Lesson summary “Genetic relationships between the main classes of organic compounds. Problem Solving"

OPTION 1

2. Calculate the amount of substance (in moles) and the mass of substance (in grams) of each product during the following transformations: ethane → bromoethane → ethanol, if ethane was taken with a mass of 90 g. The yield of the product at each stage of the synthesis is conventionally taken as 100%.

3. Draw up a diagram and equations of reactions that can be used to obtain carboxylic acid from methane.

OPTION 2

1. Write reaction equations that can be used to carry out the following transformations:

2. Calculate the amount of substance (in moles) and the mass of substance (in grams) of each product during the following transformations: benzene → chlorobenzene → phenol, if benzene was taken with a mass of 156 g. The yield of the product at each stage of the synthesis is conventionally taken as 100%.

3. Draw up a diagram and equations of reactions that can be used to obtain an amino acid from ethylene.

OPTION 3

1. Write reaction equations that can be used to carry out the following transformations:

2. Calculate the amount of substance (in moles) and the mass of substance (in grams) of each product during the following transformations: benzene → nitrobenzene → aniline, if benzene was taken with a mass of 39 g. The yield of the product at each stage of the synthesis is conventionally taken as 100%.

3. Draw up a diagram and equations of reactions that can be used to obtain an ester from coal.

OPTION 4

1. Write the reaction equations that can be used to carry out the following transformations:

2. Calculate the amount of substance (in moles) and the mass of substance (in grams) of each product when carrying out the following transformations: chloromethane → methanol → methyl acetate, if chloromethane was taken with a mass of 101 g. The yield of the product at each stage of the synthesis is conventionally taken as 100%.

3. Draw up a diagram and equations of reactions that can be used to obtain an aromatic amine from methane.

Summary lesson

Lesson objectives:

Ensure that students understand the genetic relationship between classes organic compounds;

Development of independent thinking skills;

Create conditions for developing skills of independent and team work.

Lesson objectives:

Continue to develop students’ ability to apply previously acquired knowledge;

Development logical thinking;

Development of students' speech culture;

Development of cognitive interest in the subject.

During the classes:

1. Introduction.

2. Warm up.

3. Quiz: “Guess the substance.”

4. Drawing up a genetic chain.

5. Homework.

Introduction. Knowing the chemistry of functional groups, possible ways replacing them, the conditions for their transformations, it is possible to plan organic synthesis, moving from relatively simple compounds to more complex ones. IN famous book Carroll's "Alice in Wonderland" Alice turns to the Cheshire Cat: “Please tell me where should I go?” To which the Cheshire Cat reasonably remarks: “It largely depends on where you want to come.” How can this dialogue be related to a genetic connection? We will try, using knowledge of the chemical properties of organic compounds, to carry out transformations from the simplest representatives of alkanes to high-molecular compounds.

I. Warm-up.

1. Review the classes of organic compounds.

2. What are the structure of transformation series?

3. Solving series of transformations:

1) CaC2 → C2H2 → C6H6 → C6H5Cl → C6H5OH → C6H2Br3OH

2) Al4C4 → CH4 → C2H2 → C6H6 → C6H5ONa → C6H5OCH3

3) hexane → benzene → chlorobenzene → toluene → 2.4.6-tribromotoluene

II. Quiz: “Guess the substance.”

Assignment to students: identify the substance about which we're talking about and say a few words about this substance. (The student writes down the formulas of substances at the blackboard.)

1) This substance is called swamp gas, it is the basis of natural gas, a valuable and accessible raw material for the synthesis of many substances. (Methane)

Teacher's addition: one interesting message about where methane came in handy. Specialists from one of the US Navy research laboratories managed to develop a method for producing artificial diamonds. Methane was fed to a tungsten plate heated to 2500 C, on which the resulting crystals settled.

2) This substance is called illuminating gas. This gas was initially used mainly for lighting: street lamps, theater footlights, camping and miner's lanterns. Older bikes had carbide lights. Water flowed into a vessel filled with calcium carbide, and the resulting gas flowed through a special nozzle into the lamp, where it burned with a bright flame. (Acetylene)

3) The structure of this substance took 40 years to be established, and the solution came when a snake appeared in Kekule’s imagination, biting its own tail. (Benzene)

4) Special experiments have established that when the content of this substance in the air is approximately 0.1%, vegetables and fruits ripen faster. This substance is called a plant growth regulator. (Ethylene)

Teacher's addition: It turns out that pineapples need ethylene to bloom. On plantations, fuel oil is burned, and small amounts of ethylene produced are sufficient to produce a crop. At home, you can use a ripe banana, which also releases ethylene. By the way, ethylene can transmit information. Kudu antelopes feed on acacia leaves, which produce tannin. This substance gives the leaves a bitter taste, and in high concentrations it is poisonous. Antelopes are able to select leaves from low content tannin, but in extreme conditions they eat any and die. It turns out that the leaves eaten by antelopes emit ethylene, which serves as a signal for neighboring acacias, and after half an hour their leaves intensively produce tannin, which leads to the death of the antelopes.

5) Grape sugar. (Glucose.)

6) Wine alcohol. (Ethanol)

7) Oily liquid. Which was obtained from Tolu balsam. (Toluene)

8) When there is danger, ants release this particular substance. (Formic acid)

9) An explosive substance that has several names: tol, TNT. TNT. Typically, 1 g of explosive produces about 1 liter of gases, which corresponds to a thousandfold increase in volume. The mechanism of action of any explosive is the instantaneous formation of a large volume of gas from a small volume of liquid or solid. The pressure of the expanding gases is the destructive force of the explosion. (Trinitrotoluene)

III. Drawing up a genetic chain.

Work in groups. The class is divided into groups of 4 people.

Assignment to groups: compose a series of transformations using as many more substances guessed in the quiz. The task is offered for a while. After completion, the task is checked at the board.

At the end of the lesson, evaluate the students' answers.

Let us consider the genetic series of organic substances, in which we include greatest number connection classes:

Each number above the arrow corresponds to a specific reaction equation (the reverse reaction equation is indicated by a number with a prime):

IV. Homework: Create a genetic series of transformations that includes at least five classes of organic compounds.

Alice (in Wonderland to the Cheshire cat): - Tell me, where should I go from here? Alice (in Wonderland to the Cheshire cat): - Tell me, where should I go from here? Cheshire Cat: – It depends on where you want to come? Cheshire cat: – It depends on where you want to come? 2

Synthesis Strategy “I want to sing the praises of the creation of molecules - chemical synthesis... ...I am deeply convinced that it is an art. And at the same time, synthesis is logic.” Roald Hoffman ( Nobel Prize in chemistry 1981 d) Selection of starting materials Construction of the carbon backbone of the molecule Introduction, removal or replacement of a functional group Protection of the group Stereoselectivity 5

CO + H 2 Ru, 1000 atm, C ThO 2, 600 atm, C Cr 2 O 3, 30 atm, C Fe, 2000 atm, C ZnO, Cr 2 O 3, 250 atm, C PARAFFINS ISOPARAFFINS TOLUENE, XYLENES HIGHER ALCOHOLS CH 3 OH 6

С n H 2n+2 Scheme of formation of σ-bonds in a methane molecule Models of methane molecules: ball-and-stick (left) and scale (right) CH4CH4CH4CH4 Tetrahedral structure sp 3 -hybridization of σ-bonds homolytic cleavage of the X: Y bond homolytic cleavage of the bond Radical substitution reactions ( S R) substitution (S R)CombustionDehydrogenation S – eng. substitution - replacement Reactivity forecast 7

CH 3 Cl – METHYL CHLORIDE CH 4 METHANE C – SOAR C 2 H 2 – ACETYLENE CH 2 Cl 2 – DICHLOROMETHANE CHCl 3 – TRICHLORMETHANE CCl 4 – TETRACHLOROMETHANE H 2 – HYDROGEN SYNTHESIS GAS CO + H 2 SYNTHESIS GAS CO + H 2 Cl 2, hγ Chlorination C pyrolysis H 2 O, Ni, C Conversion of O 2, Oxidation CH 3 OH – METHANOL HCHO – METHANAL solvents Benzene СHFCl 2 freon HCOOH - formic acid Synthetic gasoline SYNTHESIS BASED ON METHANE 8 CH 3 NO 2 – NITROMETHANE CCl 3 NO 2 chloropicrin CH 3 NH 2 methylamine HNO 3, C Nitration

С n H 2n Scheme of the formation of σ-bonds with the participation of sp 2 -hybrid clouds of the carbon atom Scheme of the formation of π-bonds with the participation of p-clouds of the carbon atom Model of the ethylene molecule Electrophilic addition reactions (A E) Polymerization Polymerization Oxidation OxidationCombustion Flat molecule (120 0) sp 2 – hybridization of σ– and σ– and π– bonds Eb (C = C) = 611 kJ/mol Eb (C – C) = 348 kJ/mol A – eng. addition – accession Reactivity forecast 9

C 2 H 4 Ethylene Polymerization H 2 O, H + Hydration Cl 2 Chlorination Oxidation ETHYL ALCOHOL WITH 2 H 5 OH ETHYL ALCOHOL WITH 2 H 5 OH SYNTHESIS BASED ON ETHYLENE DICHLOROETHANE ETHYLENE OXIDE ETHYLENE GLYCOL ACETALEHYDE ACEALDEHYDE O 2, KMnO4,H 2 O O 2, PdCl 2, CuCl 2 HDPE HDPE WITH MPa 80 0 C, 0.3 MPa, Al(C 2 H 5) 3, TiCl 4 SKD LDPE LDPE Butadiene-1,3 (divinyl) Acetic acid Dioxane Acetic acid 10

С n H 2n-2 Scheme of formation of σ-bonds and π-bonds with the participation of sp-hybrid clouds of the carbon atom Models of the acetylene molecule electrophilic addition reactions (A E) oxidation oxidation di-, tri- and tetramerization di-, tri- and tetramerization combustion combustion reactions involving the “acidic” hydrogen atom Linear structure (180 0) (cylindrical distribution of electron density) sp – hybridization of σ– and 2 σ – and 2π – bonds Reactivity forecast 11

C2H2C2H2 HСl, Hg 2+ H 2 O, Hg 2+ Kucherov reaction C act, C trimerization SYNTHESIS BASED ON ACETYLENE ACETALDEHYDE ACETALDEHYDE CuCl 2, HCl, NH 4 Cl dimerization ROH Acetic acid BENZENE SKD Divinyl Chloroprene SK chloroprene VINI LACETYLENE VINYL ESTERS Polyvinyl ethersPolyvinyl chloride VINYL CHLORIDE HCN, СuCl, HCl, 80 0 C ACRYLONITRILE Fibers 12

Scheme of the formation of π-bonds in a benzene molecule Delocalization of electron density in a benzene molecule Scheme of the formation of σ-bonds in a benzene molecule with the participation of sp 2 - hybrid orbitals of carbon atoms C n H 2n-6 Reactivity prediction Flat molecule sp 2 - hybridization of σ- and σ – and π – bonds Aromatic structure Electrophilic substitution reactions (S E) Radical addition reactions (A R) Radical addition reactions (A R) Combustion 14 M. Faraday (1791–1867) English physicist and chemist. Founder of electrochemistry. Discovered benzene; was the first to obtain chlorine, hydrogen sulfide, ammonia, and nitric oxide (IV) in liquid form.

BENZENE H 2 /Pt, C hydrogenation SYNTHESIS BASED ON BENZENE NITROBENZENE NITROBENZENE Cl 2, FeCl 3 chlorination HNO 3, H 2 SO 4 (concentrated) nitration CH 3 Cl, AlCl 3 alkylation CHLOROBENZENE Aniline TOLUENE TOLUENE Benzoic acid 2,4,6- trinitrotoluene STYRENE STYRENE Polystyrene 1. CH 3 CH 2 Cl, AlCl 3 Alkylation 2. – H 2, Ni dehydrogenation CH 2 =CH-CH 3, AlCl 3 alkylation CUMEN (ISOPROPYLBENZENE) CUMENE (ISOPROPYLBENZENE) CYCLOHEXANE CYCLOHEXANE Phenol Acetone HEX LORANE HEXACHLORANE 15

SYNTHESIS BASED ON METHANOL CH 3 OH VINYL METHYL ETHER VINYL METHYL ETHER DIMETHYLANILINE C 6 H 5 N(CH 3) 2 DIMETHYL ANIline C 6 H 5 N(CH 3) 2 DIMETHYL ETHER CH 3 –O–CH 3 DIMETHYL ETHER CH 3 –O–CH 3 METHYLAMINE CH 3 NH 2 METHYLAMINE CH 3 NH 2 VINYL ACETATE METHYL CHLORIDE CH 3 Cl METHYL CHLORIDE CH 3 Cl FORMALDEHYDE CuO, t HCl NH 3 METHYL THIOL CH 3 SH METHYL THIOL CH 3 SH H 2 S, t C 6 H 5 NH 2 + CO 16 H +, t

SYNTHESIS BASED ON FORMALDEHYDE METHANOL CH 3 OH METHANOL CH 3 OH PARAFORM PHENOLFORMALDEHYDE RESINS PHENOLFORMALDEHYDE RESINS TRIOXANE PRIMARY ALCOHOLS UREA RESINS UREA RESINS UROTROPINE (HEXMETHYLENTETER) AMINE) UROTROPINE (HEXMETHYLENETETRAMINE) FORMIC ACID FORMIC ACID Hexogen [O] [H] 1861 A.M. Butlerov 18

CxHyOzCxHyOz Genetic relationship of oxygen-containing organic compounds ALDEHYDES ALDEHYDES CARBOXYLIC ACIDS CARBOXYLIC ACIDS KETONES KETONES ESTERS ESTERS ESTERS ESTERS ALCOHOL hydrolysis dehydration hydrogenation oxidation, dehydrogenation esterification esterification oxidation H+, t

C n H 2n+2 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes C n H 2n-6 Arenes, benzene

C n H 2n+2 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes Primary Secondary Tertiary C n H 2n-6 Arenes, benzene 12 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes α 23

C n H 2n+2 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes Primary Secondary Tertiary C n H 2n-6 Arenes, benzene 12 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes

C n H 2n+2 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes Primary Secondary Tertiary C n H 2n-6 Arenes, benzene Polyethylene Polypropylene 12 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes Rubbers Catalyst Ziegler – Natta (1963) 25

C n H 2n+2 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes Primary Secondary Tertiary C n H 2n-6 Arenes, benzene Polyethylene Polypropylene Rubbers Fats Phenol-formaldehyde resins 12 C n H 2n Cycloalkanes Alkenes C n H 2n- 2 AlkynesAlkadienes

C n H 2n+2 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes Primary Secondary Tertiary C n H 2n-6 Arenes, benzene Polyethylene Polypropylene Rubbers Fats Synthetic dyes Phenol-formaldehyde resins 12 C n H 2n Cycloalkanes Alkenes C n H 2n-2 AlkynesAlkadienes

Application of aniline ANILINE N.N. Zinin (1812 – 1880) Medicinal substances Dyes Explosives Streptocide Norsulfazole Phthalazole Preparation of aniline - Zinine reaction Tetryl Aniline yellow Nitrobenzene p-Aminobenzoic acid (PABA) Indigo sulfanilic acid Paracetamol 28

C n H 2n+2 C n H 2n Cycloalkanes Alkenes C n H 2n-2 Alkynes Alkadienes Primary Secondary Tertiary C n H 2n-6 Arenes, benzene Polyethylene Polypropylene Rubbers Fats Synthetic dyes Phenol-formaldehyde resins Proteins 12 C n H 2n Cycloalkanes Alkenes C n H 2n-2 AlkynesAlkadienes

Tsybina Lyubov Mikhailovna Chemistry teacher Lesson notes.

Lesson summary on the topic: “Genetic connection between the main classes of organic compounds. Problem solving."

Class: Grade 11

Target: create conditions for systematizing and deepening students’ knowledge about the relationship of organic substances according to the scheme: composition – structure – properties of substances and the ability to solve calculation problems.

Tasks:

Educational:

Generalization and deepening of students’ knowledge about the relationship between the composition – structure – properties of organic substances using the example of hydrocarbons and oxygen-containing homologous series.

Expanding the general cultural horizons of students

Educational:

Development of skills to analyze, compare, draw conclusions, establish cause-and-effect genetic relationships between organic substances.

Be able to choose the right algorithm for solving a calculation problem.

Educational:

Disclosure of ideological ideas about the relationship between the composition, structure, and properties of substances; education of an intellectually developed personality; fostering a culture of communication.

Be able to work using the algorithm and additional literature.

Lesson type:

for didactic purposes: lesson on systematization of knowledge;

by method of organization: generalizing with the acquisition of new knowledge (combined lesson).

Education technology:

problem-based learning;

information and communication

Methods used in the lesson:

explanatory and illustrative:
– frontal conversation;
– teacher’s explanation.

– schema tables, algorithms

practical:
– drawing up transformation schemes and their implementation.

deductive:
– from the known to the unknown;
– from simple to complex.

Types of control:

current poll,

working with cards.

Educational technologies used:

Information

Technology for updating personal experience

Technology targeting cognitive development personalities

Form of conduct : a combination of conversation with illustrative explanatory material, independent activity students.

Equipment: computer, algorithm for solving a calculation problem.

Lesson Plan

Tasks

Organizing time

Prepare students for work in class.

Updating of reference knowledge

"Brainstorm"

(review of the material studied)

Prepare students to learn new material. Review previously studied topics to identify gaps in knowledge and to address them. Improve knowledge and skills, prepare to perceive new material.

III

Learning new material

genetic connection;

genetic series of hydrocarbons and its varieties;

genetically a series of oxygen-containing hydrocarbons and its varieties.

Develop the ability to generalize facts, build analogies and draw conclusions.

To develop students' ability to make chemical predictions and the ability to solve calculation problems using genetic relationships.

Develop environmental thinking.

Development of a culture of communication, the ability to express one’s views and judgments and rational ways of solving a calculation problem.

Consolidation of acquired knowledge

Repetition, reproduction of learned material.

Working off of this material on assignments in UNT format.

Summing up the lesson

Perception of a sense of responsibility for acquired knowledge. Assessment of students' activities in the lesson. Reflection. Making marks.

Homework

Textbook: Chemistry for grade 11 A. Temirbulatova N. Nurakhmetov, R. Zhumadilova, S. Alimzhanova. §10.6 p.119(23,26), p.150(18),

Workbook exercise 107 a), b) p.22.

Lesson stage 1

Organizational. Announcing the topic of the lesson. Updating basic knowledge.

What does the concept mean?“genetic link”?
Conversion of substances of one class of compounds into substances of other classes;

Genetic connection is the connection between substances of different classes, based on their mutual transformations and reflecting the unity of their origin, that is, the genesis of substances.
The key point of the lesson is creating problematic situation. To do this, I use problem-search conversation, which encourages students to make assumptions, express their point of view, and causes a clash of ideas, opinions, and judgments.
The main task is to point out to students the insufficiency of their knowledge about the object of knowledge, as well as the methods of action to complete the task proposed to them.

To compare means to choose, first of all, comparison criteria. Please tell me what criteria you think we should compare. Students answer:

Chemical properties substances;

Possibility of obtaining new substances;

The relationship of substances of all classes of organic compounds.

Lesson stage 2

“Brainstorming” – frontal conversation with the class:

What classes of organic compounds do you know?

What is special about the structure of these classes of compounds?

How does the structure of a substance affect its properties?

What basic formulas do you know that can be used to solve a calculation problem?

Using knowledge about the structure of organic substances and the characteristics of their general formulas, students independently write down basic formulas and predict the possible chemical properties of organic substances.

Lesson stage 3

Carrying out genetic linkage of organic compounds

First option: ethanol →ethylene →ethane →chloroethane →ethanol →acetaldehyde → carbon dioxide

second option: methane →acetylene→ethanal→ethanol→bromoethane→ethylene→carbon dioxide

Third option: acetylene→ethanal→ethanol→bromoethane→ethylene→ethanol→ethyl acetate

work at the board using cards: solving a calculation problem

Task – 1: 6 kg of methyl formate were obtained from methane. Write the corresponding reaction equations. Calculate how much methane was consumed?

Task – 2: How much ethyl acetate can be obtained by reacting 120 g of acetic acid and 138 g of ethanol if the yield of the reaction product is 90% of theoretical?

Task – 3: We oxidized 2 moles of methanol. The resulting product was dissolved in 200 g of water. Calculate the content of methanal in the solution (in%)?

Correct solution calculation tasks are designed on a smartboard.

General conclusion :

We highlight the characteristics that characterize the genetic series of organic substances:

Substances of different classes;

Different substances are formed by one chemical element, i.e. represent different shapes the existence of one element;

Different substances of the same homologous series are related by mutual transformations.

Knowledge of the genetic relationship between different classes of organic substances allows us to select convenient and economical methods for the synthesis of substances from available reagents.

Stage 4 of the lesson

Repetition, reproduction of learned material. Practicing this material using assignments in the UNT format. p.119(23); Workbook exercise 107 a), b) p.22.

Brief instructions on homework: §10.6 p.119(23,26), p.150(18),

Lesson stage 5

Summarizing. Reflection.

Students answer the questions:

What new concepts were learned in the lesson?

What questions caused difficulties? And so on.

The teacher gives grades to those students who showed good and excellent knowledge during the lesson and were active.

Target: consider the genetic relationship between the classes of inorganic and organic

substances, give the concept of the “genetic series of substances” and “genetic connections”,

consolidate skills in writing equations of chemical reactions.

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Preview:

Lesson No.___

Subject:

Target: consider the genetic relationship between the classes of inorganic and organic

Substances, give the concept of the “genetic series of substances” and “genetic connections”,

Strengthen skills in writing equations of chemical reactions.

Tasks: 1 . Educational:improve skills in conducting laboratory tests

Experiments, recording equations of chemical reactions.

2. Developmental: consolidate and develop knowledge about the properties of inorganic and

Organic substances, develop skills in working in groups and individually.

3. Educational: to develop an interest in the scientific worldview,

The desire to achieve academic success.

Equipment: multimedia projector

Reagents: alcohol lamp, matches, test tube holder, rack with test tubes, CuSO 4 NaOH

During the classes.

I. Organizational moment.

II. Explanation of new material.

You and I live in a world where thousands of reactions occur in every cell of a living organism, in the soil, air, and water.

Teacher : Guys, what do you think of unity and diversity? chemical substances, involved in the process of transformation? What is the connection between substances called? Let's remember with you who is the keeper of hereditary information in biology?

Study: Gen.

Teacher: What is a genetic link?

Study: related.

Let's formulate the topic of our lesson. (Write the topic of the lesson on the board and notebook).

And now you and I will work according to the plan that is on every desk:

Genetic series of metal.
Genetic series of a nonmetal.
Consolidation of knowledge(testing in the form of the Unified State Exam)

Let's move on to point 1 of the plan.

Genetic connection - is called the connection between substances of different classes,

based on their mutual transformations and reflecting their unity

Origin, that is, the genesis of substances.

What does the concept mean?"genetic link"

The transformation of substances of one class of compounds into substances of other classes.
Chemical properties of substances
The ability to obtain complex substances from simple ones.
The relationship between simple and complex substances of all classes of substances.

Now let’s move on to consider the concept of a genetic series of substances, which is a particular manifestation of a genetic connection.

A number of substances are called genetic - representatives of different classes of substances

Being compounds of one chemical element related

Mutual transformations and reflecting the common origin of these

Substance

Let's consider the signs of a genetic series of substances:

All substances of the genetic series must be formed by one chemical element.
Substances formed by the same chemical element must belong to different classes(i.e. reflect different forms of existence of a chemical element)
Substances that form the genetic series of one chemical element must be connected by mutual transformations.

Based on this feature, it is possible to distinguish between complete and incomplete genetic series. Let us first consider the genetic relationship of inorganic substances and divide them into

2 types of genetic series:

A) metal genetic series

b) genetic series of a non-metal.

Let's move on to the second point of our plan.

Genetic series of metal.

a) consider the series of copper:

Cu → CuO → CuSO 4 → Cu(OH) 2 → CuO→ Cu

Copper oxide sulfate hydroxide copper oxide

Copper(II) copper(II) copper(II) copper(II)

Metal base salt base base metal

Oxide oxide

2Cu + O 2 → 2CuO
CuO + H 2 SO 4 → CuSO 4 + H 2 O
CuSO 4 + 2KOH → Cu(OH) 2 + K 2 SO 4
Cu(OH) 2 → CuO + H 2 O
CuO + C→Cu + CO

Demonstration: partially from the series - equations 3.4. (Interaction of copper sulfate with alkali and subsequent decomposition of copper hydroxide)

b) genetic series of an amphoteric metal using the example of the zinc series.

Zn → ZnO → ZnSO 4 → Zn(OH) 2 Na 2

ZnCl2

2Zn + O 2 → 2ZnO
ZnO + H 2 SO 4 → ZnSO 4 + H 2 O
ZnSO 4 + 2KOH → Zn(OH) 2 + K 2 SO 4
Zn(OH) 2 +2 NaOH→ Na 2
Zn(OH) 2 + 2HCl → ZnCl 2 + 2H 2 O
ZnO + 2HCl → ZnCl 2 + H 2 O

Demonstration carrying out reactions from the series 3,4,5.

We have discussed point 2 of the plan with you. What does point 3 of the plan say?

Genetic series of a nonmetalLet's look at an examplegenetic series of phosphorus.

P → P 2 O 5 → H 3 PO 4 → Ca 2 (PO 4 ) 2

Phosphorus oxide phosphorus phosphate

Phosphorus (v) calcium acid

Nonmetal acidic acid salt

Oxide

4P + 5O 2 → 2P 2 O 5
P 2 O 5 + 3H 2 O → 2H 3 PO 4
2H 3 PO 4 + 3Ca → Ca 3 (PO 4 ) 2 + 3H 2

So, we have looked at the genetic series of metal and non-metal. Do you think the concept of genetic connection and genetic series is used in organic chemistry? Of course it is used, butThe basis of the genetic series in organic chemistry (chemistry of carbon compounds) is made up of compounds with the same number of carbon atoms in the molecule. For example:

C 2 H 6 → C 2 H 4 → C 2 H 5 OH → CH 3 CHO → CH 3 - COOH → CH 2 Cl - COOH → NH 2 CH 2 COOH

Ethane ethene ethanol ethanal acetic acid chloroethanoic acid aminoethanoic acid

alkane alkene alkanol alkanal carboxylic acid chlorocarboxylic acid amino acid

C 2 H 6 → C 2 H 4 + H 2
C 2 H 4 + H 2 O → C 2 H 5 OH
C 2 H 5 OH + [O] → CH 3 CHO + H 2 O
CH 3 CHO + [O] → CH 3 COOH
CH 3 COOH + Cl 2 → CH 2 Cl - COOH
CH 2 Cl - COOH + NH 3 → NH 2 CH 2 - COOH + HCl

We have looked at the genetic connection and genetic series of substances and now we need to consolidate our knowledge on the 5th point of the plan.

III. Consolidation of knowledge, skills and abilities.

Unified State Exam testing

Option 1.

Part A.

A) CO 2 b) CO c) CaO d) O 2

In the transformation scheme: CuCl 2 2 b)CuSO 4 and Cu(OH) 2

CO 2 → X 1 → X 2 → NaOH

A)N b) Mn c)P d)Cl

Part B.

Fe + Cl 2 A) FeCl 2
Fe + HCl B) FeCl 3
FeO + HCl B) FeCl 2 + H 2
Fe 2 O 3 + HCl D) FeCl 3 + H 2

D) FeCl 2 + H 2 O

E) FeCl 3 + H 2 O

a) potassium hydroxide (solution)

b) iron

c) barium nitrate (solution)

d) aluminum oxide

e) carbon monoxide (II)

e) sodium phosphate (solution)

Part C.

Option 2.

Part A.

a) substances forming a series based on one metal

B) substances forming a series based on one non-metal

B) substances forming a series based on a metal or non-metal

D) substances from different classes of substances related by transformations

3 (PO 4 ) 2

A) Ca b) CaO c) CO 2 d) H 2 O

In the transformation scheme: MgCl 2 2 b) MgSO 4 and Mg(OH) 2

The final product in the chain of transformations based on carbon compounds:

CO 2 → X 1 → X 2 → NaOH

Element “E” participating in the chain of transformations:

A)N b) S c)P d)Mg

Part B.

Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

NaOH+ CO 2 A) NaOH + H 2
NaOH +CO 2 B) Na 2 CO 3 + H 2 O
Na + H 2 O B) NaHCO 3
NaOH + HCl D) NaCl + H 2 O

b) oxygen

c) sodium chloride (solution)

d) calcium oxide

e) sulfuric acid

Part C.

Implement the scheme for the transformation of substances:

IV. Summing up the lesson.

D/z: §25, exercise 3, 7*

Testing on the topic"Genetic relationship between classes of inorganic and organic substances"

Option 1.

Part A. (Tasks with one correct answer)

The genetic series of a metal is:

a) substances forming a series based on one metal

B) substances forming a series based on one non-metal

B) substances forming a series based on a metal or non-metal

D) substances from different classes of substances related by transformations

Identify substance “X” from the transformation scheme: C → X → CaCO 3

A) CO 2 b) CO c) CaO d) O 2

Identify substance “Y” from the transformation scheme: Na → Y→NaOH

A) Na 2 O b) Na 2 O 2 c) H 2 O d) Na

In the transformation scheme: CuCl 2 → A → B→ Cu the formulas of intermediate products A and B are: a) CuO and Cu(OH) 2 b) CuSO 4 and Cu(OH) 2

B) CuCO 3 and Cu(OH) 2 g) Cu(OH) 2 and CuO

The final product in the chain of transformations based on carbon compounds:

CO 2 → X 1 → X 2 → NaOH

A) sodium carbonate b) sodium bicarbonate

C) sodium carbide d) sodium acetate

Element “E” participating in the chain of transformations:

E → E 2 O 5 → H 3 EO 4 → Na 3 EO 4

A)N b) Mn c)P d)Cl

Part B. (Tasks with 2 or more correct answer options)

Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

1)Fe + Cl 2 A) FeCl 2

2)Fe + HCl B) FeCl 3

3)FeO + HCl B) FeCl 2 + H 2

4) Fe 2 O 3 + HCl D) FeCl 3 + H 2

D) FeCl 2 + H 2 O

E) FeCl 3 + H 2 O

A solution of copper (II) sulfate reacts:

a) potassium hydroxide (solution)

b) iron

c) barium nitrate (solution)

d) aluminum oxide

e) carbon monoxide (II)

e) sodium phosphate (solution)

Part C. (With a detailed answer)

Implement the scheme for the transformation of substances:

FeS →SO 2 → SO 3 → H 2 SO 4 → MgSO 4 → BaSO 4

Testing on the topic"Genetic relationship between classes of inorganic and organic substances"

Option 2.

Part A. (Tasks with one correct answer)

The genetic series of a non-metal is:

a) substances forming a series based on one metal

B) substances forming a series based on one non-metal

B) substances forming a series based on a metal or non-metal

D) substances from different classes of substances related by transformations

Identify substance “X” from the transformation diagram: P → X → Ca 3 (PO 4 ) 2

A) P 2 O 5 b) P 2 O 3 c) CaO d) O 2

Determine substance “Y” from the transformation scheme: Ca → Y→Ca(OH) 2

A) Ca b) CaO c) CO 2 d) H 2 O

In the transformation scheme: MgCl 2 → A → B→ Mg the formulas of intermediate products A and B are: a) MgO and Mg(OH) 2 b) MgSO 4 and Mg(OH) 2

B) MgCO 3 and Mg(OH) 2 g) Mg(OH) 2 and MgO

The final product in the chain of transformations based on carbon compounds:

CO 2 → X 1 → X 2 → NaOH

A) sodium carbonate b) sodium bicarbonate

C) sodium carbide d) sodium acetate

Element “E” participating in the chain of transformations:

E → EO 2 → EO 3 → N 2 EO 4 → Na 2 EO 4

A)N b) S c)P d)Mg

Part B. (Tasks with 2 or more correct answer options)

Establish a correspondence between the formulas of the starting substances and the reaction products:

Formulas of starting substances Formulas of products

1) NaOH + CO 2 A) NaOH + H 2

2) NaOH + CO 2 B) Na 2 CO 3 + H 2 O

3) Na + H 2 O B) NaHCO 3

4) NaOH + HCl D) NaCl + H 2 O

2. Hydrochloric acid does not react:

a) sodium hydroxide (solution)

b) oxygen

c) sodium chloride (solution)

d) calcium oxide

e) potassium permanganate (crystalline)

e) sulfuric acid

Part C. (With a detailed answer)

Implement the scheme for the transformation of substances:

CuS →SO 2 → SO 3 → H 2 SO 4 → CaSO 4 → BaSO 4

Lesson plan:

Definition of concepts: “genetic connection”, “genetic series of an element”
Genetic series of metal.
Genetic series of a nonmetal.
Genetic relationship of organic substances.
Consolidation of knowledge(testing in the form of the Unified State Exam)

Lesson plan:

Definition of concepts: “genetic connection”, “genetic series of an element”
Genetic series of metal.
Genetic series of a nonmetal.
Genetic relationship of organic substances.
Consolidation of knowledge(testing in the form of the Unified State Exam)

Lesson plan:

Definition of concepts: “genetic connection”, “genetic series of an element”
Genetic series of metal.
Genetic series of a nonmetal.
Genetic relationship of organic substances.
Consolidation of knowledge(testing in the form of the Unified State Exam)

Lesson plan:

Definition of concepts: “genetic connection”, “genetic series of an element”
Genetic series of metal.
Genetic series of a nonmetal.
Genetic relationship of organic substances.
Consolidation of knowledge(testing in the form of the Unified State Exam)

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Lesson topic: “Genetic relationship between classes of inorganic compounds” Municipal educational institution secondary school No. 1 Chemistry teacher: Fadeeva O.S. Grachevka village, Stavropol Territory, 2011.

Lesson topic: “Genetic relationships between classes of inorganic compounds”

Lesson work plan: 1. Definition of the concepts “genetic connection”!, “genetic series of an element” 2. Genetic series of a metal 3. Genetic series of a non-metal 4. Genetic connection of organic substances 5. Consolidation of knowledge (Unified State Exam testing)

Genetic connection is the connection between substances of different classes, based on their mutual transformations and reflecting the unity of their origin.

What does the term “genetic link” mean? 1. Conversion of substances of one class of compound into substances of other classes; 2. Chemical properties of substances; 3. Possibility of obtaining complex substances from simple ones; 4. The relationship between simple and complex substances of all classes of inorganic compounds.

Genetic refers to a number of substances, representatives of different classes of substances, which are compounds of one chemical element, connected by mutual transformations and reflecting the common origin of these substances.

Signs that characterize the genetic series: Substances of different classes; Different substances formed by one chemical element, i.e. represent different forms of existence of one element; Different substances of the same chemical element are related by mutual transformations.

Genetic series of copper

Genetic series of phosphorus

Testing on the topic “Genetic relationship between classes of inorganic and organic substances” Option 1. Part A. (Tasks with one correct answer) 1. The genetic series of a metal is: a) substances forming a series based on one metal b) substances forming series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances related by transformations 2. Identify substance “X” from the transformation scheme: C → X → CaCO 3 a) CO 2 b) CO c) CaO d) O 2 3. Determine the substance “Y” from the transformation scheme: Na → Y → NaOH a) Na 2 O b) Na 2 O 2 c) H 2 O d) Na 4. In the transformation scheme: CuCl 2 → A → B → Cu the formulas of intermediate products A and B are: a) CuO and Cu (OH) 2 b) CuSO 4 and Cu (OH) 2 c) CuCO 3 and Cu (OH) 2 d) Cu (OH ) 2 and CuO 5. The final product in the chain of transformations based on carbon compounds: CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium hydrogen carbonate c) sodium carbide d) sodium acetate 6. Element “E” involved in the chain of transformations: E → E 2 O 5 → H 3 EO 4 → Na 3 E O 4 a) N b) Mn c) P d) Cl

Part B. (Tasks with 2 or more correct answer options) Establish a correspondence between the formulas of the starting substances and the reaction products: Formulas of the starting substances Formulas of the products 1) Fe + Cl 2 A) FeCl 2 2) Fe + HCl B) FeCl 3 3) FeO + HCl B) FeCl 2 + H 2 4) Fe 2 O 3 + HCl D) FeCl 3 + H 2 E) FeCl 2 + H 2 O E) FeCl 3 + H 2 O 2. A solution of copper (II) sulfate reacts : a) potassium hydroxide (solution) b) iron c) barium nitrate (solution) d) aluminum oxide e) carbon monoxide (II) f) sodium phosphate (solution) Part C. (With a detailed answer) Carry out the scheme for the transformation of substances: Fe S →SO 2 → SO 3 → H 2 SO 4 → MgSO 4 → BaSO 4

Testing on the topic “Genetic relationship between classes of inorganic and organic substances” Option 2. Part A. (Tasks with one correct answer) 1. The genetic series of a non-metal is: a) substances forming a series based on one metal b) substances forming series based on one non-metal c) substances forming a series based on a metal or non-metal d) substances from different classes of substances related by transformations 2. Identify substance “X” from the transformation scheme: P → X → Ca 3(PO 4)2 a) P 2 O 5 b) P 2 O 3 c) CaO d) O 2 3. Determine the substance “Y” from the transformation scheme: Ca → Y → Ca (OH) 2 a) Ca b) CaO c) CO 2 d) H 2 O 4. In the transformation scheme: MgCl 2 → A → B → Mg, the formulas of intermediate products A and B are: a) MgO and Mg (OH) 2 b) MgSO 4 and Mg (OH) 2 c) MgCO 3 and Mg ( OH) 2 d) Mg (OH) 2 and MgO 5. The final product in the chain of transformations based on carbon compounds: CO 2 → X 1 → X 2 → NaOH a) sodium carbonate b) sodium hydrogen carbonate c) sodium carbide d) sodium acetate 6. Element “E” participating in the chain of transformations: E → EO 2 → EO 3 → H 2 EO 4 → Na 2 EO 4 a) N b) S c) P d) Mg

Part B. (Tasks with 2 or more correct answer options) 1. Establish a correspondence between the formulas of the starting substances and the reaction products: Formulas of the starting substances Formulas of the products 1) NaOH + CO 2 A) NaOH + H 2 2) NaOH + CO 2 B ) Na 2 CO 2 + H 2 O 3) Na + H 2 O B) NaHCO 3 4) NaOH + HCl D) NaCl + H 2 O 2. Hydrochloric acid does not interact with: a) sodium hydroxide (solution) b) oxygen c ) sodium chloride (solution) d) calcium oxide e) potassium permanganate (crystalline) f) sulfuric acid Part C. (With a detailed answer) 1. Implement the transformation scheme of substances: CuS →SO 2 → SO 3 → H 2 SO 4 → CaSO 4 → BaSO 4

Homework textbook § 25, exercises 3,7