing and deepening 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., chemistry teacher 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 on performing chains of transformations are often used. They are used in 9th grade at the first stage of studying organic substances, in 10th grade when studying factual material in this course, in 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 on the materials of the Unified State Examination. Tasks for carrying out transformations are widely used in general lessons.

Target. Promote students to develop a higher level of complexity general training on the issue of genetic relationships between organic compounds; create conditions for systematizing and deepening 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 hydrocarbon derivatives, putting forward hypotheses about the chemical properties of unfamiliar organic substances); develop students’ ability to make comparisons (using the example of comparing the chemical properties of organic compounds); develop information and cognitive competence of students.

What does the term “genetic link” mean? A genetic connection is a connection between substances different classes compounds, based on their mutual transformations and reflecting their origin. Genetic relationships can be reflected in genetic series. The genetic series consists of substances that reflect 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 showing genetic connections between classes of organic substances, in chemistry lessons knowledge of the nomenclature and classification of substances is developed, Chemical properties compounds and methods for their preparation. This approach can be traced at all stages of studying this issue, only each time theoretical material deepens and expands.

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

Formula map chemical compounds(the formula map 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 – 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, ethyl acetic acid 1 2) HCOOCH 3 methyl formate, methyl formic acid.

Study of chemical properties and methods of obtaining organic compounds. When studying these questions in class, you can use reference diagrams. Support diagrams allow short form give a large amount of information. Students enjoy using them. These diagrams help them organize knowledge and develop the logic of everyone's thinking. In this way, students are prepared to perform transformation exercises. Students complete outline diagrams when studying each topic step by step. Repeated reference to the outline diagram allows you to lay a solid foundation in the acquired knowledge.

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 For fixing educational material and revitalization educational activities It is recommended to pay special attention to solving problems such as mutual transformations substances (chains of transformations).

Success in completing tasks will depend on the number of transformations performed 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 = CH - CH = 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 → mythyl 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 (ext) 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. X 4

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; replacement of one functional group with another; removal of functional groups; changing the nature of functional groups. The sequence of operations may be different, depending on the structure and nature of the starting and resulting compounds.

Reminder: Present the facts and their relationships in in a visual form. Write down, in as much detail as possible, the essence of the task 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 turn out to be correct and lead you to the right decision. Use trial and error. If there is a limited set of options, try them all.

With the help of what reactions can transformations be carried out according to the scheme: CH 3 COO Na → CH 3 – CH 3 → CH 2 = CH 2 → CH 2 Br – CH 2 Br → CH ≡ CH → KOOC – COOK Solution. 1. To obtain ethane from sodium acetate, we will 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 into ethene, we will carry out the dehydrogenation reaction: t,Ni CH 3 – CH 3 → CH 2 =CH 2 + H 2 3. To obtain a dihaloalkane from an alkene, we use the bromination reaction: CH 2 =CH 2 +Br 2 → CH 2 Br – CH 2 Br 4. To obtain ethylene from dibromoethane, it is necessary to carry out dehydrohalogenation reaction, for this purpose 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 discolors 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 put a coefficient of 8. M argan changes the oxidation state from +7 to + 4, which corresponds to the acquisition of 3 ē, so we put a coefficient of 3 in front of the organic substance. Pay attention to reaction equations 1 and 5: Kolbe synthesis and oxidation of alkynes aqueous solution potassium permanganate. In an acidic environment, the permanganate ion is reduced to Mn 2+, and ethine 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 degree of oxidation 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 Oxidizing agents KMnO 4 Mn 2+ MnO 2 MnO 4 2+ Cr 3+ Cr 3

The oxidation processes of an alkene depend on its structure and the reaction environment. When oxidizing alkenes to a concentrated solution of potassium permanganate KMnO4 in an acidic environment (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 double bond.

If in an 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

Methods for arranging coefficients in ORR 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 method electronic balance: 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 H 2 O

The coefficients can also be selected using 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 attitude to oxidizing agents of all isomeric alcohols of the composition C 4 H 10 O. 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 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 reaction of 1,3-dimethylnitrobenzene with ammonium sulfide in a neutral medium (Zinin reaction). Problems for independent solution.

Conclusion Students’ knowledge in questions about the relationships of organic substances according to the scheme: composition - structure- properties of Drawing up redox reaction equations in organic chemistry Final certification of graduates in the form and based on the Unified State Examination materials in 2012: 6 students successfully completed 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 the maximum possible five. in 2013: 4 students successfully completed tasks of parts A and B exam material, and in task C 3, three people received 5 points each, one person received 4 points in 2014. One person took the exam and completely coped with these topics in tasks of parts A, B and C 3 of the examination 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 and methodological manual. Volgograd. Teacher2008 4. L.I. Salyakhova. Pedagogical advice. Preparation technology and practical developments. Educational and methodological manual. 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 Unified State Exam. Orenburg 2011 8. N.E. Kuznetsova, A.N. Levkin Problem book in chemistry: 10th grade: − M.: Ventana-Graf, 2011. – 144 p. literature

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

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

(based on the topic “Genetic relationship of organic compounds”)

Completed by: Vidershpan Irina Petrovna

Chemistry teacher

MBOU "Klyuchevskaya Secondary School No. 1"

With. Keys 2015

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

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

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

1. Work organizationon deciphering the chains of transformations…….6
2. Sequence of actions when executing chains………..7
3. Memo…………………………………………………………………….8
4. Methods for arranging coefficients in ORR 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 on performing chains of transformations are often used. They are used in 9th grade at the first stage of studying organic substances, in 10th grade when studying factual material in this course, and in 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 on the materials of the Unified State Examination. Tasks for carrying out transformations are widely used in general lessons.

Relevance: Therefore, performing tasks with genetic connections can create problematic situation at the students. They are in search, offering various ways. Children have creativity to this issue, since they offer their chains of transformations, while showing knowledge of factual material and the logic of their thinking.

Target.

• Promote the formation of higher levels in students

difficulties in training on the issue of genetic connections between

Organic compounds;

• create conditions for systematizing and deepening students’ knowledge about the relationship of organic substances according to the scheme: composition – structure – properties of substances

Tasks .

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

2. Main part

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 link”?

Genetic connectionis 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 series.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 connections 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, development of substances, their composition, structure, and the emergence of formations capable of the origin of life took place.
In practical terms, genetic connections show from which substances and in what ways the necessary substances can be obtained. Each transition is both an expression of the chemical properties of a substance and possible ways its practical use. 1

In order to successfully complete tasks showing genetic connections between classes of organic substances, knowledge of the nomenclature and classification of substances is developed in chemistry lessons, the chemical properties of compounds and methods for their preparation are studied. This approach can be followed at all stages of studying this issue, only each time the theoretical material is deepened and expanded.

2.2 Work on studying the nomenclature and classification of organic substances

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

2.3 Work on studying 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. When studying these issues in the classroom, I use reference diagrams (see Appendix 2).

Study of the structure, chemical properties and methods of producing hydrocarbons various groups shows that they are all genetically related to each other, i.e. transformation of some hydrocarbons into others is possible.

Support diagrams allow you to provide a large amount of information in a concise form. Students enjoy using them. These diagrams help them organize knowledge and develop the logic of everyone's thinking. In this way, students are prepared to perform transformation exercises.

Students complete outline diagrams when studying each topic step by step. Repeated reference to the outline diagram allows you to lay a solid foundation in the acquired knowledge.

2.4 Organization of workon deciphering chains of transformations

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

Exercises to decipher transformation chains help to better understand the genetic relationships between organic compounds. For example, after studying the topic “Hydrocarbons”, you should pay attention to solving this particular type of problem (see.example ).

Task 1. Name the intermediate products 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
Solution. In this chain of transformations, including 4 reactions, from ethyl alcohol C 2 H 5 OH butadiene-1,3 must be obtained

CH 2 =CH–CH=CH2.

1. When heating alcohols with concentrated sulfuric acid
   H 2 SO 4 (water-removing agent) occursdehydration With

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

CH 3 - CH 2 -OH → CH 2 =CH 2 + H 2 O

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

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

CH 2 =CH 2 + HBr → CH 3 - CH 2 -Br

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

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

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 producing butadiene-1,3 CH 2 =CH–CH=CH 2 2

Success in completing tasks will depend on the number of transformations performed. The handouts for students contain individual assignments(respectively for each class and according to the level of the material studied). Developed and tested Handout for students in grades 10 (see Appendix 3), for grades 11 (see Appendix 4), for preparing for exams in the form and based on Unified State Exam materials (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. Therefore, everyone has their own ways of solving problems, but one goal has been achieved: a transformation has been carried out, showing the genetic connection between individual classes of organic compounds.

2.5 Sequence of actions when executing 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. Moreover, in some cases it is necessary to indicate specific reagents and conditions for the reactions leading to the substances that make up the chain of transformations. In others, on the contrary, it is necessary to determine what substances are formed during the action specified reagents to the original connections.

Usually in similar cases no need to indicate the fine technical details of the synthesis, the exact concentration of reagents, specific solvents,

purification and isolation methods, etc. however, it is necessary to indicate approximate conditions carrying out reactions.

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;
• replacement of one functional group with another;
• removal of functional groups;
• changing the nature of functional groups.

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

2.6 Memo

Present the facts and their relationships clearly. Write down, in as much detail as possible, the essence of the task 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 turn out to be correct and lead you to the right decision.

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

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

→ CH 3 COH → CH 3 COOH → CH 3 COOC 2 H 5

Solution.

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 sodium metal (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 a nucleophilic substitution reaction (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 into an aldehyde, it is necessary 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. What reactions can be used to carry out transformations according to the following scheme: CH 3 COONa → CH 3 – CH 3 → CH 2 =CH 2 → CH 2 Br – CH 2 Br →

→ CH≡CH→KOOC – COOK

Solution.

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

El-z

2CH 3 COONa + 2H 2 O → CH 3 – CH 3 +H 2 +2NaHCO 3

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

t,Ni

CH 3 – CH 3 → CH 2 =CH 2 + H 2

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

Bromination:

CH 2 =CH 2 +Br 2 → CH 2 Br – CH 2 Br

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

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

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

5. Ethine decolorizes an aqueous solution of KMnO 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, so before MnO 2 We set the coefficient to 8. Mn changes the oxidation state from +7 to +4, which corresponds to the acquisition of 3 electrons, so we set the coefficient to 3 in front of the organic substance.

Please 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 ethin is oxidized to oxalic acid:

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

To assign 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 electronegativities of the atoms in its immediate environment.

The degree of oxidation 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 oxidation state 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 along three σ bonds. In the carboxyl group of the benzoic acid molecule, the oxidation state of the carbon atom is “+ 3”, because electron density shifts

from a carbon atom to oxygen atoms via two σ-bonds and one π-bond (three bonds in total):

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 odds using the electronic balance method:

C -3 – 6e - → C +3 5

Mn +7 + 5е - →Mn +2 6

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

Toluene C 6 H 5 CH 3 oxidizes to benzoic acid C 6 N 5 COOH, permanganate ion MnO 4 - is 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 - +8Н + + 5е - →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 elimination or addition of hydrogen and oxygen atoms.

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

This principle is the basis for the method of arranging coefficients, which does not require determining the degree of carbon oxidation. Let us determine the number of hydrogen atoms lost by toluene and the number of oxygen atoms introduced into the molecule. Two hydrogen atoms are lost (-2e¯), two oxygen atoms are introduced (-4e¯). A total of 6e¯ was given away.

Let's consider the application in various ways arrangement of coefficients using the example of the reaction between n-methylcumene and potassium permanganate in an acidic environment.

1. Oxidation of any alkyl substituents on 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 electron 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. Eight hydrogen atoms are lost (-8e¯), four oxygen atoms are introduced (-8e¯). In addition, four oxygen atoms were included in carbon dioxide (-8e¯). A total of 24e¯ was given away. Manganese with an oxidation state of +7 is reduced to +2, then the electronic balance diagram 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, determining the oxidation states

carbon atoms. 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 the α-carbon atoms (directly bonded to the benzene ring) are oxidized to carboxyl groups, the remaining carbon atoms are oxidized to carbon dioxide.

С -1 – 4е - →С +3

C -3 – 6e - →C +3 - 24e - 5

2 С -3 – 14е - →2С +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. Let us select the coefficients using the electron-ion balance method.

In this case there is no need to depict structural formulas organic substances, therefore the formulas n-methylcumene and tere We write phthalic acid in molecular form:

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

MnO 4 - +8Н + + 5е - →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 + 56 H2 O

I offer handouts for students (see Appendix 7).

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

13

3. Conclusion

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

allowance. Volgograd. Teacher2008

4. L.I. Salyakhova. Pedagogical advice. Preparation technology and

practical developments. Educational and methodological manual. Globe. 2006

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

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

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

to the Unified State Exam. Orenburg 2011

8. N.E. Kuznetsova, A.N. Levkin Problem book in chemistry: 10th grade: − M.: Ventana-

Count, p.2011. – 144

14

Annex 1

MAP OF FORMULAS OF CHEMICAL COMPOUNDS

(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 – O CH3 ; 10) CH3 OH; 11) CH C–CH3 ; 12) CH3 – COOCH3 ; 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 – N.H.2 ;

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

Appendix 2

SUPPORT DIAGRAM

CHEMICAL PROPERTIES OF SATURATE 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) Ethane→ 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 +Ag2 O,tX2 +CH3OH,H2SO4X3

10) CH2 = CH – CH3 → X→ CH3 COCH3

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

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 IMPLEMENT 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 1500 CX1 +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) ztilen+Br2(ad)X1 +KOH, alcoholX2 +H2O, HgX3 → X4 → mythyl acetate

8) ethyl acetate → sodium acetateNaOH(fusionX1 1500CX2 400CX3 C2H5Cl,AlCl3X4

9) 1-bromopropane → hexane → benzeneCH3ClX1 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 CH3OH, H2SO4X4

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

17)CH4 1000CX1 Act,tX2 CH3Cl, AlCl3X3 KMnO4, tX4

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

19)CH4 →CH3 NO2 →CH3 N.H.2 →CH3 N.H.3 Cl → CH3 N.H.2 → N2

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

17

Appendix 5

A SET OF DIAGRAMS USED TO PREPARATION FOR COMPLETING A TASK WITH3 EXAMINATION PART IN THE FORM AND ACCORDING TO THE USE MATERIALS

1) Ethine → benzeneCH3Cl, AlCl3X1 Cl2, UVX2 CONaq.X3 HCOOH,HX4

2) 1,3-dibromobutaneZnX1 → 2-,bromobutaneNaX2 → 1,2dimethylbenzeneKMnO4 ,H2 SO4,tX3

3) C2 H5 Cl→C3 H8 t,NiX1 KMnO4,H2OX2 excess 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 OHAl2O3, 400X1 KMnO4,H2OX2 HBr(g)X3 KOH (alcohol)X4 C act, tX5

10) C2 H5 OHAl2O3, 400X1 KMnO4, H2OX2 HBr(g)X3 KOH (alcohol)X4 → C2 H4 O

11) CH4 → HCHOH2, cat.X1 NaX2 HClX1 KMnO4, H2SO4X3

12) CH4 → ethylene → vinyl acetylenehut H2,catX1 → ethanoic acidNH3X2

13)CaC2 H2OX1 KMnO4H2 C2 O4 conc. H2SO4X2 → HCOOKconc. H3PO4X3

14) H2 C2 O4 → CO → CH3 OH → CH3 COOCH3 NaOH + H2O, tX → CH4

18

Appendix 6

1. C6 N12 ABOUT6 →C2 N5 OH→CH3 –CHO→CH3 – COOH→Cl-CH2 –COOH→

→ N2 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.CH3 – 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 vinyl acetylene X3

Electrolysis +Cl2 , light +NaOH, H2 O H2 SO4 (conc.), t

6.CH3 COOH X1 WITH2 N6 X2 X3 X4

C, 400 °С Cl2 , cat. CO2 +H2 OBr2 +H2 O

7.C2 H2 X1 X2 C6 H5 OK X3 X4

19

Appendix 7

Problems 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 attitude to oxidizing agents of all isomeric alcohols of composition C4 N10 O. 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 reaction of 1,3-dimethylnitrobenzene with ammonium sulfide in a neutral medium (Zinin reaction).
• During glucose oxidation bromine water gluconic acid is formed, and upon oxidation of concentrated nitric acid– glucarova. Write down the equations for the corresponding reactions.

20

Appendix 8

Appendix 9

21

Appendix 10

Appendix 11

22|

The proposed manual presents a system of tasks of increased and high levels of complexity, similar to the tasks of control measuring tests. Unified State Exam materials in chemistry. Special attention focuses on analyzing the tasks that caused the greatest difficulties. For training and self-preparation for the Unified State Exam, assignments are offered various levels difficulties in key sections of the chemistry course.
The manual is addressed to high school students, teachers and parents. It will help schoolchildren test their knowledge and skills in the subject, and teachers - assess the degree to which the requirements have been achieved educational standards individual students and ensure their targeted preparation for the exam.

Examples.
When reducing zinc oxide carbon monoxide metal was formed. The metal reacted with a concentrated solution of potassium hydroxide to form a complex salt. Excess hydrogen sulfide was passed through the salt solution, and a precipitate formed. When this precipitate was heated with concentrated nitric acid, a 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 equations for the four reactions described.

CONTENT
INTRODUCTION
PART 1. SYSTEM OF ASSIGNMENTS OF ADVANCED AND HIGH LEVEL OF COMPLEXITY FOR THE IMPORTANT SECTIONS/TOPICS OF THE CHEMISTRY COURSE
Section 1. Redox reactions, patterns of their occurrence. Electrolysis of solutions and molten salts
1.1. Material for repetition and systematization of knowledge
1.2. Assignments 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. Interrelation of inorganic substances
2.1. Material for repetition and systematization of knowledge
2.2. Assignments with comments and solutions
2.3. Assignments 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. Assignments with comments and solutions
3.3. Assignments for independent work
Section 4. Calculation problems
PART 2. TRAINING OPTIONS, INCLUDING ADVANCED AND HIGH LEVEL OF COMPLEXITY TASKS
Option 1
Option 2
Option 3
Option 4
Option 5
PART 3. ANSWERS TO TASKS FOR INDEPENDENT WORK AND TASKS FOR TRAINING OPTIONS.

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