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

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The relationship between classes of substances is expressed by genetic chains

The genetic series is the implementation of chemical transformations, as a result of which substances of one class can be obtained from substances of another class.
To carry out genetic transformations, you need to know:
classes of substances;
nomenclature of substances;
properties of substances;
types of reactions;
nominal reactions, for example the Wurtz synthesis:

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What reactions must be carried out to obtain another from one type of hydrocarbon?
The arrows in the diagram indicate hydrocarbons that can be directly converted into each other by one reaction.

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Carry out several chains of transformations

Determine the type of each reaction:

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Checking

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Distribute substances into classes:

C3H6; CH3COOH; CH3OH; C2H4; UNSC; CH4; C2H6; C2H5OH; NSSON; C3H8; CH3COOC2H5; CH3SON; CH3COOCH3;

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Examination

Alkanes: CH4; C2H6; С3Н8
Alkenes: C3H6; С2Н4
Alcohols: CH3OH; C2H5OH
Aldehydes: НСО; CH3SON
Carboxylic acids: CH3COOH; UNDC
Esters: CH3COOC2H5; CH3COOCH3

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How can it be obtained from hydrocarbons:
a) alcohols b) aldehydes c) acids?

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Carbon Journey

C CaC2 C2H2 CH3CHO C2H5OH
CH3COOH CH3COOCH2CH3

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2C + Ca CaC2
CaC2 + 2H2O C2H2 + Ca(OH)2
C2H2 + H2O CH3CHO
CH3CHO + H2 C2H5OH
CH3CHO + O2 CH3COOH
CH3COOH + CH3CH2OH CH3COOC2H5

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For oxygen containing compounds

draw up reaction equations, indicate the conditions for the occurrence and type of reactions.

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Obtaining an ester from a hydrocarbon

C2H6 C2H5ClC2H5OH CH3CHO CH3COOH CH3COOCH2CH3

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Conclusion: Today in class - using the example of a genetic connection organic matter different homologous series We saw and proved with the help of transformations the unity of the material world.

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butane butene-1 1,2-dibromobutane butene-1
pentene-1 pentane 2-chloropentane
penten-2 CO2
Carry out transformations.

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Abstract

What is nano?�

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Demonstration of a video clip.

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What is nano?�

New technologies are what moves humanity forward on its path to progress.�

The goals and objectives of this work are to expand and improve students’ knowledge about the world around them, new achievements and discoveries. Formation of comparison and generalization skills. The ability to highlight the main thing, develop creative interest, cultivate independence in searching for material.

The beginning of the 21st century is marked by nanotechnology, which combines biology, chemistry, IT, and physics.

IN last years the pace of scientific and technological progress began to depend on the use of artificially created nanometer-sized objects. Substances and objects with a size of 1–100 nm created on their basis are called nanomaterials, and the methods of their production and use are called nanotechnologies. With the naked eye, a person can see an object with a diameter of approximately 10 thousand nanometers.

In its broadest sense, nanotechnology is research and development at the atomic, molecular and macromolecular level on a size scale of one to one hundred nanometers; the creation and use of artificial structures, devices and systems that, due to their ultra-small sizes, have significantly new properties and functions; manipulation of matter on the atomic distance scale.

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Technologies determine the quality of life of each of us and the power of the state in which we live.

The Industrial Revolution, which began in the textile industry, spurred the development of railway communication technologies.

Subsequently, the growth of transportation of various goods became impossible without new automotive technologies. Thus, each new technology causes the birth and development of related technologies.

The current period of time in which we live is called the scientific and technological revolution or information revolution. The beginning of the information revolution coincided with the development of computer technologies, without which life modern society no longer seems.

The development of computer technology has always been associated with the miniaturization of electronic circuit elements. Currently, the size of one logical element (transistor) of a computer circuit is about 10-7 m, and scientists believe that further miniaturization of computer elements is possible only when special technologies called “nanotechnology” are developed.

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Translated from Greek, the word “nano” means dwarf, gnome. One nanometer (nm) is one billionth of a meter (10-9 m). A nanometer is very small. A nanometer is the same number of times less than one meter as the thickness of a finger is less than the diameter of the Earth. Most atoms have a diameter of 0.1 to 0.2 nm, and the thickness of the DNA strands is about 2 nm. The diameter of red blood cells is 7000 nm, and the thickness of a human hair is 80,000 nm.

The figure shows a variety of objects from left to right in order of increasing size - from an atom to solar system. Man has already learned to benefit from the most different sizes. We can split the nuclei of atoms to produce atomic energy. By carrying out chemical reactions, we obtain new molecules and substances that have unique properties. With the help of special tools, man has learned to create objects - from a pinhead to huge structures that are visible even from space.

But if you look at the figure carefully, you will notice that there is a fairly large range (on a logarithmic scale) where for a long time scientists have never set foot before - between a hundred nanometers and 0.1 nm. Nanotechnology will have to work with objects ranging in size from 0.1 nm to 100 nm. And there is every reason to believe that we can make the nanoworld work for us.

Nanotechnologies use the latest achievements of chemistry, physics, and biology.

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Recent studies have proven that in Ancient Egypt Nanotechnology was used to dye hair black. For this purpose, a paste of lime Ca(OH)2, lead oxide and water was used. During the dyeing process, nanoparticles of lead sulfide (galena) were obtained as a result of interaction with sulfur, which is part of keratin, which ensured uniform and stable dyeing

The British Museum houses the "Lycurgus Cup" (the walls of the cup depict scenes from the life of this great Spartan legislator), made by ancient Roman craftsmen - it contains microscopic particles of gold and silver added to the glass. Under different lighting, the cup changes color - from dark red to light golden. Similar technologies were used to create stained glass windows in medieval European cathedrals.

Currently, scientists have proven that the sizes of these particles are from 50 to 100 nm.

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In 1661, the Irish chemist Robert Boyle published an article in which he criticized Aristotle's assertion that everything on Earth consists of four elements - water, earth, fire and air (the philosophical basis of the foundations of the then alchemy, chemistry and physics). Boyle argued that everything consists of “corpuscles” - ultra-small parts that, in different combinations, form various substances and objects. Subsequently, the ideas of Democritus and Boyle were accepted by the scientific community.

In 1704, Isaac Newton suggested exploring the mystery of corpuscles;

In 1959, American physicist Richard Feynman said: “For now we are forced to use the atomic structures that nature offers us.” “But in principle a physicist could synthesize any substance according to a given chemical formula.”

In 1959, Norio Taniguchi first used the term “nanotechnology”;

In 1980, Eric Drexler used the term.

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Richard Phillips Feyman (1918-1988) outstanding American physicist. One of the creators of quantum electrodynamics. Winner of the Nobel Prize in Physics in 1965.

Feynman's famous lecture, known as "There's Still Plenty of Room Down There," is now considered the starting point in the struggle to conquer the nanoworld. It was first read at the California Institute of Technology in 1959. The word “below” in the title of the lecture meant in “a world of very small dimensions.”

Nanotechnology became a field of science in its own right and became a long-term technical project following detailed analysis by American scientist Eric Drexler in the early 1980s and the publication of his book Engines of Creation: The Coming Era of Nanotechnology.

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The first devices that made it possible to observe nanoobjects and move them were scanning probe microscopes - an atomic force microscope and a scanning tunnel microscope operating on a similar principle. Atomic force microscopy (AFM) was developed by Gerd Binnig and Heinrich Rohrer, who were awarded the Nobel Prize for this research in 1986.

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The basis of AFM is a probe, usually made of silicon and representing a thin cantilever plate (it is called a cantilever, from English word"cantilever" - console, beam). At the end of the cantilever there is a very sharp spike ending in a group of one or more atoms. The main material is silicon and silicon nitride.

When the microprobe moves along the surface of the sample, the tip of the spike rises and falls, outlining the microrelief of the surface, just as a gramophone stylus slides along a gramophone record. At the protruding end of the cantilever there is a mirror area onto which the laser beam falls and is reflected. When the spike lowers and rises on surface irregularities, the reflected beam is deflected, and this deviation is recorded by a photodetector, and the force with which the spike is attracted to nearby atoms is recorded by a piezoelectric sensor.

The photodetector and piezo sensor data are used in the feedback system. As a result, it is possible to construct a volumetric relief of the sample surface in real time.

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Another group of scanning probe microscopes uses the so-called quantum mechanical “tunnel effect” to construct surface relief. The essence of the tunnel effect is that electricity between a sharp metal needle and a surface located at a distance of about 1 nm begins to depend on this distance - the smaller the distance, the greater the current. If a voltage of 10 V is applied between the needle and the surface, then this “tunnel” current can range from 10 pA to 10 nA. By measuring this current and maintaining it constant, the distance between the needle and the surface can also be kept constant. This allows you to build a volumetric profile of the surface. Unlike an atomic force microscope, a scanning tunneling microscope can only study the surfaces of metals or semiconductors.

A scanning tunneling microscope can be used to move any atom to a point chosen by the operator. In this way, it is possible to manipulate atoms and create nanostructures, i.e. structures on the surface with dimensions on the order of a nanometer. Back in 1990, IBM employees showed that this was possible by combining the name of their company from 35 xenon atoms on a nickel plate.

A bevel differential adorns the home page of the Institute of Molecular Manufacturing website. Compiled by E. Drexler from the atoms of hydrogen, carbon, silicon, nitrogen, phosphorus, hydrogen and sulfur total number 8298. Computer calculations show that its existence and functioning do not contradict the laws of physics.

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Classes for lyceum students in the nanotechnology class of the Russian State Pedagogical University named after A.I. Herzen.

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Nanostructures can be assembled not only from individual atoms or single molecules, but also from molecular blocks. Such blocks or elements for creating nanostructures are graphene, carbon nanotubes and fullerenes.

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1985 Richard Smalley, Robert Curl and Harold Kroteau discovered fullerenes and were able to measure an object 1 nm in size for the first time.

Fullerenes are molecules consisting of 60 atoms arranged in the shape of a sphere. In 1996, a group of scientists was awarded the Nobel Prize.

Demonstration of a video clip.

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Aluminum with a small additive (no more than 1%) of fullerene acquires the hardness of steel.

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Graphene is a single, flat sheet of carbon atoms bonded together to form a lattice, each cell resembling a honeycomb. The distance between nearest carbon atoms in graphene is about 0.14 nm.

The light balls are carbon atoms, and the rods between them are the bonds that hold the atoms in the graphene sheet.

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Graphite, what regular pencil leads are made from, is a stack of sheets of graphene. The graphenes in graphite are very poorly bonded and can slide past each other. Therefore, if you run graphite over paper, the sheet of graphene in contact with it is separated from the graphite and remains on the paper. This explains why graphite can be used to write.

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Dendrimers are one of the paths into the nanoworld in the “bottom-up” direction.

Tree-like polymers are nanostructures ranging in size from 1 to 10 nm, formed by combining molecules with a branching structure. Dendrimer synthesis is one of the nanotechnologies that is closely related to polymer chemistry. Like all polymers, dendrimers are composed of monomers, and the molecules of these monomers have a branched structure.

Cavities filled with the substance in the presence of which the dendrimers were formed can form inside the dendrimer. If a dendrimer is synthesized in a solution containing any drug, then this dendrimer becomes a nanocapsule with this medicine. In addition, the cavities inside the dendrimer may contain radioactively labeled substances used to diagnose various diseases.

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In 13% of cases, people die from cancer. This disease kills about 8 million people worldwide every year. Many types cancer diseases are still considered incurable. Scientific research shows that nanotechnology can be a powerful tool in the fight against this disease. Dendrimers – capsules with poison for cancer cells

Cancer cells need to divide and grow large quantities folic acid. Therefore, folic acid molecules adhere very well to the surface of cancer cells, and if the outer shell of dendrimers contains folic acid molecules, then such dendrimers will selectively adhere only to cancer cells. With the help of such dendrimers it is possible cancer cells made visible if some other molecules are attached to the dendrimer shell, glowing, for example, under ultraviolet light. By attaching a drug that kills cancer cells to the outer shell of the dendrimer, it is possible not only to detect them, but also to kill them.

According to scientists, with the help of nanotechnology, it will be possible to embed microscopic sensors in human blood cells that warn of the appearance of the first signs of disease development.

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Quantum dots are already a convenient tool for biologists to see various structures inside living cells. The various cellular structures are equally transparent and uncolored. Therefore, if you look at a cell through a microscope, you will not see anything except its edges. To make certain cell structures visible, quantum dots of different sizes were created that can stick to specific intracellular structures.

The smallest ones, glowing green, were glued to molecules capable of sticking to the microtubules that make up the internal skeleton of the cell. Medium-sized quantum dots can stick to the membranes of the Golgi apparatus, and the largest ones can stick to the cell nucleus. The cell is dipped into a solution containing all these quantum dots and kept in it for some time, they penetrate inside and stick to wherever they can. After this, the cell is rinsed in a solution that does not contain quantum dots and under a microscope. Cellular structures became clearly visible.

Red – core; green – microtubules; yellow – Golgi apparatus.

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Titanium dioxide, TiO2, is the most common titanium compound on earth. Its powder has a dazzling White color and is therefore used as a dye in the production of paints, paper, toothpastes and plastics. The reason is a very high refractive index (n=2.7).

Titanium oxide TiO2 has very strong catalytic activity - it accelerates the occurrence of chemical reactions. In the presence of ultraviolet radiation, it splits water molecules into free radicals - hydroxyl groups OH- and superoxide anions O2- of such high activity that organic compounds decompose into carbon dioxide and water.

Catalytic activity increases with decreasing particle size. Therefore, they are used to purify water, air and various surfaces from organic compounds, which are usually harmful to humans.

Photocatalysts can be included in the concrete of highways, which will improve the environment around roads. In addition, it is proposed to add powder from these nanoparticles to automobile fuel, which should also reduce the content of harmful impurities in exhaust gases.

A film of titanium dioxide nanoparticles applied to glass is transparent and invisible to the eye. However, such glass, when exposed to sunlight, is capable of self-cleaning from organic contaminants, turning any organic dirt into carbon dioxide and water. Glass treated with titanium oxide nanoparticles is free of greasy stains and therefore is well wetted by water. As a result, such glass fogs up less, since water droplets immediately spread along the surface of the glass and form a thin transparent film.

Titanium dioxide stops working in enclosed spaces because... There is practically no ultraviolet in artificial light. However, scientists believe that by slightly changing its structure, it will be possible to make it sensitive to the visible part of the solar spectrum. Based on such nanoparticles, it will be possible to make a coating, for example, for toilets, as a result of which the content of bacteria and other organics on toilet surfaces can be reduced several times.

Due to its ability to absorb ultraviolet radiation, titanium dioxide is already used in the manufacture of sunscreens, such as creams. Cream manufacturers have begun to use it in the form of nanoparticles, which are so small that they provide almost absolute transparency to the sunscreen.

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Self-cleaning nanograss and “lotus effect”

Nanotechnology makes it possible to create a surface similar to a massage microbrush. Such a surface is called nanograss, and it consists of many parallel nanowires (nanorods) of the same length, located at an equal distance from each other.

A drop of water falling on the nanograss cannot penetrate between the nanograss, since this is prevented by the high surface tension of the liquid.

To make the wettability of the nanograss even less, its surface is covered with a thin layer of some hydrophobic polymer. And then not only water, but also any particles will never stick to the nanograss, because touch it only at a few points. Therefore, dirt particles that find themselves on a surface covered with nanovilli either fall off it themselves or are carried away by rolling drops of water.

The self-cleaning of a fleecy surface from dirt particles is called the “lotus effect”, because Lotus flowers and leaves are pure even when the water around is cloudy and dirty. This happens due to the fact that the leaves and flowers are not wetted by water, so drops of water roll off them like balls of mercury, leaving no trace and washing away all the dirt. Even drops of glue and honey cannot stay on the surface of the lotus leaves.

It turned out that the entire surface of lotus leaves is densely covered with micropimples about 10 microns in height, and the pimples themselves, in turn, are covered with even smaller microvilli. Research has shown that all these micropimples and villi are made of wax, which is known to have hydrophobic properties, making the surface of lotus leaves look like nanograss. It is the pimpled structure of the surface of lotus leaves that significantly reduces their wettability. For comparison: the relatively smooth surface of a magnolia leaf, which does not have the ability to self-clean.

Thus, nanotechnology makes it possible to create self-cleaning coatings and materials that also have water-repellent properties. Materials made from such fabrics always remain clean. Self-cleaning windshields are already being produced, the outer surface of which is covered with nanovilli. There is nothing for wipers to do on such glass. There are permanently clean rims for car wheels on sale that self-clean using the “lotus effect”, and now you can paint the outside of your house with paint to which dirt will not stick.

From polyester coated with many tiny silicon fibers, Swiss scientists have managed to create a waterproof material.

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Nanowires are wires with a diameter on the order of a nanometer, made of a metal, semiconductor or dielectric. The length of nanowires can often exceed their diameter by 1000 times or more. Therefore, nanowires are often called one-dimensional structures, and their extremely small diameter (about 100 atomic sizes) makes it possible to manifest various quantum mechanical effects. Nanowires do not exist in nature.

The unique electrical and mechanical properties of nanowires create prerequisites for their use in future nanoelectronic and nanoelectromechanical devices, as well as as elements of new composite materials and biosensors.

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Unlike transistors, the miniaturization of batteries occurs very slowly. Size galvanic cells The power supply per unit power has decreased over the past 50 years by only 15 times, and the size of the transistor during the same time has decreased by more than 1000 times and is now about 100 nm. It is known that the size of an autonomous electronic circuit is often determined not by its electronic filling, but by the size of the current source. Moreover, the smarter the electronics of the device, the larger the battery it requires. Therefore, for further miniaturization of electronic devices, it is necessary to develop new types of batteries. And here again nanotechnology helps

In 2005, Toshiba created a prototype of a lithium-ion battery, the negative electrode of which was coated with lithium titanate nanocrystals, as a result of which the electrode area increased several tens of times. The new battery is capable of gaining 80% of its capacity in just one minute of charging, while conventional lithium-ion batteries charge at a rate of 2-3% per minute and take an hour to fully charge.

In addition to high recharging speed, batteries containing nanoparticle electrodes have an extended service life: after 1000 charge/discharge cycles, only 1% of its capacity is lost, and the total service life of new batteries is more than 5 thousand cycles. Moreover, these batteries can operate at temperatures down to -40°C, losing only 20% of their charge versus 100% for typical modern batteries already at -25°C.

Since 2007, batteries with electrodes made of conductive nanoparticles have been available for sale, which can be installed in electric vehicles. These lithium-ion batteries are capable of storing energy up to 35 kWh, charging to maximum capacity in just 10 minutes. Now the range of an electric car with such batteries is 200 km, but the next model of these batteries has already been developed, which allows increasing the range of an electric car to 400 km, which is almost comparable to the maximum range of gasoline cars (from refueling to refueling).

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In order for one substance to enter into a chemical reaction with another, certain conditions are necessary, and very often it is not possible to create such conditions. Therefore, a huge number of chemical reactions exist only on paper. To carry them out, catalysts are needed - substances that facilitate the reaction but do not participate in it.

Scientists have found that the inner surface of carbon nanotubes also has great catalytic activity. They believe that when a “graphite” sheet of carbon atoms is rolled up into a tube, the concentration of electrons on its inner surface gets smaller. This explains the ability of the inner surface of nanotubes to weaken, for example, the bond between the oxygen and carbon atoms in the CO molecule, becoming a catalyst for the oxidation of CO to CO2.

To combine the catalytic ability of carbon nanotubes and transition metals, nanoparticles from them were introduced inside the nanotubes (It turned out that this nanocomplex of catalysts is capable of launching a reaction that has only been dreamed of - the direct synthesis of ethyl alcohol from synthesis gas (a mixture of carbon monoxide and hydrogen) obtained from natural gas, coal and even biomass.

In fact, humanity has always tried to experiment with nanotechnology without even knowing it. We learned about this at the beginning of our acquaintance, heard the concept of nanotechnology, learned the history and names of the scientists who made it possible to make such a qualitative leap in the development of technology, got acquainted with the technologies themselves, and even heard the history of the discovery of fullerenes from the discoverer, Nobel Prize winner Richard Smalley.

Technologies determine the quality of life of each of us and the power of the state in which we live.

The further development of this direction depends on you.

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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) Selection of starting materials Construction of the carbon backbone of the molecule Introduction, removal or substitution functional group Group protection 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 – SOOT 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 ALCOHIDE WITH 2 H 5 OH SYNTHESIS BASED ON ETHYLENE DICHLOROETHANE ETHYLENE OXIDE ETHYLENE GLYCOL ACETALEHYDE ACETALEHYDE O 2, Ag K MnO4, 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 HCl, 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 VINYL CETYLENE 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 HEXACHLO RAS 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 (HEXMETHYLENETETERS) H) 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

74. Write equations and name the reaction products according to the scheme:

75. Write equations and name the reaction products according to the scheme:

76. Write equations and name the reaction products according to the scheme:

77. Write equations and name the reaction products according to the scheme:

78. Write equations and name the reaction products according to the scheme:

79. Write equations and name the reaction products according to the scheme:

80. Write equations and name the reaction products according to the scheme:

81. Write equations and name the reaction products according to the scheme:

82. Write equations and name the reaction products according to the scheme:

83. Write equations and name the reaction products according to the scheme:

84. Write equations and name the reaction products according to the scheme:

85. Write equations and name the reaction products according to the scheme:

86. Write equations and name the reaction products according to the scheme:

87. Write equations and name the reaction products according to the scheme:

88. Write equations and name the reaction products according to the scheme:

89. Write equations and name the reaction products according to the scheme:

90. Write equations and name the reaction products according to the scheme:

91. Write equations and name the reaction products according to the scheme:

92. Write equations and name the reaction products according to the scheme:

93. Write equations and name the reaction products according to the scheme:

94. Write equations and name the reaction products according to the scheme:

95. Write equations and name the reaction products according to the scheme:

96. Write equations and name the reaction products according to the scheme:

97. Write equations and name the reaction products according to the scheme:

98. Write equations and name the reaction products according to the scheme:

99. Write equations and name the reaction products according to the scheme:

100. Write equations and name the reaction products according to the scheme:

101. Write equations and name the reaction products according to the scheme:

Module 2. Heterocyclic and natural compounds

Five-membered heterocyclic compounds

1. Write diagrams and name the products of reactions of aziridine with the following reagents: a) H 2 O (t); b) NH 3 (t); c) HC1 (t).

2. Give a reaction scheme for the extraction of oxirane. Write the equations and name the products of the reactions of oxirane: a) with H 2 O, H +; b) with C 2 H 5 OH, H +; c) with CH 3 NH 2.

3. Give schemes for the mutual transformations of five-membered heterocycles with one heteroatom (Yur’ev’s reaction cycle).

4. What is acidophobia? Which heterocyclic compounds are acidophobic? Write the reaction schemes for the sulfonation of pyrrole, thiophene, and indole. Name the products.

5. Give diagrams and name the products of the reactions of halogenation and nitration of pyrrole and thiophene.

6. Give diagrams and name the end products of the oxidation and reduction reactions of furans and pyrrole.

7. Give a reaction scheme for the extraction of indole from N-formyl o toluidine. Write the equations for the reactions of nitration and sulfonation of indole. Name the products.

8. Give a reaction scheme for the production of 2-methylindole from phenylhydrazine using the Fischer method. Write the equations and name the products of the reactions of 2-methyl-indole: a) with KOH; b) with CH 3 I.

9. Give and name the tautomeric forms of indoxyl. Write a reaction diagram for the extraction of indigo blue from indoxyl.

10. Give diagrams and name the products of the reduction and oxidation reactions of indigo blue.

11. Write diagrams and name the reaction products of 2-aminothiazole: a) with HC1; a) with (CH 3 CO) 2 O; c) with CH 3 I.

12. What type of tautomerism is characteristic of azoles, what is it caused by? Give the tautomeric forms of pyrazole and imidazole.

13. Give a scheme for the synthesis of imidazole from glyoxal. Confirm the amphoteric nature of imidazole with the corresponding reaction schemes. Name the reaction products.

14. Give reaction schemes confirming the amphoteric nature of pyrazole, benzimidazole, nicotinic (3-pyridinecarboxylic) acid, anthranilic (2-aminobenzoic) acid.

15. Write a scheme for the synthesis of 3-methylpyrazolone-5 from acetoacetic ester and hydrazine. Give and name three tautomeric forms of pyrazolone-5.

16. Write a scheme for the synthesis of antipyrine from acetoacetic ester. Give a diagram and name the product of a qualitative reaction to antipyrine.

17. Write a scheme for the synthesis of amidopyrine from antipyrine. Specify the qualitative reaction to amidopyrine.

Six-membered heterocyclic compounds

18. Write diagrams and name the reaction products that confirm the basic properties of pyridine and the amphoteric properties of imidazole.

19. Draw and name the tautomeric forms of 2-hydroxypyridine. Write the equations and name the reaction products of 2-hydroxypyridine: a) with PCl 5 ; b) with CH 3 I.

20. Draw and name the tautomeric forms of 2-aminopyridine. Write an equation and name the products of the reaction of 2-aminopyridine and 3-aminopyridine with hydrochloric acid.

21. Give diagrams and name the reaction products that confirm the presence of a primary aromatic amino group in b-aminopyridine.

22. Give a scheme for the synthesis of quinoline using Scroup’s method. Name the intermediate compounds.

23. Give a scheme for the synthesis of 7-methylquinoline using the Scroup method. Name all intermediate compounds.

24. Give a scheme for the synthesis of 8-hydroxyquinoline using the Scroup method. Name the intermediate compounds. Use chemical reactions to confirm the amphoteric nature of the final product.

25. Give diagrams and name the products of the reactions of sulfonation, nitration and oxidation of quinoline.

26. Write diagrams and name the products of reactions of quinoline: a) with CH 3 I; b) with CON; c) with c. HNO 3, c. H 2 SO 4; d) with HC1.

27. Give diagrams and name the products of the nitration reactions of indole, pyridinine and quinoline.

28. Give diagrams and name the products of reactions of isoquinoline: a) with CH 3 I; b) with NaNH 2, NH 3; c) with Br 2, FeBr 3.

29. Give a scheme for the synthesis of acridine from N-phenylanthranilic acid according to the Rubtsov-Magidson-Grigorovsky method.

30. Give a reaction scheme for the production of 9-aminoacridine from acridine. Write equations and name the products of the interaction of 9-aminoacridine a) with HCI; b) with (CH 3 CO) 2 O.

31. Give reaction schemes for the oxidation and reduction of quinoline, isoquinoline and acridine. Name the final products.

32. Write the equations and name the products of the reaction g-Pyrone with conc. hydrochloric acid. Give the formulas of natural compounds whose structure includes the g-Pyron and a-Pyron cycles.

33. Write diagrams and name the products of reactions of pyridine: a) with HCI; b) with NaNH 2, NH 3; c) with CON.

34. Write diagrams and name the reaction products of 4-aminopyrimidine: a) with sup. NCI; b) with NaNH 2, NH 3; c) with Br 2) FeBr 3.

35. Give a scheme for the synthesis of barbituric acid from malonic ester and urea. What causes the acidic nature of barbituric acid? Support your answer with diagrams of the corresponding reactions.

36. Give a diagram of tautomeric transformations and name the tautomeric forms of barbituric acid. Write the equation for the reaction of barbituric acid with an aqueous alkali solution.

37. Give a reaction scheme for the production of 5.5-diethylbarbituric acid from malonic ester. Write the equations and name the product of the interaction of the named acid with an alkali (aq. solution).

38. Give diagrams, indicate the type of tautomerism and give names to the tautomeric forms of nucleic bases of the pyrimidine group.

39. Write a diagram of the interaction of uric acid with alkali. Why uric acid two-basic, not three-basic?

40. Give equations for the qualitative reaction to uric acid. Name the intermediate and final products.

41. Write a diagram of tautomeric equilibrium and name the tautomeric forms of xanthine. Give equations and name the reaction products that confirm the amphoteric nature of xanthine.

42. Give diagrams, indicate the type of tautomerism and give names to the tautomeric forms of nucleic bases of the purine group.

43. Which of the following compounds is characterized by lactam-lactim tautomerism: a) hypoxanthine; b) caffeine; c) uric acid? Give diagrams of the corresponding tautomeric transformations.

Natural connections

44. Write diagrams and name the products of the reactions of menthol: a) with HCI; b) with Na; c) with isovaleric (3-methylbutanoic) acid in the presence of K. H 2 SO. Name menthol according to IUPAC nomenclature.

45. Give schemes of sequential reactions for the production of camphor from a-pinene. Write reaction equations confirming the presence of a carbonyl group in the structure of camphor. Name the products.

46. Give diagrams and name the gyroproducts of the interaction of camphor: a) with Br 2; b) with NH 2 OH; c) with H 2, Ni.

47. Give a reaction scheme for the extraction of camphor from bornyl acetate. Write a reaction equation that confirms the presence of a carbonyl group in the structure of camphor.

48. What compounds are called epimers? Using D-glucose as an example, explain the phenomenon of epimerization. Give the projection formula of hexose, epimeric D-glucose.

49. What phenomenon is called mutarotation? Give a diagram of cyclo-chain tautomeric transformations of b-D-glucopyranoses into aqueous solution. Name all forms of monosaccharides.

50. Give a diagram of the cyclo-chain tautomeric transformation of D-galactose in an aqueous solution. Name all forms of monosaccharide.

51. Give a diagram of the cyclo-chain tautomeric transformation of D-mannose in an aqueous solution. Name all forms of monosaccharide.

52. Give a diagram of the cyclo-chain tautomeric transformation of a-D-fructofuranose (water solution). Name all forms of monosaccharides.

53. Write diagrams of sequential reactions for the formation of fructose in ozazone. Do other monoses form the same osazone?

54. Give reaction schemes that prove the presence of the following in a glucose molecule: a) five hydroxyl groups; b) naivacetal hydroxyl; c) aldehyde group. Name the reaction products.

55. Write the reaction schemes for fructose with the following reagents: a) HCN; b) C 2 H 5 OH, H +; c) over CH 3 I; r) Ag(NH 3) 2 OH. Name the compounds obtained.

56. Write reaction schemes for the transformation of D-glucose: a) into methyl-b-D-glucopyranoside; b) into pentaacetyl-b-D-glucopyranose.

57. Give the formula and give the chemical name of the disaccharide, which upon hydrolysis will give glucose and galactose. Write reaction schemes for its hydrolysis and oxidation.

58. What are reducing and non-reducing sugars? Of the disaccharides - maltose or sucrose, will it react with Tollens' reagent (ammonium argentum oxide solution)? Give the formulas of these disaccharides, give them names according to IUPAC nomenclature, write a reaction scheme. What disaccharides can be found in a- and b-forms?

59. What carbohydrates are called disaccharides? What are reducing but non-reducing sugars? Do maltose, lactose and sucrose react with Tollens' reagent (ammonium argentum oxide solution)? Give reaction equations and name the indicated disaccharides according to IUPAC nomenclature.

60. Write diagrams of sequential reactions of obtaining ascorbic acid from D-glucose. Indicate the acidic center in the vitamin C molecule.

61. Write reaction schemes for the preparation of: a) 4-O-a-D-glucopyranoside-D-glucopyranose; b) a-D-glucopyranoside-b-D-fructofuranoside. Name the original monosaccharides. What type of disaccharides does each of a) and b) belong to?

62. Give a reaction scheme that allows you to distinguish sucrose from maltose. Give names according to the IUPAC nomenclature for these disaccharides, give schemes for their hydrolysis.

63. Give a scheme for the synthesis of methyl-b-D-galactopyranoside from D-galactose and its acid hydrolysis.

Related information.

The structure of molecules of organic compounds allows us to draw conclusions about the chemical properties of substances and the close relationship between them. From substances of one class, compounds of other classes are obtained through successive transformations. Moreover, all organic substances can be represented as derivatives of the simplest compounds - hydrocarbons. The genetic relationship of organic compounds can be represented as a diagram:

C 2 H 6 → C 2 H 5 Br → C 2 H 5 OH → CH 3 -SON → CH 3 COOH →

CH 3 SOOS 3 H 7; and etc.

According to the diagram, it is necessary to draw up equations for the chemical transformations of one substance into another. They confirm the interrelation of all organic compounds, the complication of the composition of matter, the development of the nature of substances from simple to complex.

The composition of organic substances most often includes a small number of chemical elements: hydrogen, carbon, oxygen, nitrogen, sulfur, chlorine and other halogens. The organic substance methane can be synthesized from two simple inorganic substances - carbon and hydrogen.

C + 2H 2 = CH 4 + Q

This is one example of the fact that between all substances of nature - inorganic and organic - there is unity and a genetic connection, which are manifested in the mutual transformations of substances.

Part 2. Complete the practical task.

The task is experimental.

Prove that potatoes contain starch.

To prove the presence of starch in potatoes, you need to apply a drop of iodine solution to a cut potato. The potato cut will turn blue-violet. The reaction with iodine solution is a qualitative reaction to starch.

E T A L O N

to option 25

Number of options(packages) of tasks for examinees:

Option No. 25 from 25 options

Job completion time:

Option No. 25 45 min.

Conditions for completing tasks

Occupational safety requirements: teacher (expert) supervising assignments(safety instructions when working with reagents)

Equipment: paper, ballpoint pen, laboratory equipment

Literature for examinees reference, methodological and tables

1. Familiarize yourself with the test taker assignments, skills, knowledge, and assessment metrics being assessed. .

Option No. 25 of 25

Part 1. Answer theoretical questions:

1. Aluminum. Amphotericity of aluminum. Aluminum oxides and hydroxides.

2. Proteins are natural polymers. Structure and structure of proteins. Qualitative reactions and applications.

Part 2: Do the practice task

3. The task is experimental.

How to experimentally obtain oxygen in the laboratory, prove its presence.

Option 25 out of 25.

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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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 the 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. What do you think, in organic chemistry is the concept of genetic connection and genetic series used? 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.

Carry out the scheme of 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 the right options answer)

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)

Carry out the scheme of 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 interact:

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)

Carry out the scheme of 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