The guidelines are intended for students specializing in the field of: technology of food products based on raw materials of plant origin; environmental protection. The methodology for organizing students' independent work is outlined. A list of theoretical material for the organic chemistry course and basic concepts necessary for successful mastery of the program are presented. Theoretical questions are proposed on each topic of the course, upon completion of which students will gain practical problem-solving skills. The methodological instructions are structured taking into account the strengthening of the role of students’ independent work.
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These guidelines are compiled in accordance with the Federal Agency for Education of the Russian Federation GOSVPO organic chemistry program for technological specialties. State educational institution Guidelines are intended for students of higher professional education specializing in the field of: East Siberian State Technological Technology of food products based on raw materials of plant origin; (GOU VPO VSTU) environmental protection. The methodology for organizing students' independent work is outlined. A list of theoretical material for the organic chemistry course and basic concepts necessary for successful mastery of the program are presented. Theoretical questions are proposed on each topic of the course, upon completion of which students will gain practical problem-solving skills. The methodological instructions are structured taking into account the strengthening of the role of students' independent work. in organic chemistry for students of specialties: 260100 - food technology. 260201 - technology of grain storage and processing 260202 - technology of bread, pasta and confectionery 280201 - environmental protection and rational use of natural resources. 80202 - engineering environmental protection. Compiled by: Zolotareva A.M. Ulan-Ude, 2006 2 4 Independent work of students………………….…………..….17 CONTENTS 4.1 Types of control……………………………………………………………… ..……..17 4.2 Organization of control……………………………...………….18 5 Examples and tasks of seminar classes……….….….…… …18 Preface…………………………………………..………………..…....3 5.1 Alkanes………………………………… ……….……...……………..18 1 Subject learning chains………………………….......………........3 5.2 Alkenes . Alkadienes…………………………..…….……….……….19 2 Contents of discipline sections……………….………...4 5.3 Alkynes……………………………………………………………………..20 2.1 Introduction………………………………………………………………..20 ………….……..4 5.4 Aromatic hydrocarbons………………………………………………………20 2.2 Theoretical principles and general issues of organic chemistry..4 5.5 Halogen derivatives…………………… .…………………………...21 2.3 Classes of organic compounds……………………………...4 5.6 Heterocycles……………………………… ………………………….22 2.3.1 Hydrocarbons…………………………………………..………….4 5.7 Alcohols and phenols……………… …………………………………….22 2.3.2 Hydrocarbon derivatives……………………………………..5 5.8 Aldehydes and ketones…………………… ……………………………24 2.3.3 Oxygen-containing organic compounds……………….6 5.9 Carboxylic acids………………………………………………………………... 24 2.3.4 Nitrogen-containing organic compounds………………………. .7 5.10 Nitrogen-containing substances. Amines……………………….……………..25 2.3.5 Heterocyclic compounds…………………………………..7 5.11 Hydroxy acids……… ………………………………26 2.4 Compounds with mixed functional groups…….…7 5.12 Amino acids……………………………………………………….26 2.5 Bioorganic compounds…………………………….…………..8 5.13 Proteins……………………………………………………….……………….. ..27 3 Laboratory exercises……………………………………..…………..9 5.14 Carbohydrates……………………………………………………….…. ……………….27 3.1 Introduction to organic chemistry………………………..………………...9 6 Recommended reading……………………….……………… …….28 3.2 Methods for isolation, purification and separation of organic 6.1 Additional literature…………………………….……..…..28 compounds…………………………………………………… …………………………10 3.3 Determination of the basic physical properties of organic compounds…………………..……………………………………..10 3.4 General ideas about elemental analysis of chemical substances………………………………………………………..………………...10 3.5 Hydrocarbons………………………………………………………. …..……………11 3.6 Halogen derivatives of hydrocarbons……………………………..11 3.7 Oxycompounds……………………………………………………………...... .…………11 3.8 Oxo compounds……………………………………..……………12 3.9 Carboxylic acids……………………………………………………… ………..12 3.10 Nitrogen-containing organic compounds. Nitro compounds, amines…………………………………………………………….………....12 3.11 Heterocycles…………………………………………………… ……………………………..13 3.12 Compounds with mixed functional groups…...…..13 3.13 Carbohydrates………………………………………….……… ..…………13 3.14 Lipids……………………………………………………………..14 3.15 Protein substances…………………………… ……………..……….14 3.16 Synthesis of organic compounds…………………….…………….15 3.17 Identification of an unknown organic compound…..…....16 3 predict reactivity organic molecules from the standpoint of modern electronic concepts; identify and analyze organic compounds using chemical, physicochemical and physical methods Organic chemistry studies hydrocarbon compounds and their studies; derivatives with other elements and laws that are subject to the task of research; transformation of these substances. The special position of organic chemistry is to choose a research method. due to the fact that it is based on inorganic chemistry and is closely related to biology. These guidelines are compiled in accordance with The presented course of organic chemistry is one of the modern levels of development of organic chemistry. Particular attention to the most important disciplines of the natural science block. In connection with the general given to general patterns, the most important organic tendency of natural science disciplines to approach compounds. The course discusses in detail those organic “molecular level” compounds that represent an integral part of objects or increased requirements are presented to the course of organic chemistry, since the foundation of this “molecular production is the future specialty of students. level" is created by organic chemistry. To successfully study the organic chemistry course, independent student work is required in the organic chemistry course for students. When preparing for laboratory and specialties, much attention is paid to the consideration of chemistry and practical exercises, the student must first of all study organic substances from a modern point of view. program theoretical material: work through lecture notes with the Objective of the organic chemistry course is to formulate using recommended literature, give theoretical students correct ideas about the world around them, about the meaning of colloquia, perform home tests. and the role of organic substances in various industries. In order to organize independent work of students in Organic chemistry is a basic discipline, which in this methodological instruction, the relevant sections are carried out, determines the formation of a highly qualified specialist. programs. In preparation for defending laboratory work and passing the Study of the discipline gives a scientific and practical orientation to the theoretical colloquium, students must answer the general theoretical preparation of students. Therefore, in the field of chemistry, test questions proposed for a laboratory workshop in an engineer in this area should: for each topic. know: the theoretical foundations of organic chemistry, structure 1 SUBJECT LEARNING OBJECTIVES of organic substances, nomenclature, physical and chemical properties, distribution in nature and application; The main goal of the course is to form in scientific students the main directions of development of a theoretical and practical worldview on natural phenomena and the surrounding world, understanding of organic chemistry, mechanisms of chemical processes, principles of mechanism and purposefulness of chemical, biological and planning of organic synthesis; technological processes occurring in future objects, their methods of isolation, purification and identification of organic professional activities and their impact on the environment. connections; Organic chemistry is the basis of knowledge and experience: biological sciences. Mastering it will allow you to study such disciplines in planning and staging a chemical experiment and as biological chemistry, microbiology, physical and colloidal processing of the results obtained; chemistry, food technology, food chemistry, etc. be able to: Specialists in the field of food technology and in their activities will deal with organic compounds, 4 since many objects of future work are food products Brief information about the development of theoretical concepts in are organic substances. organic chemistry. The theory of the chemical structure of organic compounds Therefore, knowledge of the basic laws and properties of compounds. Methodological foundations of the theory of chemical structure and organic compounds, the nature of chemical bonds of atoms and their main provisions of the theory of A.M. Butlerov, as part of the objective molecules and reaction mechanisms are the main prerequisites for the truth of the laws of dialectics. The current state and significance of the theory of understanding the physicochemical, biological, technological chemical structure of A.M. Butlerov. Types of chemical bonds. processes occurring during processing, storage of raw materials and the chemical, spatial, electronic structure of organic food products, as well as their qualitative composition and biological compounds. Stereochemical representations of Van't Hoff and Le Bel. values. This necessary knowledge for students is realized more than the Concept of the quantum mechanical representation of covalent nature by an in-depth study of individual topics that are reflected in the variable connection (method of molecular orbitals - MO). Electronic structure of the part in the form of the section “Bioorganic compounds”. simple and multiple bonds: σ- and π- bonds. The nature of the carbon-carbon bond. The phenomenon of hybridization of sp3 sp2 sp-hybridization orbitals. Characteristics of a covalent bond: length, energy, 2 CONTENTS OF DISCIPLINE SECTIONS direction in space (bond angles), polarity, simple 2.1 INTRODUCTION and multiple bonds. Donor-acceptor (coordinated, semipolar) bond. Hydrogen bond. The subject of organic chemistry and its features. Place Reactions of organic compounds. The concept of the mechanisms of organic chemistry among other general scientific fundamental reactions. Bond cleavage – homologous and heterolytic. Reactions of science. The most important stages in the development of organic chemistry and its role in free radical (radical mechanism) ionic knowledge of the laws and categories of the dialectical development of nature and (electrophilic and nucleophilic) or ionic mechanism. Conditions for the formation of scientific knowledge among students about the world around them, the course of the reaction. Initiators, catalysts, inhibitors. Types of phenomena and processes encountered in the future of their professional reactions. Reactions of substitution (S), addition A), elimination (E), activity. The importance of organic chemistry in the national economy, in molecular rearrangement (isomerization). food industry. Problems of ecology, protection Reactivity of organic compounds and their environment. The problem of keeping the lake clean. Baikal and its structure. The mutual influence of atoms in a molecule is the determining basis of the pool. The main raw materials for obtaining organic reactivity substances (V.V. Markovnikov). Factors, connections. Oil and its processing. Hard and brown coals, their determining reactivity of organic compounds. usage. Gases and their applications. Gasification of Buryatia. Inductive (inductive -J) and mesoric (conjugation effect -M). Mineral deposits in Buryatia, their use. Steric (spatial) effect. Acidity and basicity. Analysis and methods for studying organic compounds. Classification and nomenclature of organic compounds. The concept of methods for isolating, purifying and identifying organics. The main classes of organic compounds. The phenomenon of homology and substances. Qualitative elemental analysis. Quantitative analysis and homology series. The law of transition of quantitative changes into the establishment of empirical formulas. Meaning and use of physical-qualitative. Functional groups. The phenomenon of isomerism. Types of chemical research methods in establishing the structure of isomerism: structural, spatial. Rotational (rotational) organic compounds (UV, IR, NMR and mass spectroscopy and dynamic (tautomerism) isomerism. Law of unity and struggle, etc.). opposites. Nomenclature of organic compounds. The concept of the equivalence of carbon atoms. The concept of radicals 2.2 THEORETICAL PROVISIONS AND GENERAL QUESTIONS (alkyls) and their names. Trivial, rational and systematic ORGANIC CHEMISTRY IUPAC nomenclature. 5 2.3 CLASSES OF ORGANIC COMPOUNDS Alkynes (unsaturated, acetylene hydrocarbons) 2.3.1 HYDROCARBONS Homologous series. Nomenclature. Isomerism. Structure of alkenes: chemical, spatial, electronic. Reactivity Alkanes (saturated hydrocarbons). Homologous series. alkynes Reactions of addition of hydrogen, halogens, hydrogen halides, General formula of the homologous series. Nomenclature. Isomerism. water, alcohols, carboxylic acids, hydrocyanic acids. Mechanism Structure of alkanes: chemical, spatial, electronic. electrophilic and nucleophilic addition reactions. Reaction Concept of conformation. Reactivity of alkanes. substitutions. Acetylenides. Basic methods of obtaining. Syntheses on Characterization of carbon-carbon, carbon-hydrogen bonds. acetylene based. Substitution reactions: halogenation, nitration, sulfoxidation, Cyclic hydrocarbons, Alicycles. Structure (chemical, oxidation. Chain mechanism of radical reaction. Spatial, electronic reactions) and stability of cycles. Theory of dehydrogenation and cracking. Reaction conditions and products. Bayer voltage. Modern interpretation of cycle stability. The most important sources and synthetic methods for the production of alkanes and Arenas (aromatic hydrocarbons). Signs of aromaticity application. Alkanes as motor fuels and raw materials in organic (aromatic character). The structure of benzene. Kekule's formula and synthesis. Ocean number. modern electronic understanding of the structure of benzene. Alkenes (unsaturated, ethylene hydrocarbons). Aromatic sextet. Hückel's rule. Homologous series. Nomenclature. Isomerism. Structure of alkenes, Benzene and its homologues, isomers. Reactivity and spatial, electronic. Reactivity of alkenes. structure. Substitution and addition reactions. Mechanism of addition reactions, mechanism of electrophilic addition of electrophilic substitution of hydrogens of the benzene ring. Rules of hydrogen, halogens, hydrogen halides, sulfuric acid, water. Substitution rule and electronic interpretation. Mutual influence of atoms in Markovnikov and electronic interpretation. Mechanism of radical molecule. Induction and mesor effects. Concordant and addition (peroxide effect), qualitative reaction to discordant orientation from an electronic point of view. double bond, oxidation of alkenes. Polymerization of alkenes and Nucleophilic substitution, reaction mechanism from the point of view of reciprocal copolymerization, Mechanism of polymerization. The most important sources and influences of atoms in a molecule. Addition reactions. Main synthetic methods of production: dehydrogenation, dehydration, sources and methods of production. Synthesis based on benzene. alcohols, dehydrohalogenation of halogen derivatives. Application. The concept of polynuclear aromatic hydrocarbons. Alkadienes. Types of diene hydrocarbons. Structure. Condensed and non-condensed systems. Concept of a conjugate system. Electronic interpretation of the nature of conjugation. carcinogenic substances and dyes. Concept of non-benzoids. Mechanism of electrophilic and radical addition reactions. aromatic systems. Cyclopentadienyl anion. Ferrocene. Qualitative reaction. Main sources, methods of preparation and tropylium cation. Azulene. use of butadiene –1,3 according to the reaction of Lebedev S.V. Rubbers and synthetic rubbers. Genetic relationship between hydrocarbons. Mutual transitions of hydrocarbons from one class to another. 2.3.2 HYDROCARBON DERIVATIVES Halogen derivatives. Classification by hydrocarbon radical and halogens. Mono-, polyhalogen derivatives. Structure, mutual influence of atoms in a molecule from an electronic point of view. Qualitative reactions. Nucleophilic substitution reactions and their mechanisms, SN1; SN2. 6 The most important reactions of production from hydrocarbons (see ketone reactions. Properties of oxo compounds. Reactions of nucleophilic halogenation of the corresponding hydrocarbons). addition of hydrogen, alcohols, hydrocyanic acid, bisulfite. Halogen derivatives of saturated, unsaturated, aromatic sodium. Ammonia, Grignard reagent. Reactions with hydrazine, series. Chloroform. Freons. Vinyl chloride. Chlorprene. hydroxylamine. Aldol-crotonic condensation. Ester Tetrafluoroethylene. Chlorobenzene. condensation. Cannizzaro's reaction. Condensation with phenols, anilines, benzoin condensation. Perkin and Claisen reaction. Hetorocycles. Classification. Aromatic five- and six-membered oxidation reactions of aldehydes and ketones. Differences between oxo compounds and heterocycles. Structure. Electronic interpretation of the aromatic fatty series from aromatic aldehydes and ketones. nature of heterocycles. Hückel's rule. Reactions and mechanism Basic methods for obtaining oxo compounds Oxidation, substitution. Reactivity and orientation. Sources: dihydrogenation of alcohols, pyrolysis of salts of carboxylic acids, preparation of five- and six-membered heterocycles. Application. Furan, hydrolysis of dihalogen derivatives, oxosynthesis of alkenes, synthesis from pyrrole, thiophene, furfural, indole. Pyridine. Vitamin RR. Alkaloids. alkynes (Kucherov reaction). Preparation of aromatic aldehydes and quinoline. Pyrimidine. ketones using the Friedel-Crafts and Gutterman-Koch reactions. Vitamin B, nucleic acids. Structure and biological role. Saturated aldehydes and ketones. Formaldehyde, acetaldehyde, 2.3.3 OXYGEN-CONTAINING ORGANIC acetone. Compaction reactions, condensation. Obtaining carbohydrates. COMPOUNDS Dialegids, diketones, diacetyl. And a role in food. Unsaturated aldehydes and ketones. Acrolein. Acetone. Hydroxy compounds (alcohols, phenols). Classification according to Methyl vinyl ketone. Aromatic oxo compounds. Benzaldehyde, hydrocarbon radical and atomicity. Homologous series. acetophenone. Vanillin. Isomerism. Nomenclature. The structure of alcohols and phenols. Mutual Carboxylic acids. Classification. Homologous series. influence of atoms in a molecule from an electronic point of view. The role of isomerism. Nomenclature. Acyls. Chemical, spatial, hydrogen bonds in OH groups. Chemical properties. Reactions electronic structure of the carboxyl group. Mutual influence of substitution atoms “OH” and “H” in the hydroxy group. Reactions with alkali metals in a molecule - the mutual influence of two functional groups in phosphorus halides, halogenated acids, carboxyl reagent. Properties of carboxylic acids. Grignard acidic character, formation of ethers and esters. The reaction mechanism of the carboxyl group. Effect of hydrogen bonding. Ethyrification reactions, reversible nature of the reaction. Metabolic processes of carboxylic acids: formation of salts, esters, anhydrides, lipids. Oxidation of alcohols. acid halides. Interaction with amines and reaction mechanism The main sources and methods for obtaining alcohols and phenols: from amidation and the reverse nature of the reaction, exchange processes in halogen derivatives, hydration of alkenes, reduction of protein molecules. Substitution reactions in the hydrocarbon radical of oxo compounds using the Grignard reagent. acids: halogenation of the α-position, oxidation to the α- and β-positions. Monohydric alcohols. Methyl, ethyl, propyl alcohol. carboxylic acids, β-oxidation in biological systems. Basic Allyl alcohol. Benzyl alcohol. Polyhydric alcohols. Glycols, sources of production and methods of synthesis: oxidation of hydrocarbons, glycerols. Xylitol, sorbitol. oxosynthesis, hydrolysis of nitriles, trisubstituted phenols, naphthols. Mono- and diatomic phenols. Ethers. halogen derivatives, esters, by the Grignard reaction. Structure. Isomerism. Properties. Antioxidants in food products. Monobasic acids. Formic, acetic, butyric acids. Thymol. Palmitic and stearic acids. Unsaturated acids: acrylic, methacrylic, crotonic, sorbic, oleic, oxo compounds (aldehydes and ketones). Homologous series. linoleic, linolenic. Aromatic acids. Benzoic acid. Isomerism. Nomenclature. Chemical, spatial, electronic Cinnamic acid. Acids are food preservatives. the structure of the oxo group, its polarity and the difference between the aldehyde group and 7 Dibasic acids. saturated, unsaturated, aromatic dyes. Structure and color. Indicators. Acid dyes. Isomerism, nomenclature. Properties. Features of the triphenylmethane, alizarin, and anthocyanidin series. dibasic acids. Reactions of formation of cyclic anhydrides, Dyes in the food industry. decarboxylation. Syntheses using malonic ether. Oxalic, malonic, adipic acids and their role in the synthesis of 2.3.5 HETEROCYCLIC COMPOUNDS of vitamins and substitutes. Maleic and fumaric acids. Their use is for the stabilization of fats, oils, and milk powder. Phthalic Definition. Classification. Nomenclature. acids. Acid derivatives. Salt. Surfactant Soap. Esters and their five-membered heterocyclic compounds. Structure and mutual use as an essence in the food industry. transformations of furan, thiophene, pyrrole. Sources of obtaining them. Acid anhydrides, acid halides, acylating agents. Aromatic character. Electrophilic substitution in furan, thiophene, pyrrole: halogenation, acylation, sulfonation, nitration. Hydrogenation and oxidation. Furfural, features of chemical behavior. The concept of chlorophyll and hemin. Indol. Heteroauxin. Tryptophan. 2.3.4 NITROGEN-CONTAINING ORGANIC COMPOUNDS The concept of five-membered heterocyclic compounds with several heteroatoms. Pyrazole, imidazole, thiazole. Nitro compounds. Classification. Isomerism. Nomenclature. Six-membered heterocyclic compounds. Pyridine. Structure. The structure of the nitro group. Semipolar connection. Tautomerism. Physical Basicity. Preparation of pyridine compounds. Physical properties. properties. Reactions of nitro compounds: reduction according to Zinin, General characteristics of pyridine. Nucleophilic and reduction reactions in various media, interaction with dilute electrophilic substitution. Recovery. alkali, reactions with nitrous acid, condensation with aldehydes. Nicotinic acid, vitamin PP. The concept of alkaloids; horsenine, Basic methods for obtaining alkanes by nitration using the reaction nicotine, anabasine. Konovalov, aromatic hydrocarbons and their mechanisms. The concept of six-membered heterocycles with two nitrogen atoms. Nitromethane, nitroethane. Nitrobenzene. Nitronaphthalenes. Pyrimidine, pyrimidine bases. Purin. Purine bases. Amines. Classification. Isomerism. Nomenclature. Structure Concept of nucleosides, nucleotides and nucleic acids. amino groups. Basic character of fatty amines and anilines. Properties of amines and anilines. Reactions: formation of salts, alkylation, 2.4 COMPOUNDS WITH MIXED FUNCTIONAL acylation. Interactions of amines and anilines with nitrous acid. Reactions of the benzene ring in anilines. Main methods of production: reduction of nitro compounds, nitriles, halogen acids. Structure. Features of halogen acids. Mono-, alkylation of ammonia (Hoffmann reaction), from amides. Monoamines. di-, trichloroacetic acid. Methylamine. Ethylamine. Diamines. Hexamethyldiamine. anilines. Hydroxy acids. Classification by functional groups and by Diazo-, azo-compounds. Aromatic diazo compounds. Structure. the structure of the hydrocarbon radical. Structural isomerism, Isomerism. The diazotization reaction and its mechanism. Properties. Reactions with nomenclature. Structure. Mutual influence of atoms in a molecule. release of nitrogen: action of water, alcohol (deamination), Properties: acidic, alcoholic. Features of α-, β-, γ-, σ-hydroxy acids. replacement of a diazo group with halogens, a nitral group (reaction. The main sources of carbohydrates produced by fermentation are synthetic Sandmeyer ones). Formation of organometallic compounds (reaction methods. Optical isomerism of hydroxy acids (Bio, L. Pasteur). Optical Nesmeyanova). Reactions without nitrogen evolution: reduction of salts, activity of organic compounds (Vant' Hoff, Le Bel). diazonium, azo coupling reaction. Nitrogen dyes. Asymmetric carbon atom. Chiral molecules. Optical antipodes of hydroxy acids, racemic mixture. Specific rotation. 8 Lactic, malic acid, their role in the production of hydroxyl products (glucosides). α-, β - anomers. Furious. Pyranose nutrition. Hydroxy acids with several asymmetric atoms rings. Cyclic structures of Colley, Tollens, Heurs. carbon. Ephidrine, tartaric, citric acids, their use in Proof of the oxide ring. Conformational forms of the food industry. Hydroxybenzoic acids, etc. Methods of monosaccharides (rotational isomerism). separating a racemic mixture. Monosaccharides. Properties of monosaccharides. Monosis reactions due to oxoacids (aldo-, ketoacids). Classification. Structure. oxo groups: reduction to polyhydric alcohols; oxidation Properties of aldonic acids and ketoacids. Mutual influence with silver or copper hydroxide, feling liquid; functional groups in the molecule. Tautomerism, keto-enol. interaction with strong acid, phenylhydrazine, Acetoacetic ester, ketone and acid cleavage, role in hydrxylamine. Reactions to the presence of hydroxyl groups: metabolic processes. alkylation, acylation. Fermentation of hexoses. Epimerization. Amino acids. Classification. Isomerism: structural, Dehydration with cyclitation of pentoses. spatial – optical. Nomenclature. Structure, properties. Obtaining monosaccharides: hydrolysis of di-, polysaccharides, aldolic amphoteric nature of amino acids. Formation of complexes with condensation. Interconversion of monosaccharides: hydroxynitrile metals. Reactions caused by the presence of a carboxyl group: synthesis (chain extension), Ruff decomposition (chain shortening). formation of salts, esters, amides, decarboxylation. Hexoses: glucose, fructose, galactose, mannose. Pentoses: ribose, Reactions to amino groups: formation of salts, acylation, arabinose, xylose. alkylation, action of nitrous acid. Polypeptides. Disaccharides. Restoring (reducing) and Specific reactions. Relation of amino acids to heat. non-reducing (non-reducing) disaccharides. Structure. The main sources of synthesis methods: protein hydrolysis, Tautomerism of reducing disaccharides. Properties of disaccharides. microbiological synthesis, amination of halogen acids, preparation of hydrolysis reactions of disaccharides, for the presence of multiatomicity in the molecule. from hydroxynitriles, unsaturated acids, nitroacids, condensation Reactions of reducing disaccharides: oxidation of aldehydes with malonic acid and ammonia by hydroxide (V.M. Rodionov). The role of silver or copper, feling liquid, the addition of hydrocyanic amino acids in the life of living and plant organisms. acids. Biozones: lactose, sucrose, maltose, cellobiose, trehalose. High molecular weight compounds. The concept of polymers. Polysaccharides. The structure of high molecular weight sugars. Classification. Substances (monomers) from which polymers are obtained. Homopolysaccharides, heteropolysaccharides. Starch, glycogen. Structure The structure of monomers and polymers. Reactions for obtaining (α-, β-anomeric glucose). Amylose, amylopectin. (α-1,4 - and 1,6-high molecular weight compounds. Polymerization and polycondensation. glycosidic bonds). Iodine reaction to starch. Application. Copolymerization. Vinyl polymers. Polyethylene, polypropylene, fiber (cellulose). Structure (β-anomeric glucose). Properties. polystyrene, polyvinyl chloride, polytetrafluoroethylene (fluoroplastic), Acylation, nitration reactions. Application of fiber and its rubbers, polyacrylic polymers. Polycondensation polymers. derivatives. Polyesters, polyamides. Lavsan. Polyptides. Nylon, nylon, Concept of pectin substances, gum, mucus. phenol aldehyde resins. Lipids. Determination of lipids. Classification. Distribution 2. 5 BIOORGANIC COMPOUNDS of lipids in nature. Simple lipids. Fats. Waxes. Glycerides. The structure of fats. Carboxylic acids that are part of fats. Carbohydrates (oxyoxo compounds, oxyaldehydes, oxyketones). Higher carboxylic acids. Saturated and unsaturated acids. Distribution in nature. Classification. Monosaccharides. Structure. Isomerism of glycerides: structural, geometric, optical. Aldoses, ketoses. Tetroses, pentoses, hexoses. Isomerism. Optical properties of fats. Reaction of glycerides: hydrolysis, transesterification, stereoisomers. Antipodes. Projection form of E. Fischer. alcoholism, acidolysis, hydrogenation, polymerization, oxidation. Tautomerism monos. Cyclo-oxotautomeric forms. Hemiacetal 9 Concept of alkyl lipids. The concept of plasmalogens. Diols The purpose of the proposed course is to expand and deepen lipids. students' knowledge in the field of carbohydrate chemistry. As part of the Wax course. Definition. Properties. Application. focuses on fundamental issues of the structure of complex lipids. Phospholipids and their role in a living organism. carbohydrate molecules, synthetic problems of this main group of phospholipids are considered. Glycerophospholipids. Main areas. The objective of the course is to describe the current state of research into structural components. Phosphatidic acids, lititin, polysaccharide regions. The special course covers phosphatidylethanolamine and phosphatidylinositol in detail. types of dietary fiber, including pectins, their Sphingolipids. Phosphorus-containing sphingolipids. classification, structures and properties. Since pectin substances are Glycosphingolipids. are considered as a means of preventing severe poisoning. Lipid analysis. Acid and iodine numbers. Saponification number. metals, this special course presents the mechanism of using chromatography. complexation. Processing of fats and oils. Margarine. Salomas. Soap. Surfactant Anionic substances. SMS. Protein substances. The role of proteins in nature. Function of protein in 3 LABOTATORY ACTIVITIES in the human and animal body. Proteins are high-molecular compounds, biopolymers. Amino acids as structural elements In laboratory classes, the student acquires the skills of a protein biopolymer. Basic amino acids included in the experimental work. When doing protein lab work. Replaceable and essential amino acids. The importance of peptides in a student should keep a laboratory workbook that studies protein chemistry. Peptide bond. Peptide synthesis. The methods are designed to record all observations during the experiment, protecting end groups for the targeted synthesis of peptides. calculations and results obtained. When making entries in the journal, you should Classify proteins. Simple (proteins) and complex (proteids) clearly state the essence of the experiment. proteins. Physicochemical properties of proteins. Amphoteric character. Qualitative reactions - color reactions. Protein hydrolysis. Precipitation Scheme of how proteins work (salting out, denaturation). History of the development of the question of the structure of proteins. The role of scientists in Work No.... research on the structure and properties of proteins: A.Ya. Danilevsky, A.D. Title of synthesis (theme) by Zelinsky, V.S. Sadikova, D.L. Talmed, N. Hofmeister, E. Fischer and Substances and reagents necessary for the experiment of others. The current state of the structure of the protein molecule. Primary, Specify the reaction conditions, secondary structure. Spatial organization Reaction equations of a macromolecular polypeptide chain. Basic types of non-valence bonds Observations in the protein chain. α-helix conformations (L. Pauling). Conclusion Tertiary, quaternary structure of proteins. Work passed _________ Globular and fibrillar proteins. Their differences. Insulin (Sanger). Collagen, keratin. Fibroin. Gelatin. Casein. 3.1 INTRODUCTION TO ORGANIC CHEMISTRY Lactoglobulin. Hemoglobin. Myoglobin. Essential oils. Bicyclic terpenes. Biterpenes. Carotenoids. Vitamin A. Purpose of the lesson: 1. To work through the basic principles, techniques and accept that after completing the basic course of organic chemistry, students will learn the rules of safe work in the laboratory. The author's course “Polysaccharides of food raw materials” is offered. 10 2. Create an idea of the content, directions and tasks 3. List the types of distillation of organic compounds and organic chemistry. determine their differences. 3. Familiarize yourself with the utensils, equipment, instruments for 4. Chromatography and its types. carrying out chemical reactions. 5. Give examples of the use of these isolation methods and Initial level of knowledge: purification of organic compounds in various industries 1. Quantum - mechanical ideas about the structure of atoms and industry. molecules; 2. Theory of molecular orbitals; Laboratory work: 3. Theory of hybridization; 1. Crystallization. 4. Butlerov’s theory of chemical structure. 2. Sublimation. Questions to prepare for the lesson: 3. Extraction. 1. The role of organic chemistry in the synthesis of professional 4. Distillation. education 5. Chromatography. 2. The main tasks of organic chemistry. 2.1. Analysis and determination of the structure of organic compounds. 2.2. Synthesis and assessment of the reactivity of organic compounds 3. 3 DETERMINATION OF BASIC PHYSICAL PROPERTIES 3. Methods for studying ORGANIC COMPOUNDS 3.1 Chemical 3.2 Physical Purpose of the lesson: 3.3 Physico-chemical 1. Introduction to methods for determining the basic physical Laboratory work: characteristics of organic substances: melting points, boiling points, 1. Chemical glassware and materials. show 2. Identification of organic compounds by physical constants; 3. 2 METHODS FOR ISOLATION AND PURIFICATION OF ORGANIC 3. Establishing the degree of purity of organic substances. COMPOUNDS Initial level of knowledge: 1. Basic physical constants of organic substances. Purpose of the work: Questions to prepare for the lesson: 1. Familiarization with the basic methods of isolation, purification and 1. Physical constants of solid, liquid and gaseous separation of organic compounds from a mixture. organic compounds. Initial level of knowledge: 2. Define the crystallization method. Basic methods of purification and isolation of organic 3. What is sublimation of organic substances. connections. 4. Distillation of organic compounds and its types. Questions to prepare for the lesson: 5. Rationale for choosing a method for purifying organic substances. 1. Theoretical foundations of methods of isolation, purification and separation 6. Give examples of the use of these methods in various mixtures of substances. industries. 2. Define the process of filtration, sublimation, distillation, Laboratory work: crystallization, chromatography. 1. Determination of melting point 2. Determination of boiling point.
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Monitoring schedule for CPC in organic chemistry
for the 5th semester 3rd year 2009-2010 academic year| month | September | October | November | December |
| weeks |
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I | ||||
| Control work “Aliphatic hydrocarbons” 13.10 14 40 -16 00 | ||||
| III | Checking notes, interview on the topic of self-study “Main sources of hydrocarbons” 16.10 14 40 | Test "Carbonyl compounds" 20.11 14 40 -16 00 | 15.12 14 40 -16 00 |
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| Control work "Halogen and nitrogen derivatives of aliphatic hydrocarbons" 30.10 14 40 -16 00 | Checking individual home activities and l/r report 25.12 14 40 -16 00 |
SRS monitoring schedule for 3rd year students, specialty “Chemistry”
in organic chemistry and fundamentals of supramolecular chemistry
VI semester 2008-2009 academic year
| month a week | February | March | April | May |
| I | Test "Carboxylic acids" 03/06/09 14 40 | Checking individual assignments on the topic “Benzene derivatives” 05/08/09 14 40 |
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| II | Individual tasks “Monosaccharides” 03/13/09 14 40 | Checking notes and solving individual assignments on the topic “Terpenes” 04/10/09 14 40 | ||
| III | ||||
| IV | Computer tests on the topic “Carbohydrates” 03/27/09 14 40 |
Types of independent work of students
- Preparation for laboratory work
- Preparation for tests
- Compilation of notes on topics submitted for independent study
- Completing coursework
- Solving individual homework
Self-study topics
Natural sources of hydrocarbons and their processing
Questions to Study
- Natural and associated petroleum gases.
- Oil and its products: physical properties and composition of oil, primary oil refining, cracking of oil products.
- Coal processing, coal tar distillation.
Thiols, thioesters
Questions to Study
- General characteristics (definition, functional groups)
- Isomerism, nomenclature
- Methods of obtaining
- Chemical properties
- Application
Terpenes
Questions to Study
- Distribution in nature
- Classification
- Monocyclic terpenes: nomenclature, properties, methods of preparation, individual representatives
- Bicyclic terpenes: nomenclature, properties, methods of preparation, individual representatives.
Nonbenzenoid aromatic systems
Questions to Study
1. Main representatives (ferrocene, azulene, etc.)
2. Structural features
3. The most important reactions
Report form - interview
Organosilicon compounds
Questions to Study
1. Classification
2. Application
Report format – abstract, seminar
1The article discusses the organization of effective research work of students, which allows students to develop the ability to independently obtain knowledge, analyze and effectively use information for maximum self-realization. The use of a taxonomic approach in preparing assignments for self-help work in the discipline “Organic Chemistry” is aimed at developing professional competencies in accordance with the needs of reality. Examples of multi-level questions on the topic “Unsaturated hydrocarbons” for an express survey are given. Using Bloom's pyramid, it is shown what results can be expected as a result of studying this topic. It is proposed to use Bloom's taxonomy when conducting experimental work in laboratory classes. To solve the problem of connecting theory with practice, the authors propose the use of the design method. This will allow you to develop competencies such as the ability to search, collect and analyze information.
Bloom's taxonomy
independent work of students (SWS)
unsaturated hydrocarbons
professional competencies
lesson planning
1. Chizhik V.P. Forms of organization of the educational process in a higher educational institution // Siberian Trade and Economic Journal. – 2011. – No. 11. – P. 119–121.
2. Nurov K. Higher education in Kazakhstan: price without quality and knowledge [Electron. resource]. – 2011. – URL: http://www.ipr.kz/kipr/3/1/44.
3. Lazareva I.N. Taxonomic approach to the design of personality-oriented intellectual developmental training // News of the Russian State Pedagogical University named after. A.I. Herzen. – 2009. – No. 94. – P. 130–136.
4. Kryukov V.F. Modern teaching methods. – M.: Norma. – 2006. – 176 p.
The most important factor in creating an innovation system and developing the country’s human capital is education.
Currently, our country has developed and adopted the State Program for the Development of Education and Science until 2020. Increasing the competitiveness of human capital and the level of training in general is the main emphasis in this program.
In many countries of the world, a personality-oriented approach is recognized as a priority, which corresponds to modern concepts of education. As a result of applying this approach, creative thinking and the ability to work with information are formed and developed. The focus is on the activities of cognition, cooperation, mutual work, i.e. The basis of this method is the independent cognitive activity of students. It is impossible to implement this approach by simply changing one system or form of training to another. First of all, it is necessary for all participants in the educational process to understand the changes taking place, and this implies a certain breaking of habits and stereotypes.
At the present stage of education, the role of the teacher has changed. It is now not so much a source of information transmission, but rather teaches the student to obtain information. The student’s task is to be able to rethink the information obtained and be able to further use the knowledge in practice. In this aspect, the implementation of all teaching functions depends on the choice of method. In a word, the effectiveness of education will depend primarily on the extent to which students have developed the ability to independently obtain knowledge, analyze, structure and effectively use information for maximum self-realization and useful participation in society.
A number of authors suggest using the organization of research work in practical classes and self-help work as one of the ways to activate divergent thinking. Research work in current practical areas allows you to develop the student’s competencies and skills in accordance with the needs of reality, which will allow the formation of competitive specialists.
We propose the use of a taxonomic approach in the preparation of assignments for SRS and SRSP for the discipline “Organic Chemistry”.
Effectively organized independent work begins with setting goals. Firstly, this will make it possible to determine the degree of progress of students towards the intended result, and secondly, it will ensure timely correction.
Long-term use of B. Bloom's taxonomic model indicates its effectiveness. It can be used as a tool for planning lessons and developing strategies and survey methods - from simple to complex.
Using the example of the topic “Unsaturated hydrocarbons” (6th hour), we wanted to show what results we expect as a result of studying this topic:
The student must know: the properties and structure of unsaturated hydrocarbons, types of organic reactions with their participation, signs and conditions of their occurrence.
The student must be able to: establish the relationship between the structure of a compound and its properties, plan and carry out a chemical experiment, analyze its results.
The student must have skills in assembling installations for conducting laboratory experiments and working with modern instruments.
As a result of studying this topic, using B. Bloom's taxonomy, the student at the initial stage (knowledge) will be able to determine the type of hydrocarbon, the features of its structure, and the presence of reaction centers. Moving from simple to complex, at the stage of applying knowledge, he will be able to interpret the stages of chemical reactions, describe transformation schemes, and at the analysis stage he will compare production methods and chemical properties of various classes of unsaturated hydrocarbons, and discuss reaction mechanisms.
Below we provide examples of multi-level questions on the topic for an express survey:
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1) Chemical formula of butadiene? 2) What is polymerization? 3) When was the theory of chemical structure discovered? |
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Understanding |
1) Compare the chemical properties of ethylene and acetylene? 2) What factors influence the halogenation of alkenes? 3) How can you call in one word the reaction of the elimination of water from alcohols? |
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Application |
1) What are the possible results of pentane isomerization? 2) What is formed during the cyclization of butadiene? 3) How can the reaction of hydration of alkenes be applied in practice? |
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1) What are the prerequisites for the emergence of the theory of chemical structure? 2) What are the results of stereochemical reactions? 3) What is the essence of Favorsky’s reaction? |
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1) How can you prove the structure of synthesized organic compounds? 2) How can you check whether the reaction took place or not? 3) How can the problem of synthesizing liquid crystalline compounds be solved? |
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Evaluation (making value judgments based on reason) |
1) Do you think the reactivity of conjugated dienes is higher than that of cumulated dienes? 2) How can one justify the low yield of radical substitution reactions? 3) How can we explain the ability of unsaturated compounds to undergo electrophilic addition reactions? |
When composing knowledge questions, question words are often used: when, what is, who, is it true, etc. Answers to such questions involve simple reproduction of information. The load is not on thinking, but on memory, for example, what is hydrohalogenation? The student simply remembers and recognizes information.
At the level of understanding, the information received is understood; formulating the problem in your own words. The student explains, transforms, i.e. information is processed, for example, how do alkenes differ from alkynes?
Application refers to the use of concepts in new situations. Application questions allow you to transfer the acquired knowledge to new settings, for example, to solve problems, for example, predict the result of the Diels-Alder reaction, what are the possible results of the hydrohalogenation of 1,3-butadiene, etc.
At the level of analysis, information is broken down into related parts. Questions for analysis require clarification of causes and consequences, isolating individual parts from the whole, for example, what is the essence of the problem, what conclusion can be drawn, what are the premises, etc.? Analysis makes it possible to understand and show how it works.
Synthesis is the compilation of information. Synthesis questions involve creative problem solving. It is not enough just the information available. It is necessary to create a new whole based on an original approach. At this level, verbs are more often used: develop, formulate, generalize, combine, modify, etc. For example, formulate Markovnikov’s rule, combine similar reactions of unsaturated hydrocarbons.
At the assessment level, the student discusses, selects and evaluates using certain criteria. At this level, verbs are more often used: prove, select, compare, draw a conclusion, justify, predict. For example, prove that the triple bond in pentine-1 is terminal, compare the methods for producing carboxylic acids by oxidation of hydrocarbons.
When preparing a description of the SRS algorithm, it is necessary to formulate questions and tasks of higher levels of thinking more often. A very important point is to teach students to independently formulate multi-level questions when completing assignments individually. Then, using Bloom's Chamomile, students will be able not only to answer questions, but also to develop certain types of questions themselves, allowing them to reveal each block of Bloom's pyramid. This principle is not typical for the traditional education system, since it was more common there for only teachers to formulate questions and ask them. The use of this method will allow the teacher to diagnose the quality of the knowledge acquired.
Too “theoretical” teaching does not allow students to develop high-quality knowledge. But knowledge that has no connection with practice causes a one-sided and very narrow understanding of the issue being studied. Additional motivation of students, aimed at activating answers to more complex questions, is possible with a differentiated system for assessing answers to questions.
The development of criteria for assessing knowledge makes the assessment process transparent and understandable for everyone, and the development of criteria together with students will help to form a positive attitude towards assessment.
When conducting experimental work on a topic, the application of Bloom's taxonomy is as follows:
It is known that the most common tasks for self-help work in most cases are essays and abstracts. Completing such tasks does not cause difficulties for students, because On the Internet you can find standard essays and abstracts on almost any discipline and topic. Consequently, in order to train competitive specialists, it is necessary to make more efforts to develop in students not only the necessary knowledge in the discipline, which was typical for the traditional system, but also it is necessary to develop skills and research competencies with reference to practical reality. This allows us to train specialists who are focused on market needs and are able to find the most effective solutions among many others. Specialists trained according to the proposed scheme will already have a clear understanding of their specialization during the training process, but at the same time will have effective tools for solving issues of a wider range. To solve this problem, the project method is widely used. A distinctive feature of this form of organization of the educational process is the fact that students receive all the necessary knowledge, skills and abilities not in the process of studying a particular discipline, but in the process of working on a particular project. The project method can be defined as a way of learning through the detailed development of a problem, which should result in a very real, tangible practical result that has a life context. In the educational process of a university, a project is understood as a set of actions specially organized by a teacher and independently carried out by students, culminating in the creation of a creative product. For experimental sciences, the use of the project method is very important.
A research project was developed for students of the Faculty of Chemistry as a project assignment in the discipline “Organic Chemistry”
Mol. Avogadro's law. Mole volume of gas
Since 1961, our country has introduced the International System of Units of Measurement (SI). The unit of quantity of a substance is taken to be a mole.
Mole - the amount of substance in a system containing as many molecules, atoms, ions, electrons or other structural units as are contained in 0.012 kg of carbon isotope12 WITH . The number of structural units contained in 1 mole of substance NA (Avogadro's number) is determined with great accuracy; in practical calculations it is taken equal to 6.02 10 23 molecules (mol -1).
It is easy to show that the mass of 1 mole of a substance (molar mass), expressed in grams, is numerically equal to the relative molecular mass of this substance, expressed in atomic mass units (amu). For example, the relative molecular mass of oxygen (Mr) is 32 amu, and the molar mass (M) is 32 g/mol.
According to Avogadro's law, equal volumes of any gases taken at the same temperature and the same pressure contain the same number of molecules. In other words, the same number of molecules of any gas occupies the same volume under the same conditions. At the same time, 1 mole of any gas contains the same number of molecules. Consequently, under the same conditions, 1 mole of any gas occupies the same volume. This volume is called the molar volume of gas (V 0) and under normal conditions (0 °C = 273 K, pressure 101.325 kPa = 760 mm Hg = 1 atm) is equal to 22.4 l/mol. The volume occupied by a gas under these conditions is usually denoted by V 0 , and the pressure by P 0 .
According to the Boyle-Mariotte law, at a constant temperature, the pressure produced by a given mass of gas is inversely proportional to the volume of the gas:
Р 0 /Р 1 = V 1 /V 0 or РV = const.
According to Gay-Lussac's law, at constant pressure, the volume of a gas changes in direct proportion to the absolute temperature (T):
V1/T1 = V0/T0 or V/T = const.
The relationship between gas volume, pressure and temperature can be expressed by a general equation combining the Boyle-Mariotte and Gay-Lussac laws:
PV/T = P0V0/T0, (1)
where P and V are the pressure and volume of gas at a given temperature T; P0 and V0 - pressure and volume of gas under normal conditions (norm).
The above equation allows you to find any of the indicated quantities if the others are known.
Example 1. At 25 °C and a pressure of 99.3 kPa (745 mm Hg), a certain gas occupies a volume of 152 cm 3. Find what volume the same gas will occupy at 0 °C and a pressure of 101.33 kPa?
Solution: Substituting these problems into equation (1) we get:
V0 = PVT0/TP0 = 99.3 152 273/101.33 298 = 136.5 cm 3.
Example 2. Express the mass of one CO2 molecule in grams.
Solution: The molecular weight of CO2 is 44.0 amu. Therefore, the molar mass of CO2 is 44.0 g/mol. 1 mole of CO2 contains 6.02 10 23 molecules. From here we find the mass of one molecule:
m = 44.0/6.02 10 23 = 7.31 10 -23 g.
Example 3. Determine the volume that nitrogen weighing 5.25 g will occupy at 26 °C and a pressure of 98.9 kPa (742 mm Hg).
Solution: Determine the amount of N2 contained in 5.25 g:
5.25/28 = 0.1875 mol, V0 = 0.1875 22.4 = 4.20 l.
Then we bring the resulting volume to the conditions specified in the problem:
V = P0V0 T/PT0 = 101.3 4.20 299/98.9 273 = 4.71 l.
Determination of molecular masses of substances in the gaseous state
To determine the molecular mass of a substance (amu), we usually find the molar mass of the substance (g/mol) that is numerically equal to it.
A. Determination of molecular mass by gas density
Example 4. The gas density in air is 1.17. Determine the molecular mass of the gas.
Solution: From Avogadro’s law it follows that at the same pressure and the same temperatures, the masses (m) of equal volumes of gases are related as their molar masses (M):
m1/m2 = M1/M2 = D, (2)
where D is the relative density of the first gas relative to the second.
Therefore, according to the conditions of the problem:
D = M1/M2 = 1.17.
The average molar mass of air M2 is 29.0 g/mol. Then:
M1= 1.17 29.0 = 33.9 g/mol,
which corresponds to a molecular weight of 33.9 amu.
Example 5. Find the nitrogen density of air having the following volumetric composition: 20.0% O2; 79.0% N2; 1.0% Ar.
Solution: Since the volumes of gases are proportional to their quantities (Avogadro’s law), the average molar mass of the mixture can be expressed not only in terms of moles, but also in terms of volumes:
Мср = (М1V1 + M2V2 + M3V3)/(V1+ V2+ V3). (3)
Let's take 100 ml of the mixture, then V(O2) = 20 ml, V(N2) = 79 ml, V(Ar) = 1 ml. Substituting these values into formula (1.2.2) we get:
Msr = (32 20 + 28 79 + 40 1)/(20 + 79 + 1),
Msp = 28.9 g/mol.
The nitrogen density is obtained by dividing the average molar mass of the mixture by the molar mass of nitrogen:
DN2 = 28.9/28 = 1.03.
B. Determination of the molecular mass of a gas by molar volume
Example 6. Determine the molecular mass of the gas if, under normal conditions, a gas weighing 0.824 g occupies a volume of 0.260 liters.
Solution: Under normal conditions, 1 mole of any gas occupies a volume of 22.4 liters, νgas = 0.26/22.4 = 0.0116 mol, and the molar mass is 0.824/0.0116 = 71 g/mol.
Therefore, the molar mass of the gas is 71.0 g/mol, and its molecular mass is 71.0 amu.
IN. Determination of molecular weight using the equation
Mendeleev-Clapeyron
The Mendeleev-Clapeyron equation (equation of state of an ideal gas) establishes the relationship between the mass (m, kg), temperature (T, K), pressure (P, Pa) and volume (V, m3) of a gas with its molar mass (M, kg/mol ):
where R is the universal gas constant equal to 8.314 J/(mol K). Using this equation, you can calculate any of the quantities included in it if the others are known.
Example 7. Calculate the molecular weight of benzene, knowing that the mass of 600 cm 3 of its vapor at 87 °C and a pressure of 83.2 kPa is 1.30 g.
Solution: Expressing these problems in SI units (P = 8.32 10 4 Pa; V = 6 10 -4 m 3; m = 1.30 10 -3 kg; T = 360 K) and substituting them into equation (1.2.3 ), we find:
M = 1.30 10-3 8.31 360/8.32 104 6 10-4 = 78.0 10-3 kg/mol = 78.0 g/mol.
The molecular weight of benzene is 78.0 amu.
Derivation of chemical formulas and calculations using reaction equations
Formulas of substances show which elements and in what quantities are included in the composition of the substance. There are simple and molecular formulas. The simplest formula expresses the simplest possible atomic composition of the molecules of a substance, corresponding to the mass ratios between the elements forming a given substance. The molecular formula shows the actual number of atoms of each element in the molecule (for substances with a molecular structure).
To derive the simplest formula of a substance, it is enough to know its composition and the atomic masses of the elements forming this substance.
Example 8. Determine the formula of chromium oxide containing 68.4% chromium.
Solution Let us denote the numbers of chromium and oxygen atoms in the simplest formula of chromium oxide by x and y, respectively. Formula of oxide CrхOy. The oxygen content in chromium oxide is 31.6%.
x: y = 68.4/52: 31.6/16 = 1.32: 1.98.
To express the resulting ratio in integers, divide the resulting numbers by a smaller number:
x: y = 1.32/1.32: 1.98/1.32 = 1: 1.5,
and then multiply both values of the last ratio by two:
Thus, the simplest formula of chromium oxide is Cr2O3.
Example 9. When a certain substance weighing 2.66 g was completely burned, CO2 and SO2 were formed with masses of 1.54 g and 4.48 g, respectively. Find the simplest formula of the substance.
Solution: The composition of the combustion products shows that the substance contained carbon and sulfur. In addition to these two elements, it could also contain oxygen.
The mass of carbon that was part of the substance can be found from the mass of CO2 formed. The molar mass of CO2 is 44 g/mol, while 1 mole of CO2 contains 12 g of carbon. Let's find the mass of carbon m contained in 1.54 g of CO2:
44/12 = 1.54/m; m = 12 1.54/44 = 0.42 g.
Calculating in a similar way the mass of sulfur contained in 4.48 g of SO2, we obtain 2.24 g.
Since the mass of sulfur and carbon is 2.66 g, this substance does not contain oxygen and the formula of the substance is CxSy:
x: y = 0.42/12: 2.24/32 = 0.035: 0.070 = 1:2.
Therefore, the simplest formula of the substance is CS2.
To find the molecular formula of a substance, it is necessary, in addition to the composition of the substance, to know its molecular mass.
Example 10. A gaseous compound of nitrogen with hydrogen contains 12.5% (mass) hydrogen. The density of the compound for hydrogen is 16. Find the molecular formula of the compound.
Solution: The required formula of the substance NxHy:
x: y = 87.5/14: 12.5/1 = 6.25: 12.5 = 1: 2.
The simplest formula for the compound NH2. This formula corresponds to a molecular weight of 16 amu.
We will find the true molecular mass of the compound based on its hydrogen density:
M = 2 16 = 32 a.m.u.
Therefore, the formula of the substance is N2H4.
Example 11. When zinc sulfate crystal hydrate weighing 2.87 g was calcined, its mass decreased by 1.26 g. Establish the formula of the crystal hydrate.
Solution: When calcined, the crystalline hydrate decomposes:
ZnSO 4 nH2O → ZnSO4 + nH2O
M(ZnSO4) = 161 g/mol; M(H2O) = 18 g/mol.
From the conditions of the problem it follows that the mass of water is 1.26 g, and the mass of ZnSO4 is equal to (2.87-1.26) = 1.61 g. Then the amount of ZnSO4 will be: 1.61/161 = = 0.01 mol, and number of moles of water 1.26/18 = 0.07 mol.
Therefore, for 1 mole of ZnSO4 there are 7 moles of H2O and the formula of ZnSO4 crystal hydrate is 7H2O
Example 12. Find the mass of sulfuric acid required to completely neutralize 20 g of sodium hydroxide.
Solution Reaction equation:
H2SO4 + 2NaOH = Na2SO4 + 2H2O.
M(H2SO4) = 98 g/mol; M(NaOH) = 40 g/mol.
According to the condition: (NaOH) = 20/40 = 0.5 mol. According to the reaction equation, 1 mole of H2SO4 reacts with 2 moles of NaOH, and 0.25 mole of H2SO4 or 0.25 98 = 24.5 g reacts with 0.5 mole of NaOH.
Example 13. A mixture of copper and iron filings weighing 1.76 g was burned in a stream of chlorine; resulting in a mixture of metal chlorides weighing 4.60 g. Calculate the mass of copper that reacted.
Solution: The reactions proceed according to the following schemes:
1) Cu + Cl2 = CuCl2
2) 2Fe + 3Cl2 = 2FeCl3
M(Cu) = 64 g/mol; M(Fe) = 56 g/mol; M(CuCl2) = 135 g/mol;
M(FeCl3) = 162.5 g/mol.
According to the conditions of the problem, the mass of a mixture of copper(II) and iron(III) chlorides, i.e. a + b = 4.60 g. Hence 135x/64 + 162.5 (1.76 - x)/56 = 4.60.
Therefore, x = 0.63, that is, the mass of copper is 0.63 g.
Example 14. When treating a mixture of potassium hydroxide and bicarbonate with an excess of hydrochloric acid solution, potassium chloride weighing 22.35 g was formed and a gas with a volume of 4.48 dm 3 (n.s.) was released. Calculate the composition of the initial mixture (ω, %).
Solution: Reaction equations:
1) KHCO3 + HCl = KCl + H2O + CO2
2) KOH + HCl = KCl + H2O
M(KHCO3) = 100 g/mol; M(KCl) = 74.5 g/mol; M(KOH) = 56 g/mol.
According to the conditions of the problem, the volume of gas (CO2) according to reaction (1) is equal to 4.48 dm 3 or 0.2 mol. Then from the reaction equation (1) it follows that the initial amount in the mixture of potassium bicarbonate is 0.2 mol or 0.2 100 = 20 g and the same amount of 0.2 mol KCl is formed or 0.2 74.5 = 14.9 G.
Knowing the total mass of KCl formed as a result of reactions (1 and 2), we can determine the mass of KCl formed by reaction (2). It will be 22.35 - 14.9 = 7.45 g or 7.45/74.5 = 0.1 mol. The formation of 0.1 mol KCl according to reaction (2) will require the same amount of KOH, that is, 0.1 mol or 0.1 56 = 5.60 g. Therefore, the content of the initial components in the mixture will be:
5.6 100/25.6 = 21.9% KOH and 20.0 100/25.6 = 78.1% KHCO3.
Calculations according to the law of equivalents
The amount of an element or substance that interacts with 1 mole of hydrogen atoms (1 g) or replaces this amount of hydrogen in chemical reactions is called equivalent of a given element or substance.
Equivalent mass (M uh ) is called the mass of 1 equivalent of a substance.
Example 15 Determine the equivalent and equivalent masses of bromine, oxygen and nitrogen in
compounds HBr, H2O, NH3.
Solution: In these compounds, 1 mole of hydrogen atoms combines with 1 mole of bromine atoms, 1/2 mole of oxygen atoms and 1/3 mole of nitrogen atoms. Therefore, according to the definition, the equivalents of bromine, oxygen and nitrogen are equal to 1 mol, 1/2 mol and 1/3 mol, respectively.
Based on the molar masses of the atoms of these elements, we find that the equivalent mass of bromine is 79.9 g/mol, oxygen - 16 1/2 = 8 g/mol, nitrogen - 14 1/3 = 4.67 g/mol.
The equivalent mass can be calculated from the composition of the compound if the molar masses (M) are known:
1) Me (element): Me = A/B,
where A is the atomic mass of the element, B is the valence of the element;
2) Me(oxide) = Me(ele.) + 8,
where 8 is the equivalent mass of oxygen;
3) Me(hydroxide) = M/n(on-),
where n(on-) is the number of OH- groups;
4) Me(acids) = M/n(n+),
where n(n+) is the number of H+ ions
5) Me(salt) = M/nmeVme,
where nme is the number of metal atoms; Vme is the valency of the metal.
Example 16. Determine the equivalent masses of the following substances Al, Fe2O3, Ca(OH)2, H2 SO4, CaCO3.
Solution: Me(Al) = A/B = 27/3 = 9 g/mol; Me(Fe2O3) = 160/2 3 = = 26.7 g/mol; Me(Ca(OH)2) = 74/2 = 37 g/mol; Me(H2SO4) = 98/2 = 49 g/mol; Me(CaCO3) = 100/1 2 = 50 g/mol; Me(Al2(SO4)3) = 342/2 3=342/6 = 57 g/mol.
Example 17. Calculate the equivalent mass of H2SO4 in the reactions:
1) Н2SO4+ NaOH = NaHSO4 + H2O
2) H2SO4 + 2NaOH = Na2SO4 + H2O
Solution: The equivalent mass of a complex substance, as well as the equivalent mass of an element, can have different values, and depend on the chemical reaction the substance enters into.
The equivalent mass of sulfuric acid is equal to the molar mass divided by the number of hydrogen atoms replaced by the metal in this reaction. Therefore, Me(H2SO4) in reaction (1) is equal to 98 g/mol, and in reaction (2) - 98/2 = 49 g/mol.
When solving some problems containing information about the volumes of gaseous substances, it is advisable to use the value of the equivalent volume (Ve).
Equivalent volume is the volume occupied under given conditions by 1 equivalent of a gaseous substance. So for hydrogen at no. the equivalent volume is 22.4 1/2 = 11.2 l, for oxygen - 5.6 l.
According to law of equivalent masses (volumes) of substances reacting with each other m 1 and m 2 are proportional to their equivalent masses (volumes) :
m1/ Me1 = m2/ M e2. (5)
If one of the substances is in a gaseous state, then:
m/Me = V0/Ve. (6)
Example 18. When a metal weighing 5.00 g is burned, a metal oxide weighing 9.44 g is formed. Determine the equivalent mass of the metal.
Solution: From the conditions of the problem it follows that the mass of oxygen is equal to the difference 9.44 g - 5.00 g = 4.44 g. The equivalent mass of oxygen is 8.0 g/mol. Substituting these values into expression (5) we get:
5.00/Me(Me) = 4.44/8.0; Mz(Me) = 5.00 8.0/4.44 = 9 g/mol.
Example 19. During the oxidation of metal(II) weighing 16.7 g, an oxide weighing 21.5 g was formed. Calculate the equivalent masses of: a) metal; b) its oxide. What is the molar mass of: c) metal; d) metal oxide?
Solution: m(O2) in the oxide will be: 21.54 - 16.74 = 4.80 g. In accordance with the law of equivalents, we obtain:
16.74/Me(Me) = 4.80/8.00,
From: Me(Me) = 27.90 g/mol.
The equivalent mass of the oxide is equal to the sum of the equivalent masses of the metal and oxygen and will be 27.90 + 8.00 = 35.90 g/mol.
The molar mass of metal (II) is equal to the product of the equivalent mass and valency (2) and will be 27.90 2 = 55.80 g/mol. The molar mass of metal(II) oxide will be 55.8 + 16.0 = 71.8 g/mol.
Example 20. From a metal nitrate weighing 7.27 g, a chloride weighing 5.22 g is obtained. Calculate the equivalent mass of the metal.
Solution: Since the equivalent mass of a metal nitrate (chloride) is equal to the sum of the equivalent masses of the metal (x) and the acidic residue of the nitrate (chloride), then according to the law of equivalents, taking into account the conditions of the problem, we obtain:
7.27/5.22 = (x + 62)/(x + 35.5).
From: x = 32.0 g/mol.
Example 21. From metal (II) sulfate weighing 15.20 g, a hydroxide weighing 9.00 g was obtained. Calculate the equivalent mass of the metal and determine the formula of the original salt.
Solution: Taking into account the conditions of the problem and equation (5), we obtain:
15.2/9.0 = (Me(Me) + 48)/(Me(Me) + 17).
From: Me(Me) = 28 g/mol; M(Me) = 28 2 = 56 g/mol.
Salt formula: FeSO4.
Example 22. What mass of Ca(OH)2 contains the same number of equivalents as Al(OH)3 weighing 312 g?
Solution: Me(Al(OH)3) is 1/3 of its molar mass, that is, 78/3 = 26 g/mol.
Therefore, 312 g of Al(OH)3 contains 312/26 = 12 equivalents. Me(Ca(OH)2) is 1/2 of its molar mass, that is, 37 g/mol. Hence, 12 equivalents are 37 12 = 444 g.
Example 23. The reduction of metal (II) oxide weighing 7.09 g requires hydrogen with a volume of 2.24 dm3 (n.s.). Calculate the equivalent masses of oxide and metal. What is the molar mass of the metal?
Solution: In accordance with the law of equivalents, we obtain:
7.09/2.24 = Me(oxide)/11.20; Me(oxide) = 35.45 g/mol.
The equivalent mass of the oxide is equal to the sum of the equivalent masses of the metal and oxygen, so Me(Me) will be 35.45 - 8.00 = 27.45 g/mol. The molar mass of metal(II) will be 27.45 2 = 54.90 g/mol.
When determining the equivalent masses of various substances, for example, by the volume of gas released, the latter is collected over water. Then the partial pressure of the gas must be taken into account. The partial pressure of a gas in a mixture is the pressure that this gas would produce if it occupied the volume of the entire gas mixture under the same physical conditions. According to According to the law of partial pressures, the total pressure of a mixture of gases that do not enter into chemical interaction with each other is equal to the sum of the partial pressures of the gases that make up the mixture. If a gas is collected above a liquid, then during calculations it should be borne in mind that its pressure is partial and equal to the difference between the total pressure of the gas mixture and the partial vapor pressure of the liquid.
Example 24. What volume will they occupy at normal conditions? 120 cm 3 of nitrogen collected above water at 20 0 C and a pressure of 100 kPa (750 mm Hg)? The saturated vapor pressure of water at 20 °C is 2.3 kPa.
Solution: The partial pressure of nitrogen is equal to the difference between the total pressure and the partial pressure of water vapor:
PN2 = P - H2O P = 100 - 2.3 = 97.7 kPa.
Denoting the required volume by V0 and using the combined Boyle-Mariotte and Gay-Lussac equation, we find:
V0 = РVT0/TP0 = 97.7 120 273/293 101.3 = 108 cm 3 .
Ministry of Education and Science of Russia
"East Siberian State University of Technology and Management"
Department of Bioorganic and Food Chemistry
Guidelines for implementation
SRS and control tasks for the course
"Organic chemistry with basic biochemistry"
specialties "Standardization and metrology" and "Quality management"
Compiled by: ,
PREFACE
The study of organic chemistry presents certain difficulties due to the large volume of factual material, a significant number of new concepts, the uniqueness of the nomenclature of organic compounds and the very close connection of one section with another. Therefore, mastering a course in organic chemistry requires systematic and consistent work. When studying, it is necessary to especially strictly observe the sequence of transition to studying each next section only after the material of the previous one has been mastered. You should not mechanically memorize formulas, constants, reaction equations, etc. It is necessary to be able to highlight the main thing, understand the essence of certain transformations, find the mutual connection of various classes of compounds and their significance and application.
SAMPLE LIST OF LABORATORY WORKS FOR CORRESPONDENCE STUDY STUDENTS (6 hours)
1. Basic rules for working in an organic chemistry laboratory, safety precautions when carrying out laboratory work.
2. Hydrocarbons.
3. Oxygen-containing organic compounds. Alcohols and phenols. Aldehydes and ketones. Carboxylic acids.
4. Carbohydrates. Monosaccharides.
5. Amino acids. Squirrels.
Control measures and distribution of points by type of work
Name of sections | Theoretical component assessment form | Practical assessment form | SRS assessment form | ||||
Section 1 (module 1) Theoretical concepts in organic chemistry |
laboratory work 1 | Self-assessment, review, public defense of the Kyrgyz Republic Assignments No. 1 | |||||
Section 2 (module 2) Hydrocarbons and their derivatives | Defense of laboratory (2, 3) and practical work | Protection of the Kyrgyz Republic, rear. No. 2 | |||||
Section 3 (module 3) Heterofunctional hydrocarbon derivatives Bioorganic compounds | Defense of laboratory (4.5) and practical work | Rear protection system No. 3 | |||||
Certification final testing | |||||||
Total: 108 (maxipoint) | |||||||
INSTRUCTIONS TO FOLLOW
CONTROL WORK
A correspondence student must complete one test according to the curriculum.
When completing and completing test assignments, the student must adhere to the following rules:
1. Design the title page of the notebook in which the test was completed according to the following model:
2. Test assignments should be completed in notebooks, leaving margins for the reviewer’s comments;
3. When performing a test, write out the conditions of the task or question in full.
4. State the answer in detail, avoiding long descriptions.
The test consists of three tasks. The student selects problems in the table as follows: problem I is found against the initial letter of his last name, task II - against the initial letter of his first name, task III - against the initial letter of his patronymic. For example, performs tasks: 7, 29, 48.
Problem numbers | |||
5. The test work completed and formatted according to the above rules should be submitted to the department of “Bioorganic and Food Chemistry” (room 8-414) for review.
TASKS AND QUESTIONS OF CONTROL TASKS
EXERCISEI.
Task algorithm:
In the given structures or formulas of your assignment:
b) give examples of possible isomers for them;
c) give names according to systematic nomenclature or trivial names;
d) indicate in which hybridization each carbon atom is located in these compounds.
Options:
1. C – C – C C – C – OH
2. C –C –C C –C –Cl
3. C –C –C C –C –C
4. C –C –C C=C –Cl
5. C –C –C–C C=C –COOH
C–C –C C–C –Br
7. C≡ C–C C – C – CN
8. C=C - C=C C - C - O - C –C
9. C - C= C - C C – CO – C
10. C =C –C C - C –N-C
11. C≡ C – C - C C – C - C
12. C - C-C C –C=O
13. C - C - C= C C –C –NH2
14. C C C - COOH
15. C=C - C C-CO - O - C - C
C - C - C - C - CONH- C -
17. S-S-S-S S-S-COON
18. S-S-S S S-S-OH
EXERCISEII.
Task algorithm:
Carry out chemical transformations for the following molecules of organic compounds, indicating the reagents with which they react. Establish their structure and give them names according to systematic nomenclature. For the final product, indicate its scope of application.
Options:
19. Halogen derivative → alkene → alcohol → alkadiene → → synthetic rubber ↓
20. Alkane →halogen derivative →alkene →dihydric alcohol →lavsan
21. Alcohol → alkene → dibromo derivative → alkyne → chloroalkene → polyvinyl chloride
22. Dichloro derivative → alkyne → ketone → oxynitrile → → hydroxypropionic acid → polyester
23. Sodium salt of carboxylic acid →alkane→halogen derivative →alkene→dihydric alcohol→polyester of ethylene glycol and succinic acid
24. Wurtz reaction → alkane → dinitro compound → diamine → polyamide → ethanediamine and adipic acid
25. Arene → aromatic nitro compound → alkylaniline → aminobenzoic acid → polyamide
26. Alkene → alkyne → oxo compound → oxynitrile → hydroxypropionic acid → polyester
27. Alkene → dichloro derivative → dihydric alcohol → polyether → ethylene glycol and succinic acid
28. Dichloro derivative → alkyne → ketone → hydroxynitrile → →hydroxyisobutyric acid
29. Chloroalkane → alkene → alcohol → alkadiene → synthetic rubber ↓
2-methylbutane
30. Alkene → dichloroalkane → dihydric alcohol → diamine → polyamide → diaminoethane and oxalic acid
31. Alkane → chloroalkane → alkene → ethylene glycol → diamine → → phthalic acid polyamide
32. Alkyne →ketone →isopropyl alcohol
hydroxynitrile→hydroxy acid→polyester
33. Alkene → alcohol → oxo compound → oxynitrile → hydroxy acid → polyester of lactic acid
34. Bromoalkane → alcohol → carboxylic acid → chlorocarboxylic acid → aminoacetic acid → polyamide
35. Alkane → alkene → alkyne → aldehyde → hydroxy acid → → α-alanine → diketopiperazine
36. Alkene→bromoalkane→alcohol→ketone→hydroxynitrile→→2-hydroxy-2-methylpropanoic acid→α-amino acid
EXERCISEIII.
Task algorithm:
a) Write the structural formulas of the tautomeric formulas of monosaccharides, mark the hemiacetal hydroxyl, and give them names. Write down the characteristic reaction equations for one monosaccharide. Obtain reducing and non-reducing disaccharides from a monosaccharide and give them names.
b) Write a scheme for obtaining isomeric triacylglycerides that are part of lipids from fatty acids. Give names to triacylglycerides. What consistency will the fat containing these acylglycerides have? How to turn liquid fat into solid fat? How to determine uncertainty? Carry out hydrolysis and saponification of the resulting triacylglycerides, give names to the resulting products.
c) Write reaction equations for an amino acid, characteristic of the amino group and carboxyl, and show amphotericity. Write the bipolar ion for the amino acid. Explain the activity using the pHi value. Synthesize isomeric tripeptides from this amino acid and two other amino acids, give names.
a) monosaccharides | b) fatty acids | c) amino acids |
|
Idose, fructose | Caprylic, erucic | ||
Altroza, | Palmitic, stearic | ||
Galactose, | Oleic, oil | ||
Linoleic, caprylic | |||
Allose, ribose | Nylon, arachidonic | ||
Ribose, thallose | Stearic, oil | Histidine |
|
Arabinose, | Kaprinovaya, linoleic | Methionine |
|
Fructose, galactose | Linolenic, caprylic | ||
Lyxose, ribose | Ricinoleic, nylon | Phenylalanine |
|
Gulose, xylose | Lauric, linoleic | Tryptophan |
|
Galactose, | Lauric, myristic | ||
Fructose, | Erukovaya, stearic | Glutamic acid |
|
|
galactose | Octadecane, ricinoleic | Aspartic acid |
|
|
fructose | Myristic, stearic | ||
Glucose, ribose | Kaprinovaya, arachidonic | ||
Mannose, idose | Arakhinovaya, palmitic | ||
Gulose, idose | Isoleucine |
||
Arabinose, altrose | Arakhinovaya, arachidonic |
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2. Grandberg Chemistry.-M.: Higher School, 1974.
3., Troshchenko chemistry.-M.: Higher School, 2002.
4. Artemenko chemistry.-M.: Higher School, 2002.
5., Anufriev on organic chemistry.-M.: Higher School, 1988.
6. Maksanov of organic chemistry in diagrams, tables and figures: Textbook. - Ulan-Ude: Publishing House of VSTU, 2007.
7. Maxanova compounds and materials based on them, used in the food industry.-M.: KolosS, 2005.- 213 p.
8. , Ayur compounds and their application. - Ulan-Ude: Publishing House of the All-Russian State Technical University, 2005. - 344 p.