m. gorky donetsk national medical university medical...

M. Gorky donetsk national medical university

Medical chemistry department

PRACTICE ACTIVITY GUIDELINES

for bioorganic chemistry practical lessons

(intended audience: first-year students of medical faculty)

Donetsk – 2009

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Practice activity guidelines were made by assistant professor Selezneva E.V., assistant

Ignatyeva V.V.

Reviewers: Head of the biochemistry department, D.B., professor Borzenko B.G. Head of the pathologic physiology department, Corresponding Member, professor Elskiy V.M. Head of the educational-methodical department Basiy R.V. Head of the department of foreign languages Pouzik A.A.

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INTRODUCTION Organic chemistry is the child of medicine, and however far it may go on its way, with its

most important achievements, it always returns to its parent. Thudichum, J.L.W. Journal of the Royal Society of Arts On the Discoveries and Philosophy of Leibig Volume 24, 1876 (p. 141) The Ability to interpret structure of organic compounds, predict their chemical properties

and qualitative determination in different biological liquids are necessary for formation of complete understanding of metabolism process in human body and practical skills development in determination of biological active substances in clinical and pharmaceutical tests.

The Practice activity guidelines are designed for one-semester course of Bioorganic chemistry. This text was written for students who have interest in biology, human medicine, pharmacy, nursing, medical technology, health science. The organization of studding course is fairly classical. The first two topics treat in sequence carbonyl compounds and carboxylic acids. The next one takes up classification and main properties of lipids. The following three topics (hydroxy- and oxo-acids; monosaccharides; di- and polysaccharides) treat compounds with two (or more) oxygen functions in the same molecule. Nitrogen-containing substances (heterocyclic compounds; amino acids) are discussed in the last two topics.

The Practice activity guideline in each topic consists of following parts: 1. Biomedical importance. 2. The Objective of the practice. 3. Academic content. 4. Chart of the logical structure. 5. Directed action base. 6. Academic exercises. 7. Methodical recommendations for student’s work. The importance of each leaning topic and connection of it with medical science is

described in first part of the practice guideline. The next two parts contain general and specific goals and theoretical topics for target. Chart of the logical structure gives ability for students to systematize the main ideas in leaning topic. The next part is the laboratory manual that contains experiments that have been done in practical course. Academic exercises appear at appropriate places within each topic to help students develop the skills.

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STRUCTURE AND PROPERTIES OF ALDEHYDES AND KETONES Biomedical importance: Aldehydes and ketones are one of the important classes of organic compounds. Carbonyl

group is a part of biological important compounds such as retinal, vitamin В6, glucose, fructose and others. Many aldehyde-containing compounds are used as medicines. Aldehyde group causes narcotic and disinfectant properties of compounds.

Formic aldehyde has ability to precipitate proteins. 40% aqeous formaldehyde solution with small amount of methanol (for inhibition of polymerization) is called formalin. It’s used as reducing, antiseptic, disinfectant and preservative agent for biological speciments. In medicine urotropin (hexamethylentetraamine) is used, it is a formaldehyde derivate. Urotropin serves as antiseptic at infectious diseases of urinary tract. Chloral hydrate is a sedative, antianxiety soporific medicine. Carbonyl- containing compound acetone is formed in metabolic pathways and excreted with urine in diabetic patients. Ketone (acetone) tests are used in clinical practice for diabetes mellitus diagnostic.

Large quantity of metabolic processes of carbonyl-containing compounds are caused by their structure and properties.

So, knowledge about structure, properties of carbonyl-containing compounds and ability to identify them in solution by specific reactions are needed to further studying students and doctor practice.

THE OBJECTIVE OF THE PRACTICE: To develop skills in interpreting of structure and chemical properties of carbonyl-

containing compounds to understand reactions that occur in human body. Specific Goals: To be able to: 1. Identify of carbonyl-containing compounds 2. Predict, identify of properties of aldehydes on basis of their electronic structure 3. Predict, identify of properties of ketones on basis of their electronic structure 4. Explain of common and specific properties of carbonyl-containing compounds 5. Interpret of medical-biological importance of carbonyl-containing compounds. 6. Identify of biological important derivates of carbonyl-containing compounds. ACADEMIC CONTENT

THEORY TOPICS FOR TARGET: 1. Classification of carbonyl-containing compounds. 2. Electronic structure of carbonyl group. 3. Nucleophilic addition reactions. The mechanism. 4. Reduction-oxidation reactions of aldehydes and ketones. 5. Reactions of α-carbon atom. 6. The difference in reactivity of aldehydes and ketones. 7. Medical-biological importance of aldehydes, ketones and their derivates.

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CHART OF THE LOGICAL STRUCTURE

Structure of oxocompounds Electronic structure of

carbonyl group

Chemical properties Chemical properties

Nucleophilic addition reactions

Reduction, oxidation reactions

Difference in reactivity of aldehydes and ketones

Biological important carbonyl compounds

CARBONYLCONTAINING CONPOUNDS

ALDEHYDES CLASSIFICATION

NOMENCLATURE

KETONES

Medical-biological importance of carbonylcontaining compounds

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THE APPLICABLE MATERIALS:

1. The synopsis of lectures. 2. Graf of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work.

Experiment 1. Oxidation of formaldehyde by copper hydroxide in alkaline solution. Pipette off 6 drops of 2n NaOH, add 6 drops of water and 1 drop of 0,2n CuSO4. Add 2-3

drops of 40% formalin to obtained Cu(OH)2. Shake solution. Heat up upper part of solution. In the warmed part of test-tube yellow and after red precipitation is forming. If test-tube is pure “copper mirror” can form on the walls.

H CO

HH C

O

OH2Cu(OH)2+ 2CuOH+ H2O+

Yellow CuOH at heating is converted into red CuO 2CuOH → H2O + Cu2O This is qualitative reaction of aldehyde group. With formalin (the most active aldehyde) this reaction can occur until pure Cu is educed.

H CO

HH C

O

OHCu2O+ 2Cu+

Experiment 2. Precipitation of proteins due to formalin action. Pour 0,5 ml of protein solution into test-tube, add several drops of formalin, mix. Observe

setting of protein under influence of formalin. Experiment 3. Color reaction of acetone with sodium nitroprusside Pipette 1 drop of Na nitroprusside ( Na2[Fe(CN)5NO] ), 5 drops of water, 1 drop of

acetone into test-tube. Add 1 drop of 2n NaOH. What do you observe? Divide solution by 2 portions. Add in first 1 drop of 2n CH3COOH. Compare color of

first and second test-tubes. Color reaction with Na nitroprusside is Legal test; it’s used in clinical practice for

detecting of acetone in the diabetic patient’s urine. ACADEMIC EXERCISES FOR ACHIEVEMENT CONTROL OF SPECIFIC

PRACTICE GOALS Exercise 1. Formaldehyde is used in producing of medical substance - urotropin. What chemical

reaction occurs at interaction between formaldehyde and prussic acid (hydrogen cyanide)? Answer: О Formaldehyde Н – С takes part in many reactions, is one of most Н reactive compounds

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Step 1. Interaction of methanal with HCN results in formation of oxinitrile this reaction

proceeds according to nucleophilic addition mechanism. For increasing reactivity of carbonyl group acidic catalysis is used, it leads to increasing

positive charge on carbon atom: О ОН+ ОН Н – С + Н+ → Н – С ↔ Н – С+ Н Н Н Step 2. Protonated methanal adds nucleophilic part of HCN molecule. The addition is

accompanied by proton releasing and formation of oxynitrile: О OH Н – С+ + СN– → H – C – СN H H Conclusion: At interaction of methanal and prussic acid oxynitrile is formed by nucleophilic addition. Exercise 2. Formation of hemiacetals is a basis of glucose cyclic form formation. What chemical reaction with 5-hydroxyhexanal takes place in acidic medium? Answer: 5-hydroxyhexanal is heterofunctional compound containing 2 functional groups: alcohol

and aldehyde, so it possesses alcohol and aldehyde properties. 6 5 4 3 2 1 O СН3 – СН – СН2 – СН2 – СН2 – С I H OH All carbon atoms in hydrocarbon radical are in sp3 hybridizated state, so carbon chain can

take various conformations, including claw-shaped. At this conformation atoms С1 – С5 or С1 – С4 draw together. In case of С1 – С5 joining in acidic medium it’s possible interaction between alcohol and

aldehyde group resulting in formation of cyclic hemiacetal. Conclusion: In molecule 5-hydroxyhexanal between aldehyde and alcohol groups intramolecular

interaction occurs (nucleophilic addition reaction). It’s resulting in formation of cyclic hemiacetal (pyranose cycle).

Exercise 3. Amines are intermediate products in many enzymatic processes, for example, in α-

aminoacid synthesis in human body. Write reaction of methylamine addition to: a) propanal, b) acetone. Why are these reactions classified as addition-elimination reactions? Exercise 4. Aldehydes are easy oxidized, so they can serve as restores. What chemical reaction occurs at interaction between acetic aldehyde and silver oxide?

What’s the name of this reaction?

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Exercise 5. Aldehydes and ketones participate in many chemical reactions. Mention reactions that are common for aldehydes and ketones. Exercise 6. Which compound: benzaldehyde or propanal can show aldol condensation reaction? Prove your answer.

METHODICAL RECOMMENDATIONS FOR STUDENT’S WORK AT THE PRACTICAL LESSON:

The lesson begins with teaching problems solving, after that it’s necessary to carry out the experiment about the structure and properties of studied compounds.

At the end of the lesson the students write final tests. After that all the students under the supervision of the teacher make analysis of class

work and summ up total results of the lesson.

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STRUCTURE AND PROPERTIES OF CARBOXYLIC ACIDS Biomedical importance: Carboxylic acids and their esters are widespread in nature. Fragrance of flowers, fruits,

berries caused by essential substances. Carboxylic acids and their derivates get into organism with food and further are converted into more simple substances. Carboxylic acids and their salts form buffers (phosphate, hydrocarbonate, haemoglobic and others) that supply pH constancy of blood and tissues.

Some carboxylic acids and their salts are medical product. For example, isovaleric acid is a part of validol, sodium benzoate is disinfectant, urethanes (esters of carbaminoacid) have psychotropic action and are used in psychiatric practice.

Theme is important for studying of biochemistry, pharmacology and other medical disciplines.

THE OBJECTIVE OF THE PRACTICE: To develop skills in interpreting of structure and predicting of properties of carboxylic

acids and their derivates that participate in many chemical processes. Specific Goals: To be able to: 1. Interpret of mechanism of nucleophilic substitution reactions. 2. Make of functional derivates of carboxylic acids: salts, amides, esters. 3. Predict of dicarboxylic acid properties, specific reactions 4. Identify of acetic acid. 5. Prepare of sodium oxalates. 6. Get of experimental proof of oxalic acid molecule structure.

ACADEMIC CONTENT

THEORY TOPICS FOR TARGET: 1. Electronic structure of carboxylic group. Chemical properties of carboxylic acids. 2. Functional derivates of carboxylic acids: salts, anhydrates, esters, amides. 3. Dicarboxylic acids: oxalic, malonic, succinic, glutaric, fumaric. 4. Carbonic acid, its derivates: urea, carbamic acid, urethanes, ureids.

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CARBOXYLIC ACIDS

MONO-CARBOXYLIC

ELECTRONIC STRUCTURE

OF CARBOXYLIC GROUP

DICARBOXYLIC

NUCLEOPHILIC SUBSTITUTION REACTIONS

ANHYDRATES ESTERS AMIDES

MEDICAL-BIOLOGICAL IMPORTANCE

SPECIAL

PROPERTIES

PROPERTIES

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THE APPLICABLE MATERIALS:

1. The synopsis of lectures. 2. Chart of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work. Experiment 1. Dissociation of acetic acid. Pipette 2-3 drops of acetic acid into test-tube. Add 2-3 drops of water. Determine medium

of solution with litmus. Write dissociation scheme. Experiment 2. Stability of acetic acid to oxidizing agents. Add several drops of 2n KMnO4 and 2n H2SO4, mix. What’s happening? Make a

conclusion. Experiment 3. Detection of acetic acid. Formation of complex iron salt and its

decomposition at boiling. Put several crystals of CH3COONa into test-tube. Look, it hasn’t smell. Add 3 drops of

water and 2 drops of 0,1n FeCl3. Yellow-red coloration of iron acetate appears. Write reaction scheme.

Heat up solution until boiling. Due to hydrolysis of salt red-brownish insoluble sediment of (CH3COO)2FeOH is formed. Solution above sediment is clear, it doesn’t contain iron ions.

This reaction is applied qualitative analysis for moving off iron oxide from solution. (This reaction is possible with acetic acid if it will be neutralizing by alkali). Experiment 4. Obtaining of sodium oxalate. Put several grains of sodium formiate into dry test-tube and heating it. Firstly salt is

melting, after it’s decomposing with releasing of hydrogen. How can you prove presence of hydrogen? Heat substance carefully so as not to make it charred.

H CO

ONa

H CO

ONa

CO

ONa

CO

ONaH2 +

sodium oxalate ACADEMIC EXERCISES FOR ACHIEVEMENT CONTROL OF SPECIFIC

PRACTICE GOALS Exercise 1.

In human body simple substances undergo transformations resulting in formation of more complex compounds. Example of such transformation is esterification reaction.

Malonic acid is metabolite of tricarboxylic acid cycle. Suggest the way of malonic acid convertion into ethylacetate.

Exercise 2. Hydrolysis of esters take place in human body. What acid is a product of ethylmalonate

hydrolysis? A. Acetic B. Oxalic

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C. Lactic D. Malonic E. Formic Exercise 3. Oxalic acid contains in sorrel. It can get into human body with food. What substance

should be added to oxalic acid to obtain its amide? A. NаОН. B. С2Н5ОН. C. NН3. D. CH3SH E. Acetic acid Exercise 4. Dicarboxylic acids participate in tricarboxylic acid cycle that takes place in human body. What is the product of succinic acid dehydration? A. Tartaric acid B. Oxyacid C. Oxoacid D. Anhydrate E. Amide Exercise 5. Butyric acid contains in rancid butter. What is the name of butyric acid according to IUPAC? A. Propanoic B. Butanoic C. Oxybutanoic D. Oxopentanoic E. 2,2-dimethylpropanoic Exercise 6. Dehydrogenation of carboxylic acid is a part of trycarboxylic acid cycle, occurs in human

body. What is the product of enzymatic dehydrogenation of syccinic acid? A. Malonic acid B. Benzoic acid C. Acrylic acid D. Maleic acid E. Fumaric acid Exercise 7. Carboxylic acids are active members of metabolic processes in human body. Indicate

dicarboxylic acid among the given compounds. A. Acetic acid B. Benzoic acid C. Butyric acid D. Oxalic acid E. Nicotinic acid Answers: 1. a) Decarboxylation of malonic acid resulting in formation of acetic acid.

t СООН – СН2 – СООН → СО2 + СН3 – СООН

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b) Esterification reaction between acetic acid and ethanol resulting in formation of ethylacetate.

СН3 – СООН + НОС2Н5 → Н2О + СН3 – С – О -С2Н5 ⎜⎜ О ethylacetate 2. D 3. С 4. D 5. В 6. E 7. D





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STRUCTURE AND PROPERTIES OF LIPIDS AND PHOSPHOLIPIDS. Biomedical importance: Lipids are large group of the biological active substances. They are present in all animal

cells and participate in different physiological and biochemical processes. Lipids are structural cell components, play protective role (for example, in skin) and they are form of stored and transported energy “fuel”. Together with lipids in human body present substances that are contained in small amount but have higher biological activity. Among them are steroid hormones, prostaglandins, fat-soluble vitamins that are the low-molecular regulators.

Lipids divide on the two groups: saponifiable and non-saponofiable. Structures of sterane and terpene underlie in the non-saponifiable lipid structure. For instance, cholesterol and bile acids those are present in animal cells. Isoprenes are important medications, for example, camphor, bromocamphor, menthol, different essential oils.

Transformation of the biological active substances in human body is the chemical essence of biological processes. Therefore ability to interpret structure of lipids, predict their chemical properties are necessary for later on studying of biochemistry, pharmacology and professional medical activity.

THE OBJECTIVE OF THE PRACTICE: General goal To be able to interpret structure and chemical properties of the physiological active

substances – saponifiable and non-saponifiable lipids. The goal achievement is supplied with solution of the specific goals.

Specific goals: To be able to: 1. Identify saponifiable and non-saponofiable lipids. 2. Identify simple lipids (waxes, fats and oils) and complex lipids (phospholipids,

glycolipids, sphingolipids). 3. Analyze chemical properties of solid fats and oils according to their structures. 4. Identify terpene carbohydrate and terpenoids (citral, menthane, limonene, menthol,

camphor). 5. Identify sterane, sterines, cholesterol, bile acids, sex hormones, carotene, fat-soluble

vitamins. ACADEMIC CONTENT THEORY TOPICS FOR TARGET: 5. Classification of lipids. Higher fatty acids (stearic, palmitic, oleic, linoleic, linolenic

and arachidonic acids). Their structure and features of biological active fatty acids. 6. Saponifiable lipids. Reaction of their formation. Chemical properties of these lipids:

hydrolysis, hydrogenation, oxidation reactions. A Notion of waxes. 7. Complex lipids. Obtaining of phosphatidic acid, phosphatidylcolamine,

phosphatidylcholine, phosphatidyl serine (lecithin). Their features. 8. Non-saponifiable lipids. Terpenes: terpene carbohydrate and terpenoids (citral,

menthane, limonene, menthol, camphor). 9. Steroids. Structure of sterane. Sterines: cholesterol, bile acids. Notions of carotenoids:

carotin, fat-soluble vitamins (A, E, K).

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CHART OF THE LOGICAL STRUCTURE LIPIDS

non-saponifiable saponifiableclassification

chemical properties

Fat hydrolysis

alkaline hydrolysis

identification

Hydrogenation

biomedical importance

Evidence of double bonds presence in HFA Soap dilution in water.

structure

acidic hydrolysis

stearic, palmitic, oleic, linoleic, linolenic and

arachidonic acids

menthol,

terpene hydrate,

isoprenes prostaglandins

simple complex

fats waxes phospho-lipids

glyco-lipids

Fat oxidation

phosphatidic acid formation

phosphatidylcolamine

fat-soluble vitamins (A,E,K), cholesterol, bile

acids

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THE APPLICABLE MATERIALS: 3. The synopsis of lectures. 4. Graf of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work. Soap preparation (saponification of fat by alkaline solution). Method principle: method is based on the alkaline hydrolysis reaction of fats. Material security: porcelain cup, pipettes, stand, gas burner, distilled water, 35% NaOH

solution, castor oil. Work progress: 1. Put in a porcelain cup one milliliter of castor oil and add four drops of 35% NaOH

solution. 2. Mix these substances using glass stick to formation of the emulsion. 3. Put the cup on the ring and heat mixture so as small flame (10-15 min) and all time stir

it. 4. When this mass begins to thicken add 2-3 milliliters of distilled water and heat it

again. 5. Stir this mixture all time until it becomes uniform. 6. Remove the cup from the flame. 7. After cooling small piece of white solid soap is formed. 8. Write saponification reaction. Dilution of the soap in water. Method principle: method is based on the soap dilution in distilled water. Material security: test-tubes, pipettes, gas burner, distilled water, piece of sodium soap

(obtained in the last experiment). Work progress: 1. Put in a test-tube small piece of sodium soap (obtained in the last experiment) and add

2-3 milliliters of distilled water. 2. Heat the test-tube and make sure that at heating soap dilutes easy. 3. Shake the test-tube solution and observe formation of soap foam. Obtaining of higher fatty acids from soap solution. Method principle: method is based on the reaction between soap (salt of HFA) and

inorganic acids and formation of sparingly soluble free higher fatty acids. Material security: test-tubes, pipettes, the sodium soap (obtained in the last experiment),

2.0M H2SO4 solution. Work progress: 1. Put in a test-tube five drops of sodium soap solution (obtained in the last experiment)

and add one drop of 2.0M H2SO4 solution. 2. Observe momentary formation of white flake sediment. 3. Write a reaction of higher fatty acid formation. Soap ability for fat emulsion. Method principle: method is based on the ability of soap is doing like emulsor (SAS). Material security: test-tubes, pipettes, distilled water, the sodium soap (obtained in the

last experiment), sun flower oil. Work progress: 1. Put in a test-tube one drop of sun flower oil and add five props of distilled water.

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2. Shake this mixture well. Oil disintegrates by small drops and forms emulsion. 3. Observe that in time small drops stick together and becomes bigger that float up. 4. Add to emulsion five drops of soap solution and shake the test-tube. 5. Observe the milky white emulsion formation that more stably in time. 6. Make a conclusion about soap action. Formation of insoluble calcium salts of higher fatty acids. Method principle: method is based on the obtaining of insoluble calcium salts of HFA. Material security: test-tubes, pipettes, the sodium soap (obtained in the last experiment),

CaCl2 solution. Work progress: 1. Put in a test-tube five drops of sodium soap solution and add a few props of CaCl2

solution. 2. Shake this mixture well. 3. Observe formation of calcium soap like white flakes. 4. Write the formation reaction of calcium salt of HFA. ACADEMIC EXERCISES FOR ACHIEVEMENT CONTROL OF SPECIFIC

PRACTICE GOALS Exercise 1. Saponifiable lipids are important energy sources in human body. Indicate compound that

relates to these class: A. vitamin A; B. cholesterol; C. sodium oleate; D. tristearine; E. palmitic acid. Exercise 2. Wax coating on the leaf and fruit surfaces protect them from drying up and bacteria

penetrations. Describe chemical nature of these compounds: A. triacylglycerol; B. ester of HFA and monoatomic alcohol; C. phosphoglycerol; D. bile acid; E. phosphatidic acid. Exercise 3. In human body neutral fats play the important role either cell structural components or

storage substance. Indicate acid that forms at acidic hydrolysis of trioleate: A. C17H35COOH; B. С15Н31СООН; C. С17Н33СООН; D. С17Н31СООН; E. С19Н31СООН. Exercise 4. In food industry liquid oils are changed into solid fats. Indicate substance that needs for

this process: A. water; B. hydrogen; C. alkali;

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CH2

CH

CH2

O

O

O

C

C

C

O

O

O

C17H29

C17H29

C17H29

CH2

CH

CH2

O

O

O

C

C

C

O

O

O

C15H31

C15H31

C15H31

CH2

CH

CH2

O

O

O

C

C

C

O

O

O

C17H33

C17H33

C17H33

CH2

CH

CH2

O

O

OH

C

C

O

OC17H29

C17H29

CH2

CH

CH2

O

O

O

C

C

P

O

O

O

C17H35

C17H33

OH

OH

D. phosphoric acid; E. alcohol. Exercise 5. Phospholipids are structural components of cell membrane. Indicate substances, which

form phosphatidic acid molecule: A. ethanolamine + 2 HFA + 1 H3PO4; B. glycerol + 3 H3PO4; C. glycerol + 2 H3PO4 + 1 HFA; D. ethanolamine + glycerol + 2 H3PO4; E. glycerol + 2 HFA + 1 H3PO4. Exercise 6. Solid fats are more energy capacious and stable than liquid oils. Indicate structure of fat: A. B. C. D. E. Exercise 7. Vitamins are biological active substances that need for normal vital functions of animals

and human. Indicate a fat-soluble vitamin: A. ascorbic acid; B. vitamin PP; C. vitamin E (tocopherol); D. vitamin B6;

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O

Br

O

OH

E. vitamin B12. Exercise 8. Menthol has antiseptic, nervine and analgetic actions. Indicate class of non-saponifiable

lipids to which it relates: A. terpenes; B. terpenoids; C. steroids; D. sterines; E. bile acids. Exercise 9. Cholesterol presents in animal tissues, pollen, sabadilla oils and participates in formation

of bile acids, vitamin D, sex hormone synthesis. Indicate substance that is base of cholesterol: A. cyclopropane; B. HFA; C. isoprene; D. sterane; E. phenanthrene. Exercise 10. Camphor is used in medicine like a stimulator of the cardiac activity. Indicate structural

formula of this compound: A. B. C. D. E.

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HO

CH3

OHCH3

COOH

OH

HO

CH3

OHCH3

C

OH

O

NH CH2R

HO

CH3

CH3

CH3

OH

CHH3C CH3

CH2

CH

CH2

O

O

O

C

C

P

O

O

O

C17H35

C17H33

O

OH

CH2 CH2 NH2

Exercise 11. Bile acids are synthesized in the liver from sterines. They emulsify food fats, make better

the lipids assimilation and catalyze fat hydrolysis. Cholic acid is the most important from them. Indicate structural formula of this compound:

A. B. C. D. E.


At the beginning of the practical lesson control of student preparation is realized. Students solve academic exercises, make out and repeat theoretical material: interpret structure and properties of lipids, write equations of the chemical reactions that illustrate chemical properties of these compounds.

After that students execute the experimental part of work. Using procedures sequence of the laboratory work they make experiments and execute a report.

The Next stages are analysis and correction of students self-work. At the end of lesson test control is realized and the work result takes up.

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HETEROFUNCTIONAL COMPOUNDS. HYDROXY- AND OXOACIDS. Biomedical importance: Heterofunctional compounds are one of the widespread classes of organic compounds

that contain different functional groups in their structure. Important representatives of this class are hydroxyl- and oxoacids. These substances take place in metabolism process, for instance: pyruvic, oxaloacetic, citric and malic acids – are participants of Creb’s cycle. Lactic acid is a product of glycolysis and it is concentrated in muscles at intensive work as a result of oxygen deficit that caused reduction of pyruvic acid under the influence of coenzyme NADH. β-hydroxybutyric acid is produced in organisms of patients with insular diabetes and precedes of the toxic acetone production. The other feature of hydroxyl- and oxoacids is their presence in fruits and vegetables (citric acid is in citrus plants, grape, gooseberries), dairy produces and products of milk fermentation.

Derivatives of hydroxyl- and oxocids provide a basis for a making of pharmaceutical substances. For example, γ-hydroxybutyric acid has weak narcotic action and its salts are used like sleeping-draught and anesthetic substances; salicylic acid has antireumatic, antipyretic and antifungal actions, its derivative substance (aspirin) is analgesic and resolvent.

Transformation of the heterofunctional compound in human body and participation of them in exchange and redox reactions are caused by their structure and chemical nature. Thus ability to interpret structure hydroxyl- and oxoacids, predict their chemical properties and qualitative determination in different biological liquids are necessary for formation of the complete understanding of metabolism process in human body and practical skills development in determination of biological active substances.

THE OBJECTIVE OF THE PRACTICE: General goal To be able to interpret structure and properties of heterofunctional compounds like

participates of metabolism process and components of drugs. The goal achievement is supplied with solution of the specific goals.

Specific goals: To be able to: 1. Interpret structure of heterofunctional compounds. 2. Analyze chemical properties of hydroxyl- and oxoacids according to presence of

hydroxyl- and carbonyl groups and their positions (α, β, γ) in chemical structure. 3. Interpret the stereoisomerism of hydroxyacids using D- and L-nomenclature and

presence of enantiomers and diastereoisomers. 4. Interpret the biological importance of heterofunctional compounds as components of

Creb’s cycle, basis of redox reaction products and drug production. ACADEMIC CONTENT THEORY TOPICS FOR TARGET: 1. Classification of heterofunctional compounds according to presence, position and

amount of functional groups (carbonyl, hydroxyl and carboxylic). 2. Nucleophilic substitutional reactions by carboxylic group. 3. Nucleophilic substitutional reactions by hydroxyl group (the aspirin obtaining) and

nucleophilic additional reactions by carbonyl group. 4. Redox reactions of hydroxyl- and oxoacids. 5. Dehydration of hydroxyacids and decarboxylation of oxoacids. 6. The Stereoisomerism of hydroxyacids. D- and L-nomenclature. 7. Medical and biological importance of heterofunctional compounds and their

derivatives.

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structure

classification according tofun. group position

classification according to amount of group

chemical properties

stereoisomerism


HETEROFUNCTIONAL COMPOUNDS

Hydroxyacids Oxoacids

α β γ β α

mono-carboxylic

di-; tri-carboxylic

di-carboxylic

mono-carboxylic

nucl

eoph

ilic

subs

titut

iona

l rea

ctio

ns

byca

rbox

ylic

grou

p

nucl

eoph

ilic

subs

titut

iona

l rea

ctio

ns

byhy

drox

ylgr

oup

oxid

atio

n re

actio

ns

dehy

drat

ion

reac

tions

D- and L- nomenclature

lactic, β-hydroxybutyric, γ- hydroxybutyric, malic, tartaric, citric

acids

pyruvic, β-ketoglutaric, acetoacetic, oxaloacetic acids

nucl

eoph

ilic

subs

titut

iona

l rea

ctio

ns

byca

rbox

ylic

grou

p

nucl

eoph

ilic

addi

tiona

l rea

ctio

ns b

y ca

rbon

ylgr

oup

deca

rbox

ylat

ion

reac

tions

redu

ctio

n re

actio

ns

23

THE APPLICABLE MATERIALS: 1. The synopsis of lectures. 2. Chart of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work. Obtaining of acid and medium salts of D-tartaric acid. Method principle: method is based on different solubility of acid and medium salts of

tartaric acid in aqueous solutions. Material security: test-tubes, pipettes, stand, spatula, distillate water, 2M solution of

tartaric acid, 0.5M solution of KOH. Work progress: 1. Put in a test-tube one drop of tartaric acid solution and add two drops of KOH

solution. 2. Shake well the obtained mixture. 3. Observe formation of white crystalline sediment. 4. Add the KOH solution excess (approximately 4-5 drops) and shake it. 5. Observe gradual disappearance of sediment. 6. Keep obtained solution for the next experiment. 7. Write equations of obtaining reactions for acid and medium salts. Formation of two different according to their solubility salts states about presence of two

carboxylic groups in structure of tartaric acid. Demonstration of hydroxyl group presence in structure of tartaric acid. Method principle: method is based on ability to form chelate salts with Cu2+ ions. Material security: test-tubes, pipettes, stand, spatula, gas burner, distillate water, solution

of medium potassium salt of tartaric acid, 0.2M CuSO4 solution, 2M NaOH solution. Work progress: 1. Put in a test-tube two drops of copper sulphate solution and two drops of potassium

hydroxide solution. 2. Observe formation of light-blue sediment. 3. Add to this sediment solution of potassium tartrate, obtained in the last experiment. 4. Observe sediment dilution with formation of deep-blue solution. 5. Heat this solution over the gas flame. 6. Observe the absence in changing in solution coloration. 7. Write a reaction of chelate salt formation and make a conclusion about an absence in

color change at heating. This chemical reaction illustrates the possibility of Felling’s liquid use for confirming of

hydroxyl group presence in structure of monosaccharide. Demonstration of presence of the phenol hydroxyl group in salicylic acid structure. Method principle: method is based on the qualitative colored reaction of salicylic acid

with Fe3+ ions. Material security: test-tubes, pipettes, stand, spatula, distillate water, 0.1M FeCl3 solution,

solid salicylic acid. Work progress: 1. Put in a test-tube couple crystals of salicylic acid, add 3-4 drops of distillate water and

shake well. 2. Add to this solution one drop of FeCl3 solution. 3. Observe the color changing. 4. Write the reaction equation.

24

Demonstration of absence of the phenol hydroxyl group in acetylsalicylic acid (aspirin) structure and its hydrolysis.

Method principle: method is based on the qualitative colored reaction of phenol hydroxyl group with Fe3+ ions.

Material security: test-tubes, pipettes, stand, spatula, gas burner, distillate water, 0.1M FeCl3 solution, solid aspirin.

Work progress: 1. Put in a test-tube couple crystals of aspirin, add 3-4 drops of distillate water and shake

well to complete solution. 2. Divide it on two parts. 3. To first test-tube add one drop of FeCl3 solution. 4. Make a conclusion about absence of coloration. 5. Boil aspirin solution in second test-tube during two minutes over gas flame. 6. Write the hydrolysis reaction. 7. Add FeCl3 solution to the cooled hydrolyzed solution 8. Observe the color changing. 9. Write the reaction equation and make a conclusion. Difference in ability of salicylic and acetylsalicylic acids react with iron (III) chloride

solution and form colored species is used for determination of aspirin purity, which in long time keeping destroys with salicylic acid formation.

ACADEMIC EXERCISES FOR ACHIEVEMENT CONTROL OF SPECIFIC

PRACTICE GOALS Exercise 1. Citric acid participates in Creb’s cycle. Indicate class of heterofunctional compounds to

which it relates: A. carbonyl compound; B. hydroxyacid; C. oxoacid; D. higher fatty acid; E. aminoacid. Exercise 2. The following hydroxyacid is the metabolism product of fatty acids:

CH3HC CH2 C

O

OHOH

Give the name to this compound: A. pyruvic; B. acetoacetic; C. lactic; D. oxaloacetic; E. β-hydroxybutyric. Exercise 3. Salicylic acid is a base for synthesis of febrifuges. Indicate the structural formula of this

compound:

25

COOH

C

CH3

O

COOH

COOHCOOH

COOH

C

CH2

O

COOH

OCH3

COOH

OH

COOH

O C CH3

O

COOCH3

OH

COOCH3

OCH3

A. B. C. D. E. Exercise 4. γ-hydroxybutyric acid has weak narcotic action and product of its react with NaOH is

used like anesthetic remedy. Indicate class of such product: A. ester; B. salt; C. ether; D. amide; E. ketone. Exercise 5. Pyruvic acid is a ketoacid thus it can participates in nucleophilic additional reactions,

including HCN. Indicate class of such product: A. ester; B. salt; C. amide; D. ketone; E. hydroxynitrile. Exercise 6. Malic acid produces in Creb’s cycle from fumaric acid and than is oxidized. Indicate the

structural formula of this oxidizing product: A. (pyruvic acid); B. (oxalic acid); C. (oxaloacetic acid);

26

CH3 C CH3

O

CH3 CO

OH

CH3 CH CO

OHOH

CH3 CO

H

CH3 CO

OH

COOH

CH3

C OHH

COOH

CH2-COOH

HO H

D. (acetic acid); E. CH3–CH2–OH (ethanol). Exercise 7. The Hydroxyl group position in structure of acid (α, β, γ) defines chemical changes at the

heating. Give the name to a hydroxyacid that in dehydration forms the next compound:

CH C

CHCCH3

CH3

O O

O

O A. β-hydroxybutyric; B. lactic; C. γ- hydroxybutyric; D. α-hydroxybutyric; E. β-hydroxypropionic. Exercise 8. Pyruvic acid is formed in the glycolysis and than its decarboxylation is going. Indicate

product of this reaction: A. (lactic acid); B. (acetaldehyde); C. (acetone); D. (acetic acid); E. CH3–CH2–OH (ethanol). Exercise 9. In muscles at intensive work is synthesized L-(+)-lactic acid that causes the typical pain.

Indicate the structural formula of this stereoisomer: A. B.

27

COOH

CH3

C HHO

COOH

CH2-COOH

H OH

CH3

C

COOH

HH

C. D. E. Exercise 10. Acetoacetic acid is formed in lipids metabolism. In the presence of insular diabetes lipid

exchange is changed that causes intoxication of human body. Please, explain this phenomenon: A. changing of the osmotic pressure; B. changing of buffer capacity of blood; C. metabolism products are concentrated and cause the typical pain; D. formation of poison “acetone bodies” from oxoacids; E. increasing of reactions in Creb’s cycle. METHODICAL RECOMMENDATIONS FOR STUDENT’S WORK AT THE

PRACTICAL LESSON: At the beginning of the practical lesson control of student preparation is realized.

Students solve academic exercises, make out and repeat theoretical material: interpret structure and properties of heterofunctional compounds, write equations of the chemical reactions that illustrate chemical properties of hydroxyl- and oxoacids as well as stereoisomer-forms of D- and L- hydroxyacids.



28

CARBOHYDRATES. STRUCTURE AND PROPERTIES OF MONOSACCHARIDES Biomedical importance: Carbohydrates are one of the most important classes of organic compounds, main

components of food. Carbohydrates are source of energy in organism, components of many biological important substances (nucleic acids, coenzymes, other biopolymers).

Glucose is a well-known monosaccharide. It enters into human body with plant food (fruits, juices) and as result of enzymatic degrading of complex carbohydrates. Glucose is a component of all known complex carbohydrates. It’s present in blood and tissues, energy source for biochemical reactions.

Carbohydrates show various chemical properties depending on their structure. Knowledge about their properties will help students to study biochemical transformations of carbohydrates that take place in human body. This theme is very important for studying of biochemistry, endocrinology, pharmacology.

THE OBJECTIVE OF THE PRACTICE: To develop skills in predicting of monosaccharide reactivity, interpreting of the most

important properties for understanding their metabolic pathways in human body. Specific Goals: To be able to: 1. Identify of monosaccharides according to functional groups and number of carbon

atoms. 2. Predict of tautomer formation of hexoses: glucose, galactose, fructose due to

aldehyde and ketone and hydroxyl groups. 3. Predict of tautomer formation of pentoses: ribose and deoxyribose (open and cyclic

forms) 4. Identify of cyclo- and oxotautomers. 5. Identify of chemical properties of glucose, galactose as aldoses such as oxidation (all

types), reduction. 6. Identify of chemical properties of fructose as ketose. 7. Interpret of formation of simple and complex glycosides as properties of cyclic forms. 8. Identify of the most important monosaccharide derivates. ACADEMIC CONTENT THEORY TOPICS FOR TARGET:

1. Carbohydrates. Classification. Monosaccharides (pentoses: ribose, deoxyribose;

hexoses: glucose, galactose, mannose, idose, fructose). 2. Stereoisomerism (D and L- forms), open and cyclic forms. Cyclooxotautomerism

(Fisher, Colly-Tollens, Haworth formulas). Furanose and pyranose cycles; α– and β–anomers. 3. Chemical properties of monosaccharides: 3.1. Reactions of hydroxyl groups: formation of esters and ethers (glycosides) 3.2. Oxidation (soft, harsh, enzymatic), formation of aldonic and uronic acids 3.3. Reduction resulting in formation of polyols (sorbitol, mannitol) 4. Derivates of monosaccharides: amino sugars (glucoseamine, galactoseamine), N-

substituted aminosugars, deoxy-derivates.

29


Reduction (formation of polyatomic alcohols -

polyols)

Derivates: 6-phosphate-

fructose, 1,6-diphos-

phate-fructose

Formation of ethers

Formation of esters

(complex glycosides)

Derivates: 6-phosphate-glucose, 2-amino-2-deoxy-6-phosphate-D-glucose

Reduction (formation of polyatomic

alcohols)

Soft oxidation, use in clinical

practice

Harsh oxidation

Enzymatic oxidation

cyclic form Haworth formulas (chemical

properties)

open form (chemical properties)

open form (chemical properties)

cyclic form Haworth formulas (chemical

properties)

CYCLOOXOTAUTOMERISM CYCLOOXOTAUTOMERISM

HEXOSES

RIBOSE DEOXY-RIBOSE

GLUCOSE GALACTOSE FRUCTOSE

CARBOHYDRATES

MONOSACCHARIDES

ALDOSES KETOSES

PENTOSES HEXOSES

30

THE APPLICABLE MATERIALS: 3. The synopsis of lectures. 4. Chart of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work. Experiment 1. Proof of OH group presence in glucose. Pipette of 1 drop of 0,5% glucose solution and 6 drops of 2n NaOH into test-tube. Add 1

drop of CuSO4 to the mixture. Cu(OH)2 precipitation is formed and quickly dissolves resulting in formation of pure light blue solution of copper saccharate. Resolution of Cu(OH)2 illustrates presence of OH groups in glucose. Save this solution for next experiment.

Write the reaction. Experiment 2. Reduction of Cu(OH)2 by glucose in the presence of alkali. (Trommer

test) Add several drops of water to copper saccharate solution (height of liquid must be equal

20 mm). Heat up upper part of test-tube until boiling (don’t boil). What do you observe? 2 Cu(OH)2 → 2 CuOH + H2O + O (for oxidation of glucose) (If you warmed test-tube more, yellow-red Cu2O precipitation could be formed 2CuOH → Cu2O + H2O) Write scheme of reaction. Trommer test is used for identification of glucose in the urine. Experiment 3. Detection of glucose according to Gainess test Pipette off 1 drop of 0,2n CuSO4 and 2 drops of 2n NaOH into test-tube. Add 1 drop of

gliserol to obtained precipitation of Cu(OH)2 and shake mixture. What’s happens? Add 1 drop of 0,5% glucose and several drops of water to obtained solution (height of solution must be 20 mm). Shake solution. Incline test-tube and heat up its upper part until beginning of boiling. Don’t shake contents of test-tube. What’s happens in upper part of test-tube? Write formula of formed compound, note its color.

Alkaline solution of copper glycerate is used for clinical detection of glucose in urine. This reagent is called Gainess reagent. It acts more quickly than Trommer test. Besides, if little amount of glucose is present, excess of Cu(OH)2 (which is in form of complex compound of glycerol) isn’t destroyed at boiling with formation of black CuO masking reaction.

Experiment 4. Selivanov reaction of fructose. Put grain of dry resorcin and 2 drops of HCL into test-tube. Add 2 drops of 0,5% fructose

and heat up until boiling beginning. Slowly liquid becomes red color caused by formation of oxymethylfurfurol:

OC

CC

CCH2

OHC

O

H

HH

Under influence of HCl latter is condensed with resorcin giving color compound.

Selivanov reaction is typical for fructose (and other ketoses) at indicated conditions without boiling! (Aldoses at long boiling can give slight red color).

31

ACADEMIC EXERCISES FOR ACHIEVEMENT CONTROL OF SPECIFIC PRACTICE GOALS

Exercise 1. Deoxyribose is a component of DNA. It’s present in nucleic acid as β,D–deoxyribose. Choose structure that matches to β,D–deoxyribose:

Exercise 2. Glucose is the most important monosaccharide, the main source of energy in human

body. What type of carbohydrate is glucose? A. Aldose, heptose B. Aldose, hexose C. Ketose, pentose D. Aldose, pentose E. Ketose, hexose Exercise 3. Sorbitol is used as sugar substitute for diabetics. It’s obtained from glucose. With what

substance does glucose react resulting in formation of sorbitol? A. methanol B. Ag2O. C. Cu(OH)2. D. CH3COOH E. H2 Exercise 4. In diabetic urine glucose and other aldehydes contain. What reaction gives opportunity to

detect glucose in urine? A. Reduction B. Soft oxidation C. Formation of ethers D. Formation of esters E. Harsh oxidation

O

O O O

OHOH2C

H

OH H

H HH

OHA) HOH2C

H

OH OH

H HH

OHB)

C)CH2OH

H

OHH

OH

OH

H

OH

H

D)CH2OH

H

OH

H

OH

OH

H

H

OH

HOH2C

H

OH H

H OHCH2OH

OHE)

32

Exercise 5. Glycosides are used as medicines for hart diseases treatment. Find out ethylglycoside of

galactose among the given compounds: Exercise 6. Carbohydrates from food get into cells. Glucose degrading begins with glucose-6-

phosphate formation. Which of the given compounds is glucose-6-phosphate? Correct answers: 1. А. 2. B 3. E

O O O

O O

A)CH2OH

H

OCH3H

OH

OH

H

OH

H

CH2OH

H

OC2H5

H

OH

OH

H

H

OH

HOH2C

H

OH H

H HH

OC2H5

C)

D)CH2OH

H

OC2H5H

OH

OH

H

OH

H

E)CH2OH

HH

OH

OH

H

OH

H OC2H5

B)

O O O

O O

A)CH2OPO3H2

H

OHH

OH

OH

H

OH

H

B)CH2OH

H

OPO3H2

H

HOH

H

H

OH

HOH2C

H

OH H

H H

H

OPO3H2

C)

D)CH2OOPO3H2

H

OHH

OH

OH

H

OH

H

E)CH2OH

H

H

OH

OH

H

OH

H OPO3H2

33

4. B 5. D 6. A





34

CARBOHYDRATES. STRUCTURE AND PROPERTIES OF DI- AND POLYSACCHARIDES

Biomedical importance: Carbohydrates are important biological active substances. Polysaccharides are natural polymers, their metabolism is chemical essence of biological

processes. Oligo- and polysaccharides in human body undergo enzymatic hydrolytic degrading up to

monosaccharides. Diabetes mellitus is accompanied by hyperglycemia. Treatment of this disease is impossible without knowledge about structure, properties, hydrolysis of homopolysaccharides.

Heteropolysaccharides also play important biological role. They supply elasticity, resiliency of organs and tissues. Chondroitin sulphate is a component of skin, cartilages, tendons; hyaluronic acid contains in base material of connective tissues and cartilages. Heparin is anticoagulant.

Knowledge about polysaccharide structure, mechanism of their metabolic reactions are needed to understanding normal and pathological carbohydrate’s metabolism.

THE OBJECTIVE OF THE PRACTICE: To develop skills in interpreting of structure and properties of di- and polysaccharides for

further usage in human biochemistry course. Specific Goals: To be able to: 1. Interpret of structure and properties of disaccharides (maltose, lactose, sucrose) and

polysaccharides (homo- and heteropolysaccharides). 2. Demonstrate of absence of reducing properties in sucrose and their presence in

maltose and lactose. 3. Identify of starch in solutions. 4. Carry out of starch and cellulose acidic hydrolysis. 5. Make resume about structure and properties of complex carbohydrates (di- and

polysaccharides). ACADEMIC CONTENT THEORY TOPICS FOR TARGET: 1. Disaccharides (maltose, lactose, sucrose). Structure, properties and biological

importance. 2. Polysaccharides. Classification, structure. 3. Homopolysaccharides: starch, glycogen, cellulose; dextrines, structure, hydrolysis,

biological role. Qualitative reaction of starch. 4. Heteropolysaccharides: structure. Glycosaminoglycanes (mucopolycaccharides) –

hyaluronic acid, chondroitin sulphate, heparin: structure, biological importance.

35

CHART OF LOGICAL STRUCTURE

STRUCTURE TYPE OF BONDS

CHEMICAL PROPERTIES

MEDICAL-BIOLOGICAL IMPORTANCE

POLYSACCHARIDES DISACCHARIDES

MALTOSE LACTOSE SUCROSE

CARBOHYDRATES

HOMOPOLY- SACCHARIDES

Cellulose Starch

Glycogen

HETEROPOLY- SACCHARIDES

Hyaluronic acid Chondroitin sulfate

Heparin

36

THE APPLICABLE MATERIALS: 1. The synopsis of lectures. 2. Chart of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work. Experiment 1. Evidence of existence of hydroxyl groups in sucrose. Put 1 drop 1% sucrose and 6 drops 2n NaOH into test-tube. Add 5-6 drops of water

(height of liquid must be 20 mm). Add 1 drop of 0,2n CuSO4. Instead of Cu(OH)2 precipitation, copper saccharate is formed. What color is it? Save it for next experiment.

What compounds do demonstrate resolution of Cu(OH)2? Experiment 2. Sucrose is nonreducing sugar. Heat up upper part of copper saccharate (from exp.1) until boiling (don’t boil). What do

you observe? What reaction does show glucose at the same conditions? Explain, why doesn’t sucrose reduce Cu(OH)2?

Experiment 3. Proof of sucrose hydrolysis Pipette 1 drop of 1% sucrose solution, 1drop of 2n HCl, 6 drops of water into test-tube.

Heat up test tube in 0,5-1 min. Hold test-tube on the angle and shake it. Divide hydrolizate to two parts.

Add 6 drops of 2n NaOH, 4-5 drops of water and 1 drop of 0,2n CuSO4 to first part. Heat up upper part of blue solution until boiling. What do you observe?

Why does solution show reducing properties? Make Selivanov reaction with second part of hydrolizate. Put grain of resorcin and 2

drops of HCl(conc.) into second test-tube. Heat up it. Red stain appears. What substance is detected by Selivanov reaction?

Write all your observations, reactions and explanations. Experiment 4. Identification of starch Starch isn’t soluble in water but forms colloidal solution – paster. Pipette off 5 drops of starch paster and 1 drop of diluted iodine solution. Starch becomes

blue color due to formation of complex compounds and adsorption. Heat up blue solution, color disappears. At cooling color restores.

Experiment 5. Lack of reducing properties in starch Pipette 1 ml of starch paster into test-tube. Add 2-3 drops of 2n NaOH and 1 drop of

CuSO4 and mix. Cu(OH)2 precipitate is formed. Heat up test-tube. Reducing of Cu(OH)2 doesn’t occur. Precipitate can become black due to formation of CuO, but reducing is absent. So starch has very small amount of reducing groups which can’t be detected by Cu(OH)2.

Experiment 6. Acidic hydrolysis of starch Pipette 1 drop of 0,5% starch paster into test-tube. Add 2 drops of 2n H2SO4 and put into

water bath for heating. In 20 min pay note that muddy solution became pure. Pipette 1 drop of hydrolizate to glass and add diluted iodine solution (yellow color). Blue is absent, so starch is hydrolyzed. Add 8 drops of 2n NaOH to hydrolizate and 1 drop of 0,2n CuSO4. What do you observe? Heat up upper part of test-tube. What’s happening?

Write scheme of starch hydrolysis.

37


Exercise 1. Maltose (malt sugar) is the main product of starch degrading under amylase action.

Amylase is produced by salivary gland. What are the products of maltose hydrolysis? А. 2 molecules of β, D-galactopyranose В. 2 molecules of β, D-glucopyranose С. α, D–glucopyranose and β, D-fructofuranose D. 2 molecules of α, D-glucopyranose E. α, D–galactopyranose and β, D-glucopyranose Exercise 2. Lactose is milk sugar, contains in milk. It’s obtained at cheese-making after separation of

curds from whey. Lactose solution gives positive Felling’s test. Which of the given reagents can be used to demonstrate lactose reducing properties? А. Copper sulfate В. Sodium hydroxide С. Copper (II) hydroxide D. Urea E. Sodium nitroprusside Exercise 3. Glycogen is functional and structural analog of starch. It’s similar to amylopectin but

more branched. Glycogen is ingredient of food and source of energy for living being. What are the products of glycogen hydrolysis? А. α, D-glucopyranose В. β, D-glucopyranose С. α, D–galactopyranose D. β, D- galactopyranose E. α, D-fructofuranose Exercise 4. Carbohydrates contain in cells and tissues of all plants and animals. They are the source

of energy in metabolic processes. Which from the given carbohydrates is nonreducing disaccharide? А. Maltose В. Sucrose С. Lactose D. Galactose E. Glucose Exercise 5. Amylose is starch fraction. It's macromolecule has helical form. Every turn conteins 6

monomeric units. 1. What type of bond is present in amylase molecule? 2. What are the products of amylose hydrolysis?

38

Exercise 6. Cellulose is plant polysaccharide. It possesses mechanical rigidity and serves as basic

material of plant cell walls. 1. What type of bond is present in cellulose molecule? 2. What are the products of cellulose hydrolysis?





39

STRUCTURE AND PROPERTIES OF HETEROCYCLIC COMPOUNDS. Biomedical importance: Heterocyclic compounds are structural components of biological active substances and

pharmaceuticals. They are parts of amino acids (tryptophane, histidine), nucleic acids, vitamins (nicotinic acid, cyanocobalamin, thiamine), enzymes and coenzymes (NAD, FAD) and other biological substances (haemoglobin, chlorophyll, ATP).

Large quantities of pharmaceuticals contain in their structure heterocyclic rings, for example, analgesics (amidopyrine, analgin); tuberculouses (phthivazide); antibiotics (derivatives of penicillin and cephalosporin); etc.

Heterocyclic compounds also are parts of metabolism products, for instance, uric acid is produced in purine bases decomposition in human body. At a dysbolism process the accumulation of heterocyclic compounds can cause pathological state formation, for example, salts of uric acid form stones in kidneys.

The heterocyclic compounds transformation in organism is conditioned by the chemical properties of these substances and their structures. Thus ability to interpret structure heterocyclic compounds, predict their chemical properties and qualitative determination in different biological liquids are necessary for formation of the complete understanding of metabolism process in human body and practical skills development in determination of biological active substances.

THE OBJECTIVE OF THE PRACTICE: General goal: To be able to interpret structure and properties of heterocyclic compounds like structural

parts of biological active substances and components of drugs. The goal achievement is supplied with solution of the specific goals.

Specific goals: To be able to: 1. Interpret structure of heterocyclic compounds. 2. Analyze chemical properties of five- and six-membered cycles according to amount of

heteroatoms and aromaticity of compounds. 3. Interpret the lactim-lactame and amine-imine tautomerism of heterofunctional

derivatives of heterocyclic compounds. 4. Interpret the biological importance of heterocyclic compounds as components of

nucleic and amino acids, vitamins and drug production. ACADEMIC CONTENT THEORY TOPICS FOR TARGET: 1. Classification of heterocyclic compounds according to amount of atoms in ring (five-

and six-membered) and according to amount of heteroatoms. Electron structure and aromaticity of general representatives: five-membered cycle (pyrrol) and six-membered cycle (pyridine).

2. Structure and chemical properties of five-membered heterocycles: with one heteroatom – pyrrol and with two heteroatoms – imidazole, pyrazole, oxazole, thiazole. Biological active derivatives (structure of the tetrapyrrol compounds, histidine, amidopyrine, analgin, thiamine).

3. Structure and chemical properties of six-membered heterocycles: with one heteroatom – pyridine and with two heteroatoms – pyrimidine and its derivatives: uracil, thymine, cytosine. Biological active derivatives of pyrimidine (nicotinamide, pyridoxal).

4. Structure of conjugated heterocyclic compound – purine and its derivatives: adenine, guanine and uric acid.

5. The Lactim-lactame and amine-imine tautomerism of pyrimidine and purine bases. 6. Medical and biological importance of heterocyclic compounds and their derivatives.

40


structure; aromaticity

classification according to

amount of heteroatoms

main representatives

chemical properties, isomerism

Biomedical importance

acidic properties

basic properties

amid

opyr

ine,

ana

lgin

thia

min

e

hist

idin

e, h

ista

min

e

nico

tinam

ide,

pyr

idox

al

urac

il, th

ymin

e, c

ytos

ine

tetra

pyrr

ol d

eriv

ativ

es

purin

e

pyrr

ol

pyra

zole

imid

azol

e

oxaz

ole

thia

zole

pyrid

ine

pyrim

idin

e

pyri

mid

in b

ases

HETROCYCLIC COMPOUNDS

FIVE-MEMBERED

SIX-MEMBERED

one hetero-atom

one hetero-atom

two hetero-atoms

two hetero-atoms

CONJUGATED

Puri

ne b

ases

lactime-lactame, amine-imine tautomerism

aden

ine,

gua

nine

41

THE APPLICABLE MATERIALS: 1. The synopsis of lectures. 2. Graf of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work. Reaction of antipyrine with iron (III) chloride. Method principle: method is based on ability of heterocyclic compounds to form colored

chelate compounds with Fe3+ ions. Material security: test-tubes, pipettes, stand, spatula, distillate water, 0.1M FeCl3 solution,

solid antipyrine. Work progress: 1. Put in a test-tube couple crystals of antipyrine and dissolve it in distillate water. 2. Add to obtained solution one drop of 0.1M FeCl3 solution. 3. Observe appearance of a coloration. 4. Write antipyrine formula. Presence of the stable light reddish coloration is caused by formation of the chelate

complex – ferropyrine. Reaction of antipyrine with nitrous acid. Method principle: method is based on ability of heterocyclic compounds in reaction with

nitrous acid to form nitrosocompounds that have characteristic coloration. Material security: test-tubes, pipettes, stand, spatula, distillate water, 1.0M H2SO4

solution, 0.5M NaNO2 solution, solid antipyrine. Work progress: 1. Put in a test-tube couple crystals of antipyrine and dissolve it in distillate water. 2. Add to obtained solution one drop of 1.0M H2SO4 solution. 3. Add couple drops of 0.5M NaNO2 solution. 4. Observe appearance of a coloration that disappears at long time keeping. 5. Write the equation of nitrosocompound formation. Appearance of the gradually vestigial emerald-green color is caused by the

nitrosoantipyrine formation. Solubility of uric acid and its medium potassium salt in water. Method principle: method is based on comparing of solubility of uric acid and its sodium

salts in water. Material security: test-tubes, pipettes, stand, spatula, distillate water, 2.0M NaOH

solution, solid uric acid. Work progress: 1. Put in a test-tube couple crystals of uric acid, add 3-4 drops of distillate water and

shake well. 2. Observe formation of heterogeneous system that suggest uric acid is sparingly soluble

substance. 3. Add to obtained mixture one drop of 2.0M NaOH solution. 4. Observe in a moment resolution of crystals owing to formation of good soluble

sodium twosubstituted salt of uric acid. 5. Keep obtained solution for the next experiments. 6. Write the reaction equation of sodium salt formation. Uric acid in lactime form has ability to form salts in reaction with alkali solutions. These

salts are called urates and have good solubility.

42

Formation of the sparingly soluble ammonium urate. Method principle: method is based on comparing of solubility of sodium and ammonium

salts of uric acids in water. Material security: test-tubes, pipettes, stand, spatula, distillate water, sodium urate

solution (from the last experiment), NH4Cl pregnant solution. Work progress: 1. Put in a test-tube couple drops of sodium urate solution that was obtained in the last

experiment. 2. Add one drop of NH4Cl pregnant solution. 3. Observe Formation of white crystalline sediment of ammonium urate. 4. Keep this mixture for the next experiment. 5. Write the reaction equation of ammonium salt formation. Different solubility of sodium and ammonium salts can be used for isolation of uric acid

from mixture of organic substances. Decomposition of urates under influence of inorganic acids (isolation of crystalline

uric acid). Method principle: method is based on the formation of uric acids crystals at acidity

changing. Material security: glass- plate, test-tubes, pipettes, stand, spatula, ammonium urate

solution (from the last experiment), 2.0M HCl solution. Work progress: 1. Using pipette put on the glass-plate one drop of sparingly soluble solution of

ammonium urate that was obtained in the last experiment. 2. In center of this drop add one drop of 2.0M HCl solution. 3. Observe formation of the typical elongated crystals of uric acid. 4. Draw the crystal shape in the laboratory copy-book. 5. Write the reaction equation of uric acid formation. In human body formation of uric acid crystals is caused to formation of stones in kidneys,

podagra joints, etc. This phenomenon takes place in cases of decreasing of biological liquids pH. Qualitative reaction on uric acid (murexide test). Method principle: method is based on the oxidation of uric acids to alloxantin and

formation of its colored ammonium salts. Material security: glass- plate, pipettes, stand, spatula, ammonium urate solution (from

the last experiment), HNO3 concentrated solution, 0.5M KOH solution. Work progress: 1. Using pipette put on the glass-plate one drop of sparingly soluble solution of

ammonium urate that was obtained in the last experiment. 2. In center of this drop add one drop of HNO3 concentrated solution. 3. Take the glass-plate under 10 cm on the flame and evaporate this mixture. 4. After complete drying of this drop stop heating and cool the glass-plate. 5. Put on the one side from the dried up stain one drop of 0.5M KOH solution. 6. Observe formation of the purple-violet color. Murexide test is used in clinic tests at determination of kidney stones and other

precipitation. Also this qualitative reaction is applied for the detection of purine bases derivatives like caffeine.

43

NH

N

NH

N

N

N

NH

N

N NH

N

OH

HO

N

NH

HO


Exercise 1. The Dibasol – a medicine, lowerring arterial pressure, is a derived of the benzimidazole:

condensed heterocycle at which structure imidazole ring is conjugated with benzene. From the following compounds indicate the benzimidazole:

A. B. C. D. E. Exercise 2. The aminoacid – histidine is a part of many proteins including haemoglobin, has in its

structure heterocyclic radical. Indicate a heterocyclic ring of this radical: N

NH

CH2 CH CO

OHNH2

A. indole; B. pyrimidine; C. pyrazole; D. imidazole; E. pyrrol. Exercise 3. The Xanthine is produced in human body like a result of nucleic acid metabolism and can

be classified like hydroxypurin. Indicate the structural formula of it: A. B.

44

N

HO OHCH2

CH3

COH

N

N

OH

HO

NH

CH2HO CH2 NH2

HN

N N

HN

H2N

O

C. D. E. Exercise 4. Phthivazide is the effective antitubercular medication:

N NH NCH2 CH OH

O CH3 Indicate a heterocyclic ring of this compound: A. pyrrol; B. pyrazole; C. pyrimidine; D. pyridine; E. purine. Exercise 5. Heterocyclic compound – pyrrol has an aromatic nature. Indicate amount of electrons that

nitrogen atom is given in closed system of π-bonds: A. one; B. electron pair; C. two electron pairs; D. two unpaired electron; E. no one electron. Exercise 6. Indicate the reactant that can be used for confirmation of acidic property of NH-group of

the pyrrol: A. CH3COOH; B. HCl; C. NaOH; D. SO2Cl2; E. H2/Ni. Exercise 7. The hydroxypurine and hydroxypyrimidine is characterized by the lactim-lactam

tautomerism. Indicate the lactam form of the guanine: A.

45

N

NH

NH2

O

N

N

OH

HO

N

N

OH

HO

CH3

N

N N

HN

H2N

OH

B. C. D. E. Exercise 8. Describe the biomedical importance of the tetrapyrrol compound – porfine: A. forms a complex compound with Fe2+, being a part of heme; B. is a coenzyme of redox processes in human body; C. is used like analgetic and sedative drugs; D. is a structural component of nucleic acids; E. is a neuromediator of the brain. METHODICAL RECOMMENDATIONS FOR STUDENT’S WORK AT THE

PRACTICAL LESSON: At the beginning of the practical lesson control of student preparation is realized.

Students solve academic exercises, make out and repeat theoretical material: interpret structure and properties of heterocyclic compounds, write equations of the chemical reactions that illustrate chemical properties of five- and six-membered cycles, write structural formulas of purine and pyrimidine bases that are parts of nucleic acids.



46

α-AMINO ACIDS. PEPTIDES. Biomedical importance: Proteins are of prime importance to the structure, function and reproduction of living

matter. Chemistry of these compounds is the special field of science that unites chemistry, biology, medicine and physic. Proteins are the material base of cells.

Proteins and peptides are building up from α-amino acids. Amount of these acids is above seventy. Thus knowing of structure and properties of α-amino acids are necessary for explanation and prediction structure and chemical nature of protein molecules and their properties. And its need for further studying of biology, biochemistry, protein functions on molecular level.

THE OBJECTIVE OF THE PRACTICE: General goal To be able to interpret structure and properties of the most important α-amino acids and

structure of protein molecules for further studying biological protein functions on the molecular level. The goal achievement is supplied with solution of the specific goals.

Specific goals: To be able to: 1. Predict reaction of α-amino acids based on properties of acidic group. 2. Predict reaction of α-amino acids based on properties of basic group. 3. Interpret lyophilic and lyophobic properties amino acids and proteins based on radical

structure. 4. Predict protein properties based on structure of α-amino acids. ACADEMIC CONTENT THEORY TOPICS FOR TARGET: 1. Classification of α-L-amino acids: - according to radical structure; - amount of amino and carboxylic groups; - water affinity. 2. Amphoteric properties of amino acids. 3. Biological important chemical reactions of amino acids. 4. Formation of dipeptides, polypeptides. Peptide bond: - formation; - structure of peptide group; - biuret test for peptide bond.

5. Levels of protein structure. Denaturation.

47



imin

e fo

rmat

ion

desa

min

atio

n

acyl

atio

n

deca

rbox

ylat

ion

form

atio

n of

am

ides

, sa

lts, e

ster

s

diss

ocia

tion

acid

ity

basi

city

Structure of protein molecules. Denaturation.

chemical properties

amino group reactions

carboxylic group reactions

amphoterism peptide formation

AMINO ACIDS

structure

classification

according to radical

structure

according to NH2; COOH

groups

according to water affinity

isomerism

L-is

omer

s

D-is

omer

s

hete

rocy

clic

aron

atic

alip

hatic

mon

oam

inom

onoc

arbo

xylic

hydr

ophi

lic

hydr

opho

bic

mon

oam

inod

icar

boxy

lic

diam

inom

onoc

arbo

xylic

48

THE APPLICABLE MATERIALS: 1. The synopsis of lectures. 2. Chart of the logical topic structure. DIRECTED ACTION BASE Procedures sequence of the laboratory work. Formation of chelate glycocol salt. Method principle: method is based on the amino acid ability to form complex chelate

compounds. Material security: test-tubes, pipettes, gas burner, spatula, 0.2M solution of glycocol,

copper (II) oxide, 2.0M NaOH solution. Work progress:

1. Put in a test-tube small amount of CuO. 2. Add three drops of 0.2M glycocol solution. 3. Heat mixture on the gas burner. 4. Wait for sedimentation of excess amount of black CuO. 5. Observe formation of deep blue solution of copper glycocol salt. 6. Add to this solution one drop of 2.0M NaOH solution. 7. Observe absence of sediment formation. 8. Make a conclusion about salt stability and write a reaction of chelate salt formation.

Copper salts of amino acids good crystallize therefore this reaction is used for separation of amino acids in pure.

Formaldehyde action on amino acids. Method principle: method is based on pH changing of amino acid solution after formalin

addition. Material security: test-tubes, pipettes, 40% formalin solution, 0.2% solution of methyl-

orange indicator, 0.2M solution of glycocol, 2.0M NaOH solution. Work progress: 1. Put in dry test-tube three drops of 40% formalin solution and add one drop of 0.2%

methyl-orange solution. 2. Observe red coloration of this mixture, which is evidence of acid presence. 3. Add one drop of NaOH solution to appearance yellow color that is evidence of neutral

medium. 4. Add obtained neutralized formalin solution to neutral glycocol solution. 5. Observe immediate appearance of red color that confirmed of acid formation. 6. Write a reaction of glycocol amino group with formaldehyde (imine formation). Blocking of amino group by formaldehyde at one time make free carboxylic group and

neutral solution of amino acid gets acidic pH that can be define by alkali titration. Presence of acidic reaction of aspartic acid solution and its absence in glycocol

solution. Method principle: method is based on amino acid dissociation. Material security: test-tubes, pipettes, 0.2M solution of glycocol, 0.1M aspartic acid

solution, 0.2% solution of methyl-orange indicator. Work progress: 1. Put in one test-tube three drops of glycocol solution and in another one three drops of

aspartic acid solution. 2. Add one drop of methyl-orange indicator solution. 3. Determine the pH media of both solution using its coloration.

49

4. Write reactions of amino acid dissociation and explain presence or absence acidic medium.

Ninhydrin test on α-amino acid. Method principle: method is based on amino acid ability to form colored compounds with

ninhydrin. Material security: test-tubes, pipettes, gas burner, 0.2M solution of glycocol, 0.1M β-

alanine solution, 0.1% ninhydrin solution. Work progress: 1. Put in dry test-tube five drops of glycocol solution. 2. Add five drops of 0.1% ninhydrin solution. 3. Boil this mixture in two minutes. 4. Observe pink-violet coloration. 5. Make the same experiment with 0.1M β-alanine solution. 6. Make a conclusion about absence of color in reaction with β-alanine. Xanthoprotein test. Method principle: method is based on a nitration of benzene ring in structure of aromatic

amino acid with formation of yellow colored nitro derivative compound. Material security: test-tubes, pipettes, gas burner, 0.2M solution of glycocol, 0.1%

tyrosine solution, 10% NaOH solution, nitric acid. Work progress: 1. Put in a test-tube five drops of tyrosine solution. 2. Add three drops of HNO3 solution. 3. Boil this mixture carefully. 4. Observe formation of yellow sediment. 5. Add drop by drop NaOH solution to appearance of orange coloration owing to

formation of sodium salt of dinitrotyrosine. 6. Make the same experiment with glycocol solution. 7. Explain absence of coloration. This is qualitative reaction on aromatic amino acids. Foil’s test. Method principle: method is based on black sediment formation of lead sulphide. Material security: test-tubes, pipettes, gas burner, Foil’s solution, rabbit wool, hair. Work progress: 1. Put in a dry test-tube rabbit wool. 2. Add five drops of Foil’s solution. 3. Boil this mixture in two-three minutes. 4. Observe formation of PbS black sediment. 5. Make the same experiment with hair. 6. Explain appearance of black sediment. Biuret test. Method principle: method is based on complex chelate salt formation. Material security: test-tubes, pipettes, protein solution, 10% NaOH solution, 1% CuSO4

solution. Work progress: 1. Put in a test-tube five drops of protein solution. 2. Add three drops of NaOH solution. 3. Add one drop of CuSO4 solution. 4. Shake this mixture well.

50

5. Observe formation of blue-violet color of obtained solution. This is qualitative reaction on peptide bond; it can be made with biuret. ACADEMIC EXERCISES FOR ACHIEVEMENT CONTROL OF SPECIFIC

PRACTICE GOALS Exercise 1. Amino acids can be synthesized in human body from other foods except eight acids that

are called essential and must be included, in the form proteins, in the diet. Indicate one of them: A. alanine; B. glycine; C. leucine; D. serine; E. glutamine. Exercise 2. More than 500 amino acids exist in nature but only 20 of them built proteins in all

organisms from bacterium to human. Amino acids classify according to radical nature. From the following compounds indicate aliphatic:

A. tryptophan; B. histidine; C. tyrosine; D. proline; E. serine. Exercise 3. Amino acids can be classified according to amount of carboxylic or amino groups. From

the following compounds indicate positive charged: A. glutamic acid; B. arginine; C. alanine; D. serine; E. phenylalanine. Exercise 4. Amino acids according to presence or absence of polar group classify to hydrophilic and

hydrophobic. Indicate amino acid that situates on the surface of protein molecule: A. alanine; B. isoleucine; C. lysine; D. leucine; E. phenylalanine. Exercise 5. Glutathione is an anti-oxidant. It present almost in all alive cells, especially in lens.

Structure of this substance is γ – Glu – Cys – Gly. Define a charge of this tripeptide: A. -2; B. -1; C. 0; D. +1; E. +2.

51

Exercise 6. According to rules in peptide formula amino acids are wrote in the special consistency.

Indicate C-ending amino acid in the following pentapeptide: Asp – Gln – Gly – Val – Lys

A. lysine; B. asparagine; C. valine; D. glycine; E. glutamine. Exercise 7. Amino acids are unique biological material. Owing to their ability to polarization α-

amino acids are monomers of proteins. Indicate bond that connect amino acids in chain of primary protein structure:

A. hydrogen; B. electrostatic (Van der Waals forces); C. peptide; D. ionic; E. disulfide. Exercise 8. Proteins are important buffers in human body. Give the explanation of their action: A. large quantity of amino acids; B. amino acid components with different pKa; C. N-end and C-end act like H+ donor and acceptor; D. peptide bond easy hydrolyzes and produces H+ and OH- groups; E. large quantity of hydrogen bonds in α-helix structure. METHODICAL RECOMMENDATIONS FOR STUDENT’S WORK AT THE

PRACTICAL LESSON: At the beginning of the practical lesson control of student’s preparation is realized.

Students solve academic exercises, make out and repeat theoretical material: interpret structure and isomerism of amino acids, write equations of the chemical reactions that illustrate chemical properties of amino acids and peptide formation.



52

CONTENTS

№ Topic Pages

1. Structure and properties of aldehydes and ketones

2. Structure and properties of carboxylic acids

3. Structure and chemical properties of lipids and

phospholipids.

4. Heterofunctional compounds. hydroxy- and oxoacids.

5. Carbohydrates. Structure and properties of

monosaccharides

6. Carbohydrates. Structure and properties of di- and

polysaccharides

7. Heterocyclic compounds.

8. α-Amino acids. Peptides.

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